Lesson: Heron’s Formula
Exercise 12.1
Question: 1
An isosceles right triangle has area 8 cm
2
. The length
of its hypotenuse is
(A)
cm32
(B)
cm16
(C)
cm48
(D)
cm24
Solution
A
Area of an isosceles triangle
1
2
(Base × Height)
1
8
2
(Base × Height) … (i)
The triangle is an isosceles triangle.
Base Height
2
Base 16 Base 4 cm
(From equation (i))
Now, in ∆
ABC,
using Pythagoras theorem,
2 2 2 2 2
4 4 16 16AC AB BC
2
32 32 cmAC AC
Hence, the length of its hypotenuse is
32 cm
.
Question: 2
The perimeter of an equilateral triangle is 60 m. The
area is
(A)
m
2
10 3
(B)
m
2
15 3
(C)
m
2
20 3
(D)
2
m100 3
Solution
D
Let each side of the equilateral triangle be
x
.
The perimeter of the equilateral triangle
60 m
(Given)
60x x x
3 60x
m
60
20
3
x
Area of an equilateral triangle
sid s e
2
3
4
( )
3
20 20
4
m
2
100 3
Thus, the area of the triangle is
m
2
100 3 ,
Question: 3
The sides of a triangle are 56 cm, 60 cm and 52 cm
long. Then the area of the triangle is
2
A 1322 cm
2
B 1311 cm
2
C 1344 cm
2
D 1392 cm
Solution
C
The sides of the triangle are
cm 56a
,
cm60b
and
cm52 .c
(Given)
Then, the semi perimeter of the triangle is given by:
cm
56 60 52 168
84
2 2 2
a b c
s
Now, the area of the triangle
( )( )( )s s a s b s c
(By Heron’s formula)
84(84 56)(84 60)(84 52)
84 28 24 32
4 7 3 4 7 4 2 3 4 4 2
6 2 2
(4) (7) (3)
cm
3 2
(4) 7 3 1344
Hence, the area of triangle is 1344 cm
2
.
Question: 4
The area of an equilateral triangle with side
cm2 3
is
2
A 5.196 cm
2
B 0.866 cm
2
C 3.496 cm
2
D 1.732 cm
Solution
A
The side of the equilateral triangle is
cm2 3
.
Hence, the area of the equilateral triangle
side
2
3
( )
4
2
3
(2 3)
4
3
4 3
4
cm
2
3 3 3 1.732 5.196
Hence, the area of the equilateral triangle is 5.196 cm
2
.
Question: 5
The length of each side of an equilateral triangle
having an area of
2
9 3 cm
is
A 8 cm
B 36 cm
C 4 cm
D 6 cm
Solution
D
The area of the equilateral triangle is
cm
2
9 3
We know that area of an equilateral triangle is
(side)
2
3
4
(side)
2
3
9 3
4
side
2
( ) 36
∴ Side = 6 cm
Hence, the length of the equilateral triangle is 6 cm.
Question: 6
If the area of an equilateral triangle is
16 3
cm
2
,
then the perimeter of the triangle is
A 48 cm
B 24 cm
C 12 cm
D 36 cm
Solution
B
The area of the equilateral triangle is
16 3
cm
2
.
∵ Area of the equilateral triangle is
2
3
( )
4
side
side
2
3
( ) 16 3
4
Side
2
( ) 64
∴ Side = 8 cm
∴ Perimeter of the equilateral triangle
side cm3 3 8 24
Hence, the perimeter of the equilateral triangle is 24
cm.
Question: 7
The sides of a triangle are 35 cm, 54 cm and 61 cm,
respectively. The length of its longest altitude is
(A) 16 5 cm
(B) 10 5 cm
(C) 24 5 cm
(D) 28 cm
Solution
C
Let
ABC
be the triangle having sides
AB
= 35 cm,
BC
= 54 cm and
CA
= 61 cm.
Now, the semi perimeter of the triangle,
cm
35 54 61 150
75
2 2 2
a b c
s
∵ Area of the triangle
ABC
( )( )( )s s a s b s c
(By Heron’s formula)
75(75 35)(75 54)(75 61)
75 40 21 14
25 3 4 2 5 7 3 7 2
5 2 2 3 7 5
cm
2
420 5 .
Now, the area of
Altitude.
1
2
ABC AB
Let the altitude be CD.
