Class 9 - Maths - Heron Formula

**Exercise 12.1**

**Question 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?

Answer:

Hence, area of the signal board = a^{2} √3/4

Now, perimeter = 180 cm

Each side of the triangle = 180/3 = 60 cm

Area of the triangle = (60)^{2}/4 * √3

= (3600 * √3)/4

= 900√3 cm^{2}

**Question 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 Fig.).

The advertisements yield an earning of Rs 5000 per m^{2} per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer:

Here, we first find the area of the triangular side walls.

a = 122 m, b = 120 m and c = 22 m

Now, perimeter = 122 + 120 + 22

=> 2s = 264

=> s = 264/2

=> s = 132 m

Now, area of triangular side wall = √{s(s - a)(s - b)(s - c)}

= √{132(132 - 122)(132 - 120)(132 -22)}

= √{132 * 10 * 12 * 110}

= 1320 m^{2}

Rent of 1 m^{2} of the wall for 1 year = Rs 5000

So, Rent of 1 m^{2} of the wall for 1 month = Rs 5000/12

So, Rent of the complete wall (1320 m^{2}) for 3 months = Rs (5000/12) * 1320 * 3

= Rs 16,50,000

**Question 3:**

There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig.).

If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer:

Here, a = 15 m, b = 11 m and c = 6 m

Now, perimeter = 15 + 11 + 6

=> 2s = 32

=> s = 32/2

=> s = 16 m

Now, area of triangle = √{s(s - a)(s - b)(s - c)}

= √{16(16 - 15)(16 - 11)(16 - 6)}

= √{16 * 1 * 5 * 10}

= 20√2 m^{2}

Hence, the area painted in colour = 20√2 m^{2}

**Question 4:**

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

Answer:

Here a = 18 cm, b = 10 cm, c = ?

Perimeter of the triangle = 42 cm

=> a + b + c = 42

=> 18 + 10 + c = 42

=> c = 42 – 28 = 14

Now, s = (a + b + c)/2

=> s = 42/2 = 21 cm

Now, area of triangle = √{s(s - a)(s - b)(s - c)}

= √{21(21 - 18)(21 - 10)(21 - 14)}

= √{21 * 3 * 11 * 7}

= 7 * 3 * √11

= 21√11 cm^{2}

**Question 5:**

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

Answer:

Let the sides of the triangle be 12x cm 17x cm and 25x cm.

Perimeter of the triangle = 540 cm

=> 12x + 17x + 25x = 540

=> 54 x = 540

=> x = 540/54 = 10

So, sides of the triangle are (12 * 10) cm, (17 * 10) cm and (25 * 10) cm i.e., 120 cm, 170 cm

and 250 cm.

Now, suppose a = 120 cm, b = 170 cm, c = 250 cm,

So, s = (a + b + c) = (120 + 170 + 250)/2 = 540/2 = 270 cm

Now, area of triangle = √{s(s - a)(s - b)(s - c)}

= √{270(270 - 120)(270 - 170)(270 - 250)}

= √{270 * 150 * 100 * 20}

= 9000 cm^{2}

**Question 6:**

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

Answer:

Here, a = b = 12 cm,

Also, a + b + c = 30

=> 12 + 12 + c = 30

=> c = 30 – 24 = 6

So, s = (a + b + c)/2 = (12 + 12 + 6)/2 = 30/2 = 15 cm

Now, Area of the triangle = √{s(s - a)(s - b)(s - c)}

= √{15(15 - 12)(15 - 12)(15 - 6)}

= √{15 * 3 * 3 * 9}

= 9√15 cm^{2}

**Exercise 12.2**

**Question1:**

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Answer:

ABCD is the park as shown in the figure.

Join BD.

In ∆DBC, we have

DB^{2} = BC^{2} + CD^{2} [From Pythagoras theorem]

=> DB^{2} = (12)^{2} + 5^{2}

=> DB^{2} = 144 + 25 = 169

=> DB = √169

=> DB = 13

Area of ∆DBC = 1/2 * base * height

= 1/2 * 12 * 5

= 6 * 5

= 30 m^{2}

In ∆ABD, a = 9 m, b = 8 m, c = 13 m

So, s = (a + b + c)/2 = (9 + 8 + 13)/2 = 30/2 = 15 m

Now, Area of the triangle ABD = √{s(s - a)(s - b)(s - c)}

= √{15(15 - 9)(15 - 8)(15 - 13)}

= √{15 * 6 * 7 * 2}

= √1260 = 35.5 m^{2} (approx.)

