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Ganit Learning Guides Intermediate Geometry Lines, Angles, Triangles, Loci, Concurrency, Quadrilaterals, Parallelograms, Areas of Plane Figures, Surface Area and Volume of Solid Figures, Graphs and Geometric Construction Author: Raghu M.D. GEOMETRY LINES, ANGLES AND TRIANGLES Geometry is a branch of mathematics which covers, points, lines, angles, areas and solids. The word geometry is derived from the Greek words ‘geo’ meaning ‘earth’ and ‘matrein’ meaning ‘measurement’. Some concepts of geometry are as follows. Postulate or Axiom: Postulates are self evident specific assumptions made to prove geometrical results or relationships. Theorem: Theorem is the result derived on the basis of postulates and common logic. Point: A point is the smallest possible dot in a plane. A point has no length, breadth or thickness. Points are the basic building blocks of lines and curves. Lines: Lines are made up of a number of points positioned side by side. A line is a geometrical figure which has length but no breadth. A line drawn between 2 points is known as a line segment. A line is considered as a straight line if the distance between any two points, is the least. Only one line can pass through two points but an infinite number of lines can pass through one point. All points on a line are called collinear points. Two distant lines on a plane cannot have more than one point in common. This point is called as the point of intersection and the lines intersecting lines. Some pairs of lines don’t intersect at all and are called parallel lines. Also from a point, not in the same line, there is only one line which passes through the point and parallel to the given line. A line with only one point defined is called a ray. A system of rays passing through points A and B A B A and B : Two points AB: Line segment. Intermediate-Geometry 1 of 96 ©2014 www.learningforkowledge.com/glg Intersecting and parallel lines. E A B O F C D A: A point in the plane BC: A line in the same plane as point A AD and BC: Intersecting lines O: Points of intersection BC and EF: A pair of parallel lines ANGLES Angle: Angle is defined as the union of two lines (two rays) having the same end point. Angles are classified according to the degree of separation of the two intersecting rays. End point where the two rays meet is known as vertex. There can be more than 2 rays emerging from a vertex resulting in the formation of more than one angle. Properties of angles are shown in figures 8.3, 8.4, 8.5 Angle ABC formed by ray AB and BC with point B as the vertex A B C Fig 8.4: Three rays AO, BO and CO emerge from the vertex O. Angles AOB and BOC are the two adjacent angles. Adjacent angles add up to equal the angle formed by the two extreme rays. A Ô B + B Ô C = A Ô C Intermediate-Geometry 2 of 96 ©2014 www.learningforkowledge.com/glg A B O C Property: Adjacent angles on a line are supplementary Fig 8.5: A ray is drawn from a point A such that it intersects the line (ray) BC at point O The adjacent angles A Ô B+A Ô C = B Ô C = 180º A B O C Property: If two lines (rays) intersect, vertically opposite angles are equal. AB and CD are line intersecting at point O. Angles A Ô C and B Ô D, B Ô C and A Ô D are pairs of vertically opposite angles. A Ô C = B Ô D, B Ô C = A Ô D A D O C B Proof: Angles A Ô D and B Ô D are adjacent angles on line AB Hence A Ô D + B Ô D = 180º Angles B Ô D and B Ô C are adjacent angles on line CD Hence B Ô C+B Ô D =180º Intermediate-Geometry 3 of 96 ©2014 www.learningforkowledge.com/glg  (A Ô D+B Ô D) – (B Ô D + B Ô C) = 180º-180º A Ô D+B Ô D - B Ô D - B Ô C = 0 A Ô D - B Ô C = 0  A Ô D+B Ô C like wise A Ô C - B Ô D TYPES OF TRIANGLES Acute angle: An angle whose measure is less than 90º is called an acute angle Right angle: An angle whose measure is exactly 90º Straight angle: If both the rays forming an angle are exactly opposite. A straight angle has a measure of 180º Obtuse angle: an angle whose measure is more than 90º but less than 180º Reflex angle: Reflex angle is also called the exterior angle of an acute, right or obtuse angle. This is an angle whose measure is more than 180º but less than 360º Angle around a point: An angle from the top of a ray to its bottom, completely encompassing the defined point of the ray is called angle around a point. By definition its measure is equal to 360º Example 1: AB is a line and C is a point on it D is a point away from the line. If A Ĉ D = 48ofind Angle B Ĉ D A Ĉ D +B Ĉ D = 180º Angles on a line are supplementary B Ĉ D =180º – A Ĉ D = 180º -48º = 132º Example 2: AB and CD are two lines intersecting of point O. If angle AOC is equal to 50º Find angle COB, BOD and AOD A D 50º C O B A Ô C = B Ô D =50º (vertically opposite angles) A Ô C+B Ô C = 180º (Angles on a line)  B Ô C =180º - A Ô C = 180º - 50º =130º B Ô C = A Ô D =130º (vertically opposite angles) Intermediate-Geometry 4 of 96 ©2014 www.learningforkowledge.com/glg Example 3: AOB is a line C is a point away from the line. Angle AOC is equal to 4xº and angle BOC is equal to 5xº Find the value of x. C 4xº 5xº A 4xº + 5xº =180º (Angles on a line) 9xº =180º  x = 20º O B Example 4: Three rays AO, BO and CO emerge from point O. Angle between any two adjacent rays (lines) are equal to any another. Find the measure of the angle. B A O C A Ô B +B Ô C+C Ô A = 360º (Angles around a point) But B Ô C = A Ô B and C Ô A = A Ô B  A Ô B+A Ô B+A Ô B = 3A Ô B = 360º  A Ô B =120º Example 5: PO and QO are line segments of opposite rays. RO and OS are two lines emerging from O such that P Ô R =Q Ô S. Also R Ô S = 90º Find P Ô R and Q Ô S Intermediate-Geometry 5 of 96 ©2014 www.learningforkowledge.com/glg R S P O Q P Ô R + R Ô S + S Ô Q = 180º -------- (Angles on a line) P Ô R + 90º + S Ô Q =180º ----------(R Ô S = 90º) P Ô R + 90º + P Ô R =180º ------- (P Ô R = S Ô Q) 2P Ô R = 90º or P Ô R = 45º = S Ô Q Example 6: If A Ô B shown in the figure if 45º, find the measure of the corresponding reflex angle. A O B Interior A Ô B + Exterior A Ô B =360º ------- (Angle around a point)  Exterior A Ô B =360º – 45º =315º =Reflex angle Example 7: Identify the following features in the figure shown below: a. Opposite rays b. Intersecting lines c. Point of intersection d. Opposite angles e. Linear pair of angle Intermediate-Geometry 6 of 96 ©2014 www.learningforkowledge.com/glg C A O B D Workings a. AO and OB, CO and OD b. AB and CD c. O d. A Ô C and B Ô D, B Ô C and A Ô D e. A Ô C and B Ô C, A Ô D and B Ô D Example 8: AO and BO are two rays meeting at point O, Another ray OC bisects the angle AOB such that A Ô C = B Ô C. If A Ô B measures 60º, find the values of A Ô C and B Ô C A C O B A Ô B = A Ô C+B Ô C A Ô B =2A Ô C 2A Ô C =60º A Ô C =30º =B Ô C Example 9: Angles A Ô C and B Ô C are vertically opposite angles. Rays (lines OE and OF are bisectors of angles A Ô C and B Ô C respectively. Show that EF is a line passing through O. Intermediate-Geometry 7 of 96 ©2014 www.learningforkowledge.com/glg A D E O F C B A Ô D + D Ô F + F Ô B + B Ô C+C Ô E + E Ô A = 360º (Angles around the point O) But C Ô B = A Ô D ----- (Opposite angles) C Ô E = A Ô E ---- (EO is the bisects or A Ô C) And B Ô F = F Ô D.  A Ô E = B Ô F Hence EF is a line Example 10: AB is a line and O is a point on it, and C is a point away from it. P and Q are points such that OP bisects A Ô C and OQ bisects B Ô C. Show that P Ô Q is a right angle. C P A Q O B Because PO and QO are angle bisectors A Ô P = P Ô C and C Ô Q = Q Ô B A Ô P + P Ô C + C Ô Q + Q Ô B =180º P Ô C + P Ô C + C Ô Q + C Ô Q =180º 2P Ô C + 2C Ô Q =180º P Ô C + C Ô Q = 90º = P Ô Q Intermediate-Geometry 8 of 96 ©2014 www.learningforkowledge.com/glg EXERCISES 1. State True or False 1. Two distinct points in a plane can form a part of straight line. 2. If two adjacent angles are equal that each has to measure 90º. 3. Two lines in a plane can have more than one point in common. 4. Angles forming a line can be both obtuse angles 5. Two adjacent angles equal in measure, adding to 180º are right angles 6. If two lines intersect vertically opposite angles are equal 7. Angles around a point add up to 180º 8. If two angles add up to 180º they are called supplementary angles 9. A point has no size 10. An angle bisector divides an angle into three equal parts. 2. O is a point on the line AB, CO is a ray drawn such that A Ô C =90º, find the measure of angle B Ô C 3. In the following figure A Ô C =3xº and B Ô C =7xo find the value of xo C 3xº A 7xº O B 4. If PO, QO and RO are three rays from the points O. If P Ô R =60º and Q Ô R =120º show that PO and QO are opposite rays. 5. In the following figure A Ô C is greater than B Ô C by 30º, find the value of both the angles C A Intermediate-Geometry O B 9 of 96 ©2014 www.learningforkowledge.com/glg 6. 3 rays AO, BO and CO emerge from a point ‘O’ such that A Ô B = B Ô C = C Ô A =120º three more rays PO, QO and RO are bisectors of the angles. Show that P Ô Q, Q Ô R and R Ô P are also equal to 120º also show that AO and QO are opposite rays. A R P O C B Q 7. Lines AB and PQ intersect at point O if A Ô P =90º find P Ô B, B Ô Q and Q Ô A 8. In the figure shown below find the value of x please note the AB is a line and O is a point on it. 2x x + 10 A x O B 9. AB is a line and O is a point one it and R is a point away from it. It A Ô R =8x and B Ô R =10x. Find the value of x and the angles A Ô R and B Ô R. 10. AB and CD are lines intersect at point E line EF bisects angle AEC. If angle AEF is 30º find all the remaining angles shown in the figure. Intermediate-Geometry 10 of 96 ©2014 www.learningforkowledge.com/glg A D 30º 1 4 E 3 2 F C B PARALLEL LINES Two lines are said to be parallel to each other, if they do not intersect any where and are in the same plane. Separation of two parallel lines is equal at all points. A transversal is a line that intersects both the parallel liner. Several properties of parallel lines can be explained with the help of a transversal. The diagram below shows a set parallel lines with a transversal. E 4 1 A B 3 2 8 5 C D 7 6 F Description: AB and CD – Parallel lines EF –Transversal 1, 5, 2, 6, 3, 7 and 4, 8 pairs of corresponding angles 2, 5 and 3, 5 –Alternate angles 3, 8 and 2, 5 –Co interior angles 1, 3, 2, 4, 5, 7 and 6, 8 – Opposite angles. PROPERTY- Corresponding angles are equal. The property when a pair of parallel lines are intersected by a transversal the corresponding angles are equal is assumed to be true. Hence this property is called the corresponding angles axiom. The converse is also true. When the corresponding angles formed by a transversal are equal, the two lines it intersects are parallel. Intermediate-Geometry 11 of 96 ©2014 www.learningforkowledge.com/glg a b Corresponding angles  a =  b Theorem If a transversal intersects two parallel lines then alternate angles are equal 8 7 4 1 3 2 5 6 Working: In the diagram shown above angles 3 and 5 are a pair of alternate angles 1 and 5 and 3 and 7 are pairs of corresponding angles.  3 =  1 (vertically opposite angles) But  1 =  5 (corresponding angles)  3 =5 Theorem If a transversal intersects two parallel lines then each pair of co interior angles are supplementary. Intermediate-Geometry 12 of 96 ©2014 www.learningforkowledge.com/glg 4 3 8 7 1 2 5 6 Working: In the diagram above angles 2 and 5 are a pair of co interior angles and 1 and 5 are a pair of corresponding angles.  