1
35 420 5
2
CD
420 2 5
35
CD
24 5CD
cm
Hence, the length of the altitude is
cm24 5 .
Question: 8
The area of an isosceles triangle having base 2 cm and
the length of one of the equal sides 4 cm, is
2
(A) 15 cm
2
15
(B)
2
cm
2
(C) 2 15 cm
2
(D) 4 15 cm
Solution
A
Let
ABC
be an isosceles triangle such that
AB
=
AC
=
4 cm and BC = 2 cm.
Let
AD
be the perpendicular bisector from A on BC.
Now, in the right angled
,ABC
AB
2
=
AD
2
+
BD
2
(By Pythagoras
theorem)
(4)
(AD)
AD cm
2 2
2
2
1
( ) 16 1
15
15
AD
AD
Area of
1
2
ABC BC AD
(∵ Area of a triangle
1
2
(Base× Height))
cm
2
1
2 15 15
2
Question: 9
The edges of a triangular board are 6 cm, 8 cm and 10
cm. The cost of painting it at the rate of 9 paise per
cm
2
is:
A 2.00 Rs
B 2.16 Rs
C 2.48 Rs
D 3.00 Rs
Solution
B
Let the edges of the triangular board be
a
= 6 cm,
b
=
8 cm and
c
= 10 cm.
Now, the semi perimeter of the triangular board is
cm.
6 8 10 24
12
2 2 2
a b c
s
Area of the triangle
( )( )( )s s a s b s c
(By Heron’s formula)
12(12 6)(12 8)(12 10)
12 6 4 2
2 2
(12) (2)
= 12 × 2 = 24 cm
2
∵ The cost of painting an area of 1 cm
2
= Rs 0.09.
∴ The cost of painting the area of 24 cm
2
= 0.09 × 24
= Rs 2.16.
Exercise 12.2
Write True or False and justify your answer:
Question: 1
The area of a triangle with base 4 cm and height 6 cm
is 24 cm
2
.
Solution
False.
Area of a triangle =
base height
1
( )
2
Base = 4 cm and height = 6 cm
∴Area of the triangle =
cm
2
1
4 6 12
2
Question: 2
The area of
ABC
is 8 cm
2
in which
cm4AB AC
and
90A
.
Solution
True.
Let
ABC
be right angled at
A.
Sides of the triangle are
cm4AB AC
∴ Area of a triangle
ABC
base height
1
2
cm
2
1
2
1
4 4
2
8
AC AB
Question: 3
The area of the isosceles triangle is
5
11
4
cm
2
, if the
perimeter is 11 cm and the base is 5 cm.
Solution
True.
Let the equal sides of the isosceles triangle be
a
.
∴ Perimeter of the triangle
5a a
11 2 5a
2 11 5
2 6
a
a
6
2
a
cm3
Hence, the semi perimeter of the isosceles triangle =
cm
3 3 5 11
2 2 2
a b c
s
Now, the area of the triangle
( )( )( )s s a s b s c
(By Heron’s formula)
This implies,
11 11 11 11
( 3)( 3)( 5)
2 2 2 2
11 5 5 1
2 2 2 2
5
11
4
cm
2
Question: 4
The area of an equilateral triangle is
20 3
cm
2
whose
each side is 8 cm.
Solution
False
The side of an equilateral triangle is 8 cm.
Area of the equilateral triangle
2
2
2
3
( )
4
3
(8)
4
16 3
side
cm
Question: 5
If the side of a rhombus is 10 cm and one diagonal is
16 cm, the area of the rhombus is 96 cm
2
.
Solution
True
The side of the rhombus
PQ
RS is 10 cm and one of the
diagonal is 16 cm.
⟹
PQ
=
QR
= RS =
SP
= 10 cm and one of the
diagonal is 16 cm, say PR = 16 cm.
Since, the diagonals of rhombus bisect each other at
right angle (say at
O
).
Hence, in
2 2 2
,POQ PQ OP OQ
(By Pythagoras theorem)
2 2 2 2 2
10 8OQ PQ OP
2
100 64 36OQ
36 OQ
cm6OQ
2 2 6 12SQ OQ cm
∴ Area of a rhombus =
1
2
(Product of diagonals)
1
2
(
QS
×
PR
) =
1
2
× 12 × 16 = 96 cm
2
.
Question: 6
The base and the corresponding altitude of a
parallelogram are 10 cm and 3.5 cm, respectively. The
area of the parallelogram is 30 cm
2
.