Now, area of the park = area of ∆DBC + area of ∆ABD

= (30 + 35.5) m^{2} = 65.5 m^{2}

**Question 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.

Answer:

In ∆ABC, we have

AB^{2} + BC^{2} = 9 + 16 = 25 = AC^{2}

Hence, ABC is a right triangle, right angled at B

[By converse of Pythagoras theorem]

So, area of ∆ABC = 1/2 * base * height

= 1/2 * 3 * 4

= 3 * 2 = 6 cm^{2}

In ∆ACD, a = 5 cm, b = 4 cm, c = 5 cm

s = (a + b + c) = (5 + 4 + 5)/2 = 14/2 = 7 cm

Now, Area of the triangle ACD = √{s(s - a)(s - b)(s - c)}

= √{7(7 - 5)(7 - 4)(7 - 5)}

= √{7 * 2 * 3 * 2}

= √84

= 9.2 cm^{2} (approx.)

So, area of the quadrilateral = area of ∆ABC + area of ∆ACD

= (6 + 9.2) cm^{2}

= 15.2 cm^{2}

**Question 3:**

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

Answer:

For the triangle marked I:

a = 5 cm, b = 5 cm, c = 1 cm

s = (a + b + c) = (5 + 5 + 1)/2 = 11/2 = 5.5 cm

Now, Area of the triangle = √{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}

= √6.1875

= 2.5 cm^{2} (approx.)

For the rectangle marked II:

Length = 6.5 cm, Breadth = 1 cm

Area of the rectangle = 6.5 * 1 cm^{2} = 6.5 cm^{2}

For the trapezium marked III:

Draw AF || DC and AE ⊥ BC

AD = FC = 1 cm, DC = AF = 1 cm

So, BF = BC – FC = (2 – 1) cm = 1 cm

Hence, ∆ABF is equilateral.

Also, E is the mid-point of BF.

Now, BE = 1/2 cm = 0.5 cm

Also, AB^{2} = AE^{2} + BE^{2} [Pythagoras theorem]

=> AE^{2} = 1^{2} – (0.5)^{2}

=> AE^{2} = 1 – 0.25

=> AE^{2} = 0.75

=> AE = √0.75

⇒ AE = 0.9 cm (approx.)

Area of the trapezium = 1/2 * (sum of the parallel sides) * distance between them.

= 1/2 * (BC + AD) * AE

= 1/2 * (2 + 1) * 0.9

= 1/2 * 3 * 0.9

= 2.7/2

= 1.4 cm^{2}

For the triangle marked IV:

It is a right-triangle

∴ Area of the triangle = 1/2 * base * height

= 1/2 * 6 * 1.5

= 3 * 1.5

= 4.5 cm^{2}

For the triangle marked V:

This triangle is congruent to the triangle marked IV.

Hence, area of the triangle = 4.5 cm^{2}

Total area of the paper used = (2.5 + 6.5 + 1.4 + 4.5 + 4.5) cm^{2}

= 19.4 cm^{2}

**Question 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.

Answer:

In the figure, ABCD is a parallelogram and ABE is the triangle which stands on the base AB

For the triangle ABE, a = 30 cm, b = 28 cm, c = 26 cm.

So, s = (a + b + c)/2 = (30 + 28 + 26)/2 = 84/2 = 42 cm

Now, area of the ∆ABE = √{s(s - a)(s - b)(s - c)}

= √{42(42 - 30)(42 - 28)(42 - 26)}

= √{42 * 12 * 14 * 16}

= √112896

= 336 cm^{2}

Now, area of the parallelogram = base * height

=> 336 = 28 * Height [Given, area of the triangle = area of the parallelogram]

=> Height = 336/28

=> Height = 12

Hence, the height of the parallelogram = 12 cm

**Question 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?

Answer:

Clearly, the diagonal AC of the rhombus divides it into two congruent triangles.

For triangle ABC, a = b = 30 m, c = 48 m.

