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Basic Geometry-2
Polygons, Triangles, Quadrilaterals
Author: Raghu M.D.
Contents
GEOMETRY ....................................................................................................... 2
POLYGONS ......................................................................................................................... 2
Triangles ........................................................................................................................... 2
Quadrilaterals ................................................................................................................ 10
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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.
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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
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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
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(angles on a line)
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∴
∟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
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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
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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
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(angles in a triangle)
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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
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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
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b) T
c) F
d) T
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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)
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∟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.
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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
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b
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∴
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.
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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
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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
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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
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