Mensuration
151
Mensuration
Introduction
In previous classes, you have learnt about the perimeter and area of closed plane figures such
as triangles, squares, rectangles, parallelograms, trapeziums and circles; the area between two
rectangles i.e. area of pathways or borders and area between two concentric circles. You have
also learnt the concept of the surface area and volume of cube and cuboid, measurement of
surface area and volume of such solids by using basic units. In this chapter, we shall review
and strengthen all these.
15.1 Perimeter and area of plane figures
The perimeter of a closed plane figure is the length of its boundary i.e. the sum of lengths of its sides.
The unit of measurement of perimeter is the unit of length.
The area of a closed plane figure is the measurement of the region (surface) enclosed by its boundary
(sides).
It is measured in square units i.e. square centimetres (abbreviated cm2) or square metres
(abbreviated m2) etc.
15.2 Perimeter and Area of Triangles
(i)Area of a triangle = 1 base × height.
Height
2
Any side of the triangle can be taken as its base, then the
length of perpendicular (altitude) from the vertex opposite
to this side is called its corresponding height.
Base
(ii) If ABC is any triangle with sides a, b and c, then
perimeter = a + b + c, and
A
area = s(s − a)(s − b)(s − c)
where s = semi-perimeter = a + b + c .
c
2
b
(This is known as Heron’s formula.)
B
a
C
15.2.1 Some special types of triangles
A
(i) Right-angled triangle:
If ABC is a triangle in which ∠ B = 90°,
then its area = 1 BC × AB
2
= 1 (product of sides containing right angle).
2
B
(ii) Equilateral triangle:
Let ABC be an equilateral triangle with side a and
AD be the perpendicular from A to BC, then D is the
C
A
mid-point of BC i.e. BD = a .
a
2
In ∆ABD, AD2 = AB2 – BD2(Pythagoras theorem)
2
 a
⇒AD2 = a2 –   = 3 a2 ⇒ AD = 3 a.
2
2
4
B
∴ Area of ∆ ABC = 1 BC × AD
C
D
a
2
2
= 1 a ×
2
3
a = 3 a2.
2
4
Perimeter of ∆ ABC = 3 a.
(iii) Isosceles triangle:
Let ABC be an isosceles triangle with AB = AC = a
and BC = b, and AD be the perpendicular from A to BC,
A
then D is mid-point of BC i.e. BD = b .
2
In ∆ABD,AD2 =AB2 – BD2
a
(Pythagoras theorem)
2
2
2
2
 b
= a2 –   = a2 – b = 4 a − b
2
4
4
B
4 a2 − b 2
.
2
⇒ AD =
1
∴Area of ∆ ABC = 2 BC × AD = 1 b ×
2
b
2
D
C
4 a2 − b 2
2
1
2
2
= 4 b 4 a − b .
Perimeter of ∆ ABC = 2 a + b.
Illustrative Examples
Example 1. Calculate the area of a triangle whose sides are 13 cm, 5 cm and 12 cm. Hence, calculate
the altitude using the longest side as base. Leave your answer as a fraction.
Solution. Since the sides of the triangle are 13 cm, 5 cm and 12 cm.
∴
∴
s = 13 + 5 + 12 cm = 15 cm.
2
Area of triangle =
s(s − a)(s − b)(s − c)
= 15(15 − 13)(15 − 5)(15 − 12) cm2
= 15 × 2 × 10 × 3 cm2 = 30 cm2.
Mensuration
2197
The longest side of the triangle is 13 cm, let h cm be the corresponding altitude, then
area of triangle = 1 base × height
⇒ 2
60
30 = 1 × 13 × h ⇒ h =
.
13
2
∴ The required altitude of the triangle = 4 Example 2. The triangular side walls of a
flyover have been used for advertisements. The
sides of the walls are 122 m, 22 m and 120 m (shown
in the adjoining figure). The advertisements yield
an earning of ` 5000 per m2 per year. A company
hired one of its walls for 3 months. How much
rent did it pay?
8
cm.
13
122 m
22 m
120 m
122 + 22 + 120
m = 132 m.
2
Solution. Here, s =
Using Heron’s formula,
area of one triangular wall =
s (s − a)(s − b)(s − c)
=
132(132 − 122)(132 − 22)(132 − 120) m2
Rent = ` 5000 per m2 per year.
=
132 × 10 × 110 × 12 m2 = 1320 m2.
\ Rent of one wall for 3 months = `
1320 × 5000 × 3
= ` 1650000.
