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Revision Question Bank

Heron's Formula

1. Find the area of an equilateral triangle having each side 4a.

Sol :

Here, side of an equilateral triangle = 2a

Area of an equilateral triangle

= 2 23 3

Side 2a4 4

= 234a

4

= 3 a2 sq units

Hence, area of an equilateral triangle is 3 a2 sq units.

2. Find the area of an isosceles triangle having base 4 cm and length of one of equal sides as

6 cm.

Sol :

In an isosceles ABC, AD is a perpendicular bisector of BC.

Here, AB = 6cm

and BD = DC = BC

2

= 4

22 cm

In ADB, using Pythagoras theorem

AB2 =AD2 +BD2



AD2 = AB2 – BD2

AD2 =(6)2 – (2)2

AD2 =36 – 4

AD2 =32

AD = 4 2 cm

[taking positive square root, because length is positive]

Area of an isosceles 1

ABC2

× BC × AD

= 1

2 × 4 × 2 8 2 cm2

Hence , area of an isosceles triangle is 28 2 cm .

3. The length of the sides of a triangle are in the ratio 4:3:5. If the perimeter of the triangle

is 96 cm. Find its area.

Sol :

Let the sides of a triangle be

a = 4x, b = 3x and c = 5x

Given, perimeter of a triangle = 96 cm

a + b + c = 96

4x + 3x + 5x = 96

12x = 96

96x 8 cm

12

Sides of a triangle, a = 4x = 4 ×8 = 32 cm.

b = 3x = 3 × 8 =24cm

and c = 5 × = 5 × 8 = 40cm

Semi-perimeter, s =a b c

2

= 32 24 40

2



= 96

2= 48 cm

Using Heron's formula,

Area of a triangle

= s s a s b s c

= 48 48 32 48 24 48 40

= 48 16 24 8

= 2147456 384 cm

Hence, area of a triangle is 384cm2.

4. Two parallel sides of a trapezium are 120 cm and 154 cm and other sides are 50 cm and

52 cm. Find the area of trapezium.

Sol :

ABCD is a trapezium in which AB = 154 cm, BC = 52 cm, CD = 120 cm and DA = 50 cm.

Through vertex C, draw CE AD, which meets AB at point E.

Since, AECD is a parallelogram and opposite sides of a parallelogram are equal.

CE = AD =50 cm, AE = DC =120 cm

and EB = AB – AE =150 – 120 = 30 cm

Let a = 50 cm, b = 52 cm

and c = 30 cm

a b c

s2

= 50 52 30 132

2 2



= 66 cm


Area of ACEB = CEB s s a s b s c

= 66 66 50 66 52 66 30

= 66 16 14 36

= 2 4 6 231

= 48 2231 cm

Now, area of 1

CEB2

× h × BE

1

48 231 h 302

96 231

h 3.2 23130

Area of trapezium ABCD

= 1

2(Sum of two parallel sides) × Height

= 1

2 (120 + 154) × 3.2 231

= 1

2 × 274 × 48.64

= 213327366663.68 cm

2

5. In the figure, find the area of the shaded region.



Sol :

For smaller BDC,

a = 22m, b =24m

and c = 26m

Then, semi-perimeter,

a b c 22 24 26s

2 2

= 72

36m2

Area of the smaller triangle

= 36 36 22 36 24 36 26

= [ s s s a s b s c by Heron’s formual]

= 36 14 12 10

= 224 105 m

For bigger ABC,

Now, 222 +1202

= 484 +14400 =14884 = 1222

It is a right triangle having sides as 22 m and 120 m.

Area of the larger ABC

= 1

2 × Base × Height =

1

2× 22 × 120

= 1320m2

Area of the shaded region = Area of larger ABC – Area of smaller BDC

= (1320 24 105 ) m2

= (1320-246) m2

= 1074m2

Hence, area of the shaded region is 1074 m2



6. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is

42 cm.

Sol :

Given, sides of a triangle, a = 18 cm, b=10 cm and perimeter = 42 cm

Hence, area of the shaded region is 1074m2.