So, s = (a + b + c)/2 = (30 + 30 + 48)/2 = 108/2 = 54 m

Now, area of the triangle = √{s(s - a)(s - b)(s - c)}

= √{54(54 - 30)(54 - 30)(54 - 48)}

= √{54 * 24 * 24 * 6}

= √(9 * 6 * 24 * 24 * 6)

= 3 * 6 * 24

= 432 m^{2}

Area of the rhombus = 2 * 432 = 864 m^{2}

Number of cows = 18

Hence, area of the grass field which each cow gets = 864/18 = 48 m^{2}

**Question 6:**

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see ig.), each piece measuring 20 cm, 50 cm, and 50 cm.

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

Answer:

First we find the area of one triangular piece.

Here, a = b = 50 cm, c = 20 cm

Now, s = (a + b + c)/2 = (50 + 50 + 20)/2 = 120/2 = 60 cm

Area of one triangular piece = √{s(s - a)(s - b)(s - c)}

= √{60(60 - 50)(60 - 50)(60 - 20)}

= √{60 * 10 * 10 * 40}

= √{6 * 10 * 10 * 10 * 4 * 10}

= 10 * 10 * 2√6

= 200√6 cm^{2}

Area of 10 such triangular pieces = 10 * 200√6 = 2000√6 cm^{2}

Hence, cloth required for each colour = 2000√6/2 = 1000√6 cm^{2}

**Question 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?

Answer:

ABCD is a square.

So, AO = OC = OB = OD and ∠AOB = 90°

[Diagonals of a square bisect each other at right angles]

Given, BD = 32 cm

=> OA = 32/2 cm = 16 cm

∆ABD is a right triangle.

So, area of ∆ABD = 1/2 * base * height

= 1/2 * 32 * 16 =

= 16 * 16

= 256 cm^{2}

Thus, area of ∆BCD = 256 cm^{2}

For triangle CEF, a = b = 6 cm, c = 8 cm.

So, s = (a + b + c)/2 = (6 + 6 + 8)/2 = 20/2 = 10 cm

Now, area of the triangle = √{s(s - a)(s - b)(s - c)}

= √{10(10 - 6)(10 - 6)(10 - 8)}

= √{10 * 4 * 4 * 2}

= √320

= 17.92 cm^{2}

Hence, paper needed for shade I = 256 cm^{2}, for shade II = 256 cm^{2} and for shade III = 17.92 cm^{2}

**Question 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 cm^{2}.

Answer:

We have lengths of the sides of 1 triangular tile are a = 35 cm, b = 28 cm, c = 9 cm.

So, s = (a + b + c)/2 = (35 + 28 + 9)/2 = 72/2 = 36 cm

Now, area of 1 triangular tile = √{s(s - a)(s - b)(s - c)}

= √{36(36 - 35)(36 - 28)(36 - 9)}

= √{36 * 1 * 8 * 27}

= √7776

= 88.2 cm^{2 }

So, area of 16 such tiles = 16 * 88.2 cm^{2}

Cost of polishing 1 cm^{2} = 50 paise = Re 0.50

Hence, total cost of polishing the floral design = Rs 16 * 88.2 * 0.50 = Rs 705.60

**Question 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.

Answer:

In the figure ABCD is the field. Draw CF || DA and CG ⊥ AB.

DC = AF = 10 m, AD = FC = 13 m

For ∆BCF, a = 15 m, b = 14 m, c = 13 m

Now s = (a + b + c) = (15 + 14 + 13)/2 = 42/2 = 21 m

∴ Area of ∆BCF = √{s(s - a)(s - b)(s - c)}

= √{21(21 - 15)(21 - 14)(21 - 13)}

= √{21 * 6 * 7 * 8}

= √7056

= 84 m^{2 }

Also, area of ∆BCF = 1/2 * base * height

= 1/2 * BF * CG

=> 84 = 1/2 * 15 * CG

=> CG = (84 * 2)/15

=> CG = 168/15

=> CG = 11.2 m

Again, area of the trapezium = 1/2 * sum of the parallel sides * distance between them

= 1/2 * (25 + 10) * 11.2

= 35 * 5.6

= 196 m^{2}

Hence, area of the field = 196 m^{2}

.