Heron’s Formula

NCERT Textual Exercise (Solved)

235

EXERCISE 8.1 1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral

triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

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 (see 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?

3. There is a slide in a park. One of its side walls has been painted in blue colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Heron’s Formula

NCERT Textual Exercise (Solved)

236

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Test Yourself – HF 1 1. Using Heron’s Formula, find the area of a triangle whose dimensions are as

follows: (i) 5 cm, 12 cm, 13 cm (ii) 10 cm, 24 cm, 26 cm (iii) 50 m, 78 m, 112 m 2. The sides of a triangular field are 65 m, 156 m and 169 m. Find the area of

the field. Also find the length of the shortest altitude. 3. Using Heron’s Formula, calculate the area of : (i) a right triangle, in which the sides containing the right angle measure

20 cm and 15 cm. (ii) an equilateral triangle whose perimeter is 60 cm. 4. The perimeter of a triangular field is 250 m and its sides are in the ratio 7 :

8 : 10. Find the area of the triangular field. 5. The base of an isosceles triangle is 30 cm and its area is 120 cm². Find the

perimeter. 6. The base of an isosceles triangle is 10 cm and one of its equal sides is 13

cm. Find its area by using Heron’s Formula. 7. In the following fig., calculate the area of shaded portion.

8. The sides of a triangular field are 90 m, 120 m and 150 m. Find the cost of leveling the field at the rate of ` 1.20 per square metre.

Heron’s Formula

NCERT Textual Exercise (Solved)

237

9. An advertisement board is in the form of an isosceles triangle with sides equal to 12 m, 10 m and 10 m. Find the cost of painting it at ̀ 3.60 per square metre.

10. ABCD is a square. F is the midpoint of AB and BE is one third of BC. If the area of DFBE is 108 sq. cm, find the length of AC.

EXERCISE 8.2 1. A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m,

BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Heron’s Formula

NCERT Textual Exercise (Solved)

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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, find the height of the parallelogram.

5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella ?

7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?

Heron’s Formula

NCERT Textual Exercise (Solved)

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8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 p per cm².

9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Test Yourself – HF 2 1. Find the area of a quadrilateral ABCD in which AB = 14 cm, BC = 11 cm,

CD = 18 cm, DA = 15 cm and AC = 17 cm. 2. Find the area of a quadrilateral whose sides are AB = 9 cm, BC = 40 cm,

CD = 25 cm and AD = 18 cm and ∠ABC = 90º. 3. The sides of a quadrilateral taken in order are 8 cm, 8 cm, 7 cm and 5 cm and

the angle between the first two sides is 60°. Find the area of the quadrilateral. 4. Find the area of a rhombus, one of whose sides is 25 m and one of whose

diagonals is 48 m. 5. The perimeter of a rhombus is 146 cm. One of its diagonals is 55 cm. Find

the length of the other diagonal and area of the rhombus. 6. The diagonals of a rhombus are 112 cm and 66 cm. Find the length of a side

and hence find the area of the rhombus.

Heron’s Formula

NCERT Textual Exercise (Solved)

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7. The area of a rhombus is 384 cm² and each side is 20 cm. Find its diagonals. 8. Two adjacent sides of a parallelogram are of length 5 cm and 7 cm, and the

diagonal connecting the ends of these sides is of length 9 cm. Find the area of the parallelogram. [ 11 = 3.317 approx.]

9. The adjacent sides of a parallelogram are 16 m and 10 m. If the distance between the longer sides is 8 cm, find the distance between the shorter sides.

10. (i) The area of a trapezium is 240 sq.m. and the sum of the parallel sides is 48 m, find its height.

(ii) The area of a field in the form of trapezium is 840 sq. cm. If the height of the trapezium is 20 cm and one of the parallel sides is 34 cm, find the other side.

(iii) The parallel sides of a trapezium are 85 mm and 63 mm and its altitude is 36 mm. Find the area of the quadrilateral.

(iv) The parallel sides of a trapezium are 24 cm and 52 cm and the other sides are 26 cm and 30 cm. Find the area.

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Heron’s Formula

NCERT Exercises and Assignments

Exercise 8.1 1. To find the area of an equilateral triangle using Heron’s formula:

If a, b, c are the lengths of sides BC, CA and AB of DABC and if s = 12

(a + b + c), then area of DABC = s s a s b s c( ) ( )( )− − −

Since DABC is given as equilateral of side = a

\ Hence a = b = c and s = 12

(a + a + a) = 32a

Now, s – a = 32a− a =

a2

s – b = 32a− a =

a2

s – c = 32a− a =

a2

\ Area = 32

×2

×2

×2

a a a a

= × =a a a2 2

3 34

2

...(1)

To find the area, when its perimeter = 180 cm Here, a + a + a = 180 cm \ a = 60 cm

\ Required area = ×3

460 2( ) cm² [On putting a = 60 in (1)]

= 900 3 cm².

