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Ganit Learning Guides Basic Geometry-2 Polygons, Triangles, Quadrilaterals Author: Raghu M.D. Contents GEOMETRY ....................................................................................................... 2 POLYGONS ......................................................................................................................... 2 Triangles ........................................................................................................................... 2 Quadrilaterals ................................................................................................................ 10 Basic-Geometry2 1 of 17 ©2014, www.learningforkowledge.com/glg GEOMETRY POLYGONS A Polygon is a closed figure with intersecting lines forming vertices. The line segments form sides of polygons. They do not cross over or extend beyond the vertices. The simplest of Polygons is a triangle with only three sides. Polygons can have any number of sides. A Polygon with infinite number of sides will have a shape which is close to a Circle. If the sides are of the same size, then the figure is called a regular polygon. Triangle Hexagon Irregular Polygon Triangles A Triangle has three sides, three angles and three vertices. It is named by the three vertices, for example as ∆ABC. A c b B C a ∆ABC has three vertices A, B and C, three sides AB, BC and CA, and three angles ∟ABC, ∟BCA and ∟CAB. The sides of the triangle can also be abbreviated as a, b and c. Here a is the side opposite to vertex A, b is the side opposite to vertex B and c is the side opposite to vertex C. The angles can also be abbreviated as ∟A, ∟B and ∟C. Basic-Geometry2 2 of 17 ©2014, www.learningforkowledge.com/glg Types of Triangles Triangles are classified according to their shapes. Scalene triangles Fig.(b) Fig.(a) These are triangles with different lengths of sides and different angles. All the three angles can be acute as shown in Figure (a) or one of them can be obtuse as shown in Figure (b). Right Angle triangle One of the angles of a Right Angle triangle is 90º. Isosceles triangle Two sides and two angles of an Isosceles triangle are equal to one another. Equilateral triangle All the three sides and three angles of an Equilateral triangle are equal to one another. Properties of Triangles B c a h A C b Basic-Geometry2 3 of 17 ©2014, www.learningforkowledge.com/glg Perimeter of a triangle The Perimeter P of a triangle of sides a, b and c is a + b + c, where a, b and c are lengths of sides opposite to the vertices A, B and C. P=a+b+c Area of a triangle Consider a triangle ABC as shown above. Side b is taken as the base of the triangle, h is the perpendicular distance of the vertex B to the base, which is also considered as the height of the triangle. Area A=½×b×h Or A = ½ bh Sum of angles The three interior angles of a triangle add up to 180º. Proof: Construction: Draw a ∆ABC. Draw a line DE parallel to BC passing through the vertex A. A D E B C Working: Hence, ∟DAB = ∟ABC (alternate angles) ∟EAC = ∟ACB (alternate angles) ∟DAB + ∟EAC = ∟ABC + ∟ACB Adding ∟BAC to both sides, ∟DAB + ∟EAC + ∟BAC = ∟ABC + ∟ACB + ∟BAC But, Basic-Geometry2 ∟DAB + ∟BAC + ∟EAC = 180º 4 of 17 (angles on a line) ©2014, www.learningforkowledge.com/glg ∴ ∟ABC + ∟ACB + ∟BAC = 