Solution
False.
Base and the corresponding altitude of a parallelogram
are 10 cm and 3.5 cm.
∴ Area of a parallelogram
= Base × Altitude
10 × 3.5
= 35 cm
2
Question: 7
The area of a regular hexagon of side ‘
a
’ is the sum of
the areas of the five equilateral triangles with side
a
.
Solution
False.
Since, a regular hexagon is divided into six equilateral
triangles
∴ Area of a regular hexagon of side a
= Sum of the area of the six equilateral triangles of
side
a
.
Question: 8
The cost of levelling the ground in the form of a
triangle having the sides 51 m, 37 m and 20 m at the
rate of Rs 3 per m
2
is Rs 918.
Solution
True.
Let the sides of the triangle be
a
= 51 m,
b
= 37 m and
c
= 20 m.
∴ Semi perimeter of the triangle,
a b c
s
2
51 37 20 108
2 2
54 m
∴ Area of the triangle
( )(s )( )a bs s cs
(By Heron’s formula)
2
54 54 51 54 37 54 20
54 3 17 34
9 3 2 3 17 17 2
3 3 2 17
306 m
∵ Cost of levelling per m
2
= Rs 3
∵ The cost of levelling 306 m
2
= 3 × 306 = Rs 918
Question: 9
In a triangle, the sides are given as 11 cm, 12 cm and
13 cm. The length of the altitude is 10.25 cm
corresponding to the side having length 12 cm. The
area of the triangle is 61.5 cm
2
.
Solution
True.
The sides of a triangle are
a
=11 cm,
b
= 12 cm and
c
=13 cm.
∴ The semi perimeter of the triangle is
a b c
s
2
11 12 13 36
18
2 2
cm
∴ Area of a triangle
s(s )(s )(s )a b c
(By Heron’s formula)
18 18 11 18 12 18 13
=
18 7 6 5
3 6 7 6 5
6 3 7 5
56 10
2
6 10.25
61.5 cm
We can also find the area using:
Area of a triangle
1
2
(base × height)
cm
2
12 10.25 6 10.25 61
1
5
2
.
Exercise 12.3
Question: 1
Find the cost of laying grass in a triangular field of
sides 50 m, 65 m and 65 m at the rate of Rs 7 per m
2
.
Solution
Let ABC be a triangular field of sides
AB
=
a
= 50 m,
BC
=
b
= 65 m and CA = c = 65 m.
∴ The semi perimeter of the triangular field is,
a b c 50 65 65
2 2
180
90
2
m
s
∴ Area of a triangle
s(s ) (s ) (sb )a c
(By Heron’s formula)
Area of the triangular field =
90 90 50 90 65 90 65
90 40 25 25
3 2 10 25
m
2
1500
∵ Cost of laying grass per 1 m
2
= Rs 7
∴ The cost of laying grass per 1500 m
2
= 7 × 1500 =
Rs 10,500
Question: 2
The triangular side walls of a flyover have been used
for advertisements. The sides of the walls are 13 m, 14
m and 15 m. The advertisements yield an earning of
Rs 2000 per m
2
a year. A company hired one of its
walls for 6 months. How much rent did it pay?
Solution
The sides of a triangular walls are
a
=13 m,
b
= 14 m
and
c
= 15 m.
∴ The semi perimeter of the triangle is:
m
13 14 15 42
2 2
21
a b c
s
2
∴ Area of the triangular side wall
s(s ) (s ) (sb )a c
(By Heron’s formula)
21 21 13 21 14 21 15
21 8 7 6
21 4 2 7 3 2
2 2
(21) (4)
m
2
21 4 84
Now, the advertisement yield earning per year for
m Rs
2
1 2000
∵ Advertisement yield earning per year on 84 m
2
2000 84 168000. Rs
As the company hired one of its walls for 6 months,
therefore the company pays the rent
1
168000 84000
2
Rs
Hence, the company paid rent of Rs 84,000.
Question: 3
From a point in the interior of an equilateral triangle,
perpendiculaRs are drawn on the three sides. 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 and the length of
each side be
a
.
O
is the interior point of the triangle
and
OQ
,
OR
and
OP
are the perpendicular lines drawn
from point
O
.