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Heron’s Formula

2. First, we find the area of the triangular side measuring 122 m, 22 m, and 120 m. Let a = 122 m, b = 22 m and c = 120 m

\ s = 12

(a + b + c)

= 12 (122 + 22 + 120) m

= ×

12

264 m = 132 m

Now, s – a = (132 – 122) m = 10 m s – b = (132 – 22) m = 110 m and s – c = (132 – 120) m = 12 m

\ Area of triangular wall = s s a s b s c( ) ( )( )− − −

= 132 × 10 × 110 × 12 m² = 11 × 12 × 10 × 10× 11 × 12 m²

= 10 × 10 × 11 × 11 × 12 × 12 m² = (10 × 11 × 12) m² = 1320 m² Rent charges = ` 5000 per m² per year

\ Rent charged from the company for 3 months = ` 5000 × 1320 × 312

= ` 16,50,000. 3. First, find the area of side wall measuring 15 m, 11 m and 6 m. Let a = 15 m, b = 11 m, c = 6 m

and s = 12 (a + b + c) =

12

(15 + 11 + 6) m = 12

×

32 m = 16 m

Now, s – a = (16 – 15) m = 1 m s – b = (16 – 11)m = 5 m and s – c = (16 – 6) m = 10 m

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Heron’s Formula

\ Area of side wall = s s a s b s c( ) ( )( )− − − = 16 × 1 × 5 × 10 m²

= 4 × 4 × 5 × 5 × 2 m² = 4 5 2× m² = 20 2 m² \ Area painted in blue colour = Area of side wall = 20 2 m² 4. Let a, b and c be the sides of a triangle such that a = 18 cm, b = 10 cm and a + b + c = 42 cm. \ c = 42 – a – b \ c = (42 – 18 – 10) cm = 14 cm

Now, s = 12

(a + b + c) = 12

× 42 cm = 21 cm.

s – a = (21 – 18) cm = 3 cm s – b = (21 – 10) cm = 11 cm and s – c = (21 – 14) cm = 7 cm

\ Area of the triangle = s s a s b s c( ) ( )( )− − −

= 21 × 3 × 11 × 7 cm²

= 3 × 7 × 3 × 11 × 7 cm²

= 3 × 3 × 7 × 7 × 11 cm² = 3 7× 11 cm² = 21 11 cm² 5. Let the sides of the triangle be a, b and c. \ a : b : c = 12 : 17 : 25

\ a b c k

12 17 25= = = (say)

\ a = 12 k, b = 17 k and c = 25 k Also, Perimeter = 540 cm \ a + b + c = 540

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Heron’s Formula

\ 12 k + 17 k + 25 k = 540 \ 54 k = 540 i.e., k = 10 Thus, a = 12 × 10 = 120 cm, b = 17 × 10 cm = 170 cm and c = 25 × 10 = 250 cm

Now, s = 12

× 540 cm = 270 cm

s – a = (270 – 120) cm = 150 cm s – b = (270 – 170) cm = 100 cm and s – c = (270 – 250) cm = 20 cm

\ Area of the triangle = s s a s b s c( ) ( )( )− − −

= 270 150 100 20× × × cm²

= 100 27 15 10 2× × × cm²

= 100 3 × 3 × 3 × 3 × 5 × 5 × 2 × 2 cm² = 100 × 3 × 3 × 5 × 2 cm² = 9000 cm² 6. Here, a = b = 12 cm and a + b + c = Perimeter = 30 cm \ c = (30 – a – b) cm

\ c = (30 – 12 – 12) cm = 6 cm

Now, s = 12

× 30 cm = 15 cm

s – a = (15 – 12) cm = 3 cm s – b = (15 – 12) cm = 3 cm and s – c = (15 – 6) cm = 9 cm \ Area of the triangle = s s a s b s c( ) ( )( )− − − = 15 3 3 9× × × cm²

= 5×3 × 3 × 3 × 3 × 3 cm² = 3 3 5× ×3 cm² = 9 15 cm²

Test Yourself – HF 1 1. (i) 30 cm² (ii) 120 cm² (iii) 1680 cm² 2. 5070 m², 60 m 3. (i) 150 cm² (ii) 100 cm²