Let the third side of a triangle be C.

Perimeter = 42

a + b + c =42

18+10+ c =42

c – 42 – 28 = 14 cm

Semi-perimeter,

a b c 18 10 14s

2 2

42s 21 cm

2


Area of a triangle

= s s a s b s c

= 21 21 18 21 10 21 14

= 21 3 11 7

= 24851 69.65cm

7. A traffic signal board, indicating 'SCHOOL AHEAD' is an equilateral triangle with side 'a'.

Find the area of a signal board using Heron's formula. If its perimeter is 180 cm, then

what will be the area of the signal board?

Sol :

Let each side of an equilateral triangle be a.

Semi-perineters =a b c

2



= a a a

2

= 3a

2


Area of the triangular signal board

= s s a s b s c

= 3a 3a 3a 3a

a a a2 2 2 2

= 3a a a a

2 2 2 2

= 23a

4

Given that, perimeter of an equilateral triangle is 180 cm.

3a = 180 cm

a = 180

3 = 60 cm

Area of the signal board = 23a

4

= 21.732

604

= 1.732 3600

4

= 1.732 × 900



= 1558.8 cm2

Hence, area of the signal board is 1558.8 cm2

8. Find the area of the cyclic quadrilateral ABCD having sides ‘a’ = 7cm, BC = 9cm, CD=12cm

and DA=6 cm.

Sol :

Let sides of a cyclic quadrilateral be

a = AB=7cm, b = BC = 9cm, c = CD=12cm

and d = DA=6cm

Semi-perimeter, s = a b c d

2

= 7 9 12 6

2

= 34

2 = 17 cm

Area of cyclic quadrilateral ABCD

= s a s b s c s d [by Brahmagupta's formula]

= 17 7 17 9 17 12 17 6

= 10 8 5 11

= 2 5 2 2 2 5 11

= 2 × 2 × 5 11

= 20 11 cm2

9. For an equilateral triangle, find.

(i) area, when its perimeter is 48 cm.

(ii) perimeter, when its area is 144 3 cm2, if required take 3 =1.732.

Sol :

(i) Given, the perimeter of an equilateral triangle = 48 cm

3a = 48



a = 16 cm

Area of an equilateral triangle = 23a

4

= 3

16 164

= 264 3 cm

(ii) Given, area of an equilateral triangle

= 2144 3 cm

23a 144 3

4

2 144 3 4a

3

a2 = 144 × 4 = 576

a = 24 cm

Perimeter = 3 × a = 3 × 24

= 72 cm

10. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours,

each piece measuring 20 cm, 50 cm and 50 cm.

How much cloth of each colour is required for the umbrella?

Solution :

For each triangular piece,

a = 50 cm, b = 50 cm

and c = 20 cm



Semi-perimeter,

a b c 50 50 20s

2 2

= 120

2 = 60 cm


Area of each triangular piece

= s s a s b s c

= 60 60 50 60 50 60 20

= 60 10 10 40

= 200 6 200 2.45

= 2490 cm

Since, the umbrella has 10 triangular pieces of clothes of two different colours,

therefore triangular pieces of cloth of each colour = 10

2 = 5

Cloth required for each colour

= 5 × 490

= 2450 cm2



Chapter Test {Heron's Formula}

M: Marks: 40 M: Time: 40 Min.

1. The triangular side walls of a flyover have been used for advertisement ‘SAVE WATER’.

The sides of the walls are 122m, 22m and 120 m. The advertisement yield an earning of

Rs 5000 per m2 per year. A company hired one of its walls for 3 months.

(i) How much rend did it pay?

(ii) What is the value of the message ‘SAVE WATER’ in the society? [4]

Sol. (i)

The length of the sides of the triangular walls are 122 m, 22 m and 120 m.