Area of ∆
OAB,
cm
2
1
2
1
14 7 (i)
2
AB OP
a a
Area of
1
2
ΔO C OQB BC
cm
2
1
10 5 (ii)
2
a a
Area of
1
2
Δ C RO OA AC
cm
2
2
6 3 iii
1
a a
∴ Area of an equilateral ∆
ABC
= Area of (∆OAB +
∆
OBC
+ ∆
OAC
)
cm
2
7 5 3 15 iva a a a
[From (i), (ii) and (iii)]
Now, the semi perimeter of the triangle
s
=
+ +
2
a a a
⇒
cm
3
=
2
a
s
∴ Area of the equilateral
( ) ( ) ( )s s sABC a b cs
(By Heron’s formula)
3
2
a a a a
a a a
3 3 3
2 2 2
2
3 3
(v)
2 2 2 2 4
a a a a
a
From equations (iv) and (v),
cm
2
3
15
4
15 4 3
20 3
3 3
a a
a
On putting
20 3
in equation (v), we get
Area of a
2
3
(20 3)
4
ABC
cm
2
3
400 3
4
300 3
Hence, the area of the equilateral triangle is
2
300 3 cm
.
Question: 4
The perimeter of an isosceles triangle is 32 cm. The
ratio of the equal side to its base is 3:2. Find the area
of the triangle.
Solution
Let
ABC
be an isosceles triangle having perimeter 32
cm.
The ratio of the equal side to its base is 3:2.
Let the sides of the triangle be
and 3 2AB AC x BC x.
∵ The perimeter of the triangle =
3 3 2 32x x x
cm
⇒ 8
x
= 32
⇒
x
= 4 cm
∴ AB = AC = 3 × 4 = 12 cm
and
cm2 2 4 8BC x
.
The sides of the triangle are
cm cm and cm.12 , 12 8a b c
cm.
+ + 12 12 8 32
16
2 2 2
a b c
s
∴ Area of the isosceles
( ) Δ ( ) ( )s s sC csAB a b
(By Heron’s formula)
16 16 12 16 12 16 8
16 4 4 8
cm cm
2 2
4 4 2 2 32 2
Hence, the area of the isosceles triangle is
cm
2
32 2
.
Question: 5
Find the area of a parallelogram given in Fig. 12.2.
Also find the length of the altitude from vertex
A
on
the side
DC
.
Fig. 12.2
Solution
Area of the parallelogram
2ABCD
(Area
of
BCD
) … (i)
Now, the sides of
BCD
are
cm cm12 , 17a b
and
cm25c
.
∴ The semi perimeter of
BCD
,
cm
2
12 17 25 54
2 2
27
a b c
s
∴ Area of
BCD
( ) ( ) ( )s s s b sa c
(By Heron’s formula)
27 27 12 27 17 27 5 2
27 15 10 2
9 3 3 5 5 2 2
cm
2
3 3 5 2 90
From equation (i), we get, the area of the
parallelogram
cm
2
2 90 180ABCD
.
Now, let the altitude of the parallelogram be
h
.
Also, the area of a parallelogram
ABCD
= Base ×
Altitude
180 DC h
180 12 h
180
12
h
= 15 cm
Hence, the area of the parallelogram is
cm
2
180
and
the length of the altitude is 15 cm.
Question: 6
A field in the form of a parallelogram has sides 60 m
and 40 m and one of its diagonals is 80 m long. Find
the area of the parallelogram.
Solution
Let
ABCD
be the field in the shape of a parallelogram
with following:
(i) Sides
m60AB CD
, and
m40BC DA
(ii) Diagonal
m80BD
Area of the parallelogram
2ABCD
(Area of
∆
ABD
) …………… (i)
Now, in ∆
ABD
,
∴ The semi perimeter of the triangle,
m
2
60 80 40 180
2 2
90
a b c
s
∴ Area of
( ) Δ ( ) ( )s s sD csAB a b
(By Heron’s
formula)
90 90 60 90 80 90 40
90 30 10 50
5100 13
m
2
15300
From equation (i), we get,
The area of the parallelogram
ABCD
= 2 × 300
15
=
600
15
m
2.
Hence, the area of the parallelogram is 600
m
2
15
.
Question: 7
The perimeter of a triangular field is 420 m and its
sides are in the ratio
6: 7: 8
. Find the area of the
triangular field.
Solution
The perimeter of the triangular field is 420 m and its
sides are in the ratio
6: 7: 8
. (Given)
Let the sides of the triangular field be
6 , 7a x b x
and
8 .c x
∴ The perimeter of the triangular field
a b c
420 6 7 8 420 21x x x x
m.