12
Example 3. The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the
smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle. Solution. Let the smallest side of the triangle be x cm, then the other two sides are
(x + 4) cm and (2x – 6) cm.
Given, perimeter of the triangle = 50 cm
⇒ x + (x + 4) + (2x – 6) = 50
⇒ 4x = 52 ⇒ x = 13
\ The lengths of three sides of the triangle are 13 cm, (13 + 4) cm and (2 × 13 – 6) cm
i.e. 13 cm, 17 cm and 20 cm.
50
Here, s = semi-perimeter =
cm = 25 cm.
2
Using Heron’s formula,
area of the triangle =
s (s − a)(s − b)(s − c)
=
25(25 − 13)(25 − 17 )(25 − 20) cm2
=
25 × 12 × 8 × 5 cm2 = 20 30 cm2.
Example 4. Find the area of a triangle whose perimeter is 22 cm, one side is 9 cm and the difference
of the other two sides is 3 cm.
Solution. Let the other two sides of the triangle be a cm and b cm, a > b.
Then 9 + a + b = 22 ⇒ a + b = 13
…(i)
and a – b = 3
…(ii)
On solving (i) and (ii), we get a = 8 and b = 5.
∴ The sides of the triangle are 9 cm, 8 cm and 5 cm.
2198
s = semi-perimeter = 22 cm = 11 cm.
2
Understanding ICSE mathematics – Ix
∴Area of the triangle = s(s − a)(s − b)(s − c)
= 11(11 − 9)(11 − 8)(11 − 5) cm2
= 11 × 2 × 3 × 6 cm2 = 6 11 cm2.
Example 5. From a point in the interior of an equilateral triangle, perpendiculars are drawn on
the three sides. If the lengths of the perpendiculars are 14 cm, 10 cm and 6 cm, find the area of the
triangle.
Solution. Let ABC be an equilateral triangle with length of each
A
side = a cm.
O is a point in the interior of DABC, OD ^ BC, OE ^ CA and
E
F
OF ^ AB such that OD = 14 cm, OE = 10 cm and OF = 6 cm.
O
Area of equilateral triangle ABC =
3 2
a cm2.
4
Also area of DABC = area of DOBD + area of DOCA + area of DOAB
C
D
= 1 BC × OD + 1 CA × OE + 1 AB × OF
2
2
2
⇒
B
3 2
a cm2 = 1 (a × 14 + a × 10 + a × 6) cm2
4
2
⇒ 3 a2 = 30a ⇒
( a ≠ 0)
3 a = 60
2
⇒ a ⇒ 20 3 .
3
× (20 3 )2 cm2 = 300 3 cm2.
4
\ Area of DABC =
Example 6. If the height of an equilateral triangle is 8 cm, calculate its area.
Solution. Let ABC be an equilateral triangle with side a cm. Let AD ⊥ BC, then D is midpoint of BC and BD = a cm.
A
2
In ∆ ABD,AD2 =AB2 – BD2
(Pythagoras theorem)
2
 a
82 = a2 –   ( height = AD = 8 cm given)
2
⇒ 2
2
⇒ 64 = a2 – a ⇒ 3 a = 64
4
⇒ a2
4
256
=
.
3
3 2
3 256
∴ Area of ∆ ABC =
a cm 2 =
×
cm2
4
4
3
=
B
C
D
64 3
cm2 = 36·95 cm2.
3
Example 7. The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its
perimeter.
A
Solution. Let ABC be the isosceles triangle with
base BC = 24 cm and area = 192 sq. cm. Let h cm be its
height and AB = AC = a cm.
Then area = 1 base × height
2
192 = 1 × 24 × h ⇒ h = 16.
⇒ In ∆ ABD,
2
AD2
= AB2
–
BD2
(Pythagoras theorem)
B
D
Mensuration
C
2199
⇒ (16)2 = a2 – (12)2
⇒ 256 = a2 – 144 ⇒ a2 = 256 + 144
⇒ a2 = 400 ⇒ a = 20.
∴ Perimeter of ∆ ABC = 2 a + b = (2 × 20 + 24) cm = 64 cm.
( BD = 1 BC = 12 cm)
2
Example 8. The base of an isosceles triangle measures 24 cm and its area is 60 cm2. Find its
perimeter (using Heron’s formula).
Solution. Let each equal side of isosceles triangle be a cm, then
s = a + a + 24 cm = (a + 12) cm.