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Heron’s Formula

4. 375 cm² 5. 64 cm 6. 60 cm² 7. 384 cm² 8. ` 6480 9. ` 172.80 10. 36 cm

Exercise 8.2

1. Area of DBCD = 12

× BC × CD

= × ×

12

12 5 m²

= 30 m². Using Pythagoras theorem, we have BD² = BC² + CD² \ BD² = 12² + 5² \ BD² = 144 + 25 \ BD² = 169

\ BD = 169 = 13 m In DABD: Let a = 13 m, b = 8 m and c = 9 m

Now, s = 12

(a + b + c) = 12 (13 + 8 + 9) m

= 12

× 30 m = 15 m

Now, s – a = (15 – 13) m = 2 m s – b = (15 – 8) m = 7 m and s – c = (15 – 9) m = 6 m

\ Area of DABD = 15 2 7 6× × × m² = 3 5 2 7 2 3× × × × × m² = 2 × 2 × 3 ×3 × 5 × 7 m²

= 2 3 35× m² = 6 × 5.9 m² (approx.) = 35.4 m² (approx.) \ Required area = DABD + DBCD = 35.4 m² + 30 m² = 65.4 m²

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Heron’s Formula

2. Since AC² = AB² + BC² (5² = 3² + 4² i.e., 25 = 9 + 16) \ ∠ABC = 90º

Area of right angled DABC = 12 × AB × BC

= 12

× 3 × 4

cm²

= 6 cm². For DACD: Let a = 5 cm, b = 4 cm and c = 5 cm. Then,

s = 12 (a + b + c) =

12

(5 + 4 + 5) cm

= 12

× 14 cm = 7 cm

Now, s – a = (7 – 5) cm = 2 cm s – b = (7 – 4) cm = 3 cm and s – c = (7 – 5) cm = 2 cm \ Area of DACD = s s a s b s c( ) ( )( )− − −

= × × ×7 2 3 2 cm² = 2 21 cm² = 2 × 4.6 cm² (approx.) = 9.2 cm² approx. Area of quadrilateral ABCD = Area of DABC + Area of DACD = (6 + 9.2) cm² = 15.2 cm² approx. 3. Area of Part I: a = 5 cm, b = 5 cm, c = 1 cm

s = 5 + 5 + 12

= =112

5.5 cm

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Heron’s Formula

\ Area of triangular part I using Heron’s formula

= s s a s b s c( ) ( )( )− − −

= 5 5 5 5 5 5 5 5 5 5 1. ( . ) ( . ) ( . )− − −

= 5 5 0 5 0 5 4 5. . . .× × ×

= 5510

× 510

× 510

× 4510

5 × 11 × 5 × 5 × 5 × 9100 × 100

=

= 75

10011

= 0.75 × 3.31 = 2.4825 cm² Area of rectangular part II = (6.5 × 1) cm² = 6.5cm² [Q area of rectangle = l × b]

Area of trapezium part III = 12

(AB + DC) × AE Q AE ^ DC

= 12

1 2 32

( )+ × In rt. DAED,

= 3 3

4 AE² + DE² = AD²

= 3 1 732

4× .

AE² = 1² – 0.5² = 1 – 0.25 = 0.75

= 1.299cm² AE = 0 75 34

32

. = = cm

Area of triangular parts IV and V = 2 × 12

1 5 6× ×

. = 9 cm²

Total area = 2.4825 + 6.5 + 1.299 + 9 Total area = 19.28 cm² 4. For the triangle: Let a = 26 cm, b = 28 cm and c = 30 cm. Then,

s = 12

(26 + 28 + 30) cm

= 12

× 84 cm = 42 cm

Now, s – a = (42 – 26) = 16 cm s – b = (42 – 28) = 14 cm s – c = (42 – 30) = 12 cm

\ Area of the triangle = s s a s b s c( ) ( )( )− − −

= 42 16 14 12× × × cm²

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Heron’s Formula

= 2 3 7 4 4 2 7 2 2 3× × × × × × × × × cm²

= 2 × 2 2 × 2× 3 ×3 × 4 × 4 × 7 × 7× cm² = (2 × 2 × 3 × 4× 7) cm² = 336 cm² For the parallelogram : Area = Base × Height

\ Height = AreaBase

Area of gm = Area of (given)