122 22 120s

2

= 264

2 = 132 m

Area of one triangular wall by using Heron's formula

= s s a s b s c

= 132 132 122 132 22 132 120

= 132 10 110 12

= 1320 m2

Now, yearly rent = Rs 5000 per m2

1 yr = 12 months



Then, monthly rent = Rs 5000 × 1

12

Company hired one of its walls for 3 months. Hence, rent paid by the company for 3

months

= Area of wall x Monthly rent × 3

= Rs 1320 × 5000

312

= Rs 1650000

(ii) Water is precious. We should not use it unnecessarily. We should conserve it for

coming generation.

2. How much paper of each shade is needed to make kite given in the figure, in which ABCD

is a square with diagonal 44 cm? [4]

Sol :

We know that, the all sides of square are equal.

AB = BC = CD = DA

In ACD , we have

AC = 44 cm, D = 900

Using Pythagoras theorem,

(AC)2 = (AD)2 + (DC)2

(44)2 = (AD)2 + (AD)2



[ AD = DC, sides of a square]

2AD2 = 44 × 44

AD = 22 44 [taking positive square root]

AD 22 2 cm

So, AB = BC = CD = DA = 22 2 cm

Area of square ABCD = Side × Side

= 22 2 22 2

= 968 cm2

Area of red portion = 2968242 cm

4

[ area of square is divided into four quadrants]

Area of green portion = 2968242 cm

4

and area of yellow portion = 2968484cm

2

In PCQ , we have

a = PC = 20cm, b = CQ = 20 cm, c = PQ = 14 cm

s = 20 20 14 54

27 cm2 2

Area of PCQ s s a s b s c [by Heron’s formula]

= 27 27 20 27 20 27 14

= 27 7 7 13 3 7 39 21 39

= 21 × 6.24 – 131.04 cm2

3. The height of an equilateral triangle measure 9 cm. Find its area, correct to 2 places of

decimal. (Take 3 = 1.732) [4]

Solution :

Height = 9 cm



Let side of the equilateral triangle be a cm.

Area of equilateral 1

ABC2

× base × height

= 1

2 × a × 9

Also, by Heron’s formula : –

Area of ABC = s s a s a s a , where s = 3a

2.

Area of ABC = 3a 3a 3a 3a

a a a2 2 2 2

1 3a 1a 1a 1a

a 92 2 2 2 2

29a a

32 4

9 3

a2 4

9 4 2 9 18

a cm.2 3 3 3

Area of ABC = 1 18

92 3

= 81 3 81 3

33 3



= 27 3

= 27 × 1.732

= 46.764 cm2.

4. A triangle and a parallelogram have the same base and the same area. If the sides of the

triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm,

then find the height of the parallelogram. [4]

Solution :

Lets be the semi-perimeter of the triangle. Then,

s = 26 28 30

42cm2

Area of the triangle = 242 42 26 42 28 42 30 cm

= 42 16 14 12 cm2 = 336 cm2

Let h be the height of the parallelogram .

It is given that the triangle and the parallelogram have the same base and between the

same parallels have the same area.

Area of the parallelogram = 336 cm2

Base × Height = 336 cm2

28 × h = 336 cm

h = 336

28cm = 12 cm.

5. A cyclic quadrilateral ABCD, in which AB = 4m, BC=9m, CD = 11 m and DA = 6m. F ind the

area of the quadrilateral ABCD. [4]

Solution :

Let a = 4 cm, b = 9 m, c = 11m & d = 6 m.

By Brahmagupta’s formula : –

Area of quadrilateral ABCD = s a s b s c s d ,

where 4 9 11 6 30

s 15m2 2

.



Area = 15 4 15 9 15 11 15 6

= 11 6 4 9

= 11 3 2 2 2 3 3

= 6 11 3 2

= 6 66 m2.

6. A rhombus shaped field has green grass for 20 cows to graze. If each sides of the

rhombus is 52 m and its longer diagonal is 96 m,. Then how much area of the grass field

will each cow be getting? [4]

Solution :

Clearly, triangles ABD and BCD are congruent.

Area of ABD = Area of BCD.

Let S be the semi-perimeter of BCD & ABD.

S = 52 52 96 200

100m2 2

Area of ABD = Area of BCD

= 100 100 52 100 52 100 96

= 100 48 48 4

= 960 m2.