180º Conclusion: The interior angles of a triangle add up to 180º. Example 1: Show that each angle in an equilateral triangle is equal to 60º. Construction: Draw a triangle with equal sides using a ruler and mark the interior angles as aº. aº aº aº Working: In an equilateral triangle all the three angles are equal. Hence, the sum of the angles, aº + aº + aº = 180º ∴ 3aº = 180º Or. aº = 180/3 = 60º (angles in a triangle add up to 180º) Answer: Each angle in an equilateral triangle is equal to 60º. Example 2: Triangle ABC is an isosceles triangle. The perimeter of ∆ABC is 13cms amd the shortest side is 3cms. Find the lengths of the remaining two sides. Construction: Draw the figure of an isosceles ∆ABC and mark the shortest side BC as 3cms. A B C 3 Basic-Geometry2 5 of 17 ©2014, www.learningforkowledge.com/glg Working: Perimeter P = AB + AC + BC Or AB + AC + BC = 13 (given P = 13) AB + AB + BC = 13 (AB = AC, ∆ABC is isosceles) 2AB + 3 = 13 (given BC = 3) ∴ 2AB = 10 or AB = 10/2 = 5 ∴ AB = AC = 5 Answer: AB = AC = 5cms. Example 3: ∆PQR is a right angle triangle, where PQ = 3cms, QR = 4cms and RP = 5cms. Find the area of ∆PQR. Construction: Draw a right angle triangle with sides PQ, QR and RP, where ∟PQR = 90º. P 5 3 Q R 4 Working: In a right angle triangle, the two mutually perpendicular sides can be considered as the base and height of the triangle. In ∆PQR, PQ ⊥ QR Area A = ½ base × height (∟PQR = 90º) A = ½ QR × PQ A = ½ ×4 × 3 = 6 Answer: Area of ∆PQR = 6 cms2 Basic-Geometry2 6 of 17 ©2014, www.learningforkowledge.com/glg Example 4: A line BC of ∆ABC is extended to point D and ∟BAC = ∟ABC = 60º. Find the value of ∟ACD. A B C D Construction: Sketch a ∆ABC and extend the side BC to a point D. Mark ∟ACD. Working: In ∆ABC, Hence, ∟ABC + ∟BAC + ∟ACB = 180º (angles in a ∆) ∟ACD + ∟ACB = 180º (angles on a line) (∟ABC + ∟BAC + ∟ACB) – (∟ACD + ∟ACB) = 180º - 180º ∟ABC + ∟BAC + ∟ACB – ∟ACD - ∟ACB = 0 60º + 60º – ∟ACD = 0 ∟ACD = 120º Answer: ∟ACD = 120º Example 5: In the figure below, the line segment DE is parallel to BC of ∆ABC. If ∟ABC = 70º and ∟BAC = 50º, find the value of ∟ACB, ∟ADE and ∟AED. A D B E C Construction: Sketch a ∆ABC and draw a line segment DE parallel to BC. Working: ∟ABC + ∟BAC + ∟ACB = 180º Basic-Geometry2 7 of 17 (angles in a triangle) ©2014, www.learningforkowledge.com/glg 70º + 50º + ∟ACB = 180º (given ∟ABC = 70º, ∟BAC = 50º) 120º + ∟ACB = 180º ∟ACB = 180º – 120º = 60º ∴ ∟ACB = 60º ∟ADE = ∟ABC (given DE ‖ BC) ∴ ∟ADE = ∟ABC = 70º (given ∟ABC = 70º) ∴ ∟ADE = 70º But, ∟AED = ∟ACB (given DE ‖ BC) ∴ ∟AED = ∟ACB = 60º (by calculation ∟ACB = 60º) ∴ ∟AED = 60º Similarly Answer: ∟ACB = 60º, ∟ADE = 70º and ∟AED = 60º. EXERCISE Triangles BasicGeTri 1. State whether True of False: a) In a ∆ABC, AB = BC + CA b) Three angles of a triangle add up to 180º. c) In a ∆PQR, ∟P is acute and ∟Q and ∟R are obtuse. d) Area of a triangle is given by the formula: A = ½ bh. 2. An equilateral triangle has a perimeter of 21cms. Find the length of each side. 3. Find the area of a scalene triangle with base 5cms and height of 4cms. 4. ∆ABC is shown in figure below. Find the value of xº and yº, if the ∟ABC = 90º. A xº B yº 2xº D C Basic-Geometry2 8 of 17 ©2014, www.learningforkowledge.com/glg 5. Find the perimeter of a ∆PQR, if the length of side PR = 8cms and ∟Q = ∟R = 60º. 6. Find the size of angles of ∆ABC shown below, given that ∟A = xº+40º, ∟B = xº and ∟C = xº−10º. A C B 7. Show that the exterior angle of a triangle is equal to the sum the two interior opposite angles. 