420
20
21
x
∴ The sides of the triangular field are
m,6 20 120a
m7 20 140b
and
m8 20 160c
.
Now, the semi perimeter (s)
420
210
2
∴ The area of the triangular field
( ) ( ) ( )s s s b sa c
(By Heron’s formula)
210 210 120 210 140 210 160
210 90 70 50
m
2
100 7 3 15 2100 15
Hence, the area of the triangular field is
m
2
2100 15 .
Question: 8
The sides of a quadrilateral
ABCD
are 6 cm, 8 cm, 12
cm and 14 cm (taken in order) respectively, and the
angle between the first two sides is a right angle. Find
its area.
Solution
ABCD
is a quadrilateral having sides
AB
= 6 cm,
BC
= 8 cm,
CD
= 12 cm and
DA
= 14 cm.
Now, we join
A
and
C
.
∵
ABC
is a right angled triangle at
B
.
Hence, by Pythagoras theorem,
2 2 2
AC AB BC
2 2 2
2
6 8
100
100
10
AC
AC
AC
AC
cm10AC
∴ Area of quadrilateral
ABCD
= Area of ∆
ABC
+
Area of ∆
ACD
Now, area of
1
2
BC BA CB A
cm
2
1
6 8
2
3 8 24
Now, in
cm cm, 10 , 12 ,ACD AC a CD b
and
cm.14DA c
∴ Semi perimeter of
ACD
cm.
a b c 10 12 14
2 2
36
18
2
s
Now, area of
( )(Δ )( )s s s sACD a b c
(By Heron’s
formula)
18 18 10 18 12 18 14
18 8 6 4
2
2
cm
(3) 2 4 2 3 2 4
3 4 2 3 2
24 6
Therefore, the area of the quadrilateral
ABCD
= 24 +
24 cm
2
Question: 9
A rhombus shaped sheet with perimeter 40 cm and one
diagonal 12 cm, is painted on both sides at the rate of
Rs 5 per m
2
. Find the cost of painting.
Solution
Let
ABCD
be a rhombus having each side equal to
x
cm.
⟹
cmAB BC CD DA x
∵ The perimeter of the rhombus = 40 cm.
40AB BC CD DA
40x x x x
4 40x
40
4
x
cm 10x
Let AC be the given diagonal.
Now, in ∆
ABC
,
Let
cm cm10 , 10a AB b BC
and
cm12c AC
∴ The semi perimeter of ∆
ABC,
a b c
s
10 10 12
2 2
32
16
2
cm
Now, by Heron’s formula,
Area of
( )(Δ )( )s s sBC csA a b
16 16 10 16 10 16 12
16 6 6 4
4 6 2
= 48 cm
2
∴ Area of the rhombus
2
(Area of ∆
ABC
)
2 48
= 96 cm
2
∵ Cost of painting of sheet of 1 m
2
= Rs 5
∴ The cost of painting of the sheet of 96 cm
2
=
96×5/10000 = Rs 0.048
Hence, the cost of the painting of the sheet for both
sides
2 0.048 . Rs 0.096
Question: 10
Find the area of the trapezium
PQ
RS with height
PQ
given in Fig. 12.3
Fig. 12.3
Solution
In the trapezium
PQ
RS, draw a line
RT
perpendicular
to
PS
.
Hence, the side,
m12 7 5 .ST PS TP
Now, in ∆
STR
,
2 2 2
SR ST TR
(By using Pythagoras theorem)
2 2 2
13 5 TR
m
2
2
169 25
144
12
TR
TR
TR
Now, the area of
1
2
STR TR TS.
(∵ Area of triangle =
1
2
(Base × Height))
m
2
1
12 5 30
2
Now, area of the rectangle
PQRT
=
PQ
×
RQ
=12×7
2
84 m
Therefore, the area of
PQ
RS =
ar(STR)
+
ar(PQRT)
= 30 + 84
= 114 m
2
The area of the trapezium is 114 m
2
.
Exercise 12.4
Question: 1
How much paper of each shade is needed to make a
kite given in Fig. 12.4, in which
ABCD
is a square
with diagonal 44 cm?
Fig. 12.4
Solution
∵ All the sides of a square are always equal.
