0

Exercise 12.1

1. A traffic signal board, indicating `SCHOOOL 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?

Sol. Triangle is an equilateral triangle of side a.
?                  $s = {{a + b + c} \over 2} = {{3a} \over 2}$
Using Heron’s formula, we get
Area of the triangle $= \sqrt {{{3a} \over 2}\left( {{{3a} \over 2} - a} \right)\left( {{{3a} \over 2} - a} \right)\left( {{{3a} \over 2} - a} \right)}$
$= \sqrt {{{3a} \over 2} \times {a \over 2} \times {a \over 2} \times {a \over 2}} = {{\sqrt 3 } \over 4}{a^2}$
If perimeter = 180 cm, then side $= {{180} \over 3} = 60cm$
?  Area of the triangle
$= {{\sqrt 3 } \over 4} \times {(60)^2} = 900\sqrt 3 c{m^2}$

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 Rs.5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Sol.  Sides of triangular region for advertisements are 122 m, 120 m, 22 m.

$= {{122 + 120 + 22} \over 2} = {{264} \over 2} = 132m$
$\sqrt {(132(132 - 122)(132 - 120)(132 - 22){m^2}}$
$= \sqrt {132 \times 10 \times 12 \times 110} {m^2}$
$= \sqrt {11 \times 12 \times 10 \times 12 \times 11 \times 10} {m^2}$
= 10 × 11 × 12 m² = 1320 m²
Rent paid for 3 months = Rs. 1320 × 5000 × ${3 \over {12}}$
= Rs. 16,50,000.

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

Sol. As the sides of the wall are 15 m, 11 m and 6 m.
$S = {{15 + 11 + 6} \over 2} = {{32} \over 2} = 16m$
∴       Area $= \sqrt {16(16 - 15)(16 - 11)(16 - 6)}$
$\sqrt {16 \times 1 \times 5 \times 10} = \sqrt {800} = 20\sqrt 2 {m^2}$

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

Sol. Perimeter = 42 cm. Third side = (42-18-10)cm = 14 cm
∴     $S = {{42} \over 2} = 21cm$
∴   Area   $= \sqrt {21(21 - 18)(21 - 10)(21 - 14)} c{m^2}$
$= \sqrt {21 \times 3 \times 11 \times 7} c{m^2} = 21\sqrt {11} c{m^2}$

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

Sol. Let the sides be 12x, 17x and 25x.
We have 12x + 17x + 25x = 540
⇒               54x = 540 ⇒ x = 10.
Therefore, sides are 120 cm, 170 cm and 250 cm.
Also, semiperimeter, $S = {{540} \over 2} = 270cm$
Applying Heron’s formula, we obtain
Area of the triangle
$= \sqrt {270(270 - 120)(270 - 170)(270 - 250)}$
$= \sqrt {270 \times 150 \times 100 \times 20}$
$= \sqrt {9 \times 30 \times 30 \times 5 \times 100 \times 2 \times 10}$
=  3 × 30  ×  10 × 10 = 9000 cm²

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

$S = {{12 + 12 + 6} \over 2} = 15$
Now, area $= \sqrt {15(15 - 12)(15 - 12)(15 - 6)}$
$= \sqrt {1 \times 3 \times 3 \times 9} = 9\sqrt {15} c{m^2}$