1+  2 =180º ------ (angles or a line) But  1 =  5 ------ (corresponding angles) Hence  5 +  2 = 180º Theorem A transversal intersects two parallel lines if the alternate angles are equal then the two lines are parallel 4 1 3 2 8 5 7 6 Working:  1 =  3 ------ (vertically opposite angles) But  3 =  5 ------- (Alternate angles)   1 =  5, because the angles 1 and 5 are corresponding, as per the axiom the two lines have to be parallel. Theorem A transversal intersects two lines. The co interior angles add up to 180º. In such a case the two lines are said to be parallel Intermediate-Geometry 13 of 96 ©2014 www.learningforkowledge.com/glg 4 1 3 3 8 7 2 2 5 6 Working:  1+  2 = 180º ------ (Angles on a line)  2+  5 =180º ------- (Co interior angles)  5 =1 Since  5 and  1 are corresponding angles, as per the theorem, the two lines are parallel EXAMPLES AND PROPERTIES Example 1: EF is a transversal cutting lines AB and CD at points P and Q respectively. If the angle EPB (marked 1) is 30º, find all the remaining angles (2 to 8) marked. Give reasons E P 4 1 3 2 A C 8 5 7 6 Q 30º B D F  1+  2=180º----- (Angle on a line) 30o +  2=180º ------  2=150º = B P̂ Q  1 =  3 =30º------ (Vertically opposite angles)  2 =  4 =150º ----- (Vertically opposite angles)  1 =  5 =30º------ (Corresponding angles)  2 =  6 =150º ----- (Corresponding angles)  4 =  8 =150º ----- (Corresponding angles)  3 =  7 =30º ----- (Corresponding angles) Intermediate-Geometry 14 of 96 ©2014 www.learningforkowledge.com/glg Example 2: AB and CD are two lines cut by a transversal EF, intersecting AB and CD at G and H. If  EGB =40º =  CHF show that lines AB and CD are parallel. E A G 40º B H C 40º D F  AGF =  EGB = 40º ---- (Vertically opposite angle) Therefore  AGF =  CHF = 40º (It is  CHF = 40º But angles AGF and CDF are corresponding angles formed by a transversal intersecting two lines. Hence by the axiom of corresponding angles being equal, the lines AB and CD are parallel. Example 3: PQ and RS are two lines intersected by a transversal TU at points X and Y Angles PXY and TYS are equal prove that PQ is parallel to RS T X P Q R S Y U  PXY =  TXQ ------- (Opposite angles) But angle  PXY =  TYS ----- (As per the problem) Hence  TXQ =  TYS Intermediate-Geometry 15 of 96 ©2014 www.learningforkowledge.com/glg But angles TXQ and TYS are a pair of corresponding angles formed by a transversal intersecting two lines. By the axiom, if corresponding angles are equal the two lines intersected by the transversal are parallel. Hence PQ is parallel to RS. Example 4: A pair of parallel lines AB and CD are intersected by a transversal EF which is also perpendicular to line AB. Show that EF is also perpendicular to CD. E A G B H C D F Let EF intersect lines AB and CD at G and H E Ĝ B = E Ĥ D ---- (Corresponding angles) But E Ĝ B = 90º  E Ĥ D = 90º Hence EH is perpendicular to CD Example 5: A transversal EF intersects a pair of parallel lines AB and CD. G and H are points of intersection of the transversal and the parallel lines AB and CD respectively. If  AGE =33º find  GHD E 33º A B G H B D F  AGE = HGB ------ (Opposite angles)   HGB = 33º ----- (  AGE = 33º) Intermediate-Geometry 16 of 96 ©2014 www.learningforkowledge.com/glg But  HGB +  GHD = 180º ----- (co interior angles) 33º +  GHD = 180º   GHD = 147º Example 6: AB and CD are a pair of parallel lines intersected by a transversal EF. G and H are points of intersection. GI and HK are bisectors of angles EGA and DHF respectively. Prove that GI is parallel to HK. E G A B K I C H D F  AGH =  GHD ------ (Alternate angles) But  AGH = 2  IGH ------- (IG is the bisector of  AGH) Also  GHD = 2  GHK ----- (HK is the bisector of  GHD)  2  IGH = 2  GHK or  IGH =  GHK  IGH and  GHK are alternate angles formed by transversal GH intersecting the lines IG and HK. They being equal IG must be parallel to HK. Example 7: PQ and RS are two parallel lines MN is a transversal line perpendicular to both PQ and RS and intersecting them at t and u from a point ‘V’ on the line PQ a line Vu is drawn so that vu bisects  MUR. Show that  TVU = 45º M V T P Q R S U N Intermediate-Geometry 17 of 96 ©2014 www.learningforkowledge.com/glg MU is perpendicular to RS   MUR = 90º VU is the angle bisector   VUT =90  2 = 45º =  VUR But Vu is also a transversal to the lines PQ and RS   TVU=  VUR = 45º (Alternate angles) Example 8: AB and CD are two parallel lines a transversal. MN intersects both of them at points P and Q. MN also intersects another line EF at point R. If  APQ =  ERN, show that CD is also parallel to EF. M A P C B Q E D R F N  APQ =  CQR ------ (Corresponding angles) But  APQ =  ERN (as given in the problem)   ERN =  CQR But  ERN and  CQR are corresponding angles formed when the transversal MN intersects lines CD and EF. By the axiom of corresponding angles we have that CD is parallel to EF. Example 9: In the figure shown below lines AB and BC originate from point B and lines DE and EF originate from point F. Also AB is parallel to DE and BC is parallel to EF. Points B and E have been joined and the line EB has been extended to the point G. If BG angle ABC prove that EB bisects angle DEF. D A B E G F Intermediate-Geometry C 18 of 96 ©2014 www.learningforkowledge.com/glg EG is a transversal to both the pairs of parallel lines AB, ED and DE, EF.   DEB =  ABG ------- (Corresponding angles) But  ABG =  CBG =½  ABC (BG is the bisector or  ABC)   DEB = ½  ABC like wise  FEB =  CBG ---- (Corresponding angles) But  CBG = ABG = ½  ABC ----- BG is bisector of  ABC)   CBG =  FEB = ½  ABC   FEB = ½  ABC =  DEB Hence EB is the bisector of  DEF Example 10: In the figure shown below AB, CD and EF are parallel lines MN is a transversal intersecting lines AB and CD. NP is a transversal intersecting lines CD and EF. If  BMN = 70º and  FPN = 140º Find  MNP. A M B 70º C N D 140º E F P  BMN +  MND = 180º----- (co interior angles)   MND = 180º -  BMN = 180º– 70º  MND = 110º  EPN +  PND = 180º ------- (co interior angles)   PND = 180º-  FPN = 180º – 140º  PND = 40º  MNP =  MND +  NPE ------ (As per Diagram) (adjacent angles)   MNP = 110º+ 40º  MNP = 150º EXERCISES 1. Which of the following are True and False a. If two parallel lines are intersected by a transversal the co interior angles are equal b. If two parallel lines are intersected by a transversal then the corresponding angles are equal c. Parallel lines will intersect at two points. Intermediate-Geometry 19 of 96 ©2014 www.learningforkowledge.com/glg d. e. f. g. h. i. 2. Two lines perpendicular to the same line are perpendicular to each other. Two lines parallel to the same line are not parallel to each other. Co interior angles are on the opposite side of the transversal Angle bisectors of alternate angles are also parallel Two transversals of a set of parallel lines never meet. Converse of the corresponding are equal if the transversal intersects two parallel lines is not true. Name the following refer to the diagram below. E A G C B H D F a. b. c. d. e. 3. Pair of parallel lines Transversal Angle corresponding to  AGE Co interior angle pairs Alternate angle to AGH PQ and RS are a pair of parallel lines OP is a transversal intersecting PQ and RS at M and N respectively. If PMO is 40º Find angles, NMQ, PNR, PMN and OMQ O M P Q N R S P 4. QR is the transversal perpendicular to both the lines MN and OP which it intersects. Prove that MN and OP are parallel to each other. Intermediate-Geometry 20 of 96 ©2014 www.learningforkowledge.com/glg 5. In the figure below two parallel lines are intersected by two transversals which are parallel to each other. If the angle marked measure 55º find the value of angle marked 2, 3, 4 and 5 1 5 6. 2 4 3 In the following figure  QRS =  PQR =70º  STU and  RST = 150º If PQ‫ ׀ ׀‬RS show that RS‫ ׀ ׀‬TU ‫׀ ׀‬PQ P Q 70º T U 150º 70º 150º R 7. S In the figure below bisectors GO and HO of co interior angles BGH and GHD meet at a point O. Show that GOH is a right angle E A B G O H C D F Intermediate-Geometry 21 of 96 ©2014 www.learningforkowledge.com/glg 8. If two lines are intersected by a transversal such that the co interior angles formed on one side add up to 180º show that the two lines are parallel. 9. In the following figure identify the pair of parallel lines, state the reason. A C 100º 100º P Q 120º R 120º S B 10. P Lines PQ and RS are parallel. T is any point between the lines. An angle QTS is formed by joining QT and TS. Show that  PQT +  RST =  QTS given  QTS is less than 180º P Q T R Intermediate-Geometry S 22 of 96 ©2014 www.learningforkowledge.com/glg TRIANGLES A triangle is the simplest geometrical figure obtained by 3 lines. A figure so formed contains 3 interior angles and hence called a triangle. They can be classified on the basis of sides or angles. 1. Scalene Triangle: A triangle with all the three sides having different length is called a scalene triangle. 2. Isosceles Triangle: A triangle with two sides having the same length is called an isosceles triangle. 3. Equilateral Triangle: A triangle with all the three sides having the same length is called an equilateral triangle. Intermediate-Geometry 23 of 96 ©2014 www.learningforkowledge.com/glg 4. Acute angle triangle: A triangle with all the three angles acute is called an acute angle triangle. 5. Right angle triangle: A triangle with one of its angle as a right angle is called a right angled triangle. 6. Obtuse angle triangle: A triangle with one of its angle more than 90º is called an obtuse angle triangle. Properties: A triangle is usually designated by three letters attached to the vertices. For example ABC indicates a triangle whose vertices are A, B and C also A, B and C denote the interior angles of a triangle. Length of sides opposite A, B and C are indicated by small care letter a and c. However the side joining any two vertices is given as AB, BC and CA. In a triangle the sum of length of any two sides will be more than the length of the remaining side. a+b>c or BC + CA > AB A c b B C a Intermediate-Geometry 24 of 96 ©2014 www.learningforkowledge.com/glg In a triangle the sum of all the length of sides is called the perimeter. Also the side opposite to the largest angle will be the biggest. Three angles of a triangle always add up to 180º and an exterior angle is equal to the sum of two opposite interior angles. Proof and explanation of these two angle properties are given in the following theorems. Theorem: Three interior angles of a triangle add up to 180º A D E Consider a ABC as shown in figure. B C Draw a line parallel to BC through the point A then AB and BC become transversals Hence  ABC =  BAD ------ (Alternate angles)  ACB =  CAE ----- (Alternate angles) But  BAD +  BAC +  CAE = 180º (Angles on a straight line)   ABC +  BAC +  CAE = 180º Hence angles in a triangle add up to 180º Theorem: If one of the sides is extended so as to form an exterior angle then the exterior angle will be equal to the sum of opposite interior angles. A B C D In the triangle ABC line BC is extended to D   ABC +  BCD = 180º ----- (Angles on a straight line) But in  ABC  ABC +  BAC +  BCA = 180º   ABC +  BAC +  ACB =  ACB +  BCD   ABC +  BAC =  BCD (Exterior angle) Hence an exterior angle is equal to the sum of opposite interior angles. Example 1: Identify which of the following figures belong to a triangle (1) a = 15cms, b = 5cms, c = 6cms Ans: ABC is not a triangle because a is greater sum of b and c Intermediate-Geometry 25 of 96 ©2014 www.learningforkowledge.com/glg (2)  A = 90º,  B = 60º and  C = 30º Ans:  A +  B +  C = 180º, hence ABC is a triangle (3) Exterior angle ACD = 100º, Opposite interior angles CBA and BAC are 50º and 30º Ans:  ACD = 100º, and  CBA +  BCA = 80º only and not 100º Hence ABC is not a triangle (4) a = 10, b = 8 and c = 6 Ans: a = 10 is less than b + c = 8+ 6 =14 hence ABC is a triangle (5)  A = 70º,  B = 60º and  C = 40º Ans:  A +  B +  C = 70º + 60º + 40º = 170º and is not equal to 180º, Hence ABC is a triangle Example 2: In the triangle shown below  A = 60º and  B = 80º Find  C A 60º 80º C B Example 3: In a triangle shown below exterior angle  ACD = 105º and  BAC = 45º. Find the remaining angles. A 45º 105º B C D  ACD =  BAC +  ABC 105º = 45º +  ABC   ABC = 60º Also  ABC +  BAC +  ACB = 180º 60º + 45º +  ACB = 180º  ACB = 75 º Intermediate-Geometry 26 of 96 ©2014 www.learningforkowledge.com/glg Example 4: In an isosceles triangle  A = 80º,  B = 50º and  C = 50º. If the side opposite A is 8cm = a and side b opposite to B = 10.3cms. Find the perimeter of the triangle. A 80º 10.3cm B 8cms C Ans: In the triangle ABC  B =  C = 50º  AB = AC  AB = 10.3cm = BC  Perimeter = AB + BC +AC = 10.3 + 8 +10.3 = 28.6cms. Example 5: In an equilateral triangle ABC show that each of the angle is equal. If the perimeter of the triangle is 15cms. Find the length of each side and each angle. A B C Â + B̂ + Ĉ =180º ---- (Angles in a triangle) But Â = B̂ = Ĉ ----- (ABC is an equilateral triangle)  Â + Â + Â = 180º or 3 Â =180º Â = 60º= B̂ = Ĉ a + b + c = 15cms But a = b = c  a + a + a = 15cms Or 3a = 15cms  a = b = c = 5cms Intermediate-Geometry 27 of 96 ©2014 www.learningforkowledge.com/glg Example 6: PQR is an isosceles triangle with PQ = QR side QP is extended to S such that QP = PS show that SRQ = 90º S P Q R Ans:  QPR =  PSR +  PRS ---- (Exterior angle theorem) But ∆PSR is also Isosceles because PS = PR   PSR =  PRS Hence  QPR = 2  PRS Similarly  SPR =  PQR +  PRQ ------ (Exterior angle theorem) But ∆ PQR is Isosceles   PQR =  PRQ   SPR = 2  PRQ   QPR +  SPR = 2  PRS + 2  PRQ But  QPR +  SPR = 180º ------ (Angles on a line)  2  PRS + 2  PRQ = 180º   PRS +  PRQ = 90º But  PRS +  PRQ =  QRS ------- (Adjacent angles)   QRS = 90º Example 7: ABC is triangle and AD is the angle bisector. If ABC = 60º and ACB = 40º, find all the remaining angles. A 1 60º B Intermediate-Geometry 2 3 4 40º D C 28 of 96 ©2014 www.learningforkowledge.com/glg Angles marked 1 =  BAD 2 =  DAC 3 =  ADB 4 =  ADC  BAC +  ABC +  ACB = 180º ------ (Angles in a triangle)   BAC + 60º + 40º = 180º   BAC = 80º But AC bisects BAC   BAD +  DAC =  BAD +  BAD = 80º or 2  BAD = 80º or  BAD = 40º   BAD =  DAC = 40º (AD is the angle bisector)  ADC is the exterior angle to ∆ ABC   ADC =  ABD +  DAB   ADC = 60º + 40º  ADC = 100º But  ADB +  ADC = 180º --------- (Angles on a straight line)  ADB + 100º = 180º  ADB = 80º Example 8: If the base of an isosceles triangle extended on both sides show that the exterior angles so formed are also equal P S Q R T Ans: Consider a triangle PQR with base QR extended to S and T on both sides PQR = PRQ ----- (PQ = PR and PQR is Isosceles)  180º – P Q̂ R = 180º – P R̂ Q But 180º – P Q̂ R = P Q̂ S ---- (Angles on a straight line) And 180º – P R̂ Q = P R̂ T  P Q̂ S = P R̂ T Intermediate-Geometry 29 of 96 ©2014 www.learningforkowledge.com/glg Example 9: PQRS is a Quadrilateral and PR is a diagonal. Show that the sum of interior angles add up to 360º P S Q R Ans: Consider ∆PSR  PRS +  RPS +  PSR = 180º ------- (Angles in a triangle) and  QPR +  QRP +  PQR = 180º ------- (Angles in a triangle)   PRS +  RPS +  PSR = 180º ---- (1) (Adding equations 1 & 2) *  QRP +  QPR +  PQR = 180º ----- (2) (and combining adjacent angles)  QRS +  QPS +  PSR +  PQR= 360º ----- (Angles in a Quadrilateral) Example 10: ABC is a triangle with side BA extended to D, AC to E and CD to F. Show that the exterior angles formed add up to 360º D A C F B E  FBA +  ABC = 180º ---- (1) ------- (Angles on a straight line) +  BAC = 180º ---- (2) ---- (Angles on a straight line)  ECB +  ACB = 180º ---- (3) ---- (Angles on a straight line) Adding equations (1), (2) and (3)  FBA +  DAC +  ECB +  BAC +  ACB+  ABC = 180º + 180º + 180º   FBA +  DAC +  ECB + 180º = 180º + 180º + 180º Because  ABC +  BAC +  ACB are interior angles of a triangle and hence add up to 180º   FBA +  DAC +  ECB = 360º  DAC Intermediate-Geometry 30 of 96 ©2014 www.learningforkowledge.com/glg 1. State True or False 1. Three angles of a triangle add up to 180º 2. An obtuse angled triangle can have two interior obtuse angles. 