⟹
AB BC CD DA
Now, in ∆
ACD
,
cm 44 , 90AC D
(∵
ABCD
is a square)
Now, in ∆
ACD
, by Pythagoras theorem, we get
2 2 2
AC AD DC
2 2 2
2
2
44
2 44 44
22 44
22 44
AD AD
AD
AD
AD
2 11 4 11AD
cm22 2 AD
Hence,
AB
= BC = CD =
DA
= 22
2
cm.
∴ Area of a square
ABCD
= Side × Side
= 22
2
× 22
2
= 968 cm
2
.
∴ Area of the red portion =
968
4
.
= 242 cm
2
.
(Since, the area of the square is divided into four
parts)
Now, area of the green portion
968
4
= 242 cm
2
.
Area of the yellow portion
968
2
= 484 cm
2
.
Now, in the triangle,
cm cm20 , 20a b
and
cm14c
.
The s ( )
2
emi perimeter
cm
20 20 1 54
27
2 2
4
a b c
s
∴ Area of the triangle =
( )( )( )s s a s b s c
(By Heron’s formula)
27 27 20 27 20 27 14
27 7 7 13
3 3 3 7 7 13
921 3
21 6.24
= 131.04 cm
2
∴ Total area of the green portion = 242 +131.04 =
373.04 cm
2
.
Hence, the papers required for each shade to make a
kite are:
Red paper= 242 cm
2
, yellow paper= 484 cm
2
and green
paper= 373.04 cm
2
.
Question: 2
The perimeter of a triangle is 50 cm. One side of a
triangle is 4 cm longer than the smaller side and the
third side is 6 cm less than twice the smaller side. Find
the area of the triangle.
Solution
Let the smaller side of a triangle be
x
cm.
According to the question, the second side is 4 cm
longer than the smaller side
i.e.,
cm,4x
the third side = 6 cm less than twice the smaller side
c.e., mi .2 6x
Given that the perimeter of the triangle = 50 cm.
4 2 6 50x x x
4 2 50x
4 52x
52
4
x
cm13x
The smaller side, .
cm.13a
The second side,
cm.13 4 17b
The third side,
cm.2 13 6 26 6 20c
Now, the semi perimeter is given by
cm
13 17 20 50
25
2 2 2
a b c
s
∴ Area of the triangle
s s a s b s c
(By Heron’s formula)
25 25 13 25 17 25 20
25 12 8 5
5 5 2 2 3 2 2 2 5
5 2 2 30
20 30
2
cm
Hence, the area of the triangle is
cm
2
20 30
.
Question: 3
The area of a trapezium is 475 cm
2
and the height is 19
cm. Find the lengths of its two parallel sides if one
side is 4 cm greater than the other.
Solution
Let one of the parallel sides of trapezium be
x
cm.
Hence, the length of the other parallel side
cm4x
(given in the question)
∵ Area of trapezium
1
2
(Sum of parallel Sides) ×distance between the
parallel sides
⇒ 475 =
1
2
(
x
+
x
+ 4) × 19
4
475
2 19
2
x
cm
2 4 475
2 19
2 2
475
2 19
475
2
19
2 25
23
x
x
x
x
x
∴ Other Side
cm4 23 4 27x
Hence, the parallel sides are 23 cm and 27 cm.
Question: 4
A rectangular plot is given for constructing a house,
having a measurement of 40 m long and 15 m in the
front. According to the laws, a minimum of 3 m, wide
space should be left in the front and back each and 2 m
wide space on each of other sides. Find the largest
area where house can be constructed.
Solution
Let
ABCD
be a rectangular plot which is 40 m long
and 15 m in the front.
∵ As per the laws, a minimum of 3 m, wide space
should be left in the front and back each.
∴ The length of the inner-rectangle
m.40 3 3 34
Similarly, the breadth of the inner–rectangle,
m.15 2 2 11
Let the rectangular plot be
ABCD
and the other
rectangle
EFGH
will be formed inside the rectangle
ABCD.
∴ Area of inner rectangle
EFGH
= Length × Breadth
(∵Area of a rectangle= length × breadth)
=34 ×11
= 374 m
2
Hence, the largest area where the house can be
constructed in 374 m
2
Question: 5
A field is in the shape of a trapezium having parallel
sides 90 m and 30 m. These sides meet the third side at
right angles. The length of the fourth side is 100 m. If
it costs Rs 4 to plough 1 m
2
of the field, find the total
cost of ploughing the field.