2
By Heron’s formula,
the area of triangle =
s(s − a)(s − b)(s − c)
∴ (a + 12)(a + 12 − a)( a + 12 − a)( a + 12 − 24) = 60 (given)
⇒ (a + 12) × 12 × 12 × ( a − 12) = 60
⇒ 12 (a + 12)(a − 12) = 60
⇒ a2 − 144 = 5
⇒ a2 – 144 = 25 ⇒ a2 = 169
⇒ a = 13
∴ The perimeter of the triangle = 13 cm + 13 cm + 24 cm = 50 cm.
( a > 0)
Example 9. Find the perimeter of an isosceles right-angled triangle whose area is 72 cm2.
Solution. Let ABC be an isosceles right-angled triangle
A
with ∠ B = 90° and AB = BC = a cm.
Then,
area of ∆ ABC = 1 BC × AB
a cm
2
1
⇒ 72 = a × a ⇒ a2 = 144
2
⇒ a = 12.
In ∆ ABC, ∠ B = 90°. By Pythagoras theorem, we have
AC2 = AB2 + BC2 = a2 + a2 = 2a2
⇒AC =
B
C
a cm
2 a cm
∴ Perimeter of ∆ ABC= AB + BC + CA = (a + a +
2 a) cm
= (2 + 2 ) a cm = (2 + 2 ) × 12 cm
= (2 + 1·414) × 12 cm = 40·97 cm.
Example 10. If the difference between the two sides of a right angled-triangle is 2 cm and the area
of the triangle is 24 cm2, find the perimeter of the triangle.
Solution. Let ABC be a right-angled triangle with
A
∠ B = 90°. Let BC = x cm, then AB = (x + 2) cm.
Area of ∆ ABC = 1 BC × AB
2
1
⇒ 24 = x (x + 2)
2
⇒ x ( x + 2) = 48
⇒ x2 + 2x – 48 = 0
⇒ (x + 8) (x – 6) = 0
3100
Understanding ICSE mathematics – Ix
B
x
C
⇒ x = – 8 or x = 6, but x cannot be negative
∴ x = 6.
∴BC = 6 cm, then AB = (6 + 2 ) cm = 8 cm.
In ∆ ABC, AC2 = AB2 + BC2
⇒AC2
=
(8)2 +
(6)2
(Pythagoras theorem)
= 64 + 36 = 100
⇒AC = 10 cm.
∴ Perimeter of ∆ ABC = (8 + 6 + 10) cm = 24 cm.
Example 11. The perimeter of a right-angled triangle is 60 cm. If its hypotenuse is 26 cm, find the
area of the triangle.
∠ B
Solution. Let ABC be a right-angled triangle with
= 90°, then its hypotenuse AC = 26 cm (given).
A
Let base BC = x cm, then
perimeter of ∆ ABC = AB + BC + CA
⇒ 60 cm = AB + x cm + 26 cm ⇒ AB = (34 – x) cm.
In ∆ ABC, ∠ B = 90°,
AB2 + BC2 = AC2
⇒ (34 –
x)2
+
x2
=
(Pythagoras theorem)
262
C
B
⇒ 1156 – 68x + x2 + x2 = 676
⇒ 2x2 – 68x + 480 = 0 ⇒ x2 – 34x + 240 = 0
⇒ (x – 24) (x – 10) = 0 ⇒ x = 24, 10.
If x = 24, then BC = 24 cm and AB = (34 – 24) cm = 10 cm.
∴Area of ∆ ABC = 1 × BC × AB = 1 × 24 × 10 cm2
2
2
= 120 cm2.
If x = 10, then BC = 10 cm and AB = (34 – 10) cm = 24 cm.
∴Area of ∆ ABC = 1 × BC × AB = 1 × 10 × 24 cm2
2
= 120
2
cm2.
Hence, the area of ∆ ABC = 120 cm2.
Example 12. Each of equal sides of an isosceles triangle is 2 cm greater than its height. If the base
of the triangle is 12 cm, find the area of triangle.
Solution. Let ABC be an isosceles triangle with
base BC = 12 cm. Let its height AD be x cm, then D is
mid-point of BC, therefore, BD = 6 cm.
A
According to given, AB = AC = (x + 2) cm.
From right angle ∆ ABD, by Pythagoras theorem, we get
AB2 = AD2 + BD2
⇒ (x + 2)2 = x2 + 62 ⇒ x2 + 4x + 4 = x2 + 36
⇒ 4x = 32 ⇒ x = 8.
1
∴Area of ∆ ABC = 2 BC × AD
B
D
C
1
= 2 × 12 × 8 cm2 = 48 cm2.