3. Each one of the interior angles of an equilateral triangle measures 60º 4. ABC is a triangle with side AB measuring 12cms BC = 5cms and CA = 6cms 5. PQR is a triangle with P̂ = 80º, Q̂ = 60º and R̂ = 40º 6. In an equilateral triangle any two sides will be equal to each other. 7. In an Isosceles triangle sides opposite to equal angles are equal 8. A triangle is not polygon 9. In a triangle the side opposite to the largest angle is the smallest 10. In a triangle the side opposite Â is denoted as ‘a’ 2. In a triangle ABC, B̂ = 2 Â and Ĉ = 3 Â Find the values of all the angles. 3. In a right angled triangle PQR show that one of the angles is equal to the sum of remaining two angles. 4. ABC is scalene triangle BO and OC are bisectors of the interior angles Bˆ and Cˆ Bˆ  Cˆ meeting at point O. Prove that  180 o  . 2 In the triangle ABC show below, B Â C is 40º and the exterior angle CAD is 100º Find the remaining two angles. 5. A 40º 100º B 6. C D In the figure shown below AB = AC = CD and remaining interior angles.  BAC = 50º Find all the values of A 50º B Intermediate-Geometry C D 31 of 96 ©2014 www.learningforkowledge.com/glg 7. 8. In the triangle PQR, P̂ = 40º and Q̂ = 60º Identify the largest and smallest sides In an Isosceles ∆ABC side AB = AC show that the bisector of exterior angle CAD is parallel to base BC. D A B 9. C Calculate the values of angle x, y and z in the following figure x 110º y 10. z 130º In the regular pentagon shown below find the sum of all the interior angles B A C E Intermediate-Geometry D 32 of 96 ©2014 www.learningforkowledge.com/glg CONGRUENCY OR TRIANGLES Any two figures which are identical, that is having the same shape and size are said to be congruent. To prove congruence of triangles is an important step in learning plane geometry. Two figures are congruent, if one figure can be super imposed on another so that they cover each other exactly. Simplest of the figures a line segment is said to be congruent if they are of equal length. Two angles are said to be congruent if both of them have the same measure. Hence, two triangles are said to be congruent if all the three sides and three angles are equal. Therefore all parts namely sides and angles have to be equal to prove that the triangles are congruent. A B P C Q R Congruent triangles In the figure above  ABC is congruent with if AB  PQ, AC  PR, BC  QR,  BAC  QPR, ACB  PRQ and ABC  PQR However, a triangle can be constructed if we know two sides and an angle or two angles and a side or all the three sides. Only three out of six parts of a triangle need to be known because the remaining three parts can be calculated. But it is important to note that two triangles can not be said to be congruent if only the three angles are found to be equal. If only the angles are equal the two triangles, will have the same shape and not necessarily the same size. Some of the important properties and theorems regarding congruency of triangles have been detailed below. Property 9.1: If ABC is congruent to DEF then DEF is congruent to ABC . Congruency is represented as ‘  ’ in the Geometric expressions If ABC  DEF then DEF  ΔABC Also if ABC  DEF and DEF  ΔPQR then ABC  PQR Theorem: Prove that two triangles are congruent if two angles and the included side of one triangle is equal to the two angles and included side of the other. Intermediate-Geometry 33 of 96 ©2014 www.learningforkowledge.com/glg Hence H should coincide with A Therefore AB  DE and AC  DF Also Aˆ  Dˆ Hence ABC  DEF If AB  DE points A, B and C will exactly coincide with DEF Therefore all the angles and sides of both the triangles are equal  ABC  DEF Reconstruct DEF such that EF is superimposed and exactly coincides with BC and D is on the opposite side of BC join AD ABC is an isosceles triangle because (Given that sides ABC are equal to sides of DEF )  BAD  BDA - - - - - - - (ΔABD is isosceles) A D H B G C E F In the triangles ABC and DEF  B =  E,  C =  F and BC = EF If we superimpose ∆ABC over ∆DEF we have three possibilities (a) AB = DE or (b) AB < DE or (c) AB > DE If AB < DE then let ∆EGF be the super imposed position of ∆ABC In the fig  HCB =  ACB  EFD =  GFD which cannot be true if G is not the same as D. Hence point G should co inside with point D Therefore we have AB = DE and DF = AC Also  A +  B +  C = 180º =  E +  F +  G   A =  E and  B = F We have  C =  G Hence ∆ABC  ∆DEF The same argument can be extended to prove ∆ABC  ∆DEF if AB > DE also. Theorem: Two triangles are congruent if two sides and their included angles are equal. Intermediate-Geometry 34 of 96 ©2014 www.learningforkowledge.com/glg Consider two triangles ABC and DFE having sides AB = DE and BC = EF and  ABC =  DEF A D B C E F A (D) B (E) C (F) Draw line BC and reconstruct ∆ ABC such that vertex E coincides with vertex B  DEF =  DEF ------- (E coincides with point C) And line EF coincide with line BC, because BC = EF Because  ABC has the same measure of  DEF line DE will coincide with line DE. Also AB and DE are if equal length and as such vertex A will coincide with vertex D. Also line BC coincides with line EF, as per construction. Because BC = EF, F will coincide with C. Therefore line AC should coincide with line DF AC = DF Hence, AB = DE, BC = EF and AC = DF ∆ ABC  ∆ DEF Example 1: ABC is an isosceles triangle AD is perpendicular to BC and D lies on BC. Show that ∆ ADB and ∆ ADC are congruent. A B Intermediate-Geometry D C 35 of 96 ©2014 www.learningforkowledge.com/glg Ans: If AD is  BC  ADB =  ADC = 90º Also AB = AC ------ (∆ ABC is isosceles) And  ABC =  ACD ------ (∆ ABC is isosceles) Since corresponding two angles and a side are equal in triangle ADB and ADC ∆ADB  ∆ADC ------- (ASA Condition) Example 2: ABC and BCD are two equilateral triangles with a common base BC vertices A and D are joined by a line AD. AD intersects BC at point E. Show that AD and BC are a pair of perpendicular bisectors. A B E C D Ans: Consider ∆ABD AB = BD ------- (∆ABC  ∆DBC)   BAD =  BDA ------ (∆ABD is isosceles) Consider ∆ACD AC = CD ------ (∆ABC  ∆DBC)   CAD =  CDA ------ (∆CAD is isosceles) ∆ABC is equilateral  AB = BC = AC And ∆BCD is equilateral  AC = BD = CD Hence AB = AC = BC = BD = CD ------ (BC is common side) ∆ ABC is equilateral   ABC =  BAC =  BCA = 60º ∆ BDC is equilateral   BDC =  BCD =  DBC = 60º   ABC =  BAC =  BCA =  BDC =  BCD =  DBC = 60º Now consider triangles BAE and EDC BA = DC  ABC =  ABE =  ECD =  BCD = 60º  AEB =  DEC --------------- (vertically opposite angles)  ∆ BAC  ∆ EDC ------------- (ASA criteria)  AE = ED Intermediate-Geometry 36 of 96 ©2014 www.learningforkowledge.com/glg and BE = EC Hence AD and BC are a pair of bisectors intersecting of point E Consider triangles ABE and DBE  ABE =  DBE = 60º  BAE =  BDE - - - - - - - (∆ BAD is isosceles)  The remaining pair of corresponding angles  BEA =  BED But  BEA +  BED = 180º------------ (Angles on a line)  2  BEA = 180º  BEA = 90º =  BED Hence BE and ED or BC and AD are per perpendicular Example 3: ∆ABC and ∆PQR are congruent scalene triangles D and E are mid points of AB and AC. S and T are mid points of PQ and PR. Show that ∆ADE is congruent to ∆PST A D B P E S C T Q R AB = PQ -------------- (∆ABC  ∆PQR) But AB = 2AC and PQ = 2PR ----------- (D and S are mid points)  2AD = 2PR AD = PS AC = PR -------------- (∆ABC  ∆PQR) But AC = 2AE and PR = 2PT --------- (E and T are midpoints)  2AE = 2PT AE = PT Consider the included angle  DAE is the same as  BAC and  SPT is the same as  QPR But  BAC =  QPR -------------- (∆ABC  ∆PQR)   DAE =  SPT  Since two sides and their included angle are identical in ∆DAE and ∆SPT ∆DAE  ∆SPT Intermediate-Geometry 37 of 96 ©2014 www.learningforkowledge.com/glg Example 4: In two triangles ABC and DEF the exterior angles and interior angles at points B and E are equal. Also side BC is equal to side EF. Are the triangles congruent? Explain your answer. No, the exterior angle is equal to sum of the opposite angles. Hence the interior angles of ∆ABC may not be equal to those of ∆DEF. Example 5: ∆ABC and ∆PQR are two scalene triangles. BC is extended to D such that BC = CD and QR is extended to S such that QR = RS given ∆ABC is congruent to ∆PQR show that ∆ABD is also congruent to ∆PQS A B P C D Q R S BD = 2BC ----------- (Given BC = CD) And QS = 2QR ----------- (Given QR = RS) But BC = QR ------------- (∆ABC ∆PQR)  BD = QS Also AB = PQ -------------- (∆ABC  ∆PQR)  ABC =  ABD =  PQR =  PQS ------------- (∆ABC ∆PQR)  Two corresponding sides and included angles are equal in triangles ABD and PQS  ∆ABD  ∆PQS -------------- (SAS criteria) Example 6: ABC and PQR are two triangles in which (AB +BC) = 7cm and CA = 5cms. Lengths of all the sides are integers. Also PQ = 3cm and (PR + QR) = 9cm Check if ∆ABC, ∆PQR if they are both right angled triangles. Check (AB+BC) +CA = PQ + (QR +PR) = 7+5=3+9 Assuming the triangles are congruent (QR +PR) – (AB +BC) (AC + BC) – (AB +BC) = AC – AB = 9-7 =2 - - - - - - - - (Assumed AC = QR and BC = PR) But AC = 5cms  AB = 5-2 = 3cm and AB + BC = 7cm  BC = 4cm  AB = PQ = 3cm AC = PR = 5cm BC = QR = 4cm Intermediate-Geometry 38 of 96 ©2014 www.learningforkowledge.com/glg Triangles with sizes of sides mentioned above can be verified to be right angled triangles as given from Pythagoras theorem 32 +42 =52 Hence no other combination of side length is possible  ∆ABC  ∆PQR ----------- (S.S.S criteria) Example 7: ABCD is a square AC and BD are two diagonals. Prove that ∆ABC is congruent to ∆DCB Hence show that AC = BD A B D C AB = DC ------------------- (ABCD is a square) And AD = BC ---------------- (ABCD is a square)  BAC =  BCD ---------------- (ABCD is a square) Since two corresponding sides and included angles are equal ∆ADC  ∆BDC The remaining pair of corresponding sides AC = BD Example 8: ABC and DEF are two triangles. Congruent triangles as shown below. Name all the corresponding parts and find their value. A D 70º 5cm 4cm 70º 60º B Corresponding parts Side AB – side DE Side BC –side EF Side AC – side DE Intermediate-Geometry 50º 6cm ∆ ABC 4cms 6cms 5cms C E ∆ DEF 4cms 6cms 5cms 39 of 96 6cm F Remarks AB = DE = 4 Given DF = AC = 5 ©2014 www.learningforkowledge.com/glg Â  D̂ 70 o 70 o Given B̂  Ê 60 o 60 o B̂  Ê  60 o Ĉ  F̂ 50 o 50 o B̂  Ĉ  50 o Example 9: Show that two right angled triangles are congruent if the hypotenuse and a side are equal. Working: Let ABC and DEF be two right angled triangles A B A (D) C B C (E) F Given Hypotenuse AB = DF Side AC = DE Cˆ  Eˆ  90 o Reconstruct ∆DEF such that AC coincides with DE and F is on the other side of B Now  BCA =  ACF = 90º ------------- (  DEF and  ACF are same because DE coincides with AC)   BCA +  ACF =  BCF 90º + 90º = BCF or  BCF = 180º  B, C and F are on the same line Hence ABF is a triangle  AB = AF = (DF) ABF is isosceles   ABC =  AFC =  DFC Hence in triangle ABC and DEF we have two angles and the included side are equal  ∆ABC  ∆DEF Example 10: Two lines AB and CD bisect each other at S. Prove that ∆ADS is congruent to ∆CSB A D S C Intermediate-Geometry B 40 of 96 ©2014 www.learningforkowledge.com/glg S is the mid point of AB  AS = SB S is also the mid point of CD  CS = SD Also  ASD =  CSD ------------------ (Vertically opposite angles) Since two sides and the included angle are equal ∆ASD = ∆CSB Example 11: AB and CD are a pair of parallel lines cut by a transversal MN at points P and Q. Bisectors of  APQ and  CQP meet at R and bisectors of  BPQ and  DQP meet at S. Prove that ∆PRQ is congruent to ∆PSQ Working: AB‫׀׀‬CD and PQ is a transversal   APQ =  DQP ---------------- (alternate angles)  BPQ =  DQP ---------------- (alternate angles) M P A B R C S Q D N Since PR bisects  APQ and SQ bisects  DQP  APR = 1  APQ = 1  DQP =  PQS 2 2 2 2 since QR bisects  CQP and PS bisects  BPQ  PQR = 1  CQP = 1  BPQ = 1  SPQ 2 Also side PQ is common to both the triangles PQR and PQS Hence ∆PQR  ∆PQS -------------- (ASA criteria) EXERCISES 1. ABC and DEF are two isosceles triangles prove that ∆ABC  ∆DEF if any one side and an angle are equal. 