Exercise 10.1

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?

Solution:

Given,

Side of the singal board = a

Perimeter of the signal board(s) = cm

∴ a = cm

Semi perimeter of the signal board(s) = 3a2

By using Heron's Formula,

Area of the triangular signal boards will be = ss−as−bs−c

= 3a2×3a2−a×3a2−a×3a2−a

= 3a2×a×a2×a

= 3a4

= 3a243a24

Substituting the value of a, to find the area:

Area of triangle = 34 × × = 90036003 cm2.

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. 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?

Solution:

The sides of the triangle ABC are m, m and m respectively.

Now, the perimeter will be (122 + 22 + 120) = m

Also, the semi perimeter(s) = 2642 = m

Using Heorn's formula,

Area of the triangule = ss−as−bs−c

= 132−132−132−m2

= 132×××m2

= m2

We know that the rent of advertisting per year = ₹ 5000 per m2.

∴ The rent of one wall for 3 months = RS 1320×5000×312 =

Continue

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

Solution:

It is given that the sides of the wall as m, m and m respectively.

So, the semi perimeter of trianglar wall (s) = 15+11+62 = m

Using Heorn's formula,

Area of the message = ss−as−bs−c

= 1616−×16−×16−m2

= 16×××m2

= 202162 m2

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

Solution:

Assume the third side of the triangle to be "x".

Now, the three sides of the triangle are cm, cm and "x" cm.

It is given that the perimeter of the triangle = cm

So, x = 42 - ( + )cm = cm

∴ The semi perimeter of the triangle = 422 = cm

Using Heorn's formula,

Area of the triangular = ss−as−bs−c

= 2121−×21−×21−cm2

= 21×××m2

= cm2

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

Solution:

The ratio of the sides of the triangle are given as 12 : 17 : 25

Now, let the common ration between the sides of the triangle be "x"

∴ The sides are x, x and x

It is also given that the perimeter of the triangle = cm

12x + 17x + 25x = cm

x = 540 cm

So, x =

Now, the sides of triangle are 120 cm, 170 cm, 250 cm

So, the semi perimeter of the triangle(s) = 5402 = cm

Using Heorn's formula,

Area of the triangular = ss−as−bs−c

= 270270−×270−×270−cm2

= 270××cm2

= cm2

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

Solution:

First, let the third side be x.

It is given that the length of the equal sides is cm and its perimeter is cm.

So, 30 = 12 + 12 + x

The length of the third side = cm

Thus, the semi perimeter of the isoscles triabgle(s) = 302 = cm

Using Heorn's formula,

Area of the triangule = ss−as−bs−c

= 1515−×15−×15−cm2

= 15×××cm2

= 12151315

= cm2

Thus, area of triangle is equal to 915 cm2.