So, area of rhombus = 2 × 960

= 1920 m2.

Area of grass field each cow will graze = 1920

20 = 96 m2

7. If a field is in the shape of trapezium whose parallel sides are 25m and 10m. The non-

parallel sides are 14 m and 13 m, then find the area of the field. [4]

Solution :

From C, draw CE DA. Clearly, ADCE is a parallelogram having AD CE and DC AE such

AD = 13 m and DC = 10 m.



AE = DC = 10 m and CE = AD = 13 m

BE = AB – AE = (25 – 10) m = 15 m

Thus, in BCE, we have

BC = 14 m, CE = 13 m and BE = 15 m

Let s be the semi-perimeter of BCE. Then,

2s = BC + CE + BE = 14 + 13 + 15 = 42

s = 21

Area of BCE = 21 21 14 21 13 21 15

Area of BCE = 21 7 8 6

Area of BCE = 2 2 2 27 3 4 84 m

Also, Area of BCE = 1

2 (BE × CL)

1

842

× 15 × CL

168 56

CL15 5

Height of parallelogram ADCE = CL = 56

5m

Area of parallelogram ADCE = Base × Height = AE × CL = 10 × 56

5 = 112 m2



Hence, Area of trapezium ABCD = Area of parallelogram ADCE + Area of BCE

= (112 + 84) m2 = 196 m2

8. Anita has piece of land which is in the shape of a rhombus. She has two

children, one daughter Preeti and one son. Narendra to work on the land produce

different types of crops. She divided the land in two equal parts. If the perimeter of the

land is 200 m and one of the diagonal is 70 m, then solve the following questions.

(i) How much area each of them will get for their crops?

(ii) Anita covered the land by wires. How much wire is required to cover the land?

(iii) As Anita divided the land into two equal parts. Justify her decision. [4]

Solution :

(i) Perimeter = 200 m.

Length of each side = 200

4

= 50 m

Let a = 50 m, b = 50 m & c = 70 m

Area of CBD = s s a s b s c , where s = 50 50 70 170

2 2

= 85m

Area of CBD = 85 85 50 85 50 85 70

= 85 35 35 15

= 5 17 5 7 5 7 5 3



= 5 5 5 5 7 7 17 3

= 25 7 17 3

= 2175 51 m

Area each of them will be getting = 2175 51 m

(ii) The wire required to cover the land = Perimeter = 200 m.

(iii) Anita believes in gender equality, there is no difference between son and daughter.

9. A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10m long, is

pained on both sides at the rate of Rs. 5 per m2. Find the cost of painting. [4]

Solution :

Perimeter = 32 m

4 × Side = 32

Side = 32

8m4

Let AC be the diagonal whose length = 10 m

By Heron’s formula,

Area of ABC s s a s b s c , where s = 10 8 8

2

= 26

2 = 13m.



= 13 13 8 13 8 13 10

= 13 5 5 3

= 25 13 13 5 39 m

Area of rhombus ABCD = 2 × 5 39

= 10 × 6.245

= 62.45 m2

Cost of 1 m2 area = Rs 5

Cost of 62.45m2 area = 5 × 62.45

= 312.25 Rs

Since rhombus is to be pointed on both the sides

Total cost = 2 × 312.25

= 624.5 Rs.

10. Find the area of a triangle whose sides are 9 cm and 5 cm and perimeter 21 cm. [4]

Solution :

Perimeter = 21 cm

Let a = 9 cm, b = 5 cm

Let c be the third side of a triangle.

Now, 9 + 5 + c = 21

c = 21 – 14

= 7 cm

Semi-perimeter, S = 21

2 = 10.5 cm

By Heron’s formula.

Area of triangle = s s a s b s c

= 10.5 10.5 9 10.5 5 10.5 7



= 10.5 1.5 5.5 3.5

= 1

105 15 55 35100

= 1

5 3 7 5 3 5 11 5 7100

= 1

100 × 5 × 5 × 3 × 7 11

= 3 7

114

= 22111 cm

4.

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