Mensuration For Class 9
Mensuration
1. Find the area of a triangle with following sides ( All measures area in cm)
i) 10, 17, 21
ii) 17, 25, 28
Solution :
i) Let a =10, b = 17, c = 21
s = a + b + c / 2 = 10 + 17 + 21 / 2 = 48/2 = 24
Using Heron's formula
Area of triangle = √ s( s- a) (s - b ) ( s - c)
= √ 24( 24 - 10) ( 24 - 17) ( 24 - 21)
= √ 24 * 14 * 7 * 3
= √ 2 * 2 * 2 *3 * 2 * 7 * 7 * 3
= 2 * 2 * 3 * 7
= 84 cm2
ii) 17, 25, 28
Let a =17, b = 25, c = 28
s = a + b + c / 2 = 17 + 25 + 28/ 2 = 70 /2 = 35
Using Heron's formula
Area of triangle = √ s( s- a) (s - b ) ( s - c)
= √ 35( 35 - 17) ( 35 - 25) ( 35 - 28)
= √ 35 * 18 * 10 * 7
= √ 5 * 7 * 2 *3 * 3 * 2 * 5 * 7
= 5 * 7 * 2 * 3
= 210 cm2
2. The sides of a triangle are in the ratio 5 : 12 ; 13. If its perimeter is 90 cm, find the area of the triangle.
Solution.
Given ratio in sides = 5 : 12 : 13
Perimeter = 90 cm
and sum of ratios terms = 5 + 12 + 13 = 30
first side = 90/30 * 5 = 15 cm = a
second side = 90/30 * 12 = 36cm = b
third side = 90 /30 = 13 = 39 cm = c
s = 90 /2 = 45
Area of triangle = = √ s( s- a) (s - b ) ( s - c)
= √ 45( 45 - 15) ( 45 - 36) ( 45 - 39)
= √ 45 * 30 * 9 * 6
= √ 3 * 3 * 5 * 2 * 3 * 5 * 3 * 3 * 2 * 3
= 2 * 3 * 3 * 3 * 5
= 270 cm2
3. In △ABC, ∠ A = 90° , AB = 14 cm , AC = 48 cm. Find
i) Area of △ABC
ii) length of perpendicular from A to BC
Solution,
In △ABC, ∠ A = 90° , AB = 14 cm , AC = 48 cm.
AD ⊥ BC
= 1/2 * 14 * 48
= 336 cm2
Using pythagoras theorem, we have
BC = √ AB 2 + BC 2
= √ 14 2+ 48 2
= √196 + 2304
= √2500
= 50 cm
AD = 2 * Area of △ABC /BC
= 2 * 336/ 50
= 672 / 50
= 13.44 cm
4. Find the area of an isosceles right triangle with hypotenuse 40 cm.
Solution .
ABC is an isosceles right triangle in which AB = BC and AC = 40 cm.
Let each equal side be = x
Thus by pythagoras theorem, we have
x 2 + x 2 = 402
2 x 2 = 1600
x 2 = 1600 / 2
x 2 = 800
x = 20√ 2
Area of △ABC = 1/2 * AB * BC
= 1/2 * x * x
= 1/2 x 2
= 1/2 * 800
= 400 cm2
Q 5. The base of an isosceles triangle is 40 cm and its area is 420 cm2. Find the length of its equal sides.
Solution :
Here ABC is an isosceles triangle and AB = AC and BC = 40 cm.
Area of △ABC = 420 cm2.
Draw AD ⊥ BC which bisects BC at D
BD = DC = 40/2 = 20cm
Area of △ABC = 1/2 * AD * BC
420 = 1/2 * AD * 40
AD = 420 *2 /40
AD = 21 cm
In △ABD ,
Using pythagoras theorem, we have
AB = √ AD 2 + BD 2
= √ 21 2+ 202
= √441 + 400
= √841
= 29 cm
Each equal side = AB = AC = 29 cm.
Q 6. In an isosceles triangle, the unequal side is 22 cm and perimeter is 144 cm. Find its area.
Solution. 
Perimeter of an isosceles triangle = 144 cm
Unequal side = 22 cm
Let equal side be x
Perimeter = sum of three side
144 = 22 + x + x
144 - 22 = 2x
122 = 2x
x = 122 / 2
x = 61 cm
Each equal side = 61 cm
s = Sum of sides / 2 = 61 + 61 + 22/ 2 = 144 /2 = 72
Using Heron's formula
Area of triangle = √ 72( 72- 61) (72 - 61 ) ( 72 - 22)
= √ 72( 11) ( 11) ( 50)
= √ 2 * 2 * 2 * 3 * 3 * 11 * 11 * 2 * 5 * 5
= 2 * 2 * 3 * 11 * 5
= 660 cm2
Q 7. If the area of an equilateral triangle is 25√3 m2, find the perimeter.
Solution :
Given , area of equilateral triangle = 25√3 m2,
let side be x
area of equilateral triangle = √3/4 * x2
25√3 = √3/4 * x2
x2 = 25 * 4
x2 = 100
x = 10
side = 10 cm
Perimeter = 10 + 10 + 10 = 30 cm
Q 8. In the given figure, PQR is an equilateral triangle of side 20cm. △ QSR is inscribed in it .
∠ QSR = 90°, QS = 16 cm. Find i) SR , ii) the area of the shaded portion ( Take √3 = 1.732)
Solution :
Given
△ QSR is inscribed in △PQR
∠ QSR = 90°,
QS = 16 cm.
In △ QSR ,
Using pythagoras theorem,
QR 2 = QS 2 + SR 2
20 2 = 16 2+ SR2
400 = 256 + SR2
SR2 = 400 - 256
SR = √144
SR = 12 cm
Area of △PQR = √ 3/4 * side2
= √ 3/4 * 20 * 20
= √ 3/4 * 400
= √ 3 * 100
= 1.732 * 100
= 173.2 cm2
Area of △QSR = 1/2 * QS * SR
= 1/2 * 16 *12
= 16 * 6
= 96 cm2
Area of shaded portion = 173.2 - 96 = 77.2 cm2
Q 9 ) In triangle PQR, ∠ Q = ∠ R = 45° and QR = 10 cm. Find its area.
Solution.
In △ PQR,
∠ Q = ∠ R = 45°
QR = 10 cm
Draw PL ⊥ QR which bisects QR at L 
QL = LR = 10 / 2 = 5 cm
In △ PQL,
∠ Q = 45°
∠ L = 90°
∠ QPL = 90° - 45° = 45°
PL = QL = 5 cm
Area of △ PQR = 1/2 * QR * PL
= 1/2 * 10 * 5
= 25 cm2
Q 10) A triangular advertising board is 13dm, 14dm and 15dm in length. Find the cost of painting
it at ₹10 per dm2 .
Solution :
Given Sides of triangle
a = 13dm , b = 14dm , c = 15dm
s = 13 + 14 + 15 / 2 = 42 / 2 = 21dm
Area of triangle = √ 21( 21- 13) (21 - 14 ) ( 21 - 15)
= √ 21( 8) ( 7) ( 6)
= √ 3 * 7 * 2 * 2 * 2 * 7 * 2 * 3
= 2 * 2 * 3 * 7
= 84 cm2
Rate of painting the board =₹10 per dm2
Total cost of painting = 84 * 10 = ₹ 840
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