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

.