8. In the figure shown below, Line CE is parallel to side AB. Find the size of ∟ACD and ∟ECD. A E 60º B 40º C D 9. Using a ruler and protractor, draw the figure of ∆ABC such that ∟B = 40º, ∟C = 60º and BC = 6cms. Measure ∟A using the protractor. 10. Find the sizes of angles marked xº and yº. xº 60º xº 80º yº ANSWERS 1. a) F 2. 7cms Basic-Geometry2 b) T c) F d) T 9 of 17 ©2014, www.learningforkowledge.com/glg 3. 20 sq.cms 4. x = 30º, y = 120º 5. Perimeter = 24cms 6. ∟A = 90º, ∟B = 50º and ∟C = 40º 7. Hint: Angles in a triangle and angles on a line add up to 180º. 8. ∟ACD = 100º and ∟ECD = 40º. 9. 80º ± 1º 10. x = y = 40º Quadrilaterals A Quadrilateral is a four-sided polygon, with four sides, four angles and four vertices. D A B C In this figure A, B, C and D are the vertices. AB, BC, CD and DA are the sides. Lines joining A, C and B, D are diagonals of the quadrilateral. ∟A, ∟B, ∟C and ∟D are interior angles. In any quadrilateral, the sum of interior angles is equal to 360º. Proof: Consider two triangles ∆ABC and ∆ADC in the quadrilateral shown above. ∟ABC + ∟BAC + ∟ACB = 180º (angles in a triangle) ∟ADC + ∟CAD + ∟DCA = 180º (angles in a triangle) Basic-Geometry2 10 of 17 ©2014, www.learningforkowledge.com/glg ∟ABC + ∟ADC + ∟BAC + ∟CAD + ∟ACB + ∟DCA = 180º + 180º ∟ABC + ∟ADC + (∟BAC + ∟CAD) + (∟ACB + ∟DCA) = 360º ∟ABC + ∟ADC + ∟BAD + ∟BCD = 360º Conclusion: The interior angles of a quadrilateral add up to 360º. Special Quadrilaterals Quadrilaterals can be classified according to their shapes. Some of them are special because of their unique properties such as parallel sides, equal angles, etc. Using these properties, the lengths of sides, size of angles, perimeter and area can be calculated. l l l l l l l l l b Square Properties Equal sides of length l. Equal interior angles of size 90º. Perimeter = 4l Area = l2 Diagonals bisect the interior angles and are of equal length. Rhombus Properties Equal sides. Pairs of equal opposite angles. Perimeter = 4l Area = Product of the lengths of diagonals Diagonals intersect at their midpoints. Rectangle Properties Opposite sides are equal and parallel. Equal interior angles of size 90º. Perimeter = 2l + 2b Area = l × b Parallelogram Square, rectangle and rhombus are all types of parallelograms. Properties Opposite sides are equal and parallel. Pairs of equal opposite angles. Perimeter = Sum of the lengths of all sides Area = Product of the height (distance between two sides) and the length of one of the remaining side. Basic-Geometry2 11 of 17 ©2014, www.learningforkowledge.com/glg Trapezium A trapezium has a pair of parallel sides joined by unequal parallel sides. Properties Area = Product of the height and the average length of the two parallel sides. Kite Properties Two pairs of adjacent equal sides. One pair of equal opposite angles. Area = Product of the lengths of the diagonals Example 1: A side of a square measures 5cms. Find its perimeter and area. Construction: Draw a square of side 5cms. l=5 Working: Perimeter P=4l=4×5 (give l = 5cms) P = 20cms A = l2 = 5 × 5 = 25 Area A = 25 sq.cms Answer: Perimeter = 20cms, Area = 25cms. Example 