Area of gm = 336cm²and its base =

||||

∆∴

228cm

= 33628

cm = 12 cm

5. We know that the diagonals of the rhombus bisect each other at right angles.

\ AO = 12 AC =

12

× 48 = 24 m

Using Pythagoras theorem, we have In DAOD,

OD = AD AO2 2 2 230 24− = −

= ( ) ( )30 24 30 24+ −

= 54 6× = 9 6 6× × = (3 × 6) m = 18 m

Area of DAOD = 12

24 18× ×

m² = 216 m²

Area of rhombus = 4 × DAOD = (4 × 216) m² = 864 m² Grass area for 18 cows = 864 m²

Grass area for 1 cow = 86418

m² = 48m²

6. In one triangular piece, let a = 20 cm, b = 50 cm and c = 50 cm.

Now, s = 12 (a + b + c) =

12 (20 + 50 + 50) cm

= 12 × 120 cm = 60 cm

Now, s – a = (60 – 20) cm = 40 cm s – b = (60 – 50) cm = 10 cm s – c = (60 – 50) cm = 10 cm

\ Area of the triangle = s s a s b s c( ) ( )( )− − −

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Heron’s Formula

= 60 40 10 10× × × cm² = 200 6 cm²

\ Area of 5 red triangles = 1000 6 cm²

and Area of 5 green triangles = 1000 6 cm² 7. ABCD is a square such that AC = BD = 32cm and AEF is an isosceles triangle in which AE = AF = 6cm. Clearly, from the figure Area of shaded region I = Area of shaded region II = Area of DCDB

= 12

× DB × CM

= 12

× 32 × 16 sq.cm = 256 sq.cm.

For the area of shaded region III (i.e., DAEF)

EL = LF = 12

EF

= 12

× 8 = 4 cm

and AE = 6 cm (given) \ AL = AE EL2 2−

= 36 16− = 20 2 5= cm

Area of shade region III = 12 × EF × AL

= 12 × 8 × 2 5 sq. cm. = 8 5 sq. cm.

= 8 × 2.24 sq. cm. (approx.) = 17.92 sq.cm. (approx.) 8. For one triangular tile: Let a = 9 cm, b = 28 cm and c = 35 cm

Now, s = 12

(a + b + c) = 12 (9 + 28 + 35) cm

= 12

× 72 cm = 36 cm

Now, s – a = (36 – 9) cm = 27 cm s – b = (36 – 28) cm = 8 cm s – c = (36 – 35) cm = 1 cm

\ Area of the triangle = s s a s b s c( ) ( )( )− − −

= 36 27 8 1× × × cm² = 9 4 9 3 4 2× × × × × cm²

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Heron’s Formula

= 2 3× × 4 × 4 × 9 × 9

= 4 9 6× cm² = 36 6 cm² = 36 × 2.45 cm² approx. = 88.2 cm² approx. \ Area of 16 such tiles = (16 × 88.2) cm² = 1411.2 cm² approx.

Cost of polishing @ 50 p per cm² = ` 1411 2 50100

. ×

= ` 705.60 approx.

9. Let ABCD be a field in the shape of a trapezium. Parallel sides, AB = 10 m and CD = 25 m. Non - parallel sides, AD = 13 m and BC = 14 m Draw BM ^ DC, BE || AD. \ oABED is a parallelogram \ BE = AD = 13m and DE = AB = 10m \ EC = 25 – 10 = 15m In DBEC,

Semi perimeter: S = 13 14 15

2+ +

= 422 m

= 21 m Using Heron’s Formula,

Area of DBEC = 21 21 13 21 14 21 15( ) ( ) ( )− − − = 21 8 7 6× × ×

= 7056 = 84 m²

Also, Area of DBEC = 12 × b × h

\ 84 = 12

× EC × BM

84 = 12 × 15 × BM .... [Q EC = DC – DE = (25 – 10) m = 15 m]

\ BM = 84 2

15×

\ BM = 11.2 m

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Heron’s Formula

Now, Area of trapezium ABCD = 12

× (AB + CD) × BM

= 12

× (10 + 25) × 11.2

= 12

× 35 × 11.2

= 196 m²

Test Yourself – HF 2

1. 20 35 + 14 30( ) cm² 2. 12 119 + 15( ) cm² 3. 26 3 cm² 4. 336 m² 5. 1320 cm², 48 cm 6. 65 cm, 3696 cm² 7. 32 cm and 24 cm 8. 34.83 cm² 9. 12.8 m 10. (i) 10 m (ii) 50 cm (iii) 2664 mm² (iv) 912 sq. cm.