2. AX and BY are two lines of equal length, drawn from the end points A and B of line AB. AX and BY are parallel and are on the opposite sides of line AB. XY is joined and intersects AB at Z. Show ∆AXZ  BYZ Intermediate-Geometry 41 of 96 ©2014 www.learningforkowledge.com/glg X Z A B Y 3. ABC and DEF are two congruent triangles P is the midpoint of BC and Q is the mid point of EF given BC = EF. Show that ∆APB  ∆DQE and ∆APC  ∆DQF. 4. ∆ABC is congruent to ∆PQR List all the criteria to be satisfied for the triangles to be congruent. List all the combination of corresponding sides and angles to be equal for the triangles to be congruent. 5. ABC is an isosceles triangles with  B =  C. CD and BE are perpendicular to lines BC and AB respectively. Prove that ∆BCD is congruent to ∆BCE and hence show that BE = CD. 6. ABC and DEF are two triangles having a perimeter of 12cms each. If the longest sides AB and DE are 5cms each and side BC is 4cms where as side DF is 3cms. Show that both the triangles are congruent. 7. ABC is an equilateral triangle D, E and F are mid points of AB, BC and CA. Show that ∆ADF  ∆BDE  ∆CEF  ∆DEF. A D F B C E 8. AB and CD, PQ and RS are two sets of perpendicular lines. AB meets CD at point E and PQ meets RS at T. Also AE = PT, and DE = RT. Show that two triangles ACD and PRS are congruent. Intermediate-Geometry 42 of 96 ©2014 www.learningforkowledge.com/glg C R E T A B P Q D 9. S ABC and DEF are two congruent triangles placed such that B, C, E and F are on the same line. Show that AB is parallel to DE and AC is parallel to EF. AD when joined will be parallel to BC or EF. A B Intermediate-Geometry D C E F 43 of 96 ©2014 www.learningforkowledge.com/glg LOCI AND CONCURRENCY Loci: Locus of a point is the path traced by it under some given mathematical conditions. Plural of locus is loci, simplest of the examples of a locus is a circle. A circle is traced by a moving point which moves in such a way that it i.e., at constant distance from a fixed point called the center of the circle. In terms of practical Geometry the fixed point is the sharp point of the compass and the moving point is the sharp end of the pencil. r O Locus of a line at equidistant from another line is a parallel line. Converse of a loci is to find the mathematical relationship given the locus and the reference point or a feature. There are many practical application for the study of loci. For example trace a specific path of a planet around the sun with the sum as the reference point. Study of such path helps in predicting their location at any time. Theorem: Locus of a point which is at an equal distance form two fixed points is the perpendicular bisector of the line joining the fixed points. Let C, D be the positions of moving point and A B the fixed points. C A D B Let D be the position of the moving point When it is on the line AB Then it is given that AD = DB and hence D is the mid point of line AB Consider another point on the locus away from D Then it is given AC = CB Hence  CAB =  CBA In triangles ACD and ADB  CAB =  CBA Intermediate-Geometry 44 of 96 ©2014 www.learningforkowledge.com/glg AC = DB and AD = DB  ∆ADC  ∆BDC -------------- (SAS criterion)   CDA =  CDB But  CDA +  CDB = 180º ---------------- (Angles on a line) 2  CDA = 180º  CDA = 90º =  CDB Since AD = AB, CD is the perpendicular bisector of AB and the locus is the perpendicular bisector Theorem: Prove that the locus of a point equidistant from two lines meeting at a point is angle bisector of the angle formed by the two lines A P B C Let BP is the locus of the point Then AP = PC -------------- (Given) Since AP and CP are the shortest distance form P to line AB and AC  BAP = 90º =  PCB ------------ (perpendicular is the shortest distance form a point to a line)   ABP =  PBC, Hence PB is the bisector  ABC  ∆BAP and ∆BCP are right angled triangles form Pythagoras theorem we have BP2 = BA2 + AP2 or AB  BP 2 - PA 2 also BP 2  BC 2  BP 2  BC  BP 2 - PB 2  PA  PB AB  BP 2 - PA 2  BP 2 - PB 2  BC Since three sides ABP is equal to the three sides of BPC, ABP  BDC  ABP  BPC hence BP is the bisector of ABC Example 1: What is the locus of a point P always at a distance of 1.5cms from a fixed point O. Draw the locus. O 1.5cms P Intermediate-Geometry 45 of 96 ©2014 www.learningforkowledge.com/glg Working: The locus P will be a circle of radius 1.5cms and center O- Figure 10.1.1 shows the locus of P and a point on the locus P and the center O Example 2: A and B are two fixed points on a line CD. P and Q are two points on the loci moving such that  PAB = 60º =  QBD. Show that the Loci PA and QB are parallel. P C Q A B D Working:  PAB =  QBD = 60º --------------- (Given) But  PA and  QBD are corresponding angles  PA‫׀׀‬QB ---------------- (corresponding angles axiom) This axiom is true of all length of PA or QB which represent the locus of P and Q Example 3: Show that the locus of the vertices formed by a moving point P with a fixed base QR of isosceles triangles is the perpendicular bisector of QR, lets be the point of the locus on QR. PQ = QR  PQR =  PRQ or  ∆PQS  ∆PRS  QS = RS Also  PSQ =  PSR = 90o Hence the locus is the perpendicular bisector P1 P2 Q R S P3 Working: Because P1QR is isosceles PQ = QR Intermediate-Geometry 46 of 96 ©2014 www.learningforkowledge.com/glg  P1QR =  P1RQ or  P1QS =  P1RS Also P2QR is isosceles  P2 QS  P2 RS  P1QS - P2 QS   P1 RS - P2 RS  PQP2  P1 RP2 - - - - - - - - - (Adjacent angles) Also in tringle s P1QP2 and P1 RP2 we have P1Q  P1 R P1QP2  P1 RP2 P1 P2 is common  P1QP2  P1 RP2 - - - - - - - - - - - - (SAS criteria)  P1 P2 Q  P1 R 2 R But P1 P2 Q  QP2 S  P1 P2 R  RP2 S  180 o (Angles on a line)  QP2 S  RP2 S In trianlges P2 QS and P2 RS P2 Q  P2 R QP2 S  RP2 S P2 S is common  P2 QS  P2 RS  P2 SQ  P2 SR But P2 SQ  P2 SR  180 o - - - - - - - - - - - (Adjacent angles) P2 SQ  90 o  P2 SR and SQ  SR hence P2S is the perpendicular bisector. This result can be shown for any point P1, P3 or any other vertex of the isosceles triangle formed by base QR. Hence the perpendicular bisector of QR is the locus. Example 4: Find the locus of the mid points of the parallel chords of a circle. A B C P2 P1 E D F O Intermediate-Geometry 47 of 96 ©2014 www.learningforkowledge.com/glg Working: Let EF be the diameter (longest chord) of a circle passing through the center O let AB be a chord parallel to EF with mid point P2 and CD be another chord parallel to EF with mid point P1. Consider triangles OP2B and OP2A OA = OB ----------- (radii of a circle) AP2 = P2B ---------- (Given P2 as mid point of AB) OP2 is common  OP2   B OP2 A  OP2 A  OP2 B But OP2 A  OP2 B  180 o - - - - - - - - - - - (Angles on a line)  OP2 A  OP2 B  90 o Since AB‫ ׀׀‬CD and OP2 is a transversal cutting CD at P1  CP1O =  DP1O = 90o - - - - - - - - - - - - (Corresponding angles) Consider triangles CDO and DPO OC = OD - - - - - - - - - - - - - (radii of a circle)  CP1O =  DP1O P1O is common  CPO  DPO  CP1  DP1 Hence P1 is the mid point of CD and P1O is perpendicular to CD and P1 is a point on the line OP2. Hence the locus of midpoints of chords parallel is their perpendicular bisector which is also a diameter of the circle. Example 5. Find the locus of points which are equidistant from three given points. These points are not in a line A R P B C S A, B and C are three points AB and BC are joined. Let R be the mid point of AB and S the mid point of BC. Locus of a point equidistant form AB is perpendicular bisector BP. Locus of a point equidistant form BC is its perpendicular bisector PS. P the point of intersection is the only point that is equidistant form A, B and C. Hence ‘P’ is the locus. Example 6: BC is a fixed line AC is a line of variable length and its length varies such that AC2 = BC2 + BA2  BAC and  CAB are variable. Show that  ABC is a right angle and the locus of point A(P) is perpendicular to BC. Intermediate-Geometry 48 of 96 ©2014 www.learningforkowledge.com/glg A1 (P)A B C Working: From Pythagoras theorem we have AC2 = AB2 + BC2 also given AC2 = AB2 + BC2 AC is hypotenuse of the right angled triangle.  AB is  BC Let A another position of the moving point P Again A1C2 = A1B2 + BC2  A1BC is right angled triangle AC is the hypotenuse A1B  BC Hence the locus P is perpendicular to BC Example 7:  ABC is equal to 60o moving point P has a locus such that it is equidistant from AB and BC line DE is drawn perpendicular to BP at B. Show that at any position of P ∆DBE is equilateral. A1 A Q o P1 P 60 B C1 R C Working: Since P in any position on the locus BP must be the angular bisector or ABC ABP  PBC Also BP  AC - - - - - - - - - - - - (given)  APB  90 o  BPC and BP is common  ABP  PBC - - - - - - - - - (SAS criterion)  AP  PC and BAC  CAB But ABC  BAC  BCA  180 o - - - - - - - - - (Angles in tringle ) ABC  2BAC  180 o 60 o  2BAC  180 o or BAC  60 o  BCA  60 o Intermediate-Geometry 49 of 96 ©2014 www.learningforkowledge.com/glg  ABC is equilateral triangle In the new position of ‘P’ also we can prove using the same steps that A ‘BC’ is equilateral. Hence any line perpendicular to the locus of P will form an equilateral triangle with line AB and BC. Example 8: O is the center of a circle which is the locus of a point P such that OP = 2cms.Find the locus of a point which is 1cm away from the circle. Find the total length of the locus of Q. 2 O P 1 Q Working: Let P be a point on the circle with O as center and radius of 2cms. Extend OP to Q such that OQ is 2 + 1 = 3cms. If a circle is drawn with O as center and a radius of 3cms. It will represent the locus of Q Since OQ – OP = 3 -2 = 1cm Example 9: In the ∆ABC lines AB and AC are extended to D and E. Loci of angles CBD and BCE meet at G. Show that line BG is the locus of a point equidistant from AB and AC. A B C D E G Working: BG is locus of a point G that is equidistant from lines BD and BC similarly G is a point on the locus of a point which is equidistant form BC and CE. Hence G is equidistant form BD and CE BD and CE are extent ion of lines AB and AC. Example 10: AB and CD are a pair of parallel line. Show that the locus of a point P which is the mid point of any transversal across the parallel lines is a line equidistant form AB and CD. Intermediate-Geometry 50 of 96 ©2014 www.learningforkowledge.com/glg E G A B X p Q Y C S F H Working: Let EF and GH be any two transversals to AB and CD. Let B and Q be their mid points. RS is a line through P which is perpendicular to both AB and CD. Consider triangles PER and PSF PE = PF - - - - - - - - - - - - - - (given)  EPF =  SPE - - - - - - - - - - - - (Vertically opposite angles)  PRE =  PSF = 90o  ∆PER  ∆PSF  PR = PS and RS  AB and CD Hence P is equidistant from AB and CD similarly we can prove Q as equidistant form AB and CD Since the distance between AB and CD is constant and PQ is half this distance away form AB and CD we have AB‫ ׀׀‬PQ‫׀׀‬CD. EXERCISES 1. 2. 3. 4. 5. 6. 7. 8. Define the term Locus. What is Loci? Select the best answer. What is locus of a point which is equidistant form three points not in a line? a) A point b) A circle c) A line d) A pair of lines Find the locus of centers of all circles packing through two given points A and B show that the locus is the perpendicular bisector of AB What is the loci of the vertices of a right angled triangle given that the vertex forming the right angle remains unchanged? Find the locus of mid points of equal chords of a circle. ABC is an isosceles triangle given Bˆ  Cˆ , BAD is an exterior angle at A formed by extending CA. Show that the locus of a point equidistant from BA and AD is parallel to BC. Find the loci of points P and Q which are equidistant from adjacent points A and B or B and C given A, B and C are on a line. Q and R are centers of two sets of concentric circles. Show that the locus of the point of intersection ‘P’ of any two circles with the same radius is the perpendicular bisector of line AB. Intermediate-Geometry 51 of 96 ©2014 www.learningforkowledge.com/glg 9. 