Solution
Let the trapezium be
ABCD
.
Now, let us draw a perpendicular line
CE
to the line
AB.
We have,
m.30DC AE
Now,
m.90 30 60BE AB AE
In the right angled Δ
BEC,
2 2 2
BC BE EC
(Using Pythagoras theorem)
2 2 2
2
2
100 60
10000 3600
6400
EC
EC
EC
m EC 6400 80
∴ Area of a trapezium
1
2
(Sum of parallel Sides) ×distance between
parallel sides
1
2
(
AB
+
CD
)×
EC
=
1
2
(90+30) × 80
m
2
120 80 48
1
0
2
0
∵ Cost of ploughing the field of 1 m
2
= Rs 4
∴ The cost of ploughing the field of 4800 m
2
= 4800 ×
4 = Rs 19200
Hence, the total cost of ploughing the field is Rs
19200.
Question: 6
In Fig. 12.5,
ABC
has
sides
cm cm7.5 , 6.5AB AC
and
cm.7BC
On
the base
BC
a parallelogram
DBCE
of same area as
that of
ABC
is constructed. Find the height
DF
of
the parallelogram.
Fig. 12.5
Solution
The sides of the triangle are
cm cm7.5 , 7AB a BC b
and
cm.6.5CA c
∴ The semi perimeter of the triangle
ABC
,
10
.
cm
Area of
+ + 7.5+7+6.5 21
.5
2 2 2
the ΔABC
a b c
s
s s a s b s c
(By Heron’s formula)
2
10.5 10.5 7.5 10.5 7 10.5 6.5
10.5 3 3.5 4
441 21 ..................... icm
Now, the area of the parallelogram
BCED
= Base ×
Height
7 iiBC DF DF
According to the question,
Area of a Δ
ABC
=Area of parallelogram
BCED
21 7 DF
(From Equations (i) and (ii))
cm
1
7
3
2
DF
Hence, the height of the parallelogram is 3 cm.
Question: 7
The dimensions of a rectangle
ABCD
are 51 cm × 25
cm. A trapezium
PQCD
with its parallel sides
QC
and
PD
in the ratio 9:8 is cut off from the rectangle as
shown in the Fig. 12.6. If the area of the trapezium
PQCD
is
th
5
6
part of the area of the rectangle, find
the lengths
QC
and
PD
. Fig. 12.6
Fig. 12.6
Solution
Given, the dimensions of the rectangle
ABCD
as 51
cm × 25 cm.
Also, in trapezium
PQCD
, the parallel sides
QC
and
PD
are in the ratio 9:8.
⟹
: 9: 8 QC PD
Let, the length of
be 9xQC
and
8 .PD x
∵ The area of the trapezium
PQCD
5
6
area of the
rectangle
ABCD,
1
2
(Sum of the parallel Sides × distance between
the parallel sides )
5
6
BC CD
⇒
1
2
(8
x
+ 9
x
) × 25 =
5
6
× 51 × 25
⇒
1
2
× 17
x
× 25 =
5
6
× 51 × 25
⇒
x
=
5 51 25 2
25 17 6
∴
5x
cm
So,
cm9 5 45QC
cm8 5 40PD
Question: 8
A design is made on a rectangular tile of dimensions
50 cm × 70 cm as shown in Fig.12.7. The design
shows 8 triangles, each of sides 26 cm, 17 cm and 25
cm.
Find the total area of the design and the remaining
area of the tile.
Fig.12.7
Solution
The dimension of rectangular tile is 50 cm × 70 cm.
∴ The area of the rectangular title = 50 ×70 = 3500
cm
2
.
The sides of the triangle are
a
= 25 cm,
b
= 17 cm and
c
= 26 cm.
Hence, the semi perimeter of the triangle is
25 1 26
34
.
cm.
Area of the triangle
7 68
2 2 2
a b c
s
s s a s b s c
(By Heron’s formula)
(34 25) (34 17) (34 26)
2
34
34 9 17 8
17 2 3 3 17 2 2 2
17 3 2 2 204 cm
∵ The design is made of eight triangles.
∴ The area of the design
cm
2
204 8 1632
.
Also, the remaining area of the tile = Area of the
rectangular tile–Area of the design.
= 3500 – 1632 = 1868 cm
2
.
Hence, the total area of the design is 1632 cm
2
and the
remaining area of the tile is 1868 cm
2
.