Mensuration
3101
Example 13. If the area of an isosceles triangle is 120 cm2 and the length of each of its equal sides
is 17 cm, find its base.
Solution. Let ABC be an isosceles triangle with
AB = AC = 17 cm and its area = 120 cm2.
A
Let base BC = 2x cm.
Draw AD ⊥ BC, then D is mid-point of BC.
∴BD = DC = x cm.
In ∆ ABD, ∠ D = 90°,
∴AD2 + BD2 = AB2
(Pythagoras theorem)
B
C
D
⇒AD2 = AB2 – BD2 = 172 – x2
⇒AD =
289 − x 2 cm.
Area of ∆ ABC = 1 × BC × AD
2
⇒ 120 = 1 × 2x ×
289 − x 2 = x 289 − x 2
2
⇒ (120)2 = x2(289 – x2) ⇒ x4 – 289x2 + 14400 = 0
⇒ (x2 – 225) (x2 – 64) = 0 ⇒ x2 = 225, 64
⇒ x = 15, 8.
( x cannot be negative)
∴Base = BC = 2x cm = 30 cm or 16 cm.
Example 14. In the adjoining figure, ABC is an isosceles
triangle with base BC = 8 cm and AB = AC = 12 cm. AD
is perpendicular to BC and O is a point on AD such that
∠ BOC = 90°. Find the area of the shaded region.
A
Solution. As AD ⊥ BC, D is mid-point of BC.
O
∴BD = 1 of 8 cm = 4 cm.
2
From ∆ ABD, by Pythagoras theorem,
AD2
= AB2
–
BD2
=
122
–
42
= 144 – 16
B
D
C
⇒AD = 128 cm = 8 2 cm.
∴Area of ∆ABC= 1 BC × AD
2
= 1 × 8 × 8 2 cm2
= 32 2 cm2.
2
∆ OBD ≅ ∆ OCD
⇒ OB = OC.
Let OB = OC = x cm.
From ∆ OBC, by Pythagoras theorem,
OB2 + OC2 = BC2 ⇒ x2 + x2 = 82
⇒ 2x2 = 64 ⇒ x2 = 32.
3102
Understanding ICSE mathematics – Ix
(SAS rule of congruency)
∴Area of ∆ OBC= 1 OB × OC = 1 x × x cm2
2
2
= 1 x2 cm2 = 1 × 32 cm2 = 16 cm2.
2
2
∴Area of the shaded region= area of ∆ ABC – area of ∆ OBC
= 32 2 cm2 –­16 cm2 = 16 (2 2 – 1) cm2
= 29·25 cm2.
Exercise 15.1
1. Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.
2. Find the area of a triangle whose sides are :
(i) 3 cm, 4 cm and 5 cm.
(ii) 29 cm, 20 cm and 21 cm.
(iii) 12 cm, 9·6 cm and 7·2 cm.
3. Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the
length of the altitude corresponding to the shortest side.
4. The sides of a triangular field are 975 m, 1050 m and 1125 m. If this field is sold at the rate
of ` 1000 per hectare, find its selling price.
[1 hectare = 10000 m2]
5. The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its
area and the perimeter.
6. Find the area of an equilateral triangle whose side is 8 m. Give your answer correct to
two decimal places.
7. If the area of an equilateral triangle is 81 3 cm2, find its perimeter.
8. If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.
9. (i) If the lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is
48 cm, find its area.
(ii) The sides of a triangular plot are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find
its area. Take 3 = 1·732.
10.ABC is a triangle in which AB = AC = 4 cm and ∠ A = 90°. Calculate the area of ∆ ABC.
Also find the length of perpendicular from A to BC.
Hint:By Pythagoras theorem, BC2 = AB2 + AC2 = 42 + 42 = 32 ⇒ BC = 4 2 cm.
11. Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter
is 30 cm.
12. Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.
13. The sides of a right-angled triangle containing the right angle are 5x cm and (3x – 1) cm.
Calculate the length of the hypotenuse of the triangle if its area is 60 cm2.
14. In ∆ ABC, ∠ B = 90°, AB = (2x + 1) cm and BC = (x + 1) cm. If the area of the ∆ ABC is
60 cm2, find its perimeter.
15. If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its
area.
16. In ∆ ABC, ∠ B = 90° and D is mid-point of AC. If AB = 20 cm and BD = 14·5 cm, find the
area and the perimeter of ∆ ABC.
Hint:BD = AD = DC ⇒ AC = 2 BD = 29 cm. Use Pythagoras theorem to find BC.
Mensuration
3103