10. P and Q are centers of circle of r1 and r2. Given that circle with center Q just touches the circle with center P, find the locus of P such that the two circles just touch each other. AB and CD are two intersecting lines, meeting at O. Prove that loci of points P and Q which are equidistant form the lines AB and CD, are mutually perpendicular. Concurrency When three or more lines pass through the same point they are said to be concurrent and the common point is called the point of concurrency of the given lines. There are several properties of triangles which are associated with concurrency. Theorem 1: Bisectors of interior angles of a triangle meet at a point. A F O E B D C Working: consider a triangle ABC let BO and CO be the angle bisectors pr  B and  C meeting at point O. Let OF and OE be the perpendiculars to AB and BC also draw OD  BC. If OA can be proved to be the bisector of  BAC then all the angle bisectors are concurrent. Consider triangles ODB and OBF OB is common  OBD =  OBF ---------- (Construction)  ODB =  OFB = 90o  ∆ODB  ∆OBF Hence OD = OF Using the same procedure OD = OE . Therefore O is equidistant form AB and AC. Hence OA dissects angle A. Therefore O is the point of concurrency of all three angular bisectors. This point is also the center of the in circle which is a circle with all the three sides as tangent. Theorem 2: In a triangle the perpendicular bisectors of all the three sides are concurrent. A D F O B Intermediate-Geometry E C 52 of 96 ©2014 www.learningforkowledge.com/glg Consider a ∆ ABC in which OD and OF are perpendiculars to AB and AC and O is the point of intersection of the perpendicular bisectors. The perpendicular to BC from point O, OE should also bisect OE to prove that all the perpendicular bisectors are concurrent. Join OB, OA and OC consider triangles OBD and OAD  OD is common DA = DB and  ODB = 90o =  ODA  ∆OBD = ∆OAD  OB = OA Similarly we can show OA = OC by considering triangles OAF and OFC  OB = OA = OC Or OB = OC Consider triangles OEB and OEC OE is common  OEB =  OEC = 90o OB = OC ------------ (SAS Criterion) Hence BE = CE and  BEO =  CEO But  BEO +  CEO = 180º ---------- (Angles on a line)   BEO =  CEO = 90º Hence OE is the perpendicular bisector of BC Therefore O is the point of concurrency of the three perpendicular bisectors Also OB = OA = OC Therefore we can draw a circle with a radius equal to OA or OB or OC and center O the circle will pass through all the three vertices. This circle is called the circum circle and O is the circum center. Theorem 3: Altitude of a triangle is the perpendicular to a side of a triangle from a vertex opposite to it. The altitudes of a triangle are concurrent. D A E H I O B C G F Intermediate-Geometry 53 of 96 ©2014 www.learningforkowledge.com/glg Construction: Draw a triangle ABC and Altitudes AG, BH and CI. Let O be the point of intersection of BH and CI. Draw ED‫ ׀׀‬BC, DF‫ ׀׀‬AC and EF‫ ׀׀‬AD. To prove the concurrency AG should also pass through O. Consider triangle ABC and ACE  BAC =  ACE --------- (BA‫ ׀׀‬EC, Alternate angles)  BCA =  CAE --------- (BA‫ ׀׀‬EC, Alternate angles) enough AC is common Hence, ∆BAC  ∆CAE  AE = BC Consider triangles BAD and BAC  DBA =  BAC (AC‫ ׀׀‬DB Alternate angles)  DAB =  ABC (AC‫ ׀׀‬DB Alternate angles) AB is common Hence ∆BAD  ∆BAC  DA = BC  AE = BC = DA  A is the mid point of DE Also DE ‫ ׀׀‬BC Hence AG is also perpendicular to DE AG is also perpendicular to DE  AG is the perpendicular bisector of the side DE of ∆DEF Similarly we can prove that BH is the perpendicular bisector of DF and CI is the perpendicular bisector of CI. Theorem states that perpendicular bisectors of a triangle sides meet at a point. Therefore the three altitudes AG, BH and CI meet at the point O. This point is called as the orthocenter of a triangle. Theorem 4: A line drawn form the vertex of a triangle to the mid point of the side opposite to it is called the median of the triangle. The medians of a triangle are also concurrent. A property of the point of intersection is that it divides the median in the ratio 2:1. Construct a triangle ABC in which the medians BE and CF intersect at a point G. Extend AG to intersect BC at D and further to point K such that AG = GK . This theorem can be proved by showing AS BG CG 2 BD = DC and    GD GE GF 1 Proof: A F B G D E C K Intermediate-Geometry 54 of 96 ©2014 www.learningforkowledge.com/glg Consider triangle AFG and ABK ∆AGK is similar to ∆ABK since all the three angles have the same value.  FG‫ ׀׀‬GK ---------- (Corresponding angles are equal) and FG = ½ BK Similarly we can prove GE‫׀׀‬KC And GE = ½ KC Consider triangles BDK and GDC since all the three angles are equal and because BG is parallel to KC GD BD We have  DK DC Also because GC is parallel to BK  BC and GK intersect at D (Detailed proof for this is provided in the unit on parallelograms) Hence BD = DC AD is a median and all three medians of ∆ABC meet at G 1 Becuase GD  DK  GK 2 1 But GK  1 AG - - - - - - - - - - - - - (Given GK  AG) 2 AG 2 or AG  2GD or  GD 1 AG BG CG 2 similarly we can prove    GD GE GF 1 Example 1: ∆ABC is an equilateral triangle show that the in center, circum center, the centroid are all the same point O. Construction: Draw an equilateral triangle ABC. Mark D, E and F as the mid points of AB, BC and CD. Let O be the points of intersection of the medians. Now deduce O is also the in center, circum center and the orthocenter A F E O B D C Working: Since D is the mid point of BC BD = DC Also  ABD =  ACD = 60o ----------- (Given ∆ABC is equilateral) Intermediate-Geometry 55 of 96 ©2014 www.learningforkowledge.com/glg AB = AC -------------- (Given ABC is equilateral)  ∆ABD  ∆ACD Hence AD is also the angle bisector of angle A Similarly BE and CF are also angle Bisectors or  B and  C Hence O is the point of intersection of the angle or the in center of the triangle. Since ∆ABD  ∆ACD  ADB =  ADC But  ADB +  ADC = 180o ------------ (Angles in a line)   ADB = 90o =  ADC Hence AD is  BC Similarly CF  AB and BE  AC Hence O is the point of intersection of perpendiculars from the mid points of sides BC, AB and AC. There O is the circum center. The perpendiculars from the vertices of the equilateral triangle ABC bisect the sides ∆ABD  ACD And  ADB =  ADC = 90o And BD = DC Hence O is the point of intersection of the three altitudes or the or two center. Example 2: Show that in a right angled triangle is orthocenter is concurrent to a vertex. A B C Consider a right angled triangle ABC, with vertex B or the right angle.  Side AB will be an altitude since AB  BC Side BC will be an altitude since BC  AB Let BD be the perpendicular to AC Hence B becomes the point of intersection of all the three latitudes Example 3: D, E and F are the mid points of sides AB, BC and CA of ∆ABC. If D, E and F ate the mid points of sides. AB, BC and CA of a ABC. IF O is the point of intersection of AD, AE and AF. Show that it is also the point of intersection of the medians of the ∆DEF. Assume DE‫ ׀׀‬BC and PE =½ BC, DE =½ AB and DE = ½AB FD‫ ׀׀‬AC and FD‫ ½ ׀׀‬AC A F P R Intermediate-Geometry B E O D 56 of 96 C ©2014 www.learningforkowledge.com/glg Construction: Draw a triangle ABC with D, E and F as mid points of sides AB, BC and CA. join AD, CF and BE. Let O be the point of intersection. P, Q and R be the points where DE, EF and DF intersect with the medians. Given DE ‫׀׀‬AB and DE = FA =½ AB Similarly AE = FD And FE is common  ∆AFE  ∆FDE Now consider triangles AFP and PDF We have AF = ED -------- (Given)  FAP =  PDE ---------- (Alternative angles)  FPA =  DPE ---------- (Opposite angles)  ∆FAP  ∆ PDE Hence FP = PE And AP = PD  P is the mid point of FE also lies on AD similarly we can prove Q and R as mid points of FD and EF. Since B, Q and O, C, R, O are also collinear O must be the point of intersection of the median of ∆DEF Example 4: point O. a. b. c. PQR is a triangle with PS, QT and RU as its medians intersecting at the If QO = 10cm find QT If RO = 7cmx find OU If PS = 15.6cm find QD P U Q O T S R Answer : QO 2 3   QT   QO  15cms QT 3 2 RO 2 1 1 b)   OU   RO   7  3.5cms OU 1 2 2 OP 2 2 2 c)   OP   PS   15.6  10.4cms PS 3 3 3 a) Example 5: If O is the orthocenter of ∆ABC show that A is the orthocenter of ∆ABC Construct a ∆ABC and draw perpendicular to the opposite sides from A, B and C. Mark the point of intersection O. Intermediate-Geometry 57 of 96 ©2014 www.learningforkowledge.com/glg A E F O B C D Consider ∆OBC OD  BC - - - - - - - - - - (Given AD  BC and O is a point on AD) OB  CE - - - - - - - - - - (Given BE  CE and O is a point on BE) OC  BF - - - - - - - - - - (Given CE  BF and O is a point on CF) Hence AO  BC AE  OB AF  OC Therefore A is the point of intersection of perpendiculars opposite sides from the vertices of triangle OBC. Hence A is the orthocenter of triangle ABC. EXERCISES 1. 2. 3. 4. 5. 6. 7. Explain with sketches the following terms a) Orthocenter b) Circum center c) Incenter d) Centroid If the two medians are equal show that the triangle is isosceles and that the centroid lies on the perpendicular to the remaining side, which is not equal to any other side. Prove that the angle bisectors of a triangle pass through the same point (Theorem 1). Perpendicular bisectors of two sides AB and Ac meet at a point O. Show that the perpendical OD to line BC, bisects the line BC at D. If the medians of a triangle are equal show that the triangle is equilateral. D, E and F are mid points of BC, BA and AC of ABC prove that the orthocenter of ∆DEF is the circumcenter of ∆ABC. Given sum of two sides of a triangle is always more than the third side show that sum of any two medians is greater than the third median. Intermediate-Geometry 58 of 96 ©2014 www.learningforkowledge.com/glg QUADRILATIERALS AND PARALLELOGRAMS QUADRILATERALS A quadrilateral is a four sided figure. It has four vertices and four interior angles. The figure below shows a quadrilateral and its features. A B C D Sides: AB, BC, CD and AD Vertices: A, B, C and D Interior Angles: Â, B̂, Ĉ and D̂ Diagonals: AC and BD Quadrilaterals are of different types A quadrilateral with two sides parallels is called Trapezium A quadrilateral in which two adjacent sides are equal A quadrilateral in which opposite sides are parallel is called as a parallelogram Intermediate-Geometry 59 of 96 ©2014 www.learningforkowledge.com/glg A quadrilateral parallelogram with each of the interior angles equal to 90º is called a Rectangle A square a rectangle with equal sides interior angles of square is equal to 90º A Rhombus is a parallelogram with opposite sides equal. Interior angles of Rhombus is not equal to 90º Theorem 1: Sum of interior angle of a quadrilateral is 360º Construction: Construct a quadrilateral ABCD join the Diagonal AC and BD A B D C Proof: consider ∆ADB  D Â B +  ADB +  ABD = 180º ----------- (Angles in a triangle) consider ∆BCD Intermediate-Geometry 60 of 96 ©2014 www.learningforkowledge.com/glg  BCD +  BDC +  DBC = 180º ----------- (Angles in a triangle) Therefore  D Â B +  ADB +  ABD =180º +  (BCD +  BDC +  DBC) = 180º  DAB +  BCD + (  ADB +  BDC) + (  ABD +  DBC) = 360º But  ADB +  BDC =  ADC ---------------- (Adjacent angles)  ABD +  DBC =  ABC -------------- (Adjacent angles)   DAB +  BCD +  ABD +  DBC = 360º Hence the interior angles of a quadrilateral add up to 360º Example 1: Four angles of a quadrilateral are 110º 70º 60º and 120º name the type of quadrilateral given that the angles are in the order Â, B̂, Ĉ and D̂ Construction: construct a quadrilateral ABCD with Â = 110º, B̂ = 70º, Ĉ = 60º and D̂ = 120º Select convenient length for sides AB and BC A B D C  A +  B = 180º = (110º + 70º)  AD ‫ ׀׀‬BC ------------- (cointerior angles add up to 180º) Similarly  C +  D = 180º Confirms AD‫ ׀׀‬BC However  A +  D = 110º + 120º = 230º Hence  A +  D  180º  AB is not parallel to CD Therefore two sides of the quadrilateral are parallel and it is a Trapezium. Example 2: In a kite shaped quadrilateral show that the opposite angles formed by pairs of unequal sides are equal Intermediate-Geometry 61 of 96 ©2014 www.learningforkowledge.com/glg B A Construction: Draw a kite ABCD such that AB = BC and AD = DC Join AC To show  BAC =  BCD C D Working: Consider ∆ABC We have AB = BC   BAC =  BCA ------------------- (∆ABC is isosceles) Consider ∆ACD We have AD = CD   DAC =  DCA ------------------ (∆ACD is isosceles) Therefore  BAC +  DAC +  BCA +  DCA But  BAC +  DAC =  BAD ------------ (Adjacent angles) And  BCA +  DCA =  BCD ----------- (Adjacent angles) Hence  BAD =  BCD Example 3: In a quadrilateral PQRS the opposite angles P̂ and R̂ are equal. If angles Q = 70º and S = 50º find all the angles of quadrilateral. Given  P +  Q +  R +  S = 360º   P + 70º +  R + 50º = 360º Or  P +  R = 240º But  P =  R ------------ (Given)  2  P = 240º  P = 120º =  R Hence  P = 120º,  Q = 70º,  R = 120º and  S = 50º Example 4: Show that the exterior angles of a quadrilateral also add up to 360º H Construction: Draw a quadrilateral ABCD Extend side AB to E, BC to F, CD to G and GA to H. To prove:  HAB +  CBE +  FCD +  ADG = 360º A B E G D Intermediate-Geometry C 62 of 96 ©2014 www.learningforkowledge.com/glg Working:  HAB +  BAD = 180º ------------ (Angles on a straight line)  CBE +  ABC = 180º ------------ (Angles on a straight line)  FCD +  BCD = 180º ------------ (Angles on a straight line)  ADG +  ADC = 180º ------------ (Angles on a straight line)   HAB +  CBE +  FCD +  ADG +  BAD +  ABC +  BCD +  ADC + 720º But  BAD +  ABC +  BCD +  ADC = 360º ---------- (Angles in a quadrilateral)   HAB +  CBE +  FCD +  ADG + 360º = 720º   HAB +  CBE +  FCD +  ADG = 360º Hence exterior angles add up to 360º Example 5: Any pair of adjacent angles in a quadrilateral add up to 180o. Find all the four angles of this quadrilateral ABCD given  A = 80o Given  A +  B = 180o  C +  B = 180o A – C = 0 A = C  A = 80o =  C  A +  B = 180o   B = 180o – 80o = 100 o  C +  D = 180o   D = 180o -80o = 100o   A = 80o,  B = 100o,  C = 80o,  D = 100º EXERCISES 1. 