2: The sum of lengths of two adjacent sides of a rectangle is 8cms. Its length l is 5cms. Find its perimeter and area. Construction: Draw a rectangle and mark its length as l and breadth as b. l=5 Working: l+b=8 5+b=8 b=8–5 Basic-Geometry2 b 12 of 17 ©2014, www.learningforkowledge.com/glg ∴ b=3 Perimeter P = 2 (l + b) P = 2 × 8 = 16 Area A = l × b = 5 × 3 = 15 Answer: Perimeter = 16cms, Area = 15sq.cms Example 3: The perimeter of a rhombus is 20cms and one of its angles is 120º. Find the values of remaining angles and length of the sides. A D 120º l B C Construction: Draw a rhombus ABCD. Mark ∟A = 120º. Working: Perimeter P=4l 20 = 4 l ∴ l = 20/4 = 5 ∟A = 120º ∴ But ∟A = ∟C = 120º (pair of opposite angles) ∟A + ∟B + ∟C + ∟D = 360º (angles in a quadrilateral) 120º + ∟B + 120º + ∟D = 360º ∟B + ∟D = 360º - 120º - 120º = 120º But ∟B = ∟D ∴ (pair of opposite angles) ∟B + ∟B = 120º 2 × ∟B = 120º ∟B = 120/2 = 60º = ∟D Answer: Length of each side l = 5cms, ∟B = 60º, ∟C = 120º, ∟D = 60º. Basic-Geometry2 13 of 17 ©2014, www.learningforkowledge.com/glg Example 4: Following is a diagram of a kite. Given ∟A = 110º and ∟B = 80º, find angles C and D. C B D 80º 110º A Working: ∟A = ∟C = 110º (pair of equal opposite angles) ∟A + ∟B + ∟C + ∟D = 360º (angles in a quadrilateral) 110º + 80º + 110º + ∟D = 360º 300º + ∟D = 360º ∴ ∟D = 360º - 300º - 60º Answer: ∟C = 110º, ∟D = 60º. EXERCISE Quadrilaterals BasicGeQuad 1. State whether True or False: a) In a quadrilateral, diagonals intersect and cross over. b) In a quadrilateral, sides intersect and cross over. c) Sum of interior angles of a quadrilateral add up to 360º. d) In a kite, opposite sides are parallel. 2. Two pairs of opposite sides are parallel. Name the quadrilateral. 3. ABCD is a square. Show that the diagonal AC divides the ∟BAC into two equal parts. 4. Find the size of xº in the following figure. A B xº Basic-Geometry2 ∟A = x + 20 ∟B = x ∟C = x – 20 ∟D = x + 40 D C 14 of 17 ©2014, www.learningforkowledge.com/glg 5. A trapezium ABCD has two sides AB and CD equal to 5cms. The lengths of two parallel sides AD and BC are 8cms and 4cms respectively. Find its perimeter. 6. PQRS is a parallelogram. Interior ∟P = 60º and ∟P = 120º. Find the angles R and S. 7. The following figure shows a kite ABCD. Its diagonals AC and BD intersect at point E. Given AE = 2cms, AC = 3cms and BD = 2cms, find the area of ABCD. A B E D C 8. ∆PQR is an isosceles triangle. Line ST is parallel to base QR. ST divides the triangle into a triangle PST and trapezium QRTS as shown in the figure. If ∟P is 30º, find the angles Q, R, S and T of the trapezium. P 30º S T Q R 9. A rectangular flag of length 30cms and width 20cms has three coloured vertical stripes of the same size. Find perimeter and area of each stripe. 10. A fenced land has a walking path all around inside. The path is adjacent to the fence and is 1m wide. This rectangular fence is of size 16m by 12m. Find the area of the land surrounded by the path. ANSWERS 1. a) T 2. Parallelogram 3. Hint: ∆ABC is a right-angle isosceles triangle Basic-Geometry2 b) F c) T d) F 15 of 17 ©2014, www.learningforkowledge.com/glg 4. 80º 5. 22cms 6. ∟R = 60º, ∟S = 120º 7. 10 sq.cms 8. ∟S = 105º, ∟T = 105º, ∟Q = 75º, ∟R = 75º 9. 60cms, 200sq.cms 10. 140sq.m Basic-Geometry2 16 of 17 ©2014, www.learningforkowledge.com/glg