2. 3. Draw examples of following types of quadrilaterals. a. Trapezium b. Kite c. Rectangle d. Rhombus Three angles of a quadrilateral are 100o, 90o and 80o find the remaining angle. Oppositeangles of a quadrilateral (B̂  D̂)  (Â  Ĉ) upto180o. One pair of adjacent  4. 5. 6. 7.  angles (Ĉ  D̂) add upto170 and 150 . Find the angles Â, B̂, Ĉ and D̂. Four angles of a quadrilateral measure x o, 2x o, 3x o and 4x o. Find the values of Show that the diagonal QS bisects angles Q̂ and Ŝ . Two angles of a quadrilateral are equal to 70o. Remaining two angles are also equal. Find their values. Sides AB, BC, CD and DA are extended to points E, F, G and H. If three of the exterior angles are 100o, 90o and 80o find all the interior angles of the quadrilateral. A Trapezium ABCD shown below is divided to form two quadrilaterals some of the angles have been marked. Find all the remaining angles. o Intermediate-Geometry o 63 of 96 ©2014 www.learningforkowledge.com/glg A 100o 5 E 1 8. 2 85o D B 3 60o 4 F C PQRS is a quadrilateral. Bisectors of angles  P and  Q meet at T. Show that  PRS +  SRP = 2  PTQ R S T Q P 9. ABCD is an isosceles Trapezium because  D =  C and AB ‫ ׀׀‬CD show that AD = BC PARALLELOGRAMS Theorem 1: In a parallelogram show that opposite sides and angles are equal A B C D Construction: Draw a parallelogram with AB‫׀׀‬CD and AC‫ ׀׀‬BD join Proof:  DCB =  ABC ------------ (Given AB‫׀׀‬CD) Also  ACB =  CBD ------------- (Given AC‫׀׀‬BD) BC is common  ABC  ΔACD ---------- (ASA criteria) Intermediate-Geometry 64 of 96 ©2014 www.learningforkowledge.com/glg Hence AB = CD And AC = BD Also  BAC =  BDC Similarly we can prove  ACB =  ABD Converse of this theorem is also true if the opposite sides and angles are equal quadrilateral is a parallelogram Theorem 2: Diagonals of parallelogram bisect each other. A B O D C Construction: Draw a parallelogram ABCD with sides AB‫ ׀׀‬CD and AD‫ ׀׀‬BC join the diagonals AB and CD. Let O be their point of intersection. Proof: consider triangle OAB and OCD  OAB =  OCB ------------- (Alternative angles AB‫ ׀׀‬CD)  OBA =  ODC ------------- (Alternative angles AB‫ ׀׀‬CD) AB = CD ------------ (Given ABCD is a parellogram)  ∆AOB  ∆COD ------------ (ASA criteria) Hence OA = OC And OB = OD ------- (corresponding parts in corresponding triangles) Converse of this theorem is also true. If the diagonals of a quadrilateral bisect each other then it is parallelogram If the parallelogram is a rectangle the diagonals bisect each other and the diagonals are also equal in length. Theorem 3: In a Rhombus show that the diagonal are perpendicular to each other. A O B C D Intermediate-Geometry 65 of 96 ©2014 www.learningforkowledge.com/glg Construction: construct a Rhombus such that AB = BD = DC = CA. Join the diagonals AD and BC Let them intersect at point O. Prove AO  BC Consider the two triangles AOB and AOC AB = AC ---------------- (Given ABCD is a Rhombus) BO = OC ---------------- (Because ABCD is a parallelogram its diagonals bisect each other) AO is common  ΔAOB  ΔAOC  AOB  AOC But  AOB +  AOC = 180o 2  AOB = 180o or  AOB = 90o =  AOC Hence AO  BC The converse of this theorem is also true therefore if the diagonals of a quadrilateral bisect at right angles, it will be a Rhombus. In the case of a square which can be considered as a Rhombus with equal sides, the diagonals are equal and bisect each other at right angles. Example 1: State any four important properties of a parallelogram Answer: a. Opposite sides of a parallelogram are equal b. Opposite angles of a parallelogram are equal c. Diagonals of a parallelogram bisect each other d. Any two adjacent angles of a parallelogram add upto 180o Example 2: Given one of the angles Â = 70º in a parallelogram ABCD find the remaining three angles. Working: Adjacent angle B̂ = 180º - Â ------- (AB ‫ ׀׀‬CD and BC is a transversal)  B̂ =180º -70º = 110º Ĉ  Â = 70º --------- ABCD is a parallelogram and Ĉ is opposite to Â ) Also D̂  B̂ = 110º ------------- (ABCD is a parallelogram and Ĉ is opposite to Â ) Example 3: Shown below is rectangle ABCD with diagonals AC and BD intersecting at O if CO measures 3cm find the length of diagonal BD A B 3cm O D Intermediate-Geometry C 66 of 96 ©2014 www.learningforkowledge.com/glg Working: Consider triangle ABCD and ACD We have AD = BC ------------------- (ABCD is a rectangle and AD and BC are opposite sides) DC is common  ADC =  BCD = 90º ------------- (ABCD is a triangle)  ∆AOC  ∆BCD Hence AC = BD Also AC and BD bisect at O Hence AC = BD = 2CO Or AC = BD 2×3 = 6cms Example 4: PQRS is a Rhombus if  PQR = 60º show that ∆PQS is equilateral Q P R S Consider ∆PQR  PQR = 180º -  QPR –  QRP ------------- (Angles is a triangle) But  QPR =  QRP ------------- (because QP = QR)   PQR = 180º – 2  QRP  2  QRP = 180º –  PQR = 180º- 60º =120º   QRP = 60º =  QPR PQR is an equilateral triangle since all the three interior angles are equal to 60o Example 5: ABCD is a parallelogram E and F are the mid points of parallel sides AB and CD. Show that EC‫ ׀׀‬AF and AECF is a parallelogram. A E D B F C Construction: Draw a parallelogram ABCD with AB‫ ׀׀‬CD and AD‫ ׀׀‬BC. Mark mid points of AB and CD as E and F. Join AF and EC Intermediate-Geometry 67 of 96 ©2014 www.learningforkowledge.com/glg Working: Consider triangles ADF and BEC We have  ADF =  EDB ---------- (opposite angles of a parallelogram) AD = BC --------------- (opposite sides of a parallelogram) 1 1 DF = DC = AE = (AB) ------------ (AB and DC opposite sides of a parallelogram) 2 2  ∆ADF  ∆ADB --------- (SAS criteria)  AF = EC And  AC = FC -------- (E and F are mid points of AB and DC)  AEFC is a parallelogram since opposite sides are equal Example 6: Two angles of a parallelogram are x + 30º and x - 30º. Find the value of x and all the angles of a parallelogram. A B x + 30º x – 30º D C Construction: Draw a parallelogram ABCD. Let  A = x + 30º and  D = x – 30º Working: Since AB‫ ׀׀‬DC We have  A +  D = 180º ------------- (cointerior angles)  x + 30º + x -30º = 180º 2x = 180º x = 90º   C =  A = x + 30º = 60º + 30º  C = 90º  B =  D = x – 30º = 60º – 30º  B = 30º  A = x + 30º = 90º + 30º = 120º  D = x – 30º = 90º – 30º = 60º Example 7: ABCD is a square AC and BD are two diagonals intersecting at point O. Prove that  OAB =  OBA = 45º A B O D Intermediate-Geometry C 68 of 96 ©2014 www.learningforkowledge.com/glg Construction: Draw a square ABCD join diagonals AC and BD let O be their point of intersection. Working: Consider triangle AOD and AOB AD = AB -------------- (sides of a square) DO = OB ------------- (ABCD is also a parallelogram and diagonals bisect each other) AO is common  ∆AOD  ∆DOC (similarly ∆AOB  ∆AOD  ∆ODC  ∆OBC)   AOD =  AOB But  AOD +  AOB = 180º Or 2  AOB = 180º   AOB =  AOD = 90º Consider ∆AOB  OAB +  OBA +  AOB = 180º ---------- (Angles in a triangle)   OAB +  OBA = 90º ------------ (  AOB = 90º) But OA = OB -------------- (∆AOB  ∆AOD  ∆AOC)  AOB is isosceles Hence  OAB =  OBA   OAB +  OBA = 90º Or 2  OAB = 90º  OAB = 45º Hence  OAB =  OBA = 45º Example 8: ABCD is a rectangle OA and OB are angle bisectors or A and B. Show that  AOB = 90º A B O D C Working:  DAB =  CBA = 90º --------------- (ABCD is a rectangle) 1 1  OAB =  DAB =  CBA =  OBA --------------- (OA and OB are bisectors) 2 2   OAB +  OBA =  DAB = 90º But  OAB  OBA +  AOB = 180º   AOB =180º – 90º = 90º Example 9: Following figure shows a parallelogram ABCD with a diagonal AC. Find all the angles of the parallelogram given  CAB = 40º and  CAD = 80º Intermediate-Geometry 69 of 96 ©2014 www.learningforkowledge.com/glg D C 80º 40º A B Given  CAB = 40º and  CAD = 80º We have  CAB +  CAD = 40º + 80º = 120º  DAC = 120º  ADC +  DAC = 180º ------------ (cointerior angles AD‫ ׀׀‬BC)  ADC + 120º = 180º   ADC = 60º  DCB = 120º =  DAC ----------- (Opposite angles in a parallelogram) And  ADC = 60º =  ABC --------- (Opposite angles in a parallelogram) Example 10: ABCD is parallelogram. EF is a line joining mid points of sides AD and BC. GH is a line joining any two points on lines AB and CD. O is the point of intersection of EF and GH show that DG = OH Construction: Draw a parallelogram ABCD mark E and F as mid points of AD and BC. Mark any point G on AB and H on DC join EF and GH. Mark O as the point of intersection. Draw IJ ‫׀׀‬AD and BC passing through O. A I G B O E F D H J C Working: Consider Triangles ABE and BEF 1 1 AD = AE = BF = BC ---------- (Given E and F are mid points) 2 2 BE is common  ∆ABE  ∆EBF Hence AD = EF Also AB‫ ׀׀‬EF‫ ׀׀‬DC Consider the quadrilateral AIOE Intermediate-Geometry 70 of 96 ©2014 www.learningforkowledge.com/glg AE‫׀׀‬IO -------- (Given) AI EO ---------------- (AB‫ ׀׀‬EF and I and O are points on AB and EF)  AIOE is a parallelogram 1 1  AE = IO = AD = BC 2 2 Similarly we can show 1 1 ED = OJ = AD = BC 2 2  IO = OJ Consider triangle IOG and HOJ IO = OJ  OIJ =  OJH ---------- (opposite angles)  OHJ =  OGI ----------- (Alternate angles)  ∆OHJ  ∆OGI Hence EF bisects GH  GO = HO EXERCISES 1. 2. 3. 4. 5. State True or False a. In a parallelogram opposite sides are equal b. In a parallelogram opposite angles add upto 180º c. Diagonals of a Rectangle bisect the vertex angles d. Interior angles of a square is equal to 90º e. Sum of exterior angles in a parallelogram is 360º f. Interior angles of a Rhombus are all equal g. A parallelogram is a quadrilateral h. Trapezium is a type of parallelogram i. Diagonals of a Rhombus are mutually perpendicular Interior angles  A of parallelogram is 55º side AB measures 3cms the perimeter of ABCD is 16cms. Find all the angles and length of all the sides. PQRS is a Rhombus length of the diagonal PR is equal to the length of a side. Show that Pr divides the Rhombus into two equilateral triangles. ABCD is a rectangle. Angle bisectors of  A and  B meet at O. Show that AO is Perpendicular to BO. ABCD is a parallelogram and side AB is equal to diagonal AC. If side DA is extended to E show that AB bisects  EAC. 6. E A B D Intermediate-Geometry F C 71 of 96 ©2014 www.learningforkowledge.com/glg s 7. 8. 9. In the figure shown above ED and FB bisect angles D̂ and B̂ of a parallelogram show that DE BF and DE = BF. Show that a parallelogram is a square if its diagonals are equal and intersect at right angles. E is the mid point of side AB of a parallelogram ABCD. If CE bisects  BCD show that DE also bisects  ADC. Find the value  DEC. ABCD is a parallelogram side DA is extended to E such that DA = EA and EC is 1 1 joined. Show that AF = DC = AB and F is the mid point of EC. 2 2 E A D 10. B C E and F are mid points of AB and CD of parallelogram. AF and CE, and DB (diagonal) Joined AF intersects DB at x and CE intersects DB at y. Show that Dx 1 = xy = yB = DB 3 Intermediate-Geometry 72 of 96 ©2014 www.learningforkowledge.com/glg AREAS OF PLANE AND SURFACE AREA OF SOLID FIGURES AREA OF PLANE FIGURES Consider a plane figure such a polygon or a closed curve. Area is a measure of the size interior of this figure. The figure 12.1 shows a Rectangle made from flat square tiles. Simplest way to express the area is to total the number of square tiles involved in making this rectangle. If the square tiles have sides equal in length to a unit of measurement for e.g. cm or mm, the total number of squares will be the area of the figure in mm mm or cm cm (mm2 or cm2) 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 Rectangle from 15 square tiles Figure above shows a rectangle made from 15 square tiles and therefore can be side to be of 15sq units size. Since the layout is 5 tiles long and 3 tiles broad area can also be expressed as lb or 5×3 = 15sq units Irregular shapes for e.g. Scalene triangle may also need a number of fraction or part tiles to build. The whole number of tiles and the sum of part tiles can be added to obtain the area of this figure. Figures can be graph sheet and squares counted. However formulas can be to calculate the areas of polygons, circles and similar shapes. Every polygon has a part of the plane enclosed by it. This part with the polygon is defined polygon region. Every polygon region has an area which is positive and real and is measured in square units. However, the sides of the polygon has only length is its dimension and has no width. If two polygons are congruent, their areas are also equal. Theorem 1: A parallelogram can be divided into two triangles having the same area by its diagonal. A B D Intermediate-Geometry C 73 of 96 ©2014 www.learningforkowledge.com/glg Construction: Draw a parallelogram ABCD and join BC. Proof: Consider triangles ABD and BCD AB = CD - - - - - - - - - - (ABCD is a parallelogram) AD = BC - - - - - - - - - - (ABCD is a parallelogram) BD is common  ∆ABD  ∆BCD - - - - - - - - - - (SSS criterion) Hence Area of ∆ABD = Area of ∆BCD BC divides a parallelogram into two parts. Theorem 2: Parallelogram with a common side and between the same parallel lines have the same area. E C F D A B Construction: Draw a parallelogram ABCD. Use AB as the common side or the base and draw another parallelogram ABFE. Points E and F should be on the line AB or extention of AB. Proof: Consider triangles AED and BCF AE = BF - - - - - - - - - - - - (Given ABEF is a parallelogram) AD = BC - - - - - - - - - - - - (Given ABCD is a parallelogram)  ADE =  BCF - - - - - - - - - - (Corresponding angles AD BC)  AED =  BFC - - - - - - - - - - (Corresponding angles AE BE) ∆AED  ∆BCF - - - - - - - - - - - (ASA criterion) Hence area of ∆ADE = area of ∆BCF Consider parallelogram ABCD Area of ABCD = Area of Trapezium ABDF + Area of ∆BCF similarly Area of ABEF = Area of Trapezium ABDF + Area of ∆AED Given area of ∆AED = Area of ∆BCF and that ABCD is common to both the parallelograms Area of ABEF = Area of ABCD Theorem 3: Triangle with their base as a common side and with their remaining vertices on a line parallel to the base have the same area. E A D F B Intermediate-Geometry C 74 of 96 ©2014 www.learningforkowledge.com/glg Construction: Draw two triangles ABC and DBC such that AD is parallel to BC. Extend DA to E such that BE‫ ׀׀‬AC and AD to F such that BD‫׀׀‬CF Proof : Consider parallelograms ACBE and DBCF Area of ACDE = Area of DBCF Given that two parallelograms have a common base BC and the sides opposite to BC are on the same line EF. Area of ACBE = Area of ∆ACB + ∆ABE But area of ∆ACB = area of ∆ABE Given ACBE is a parallelogram and AB is a diagonal  Area of ACBE = 2 ×area of ∆ABE Similarly Area of DBEF = 2× area of ∆BCD  2× area of ∆ABE = 2× area of ∆BCD  area of ∆ABE = area of ∆BCD AREA FORMULAE – 1. RECTANGLE AXIOM: Area of a rectangle is defined as l ×b units. Where l is the length (measure of longest side) and b is the breadth (measure of the shortest side) PARALLELOGRAM Area of a parallelogram can be calculated by comparing its area to that of a rectangle. Consider the following figure. ABCD is a parallelogram. A rectangle is drawn with one of its sides common to one of the sides of the parallelogram. Base of the parallelogram AB can be selected as the common side perpendiculars from A and B are drawn and points F of intersection of these perpendiculars with line CD or extention of it are marked as E and F. ABEF is the rectangle contained between the same parallel lines of the parallelogram. D E C F A B Area of parallelogram ABCD = Area of rectangle AEFB Area of parallelogram ABCD = AE×AB - - - - - - - - - - ( l ×b AEFB is a rectangle) AB = Base of the parallelogram and AE = height of distance between the opposite sides  Area of parallelogram = base× height = b× h Intermediate-Geometry 75 of 96 ©2014 www.learningforkowledge.com/glg TRIANGLE Area of a triangle is half the area of a parallelogram formed by two sides and their parallel lines with the remaining side forming a diagonal. A B D E C F Consider a ∆ABC drawn as a part of the parallelogram ABCD. 1 Area of ABC  Area of parallelogram ABCD 2 - - - - - - (AC is diagonal of ABCD) But Area of ABCD  Area of Rectangle  ADEF  AE  EF AE  height of triangle - - - - - - - (AE  BC) Also EF  AD  BC - - - - - - - (Oppositeof parallelograms)  Area of ABC  1  AE  EF 2 1  height  base 2 1  bh 2  Example 1: Show that diagonal of a parallelogram divides it to four triangles of equal area. A B F O E D Intermediate-Geometry C 76 of 96 ©2014 www.learningforkowledge.com/glg Construction: Draw a parallelogram ABCD. Join diagonals AC and BD. Draw AE  BD and CF  BD, AC and BD intersect at O. Answer : Consider triangles AEO and AOB 1 1 Area of AOD  base height  OD  AE 2 2 1 1 Area of AOB  base height  OB  AE 2 2 But OD  OB - - - - - - - - - (BD is the diagonal of ABCD) 1 1  OD  AE  OB  AE 2 2  Area of AOD  AOB (area) Similarly we can prove that Area of AOB = Area of ∆BOC = Area of ∆COD Example 2: Show that the area of a Rhombus is equal to the product of 1 lengths of two 2 diagonals. B O A C D Construction: Draw a Rhombus ABCD join AC and BD. O be the point of intersection. 1 1 Answer : Area of ABC   BO  AC  height  base 2 2 1 1 Area of ADC   DO  AC  hight  base 2 2 1 1 Area of Rhombus ABCD  ABC (area)  ADC (Area) 2 2 1 1 Area of ABCD  BO  AC  DO  AC 2 2 1  (BO  DO)  AC 2 1  BD  AC 2 Intermediate-Geometry 77 of 96 ©2014 www.learningforkowledge.com/glg Example 3: ABCD is a quadrilateral diagonal AC measure 5cms. BE and DF are perpendiculars to AC. Given BE = 3cms and DE = 2cms. Find the area of the quadrilateral. B A F E C D Construction: Draw a quadrilateral. Draw AC and BE and FD (Quadrilateral will not be to any scale) Answer : Consider ABC 1 1 Area of ABC  base  height   AC  BE 2 2 1 Area of ABC   5  3  7.5cm 2 2 1 1 Area of ACD  Base  height  AC  FD 2 2 1   5  2  5cm 2 2 But area of quadrilateral ABCD  Area of ABC  Area of ACD  Area of ABCD  7.5  5  12.5cm 2 Example 4: ABCD is a parallelogram. E and F are mid points f AB and CD. Show that area of AEFD is half of area of ABCD. Construction: Draw a parallelogram ABCD. Mark E and F as mid points of AB and CD join EF Draw EG  CD. A D E F B G C Working: Consider the quadrilateral AEFD Intermediate-Geometry 78 of 96 ©2014 www.learningforkowledge.com/glg AE‫ ׀׀‬FD - - - - - - - - (Given AB‫׀׀‬CD) 1 1 AE  AB  CD  DF - - - - - - - (Given ABCD is a parallelogram) 2 2 AB  CD  AEFD is a parallelogram since oppositesides are equal and parallel Area of parallelogram ABCD  CD  EG - - - - - - - - (base  height) Area of parallelogram AEFD  DF  EG - - - - - - - (base  height) 1  CD  EG 2 1   Area of parallelogram ABCD 2 Example 5: ABCD is a trapezium. Prove that area of a trapezium is its height times the average length of the two parallel lines. Construction: Draw a trapezium ABCD. Draw perpendiculars form A and B to CD. Let AE and BF be the perpendiculars. A B C E F D Area of trapezium  Area of ACE  Area of BDF and Area of Rectangle ABFE 1 Area of ACE  CE  AE 2 1 1 Area of BFD  FD  BF  FD  AE - - - - - - - - - -(AE  BF) 2 2 Area of Rectangle ABFE  AE  EF Area of trapezium ABCD  1 1 CE  AE  FD  AE  AE  EF 2 2 Intermediate-Geometry 79 of 96 ©2014 www.learningforkowledge.com/glg   1 1   AE CE  FD  EF  2  2   1 1 AB  CD But CE  FD  EF  2 2 2 AB  CD AB  AD  CE  FD Becasue  2 2 2EF CE FD    - - - - - - - - - - - ( AB  EF ) 2 2 2  AB  CD   Area of Trapezium  AE    2   EXERCISES 1. 2. 3. Find the area of the following figures a. A triangle with base 10cms and height 5cms. b. A parallelogram with base 10cms and height 7cms. c. T trapezium with its parallel sides equal to 5 and 7cms and height 8cms. d. A right angled triangle with sides 3cm, 4cm and 5cms. e. A Rhombus with diagonals measuring 12cms and 15cms. Show that a median divides a triangle into two parts having the same area Following figure shows a parallelogram ABCD, and AE  CD and CF  AD. Given AE = 5cms, CD = 7cms and CF = 6cms Find AD. A B 5cms 6cms 4. 5. D 7cms C ABCD is a quadrilateral and AC is a diagonal. If O is mid point of AS show that area of quadrilateral AOCD = area of equilateral AODB. PQRS is a parallelogram. T and U are mid points of PQ and RS. RT, TU and RS are joined to divide the parallelogram into four triangles given ∆PST has an area of 5sq cms find the areas of remaining triangles and the parallelogram. S U R P Intermediate-Geometry T Q 80 of 96 ©2014 www.learningforkowledge.com/glg 6. 7. 8. P, Q and R are the mid points of an equilateral triangle ABC. Show that the area of ∆PQR is quarter the area of ∆ABC. To isosceles triangles ABC and BCD are drawn with base BC as common and A and D are equidistant from BC and on opposite sides of it. F in the area of ∆ABD, ∆ACD and Rhombus ABDC given BC = 2cms and AD = 3cms. ABCD and ABEF are two parallelograms having AB as common side and C, D, E and F are in the some line. FB is joined given AB = 12cms and ∆FAB = 120cm2 find the following: D F C E A B a) Area of both the parallelogram b) Height of the parallelogram c) Length of side AB 9. 10. ABCD is a rectangle DC is extend to E such that ABCE is a parallelogram. If the dimensions of the rectangle are 8cms = AB and CD = 6cms find a) Area of the parallelogram b) Areas of ∆ABC, ∆ADC and ∆CBE D C A B E ABCDEF is a regular Hexagon. Two of its diagonal AD and BF measures 4 and 2 3 cms. Find the area of the Hexagon and length each side given all the sides are equal. B C A B F Intermediate-Geometry E 81 of 96 ©2014 www.learningforkowledge.com/glg PERIMETER AND AREA Perimeter of a plane figures like polygons is sum of the length of all the sides. In the case of closed curves it is the length of the bounding perimeter is expressed in the same units of length for e.g. cm or mm. Area of polygon and closed curves like circles can be found if the perimeter and some of their properties are known. Consider a triangle ABC a, b and c denote the sides BC, AC and AB of the triangle  The perimeter P = a + b + c Semi perimeter of half the value of perimeter S = ½(a + b + c) Heros formula: A formula named after ‘Hero’ connects the dimension pf the sides of a triangle to its area. AreaA  s(s )(s  b)(s  c) Quadrilateral: Area of a quadrilateral can be found from the length of all the four sides and one of its diagonals. B A C D Diagonal AC divides the quadrilateral into two triangles. Area of each triangle can be found using Hero’s formula. Adding the areas of the triangle gives the area of the quadrilateral special quadrilaterals for e.g. Rectangles area and perimeter can be linked by other formulae also. Rectangle Area  l  b - - - - - - - - (by definition ) Perimeter  2l  2b - - - - - - - - (oppositesides are equal) Square Area  l  l  l 2 - - - - - - - - (lengths of all sides are equal) Perimeter  4l Rhombus Area  d 1  d 2 , where d 1 and d 2 are length of diagonals Perimeter  4l - - - - - - (lengths of all sides are equal) Circle Perimeter of a circle is known as circumference by definition of A The circumference ' C' of a circle  π  D  2π r - - - - - (D = 2r) Intermediate-Geometry 82 of 96 ©2014 www.learningforkowledge.com/glg Where D is the diameter or the longest chord (line) intersecting the circle at points and passing through its center. πD 2 Area of a circle   πr 2 - - - - - - (D  2  radius) 4 Part of the circumference is called as the are of a circle. When the ends of an arc are joined with center of the circle by two radii sector of the circle is formed. Perimeter of an arc is given as r  θ o  2π S where is the angle seperating the radii 360 o Since, if we consider the entire circumference which has a perimeter of 2π r as the arch the radii are separated by 360o Hence perimeter of a sector P  2r  r  2 360 o  r Area of a sector  πr 2   360 o Segment of a circle is the part of a circle bound by a line intersecting the circle at two points. The line divides a circle in to major and minor segments. This line is called as a chord. When the circle is exactly divided into two parts this line or chord will be diameter. Half of a circle is called a semi circle. O  A C B Perimeter of a segment  r   C, where is the angle formed 360 o by tworadii OA and OB and C is the length of the chord Area of segment : Area of sector OAB - Area of OAB Intermediate-Geometry 83 of 96 ©2014 www.learningforkowledge.com/glg Example 1: Figure below shows shaded cross B and of a national flag. Find the area of shaded and unshaded part. Find the perimeter of the flag. 10-50cms 2 1 10cm 30cm 3 10cm Total area of flag  50  30  1500cm 2 Area of shaded region excluding any repition 10  50  10 10  10 10  700cm 2 (Adding regions (1), (2) and (3))  Area of unshaded part  1500 - 700  800cm 2 Parimeter of the flag  2l  2b  2  50  2  30  160cms Example 2: ABC is an equilateral triangle, side AB is 2cms. Find the perimeter and the area of the triangle. From the area find the height of the triangle. A B D C Construction: Draw a triangle ABC with all the sides equal to 2cms. Draw AD  BC Perimeter of ∆ABC = AB + BC + CA = 2 + 2 + 2 = 6cms 6 Semi perimeter S =  3cm 2 Area of the triangle as per Hero’s formula Intermediate-Geometry 84 of 96 ©2014 www.learningforkowledge.com/glg A  s(s  a)(s  b)(s  c) where s  Semi perimeter  3cms a  Side BC  2cms b  side AC  2cms c  side AB  2cms  A  3(3 - 2) (3 - 2) (3 - 2) A 3  1.73cm But A  1 base  Height 2 1 1.73   2  Height 2  Height  1.73cm Example 3: PQRS is a parallelogram with PQ and QR measuring 34 and 20cms. The diagonal QRS measures 42cms. Find the perimeter and the area of the parallelogram. P 34cms Q 42cms S 20cms R Perimeter = 2 (20 + 34) - - - - - - - - (PQRS is a parallelogram) = 108cms Area of PQRS  2  Area of PQR Area of PQS  s (s - a) (s - b) (s - c) where s  semi perimeter of PQR 20  34  42 97  cms  48cms 2 2 a  QS  42cms b  PS  20cms c  PQ  34cms  Intermediate-Geometry 85 of 96 ©2014 www.learningforkowledge.com/glg  Area of PQS  48 (48 - 42)  (48 - 20)  (48 - 34)  48  6  28  14  8  6  6  2  14  14  16  36  14  14  4  6 14  336cm 2 Example 4: ABCD is a rectangle sheet of size 30cms 10cms. A semicircle of Diameter 10cms has been cutout from both the smaller edges. Find the remaining area of the sheet. Also find the perimeter. Construction: Draw a rectangle (not to scale) with sides in the ratio of 3:1. Mark the center of the shorter sides BC and DA. Draw semi circles. A B D C Area of uncut rectangle ABCD  l  b  AB  BC  30  20  600cm 2 Area of semicircle     r2 2 2 22 10  - - - - - - - - - (BC  20cms hence r  10cm) 7 2 22  50  155cm 2 7  Area of ABCD ofter cutting   600 - 2 155  290cm 2 Example 5: Find the area of the shaded part of the sector given inside radius = 15cms, outside radius = 2.5cms and the angle of separation of the two radii = 30o. Also find its perimeter. Intermediate-Geometry 86 of 96 ©2014 www.learningforkowledge.com/glg Construction: Draw a sector of a circle with out radius of 2.5cms and 30o as angle of separation. Draw the inner arc with a radios of 15cm shade the portion enclosed by the two arcs and the radii. A D O C B πr 2  θ 360 o 22 30   2.5 2  7 60 2  1.64cm Area of sector AOB  22 30  1.5 2  7 360 2  0.59cm Area of ABCD  1.64 - 0.59 Area of sector ODC   1.05cm 2 Perimeter of ABCD  Arc AB  BC  Arc CD  DA 30 360  1.31cm Arc AB  2   2.5  30 360  0.79cm BC  CD  2.5 - 1.5  1cm Arc CD  2  1.5   Perimeter of ABCD  1.31  0.79  4.1cm EXERCISES 1. Area of a rectangle is 9.8cm2. Given its length equals twice breadth, find the dimentions of the rectangle. 2. Sides of a triangle are in the ratio x:2x:3x. If the perimeter of the is 36cms find its dimentions and area. Find the shortest height or altitude of this triangle. Intermediate-Geometry 87 of 96 ©2014 www.learningforkowledge.com/glg 3. A right angle triangle has a perimeter of 24cms and an area of 24cm2. Given the length of an adjacent side of the right angle as 6cms. Find the lengths of remaining two sides. 4. Diagonal of a quadrilateral measures 8cms. Perpendiculars drawn from the vertices opposite to the diagonal measures 6cms and 4cms. Find the area of the quadrilateral. 5. Find the perimeter and area of the following figure. 35 10 30 10 45 1-15 6. The flag shown below comprises of the shaded circle of 20cms diameter at the center of 30cms by 90cms rectangle. Find the areas of shaded and unshaded part. Also find the perimeter of the flag. 7. Circumference of a circle exceeds the diameter by 16.8cms. Find the circumference and the area of the circle. 8. Perimeter of a semicircle is 90cms. Find its radius and the area. 9. A cyclist goes around a track of 100m diameter. Given the diameter of the cycle wheel as 1m and that the cycle travels a distance of or meters per revolution find the number of revolution of the cycle wheel to go round the track once. Also find the length of the track. 10. Following figure shows two circles of 6cms diameter intersecting at two points. Given that the centers of the circles are 4cms apart. Find the shaded common area. Intermediate-Geometry 88 of 96 ©2014 www.learningforkowledge.com/glg SURFACE AREA AND VOLUMES OF SOLIDS Exterior of a solid comprises of a number of plane or curved surfaces. Area such surfaces can be found using formulae in the case of surfaces having a regular Geometric shape for example prisms and pyramids. A prism has two identical polygons as base and top joined by edges forming rectangular lateral surfaces. TRIANGULAR PRISM TRIANGUALR PYRAMID A pyramid has a polygon as its base. Sides of polygon form the base of triangles. Vertices of these triangles meet at a common point which becomes the vertex of the pyramid. The side faces or the triangles become the lateral surfaces. Volume of a solid is the total space with in the surfaces. 1. Write the formulae for finding L.S.A, T.S.A and volume of Pyramids and Prisms. 2. Find the lateral and total surface area of a prism whose height is 4cm and perimeter of the equilateral triangle base is 6cm. 3. Height of a right angle triangle base and top prism is 15cms. Its base has dimentions of 13cm, 14cm and 15cm. find its lateral surface area, total surface area and volume. (Hint: Use Quadratic equation) 4. Total surface area of a square based prism is 1000cm2. Given its height as 20cm. find the dimentions and volume of the prism. 5. Base and top of a prism are congruent isosceles triangles of area 192cm2, perimeter 64cm and the side which forms the base of the triangle 24cms. If the height if 30cms find its L.S.A, T.S.A and volume. 6. A tetrahedron (triangle based pyramid) consists of four equilateral triangles of 6cms each side. Find its L.S.A and T.S.A (Hint: Add area of triangles) Intermediate-Geometry 89 of 96 ©2014 www.learningforkowledge.com/glg 7. A square based pyramid has an on area of 576 units. If it is vertical height is 16cms. Find it L.S.A, T.S.A and Volume. 8. A fish tank side walls are made from 2 sq 10cms square glass panels and two 10 2cms rectangular glass panels. Top and bottom are made from steel plates. Find the area of glass sheets and steel plats required. Find the volume of water the tank ban hold in liters. 9. A tetrahedron is formed by 4 equilateral triangles of sides 2a. Given median of the equilateral triangle is 3a find the L.S.A and T.S.A of the pyramid. 10. A sky scraper is modeled on a pyramid at the top of a square based prism. Area of base if 576m2. If the total height is 180m to the top most point, find the L.S.A, T.S.A and the volume of the sky scraper. Intermediate-Geometry 90 of 96 ©2014 www.learningforkowledge.com/glg GRAPHS AND GEOMETRIC CONSTRUCTION Graphs: Scientific data is represented as graphs show the relationship between two groups. First group of data where the variable is controlled is represented on the x – axis. The second group of data which is dependent on the y –axis. A point on the graph sheet (x –y plane) represent a set of values of both the variable measured as the co ordinates of the point. For example consider a water tank being filled by water from a tap. A set of x – values gives the time elapsed after opening the tap and y the amount of water filled. X 1 2 3 4 5 6 Y 1.5 3 4.5 6 7.5 9 Y Time vs. volume of water 7 6 Volume in liters Note: A graph should have a title x axis and y-axis marked and indented x –y axis to be labeled 5 6 4 3 2 1 O 1 7 2 8 3 4 5 6 X 9 10 Time in minutes Example 1: A cup containing boiling water at 100oC is cool till the water is close to room temperature. Readings of temperature against time is given below. Draw a graph of temperature vs. time for the first 10 minutes of cooling. X0 time Y100o temp Intermediate-Geometry 1 2 4 6 8 10 90o 82o 67o 55o 45o 36o 91 of 96 ©2014 www.learningforkowledge.com/glg Working: (1) Draw x and y axis of 10cms each. (2) Indent x axis (scale markings) 1cm = 1minute. (3) Indent y –axis (scale markings) 1cm = 10oc (4) Label x and y axis (5) Mark each point Point 1 Mark 100oc an the y –axis (x = 0) 2 Draw a horizontal line from 82o on the y –axis and a vertical line from t minutes. Mark the point of intersection and label the co ordinates 3 -10. Mark all the remaining points in the same way. (6) Join the points and title the graph Temp vs. time 100 (0.100) 80 (2.82) o Temp in C 60 40 20 2 4 6 8 10 Time in minutes Example 2: A cyclist goes on a straight road. He accelerates from 0 – 10km/hr in 10 minutes time. He then slows down at an uniform rate and halts after an additional period of 10 minutes D. Draw the velocity time graph Y 10 (10.10) A B (20.10) Km/ph C (30.10) 0 10 Intermediate-Geometry 20 Minutes 92 of 96 30 X ©2014 www.learningforkowledge.com/glg Construction: Draw x-y axis Mark 10, 20 and 30 minutes on the x-axis at equal intervals Mark 5 and 10km/ph at equal intervals on the y-axis Mark point A coordinates (10, 10).Point B (20.10) and point C (30.0) Explain the graph line OA shows the accelerating part of the cyclists journey. Velocity of the bike can be obtaining by drawing a vertical line from a point an the x- axis corresponding the time. Mark the point of intersection of the graph and the vertical line. Y – coordinate of this point gives the velocity of bike. Horizontal line AB is the graph of velocity vs. time during the next 10 minutes of the journey when the velocity was constant. Line BC shows the decelerating or slowing down to rest part of the journey. GEOMETRIC CONSTRUCTION Geometric shapes can be drawn on a sheet of paper given dimentions or sizes of parts of a line, triangle or a polygon. An instrument box (Geometry box) comprising of a compass, divider, scale, protractor and set squares is the required tool kit. Unknown dimensions of a shape can be measured from an accurate construction of that shape. Example 3: Construct a ∆ABC with sides equal to 6cms, 7cms and 8cms. Steps: 1. Draw a horizontal line. Mark point B. Mark point C using a compass or a rules such that the line segment BC = 7cms. 2. With B as the center draw an arc with a radius of 6cms. 3. With C as the center draw an arc with a radius of 8cms. 4. Mark the point of intersection of two arcs as A. Join AB and AC. A 6cms 8cms B C 7cms Intermediate-Geometry 93 of 96 ©2014 www.learningforkowledge.com/glg Example 4: Draw a right angled triangle PQR given hypotenuse QR = 5cms, side PR = 3cms and  P = 90o P Q R Construction: 1. Draw a horizontal line segment PR = 3cms. 2. Draw a line perpendicular to PR at the point P (a) Draw semi circle of point P (b) Using the same radius draw two consecutive arcs with the point of intersection of the semi circle and line PR as the center. Mark the points of intersection of the semi circle with the arcs. Use these points as centers and draw a pair of intersecting arcs. (c) Join the point of intersection of the arcs with Q and extend the same. 3. With R as center draw an arc, to intersect the perpendicular to QR, having Q radius of 5cms. 4. Mark this point as P and join PQ and PR Example 5: Construct a parallelogram with adjacent sides equal to 4cms and 6cms and an included angle of 60o Construction: (1) Draw a line DC = 6cms (2) Draw an arc with D as center and intersecting DC (3) Use the above point of intersection as center and with the radius unchanged draw another arc (4) Mark the point of intersection of the two arcs. Join D with this point and extend it to a length of 4cms. Mark the point as A. (5) Draw an arc of radius 6cms with A as the center. (6) Use C as center and draw an arc with a radius of 4cms. (7) Mark the point of intersection of the two arcs as B. Join AB, BC, CD and DA A B 4 D Intermediate-Geometry 6 C 94 of 96 ©2014 www.learningforkowledge.com/glg EXERCISES 1. Construct the graph of area of a square against length of its side from the table given below. X 0 0.5 1 1.5 2 2.5 3 y 0 0.25 1 2.25 4 6.25 9 2. A car accelerates from 0-30kmph on a straight road in 60secs. It continues to move for a further period of 2minutes but its speed comes down 25kmph due to friction. The driver then applied the brakes and brings the car to a halt in 45seconds. Draw the speed time graph. 3. Construct a triangle with three sides of lengths 8, 9 and 10cms. 4. Draw a right angled triangle given the hypotenuse = 10cms and the shortest side = 6cms. 5. Draw a triangle given its base = 5cms and two of the angles adjacent to the base = 45o and 60o . 6. Divide a line segment AB (10cms) into 3 (ratio of parts) 2:3:4. (Hint- Draw a line AC and use a ruler to mark 2, 2 + 3 and 2 + 3 + 4cms. Join C to B and draw parallel lines to it to intersect AB at D and E. Now AD: DE: EB = 2:3:4) 7. Draw a triangle ABC given BC = 5cms AB + AC = 8cms and ABC = 60o (Hint: draw two lines DB and BC = equal to 8 and 5cms and  DBC = 60o Join DC and draw a perpendicular bisector to CD to intersect DB at A. 8. Construct a triangle ABC with BC = 10cms, B = 60o, C = 45o. Draw a circum circle passing through points A, B and C. (Hint: Draw ABC, Draw perpendicular bisectors to AB and BC. Point of intersection of the bisectors is the circum center). 9. Construct a Rhombus of side 4cms and an include angle of 45o. 10. Triangle ABC has perimeter of 14cms and  B = 80o and  C = 60o. Draw ABC. (Hint: Draw a line PQ = 14cms. Draw P = 40o and Q = 30o, two lines intersecting at point A. Draw perpendicular bisectors to PA and QA intersecting PQ at B and C. Join AB. BC and CA). ----------------------------------------------------* * * ---------------------------------------------- Intermediate-Geometry 95 of 96 ©2014 www.learningforkowledge.com/glg

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