NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula

Q1. 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:

Given the perimeter of an equilateral triangle is 180 cm.

So, $3a = 180$
$⇒a = 60$ cm

Hence, the length of the side is 60 cm.

Now,

Calculating the area of the signal board by Heron's Formula:

$A = \sqrt{s(s-a)(s-b)(s-c)}$

Where s is the half-perimeter of the triangle and a, b, and c are the sides of the triangle.

Therefore,

$s = \frac{1}{2}\times$ Perimeter $= \frac{1}{2}\times180 = 90$ cm

$a =b=c = 60$ cm as it is an equilateral triangle.

Substituting the values in Heron's formula, we obtain

$\implies A = \sqrt{90(90-60)(90-60)(90-60)} = 900\sqrt{3}$ cm2

Q2. 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. 12.9). 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:

From the figure,

The sides of the triangle are:

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

The semi-perimeter, s, will be:

$s = \frac{a+b+c}{2} = \frac{122+120+22}{2} = \frac{264}{2} = 132$ m

Therefore, the area of the triangular side wall will be calculated by Heron's Formula,

$A = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{132(132-122)(132-120)(132-22)}$

$= \sqrt{132(10)(12)(110)}$

$= \sqrt{(12\times11)(10)(12)(11\times10)} = 1320$ m2

Given the rent for 1 year (i.e., 12 months) per meter square is Rs. 5000.

Rent for 3 months per meter square will be:

Rs. $5000\times \frac{3}{12}$

Therefore, for 3 months for 1320 m2 :

Rs. $5000\times \frac{3}{12}\times 1320 =$ Rs. $16,50,000$

Q3. There is a slide in a park. One of its side walls has been painted in some colour with the message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer:

Given the sides of the triangle are:

$a = 15 m,\ b= 11m\ and\ c= 6m.$

So, the semi-perimeter of the triangle will be:

$s = \frac{a+b+c}{2} = \frac{15+11+6}{2} = \frac{32}{2} = 16m$

Therefore, Heron's formula will be:

$A = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{16(16-15)(16-11)(16-6)}$

$= \sqrt{16(1)(5)(10)}$

$= \sqrt{(4\times 4)(1)(5)(5\times 2 )}$

$= 4\times 5 \sqrt2 = 20\sqrt2\ m^2$

Hence, the area painted in colour is $20\sqrt2$ m2.

Q4. Find the area of a triangle, two sides of which are 18 cm and 10 cm, and the perimeter is 42 cm.

Answer:

Given the perimeter of the triangle is $42$ cm
and the sides length $a= 18$ cm and $b= 10$ cm

So, $a+b+c = 42$

$⇒c = 42 - 18-10 = 14$ cm

So, the semi-perimeter of the triangle will be:

$s = \frac{P}{2} = \frac{42}{2} = 21$ cm

Therefore, the area given by Heron's Formula will be,

$A = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{21(21-18)(21-10)(21-14)}$

$= \sqrt{(7\times3 )(3)(11)(7)}$

$= 21\sqrt{11}$ cm2

Hence, the area of the triangle is $21\sqrt{11}$ cm2.

Q5. Sides of a triangle are in the ratio of 12 : 17 : 25, and its perimeter is 540 cm. Find its area.

Answer:

Given the sides of a triangle are in the ratio of $12:17:25$ and its perimeter is $540cm$

Let us consider the length of one side of the triangle to be $a = 12x$

Then, the remaining two sides are $b = 17x$ and $c = 25x.$

So, by the given perimeter, we can find the value of x:

Perimeter $= a+b+c = 12x+17x+25x = 540$ cm

$\implies 54x = 540$ cm

$\implies x = 10$

So, the sides of the triangle are:

$a = 12\times10 =120$ cm

$b = 17\times10 =170$ cm

$c = 25\times10 =250 $ cm

So, the semi-perimeter of the triangle is given by

$s = \frac{540cm}{2} = 270$ cm

Therefore, using Heron's Formula, the area of the triangle is given by:

$A = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{270(270-120)(270-170)(270-250)}$

$= \sqrt{270(150)(100)(20)}$

$= \sqrt{81000000} = 9000$ cm2

Hence, the area of the triangle is $9000$ cm2.

Q6. An isosceles triangle has a perimeter of 30 cm, and each of the equal sides is 12 cm. Find the area of the triangle.

Answer:

The perimeter of an isosceles triangle is 30 cm (Given).

The length of the sides, which are equal, is 12 cm.

Let the third side length be 'a' cm.

Then, the Perimeter $= a+b+c$

$\Rightarrow 30= a+12+12$

$\Rightarrow a = 6cm$

So, the semi-perimeter of the triangle is given by,

$s= \frac{1}{2}\times $ Perimeter $=\frac{1}{2}\times30 = 15$ cm

Therefore, using Heron's Formula, calculating the area of the triangle

$A = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{15(15-6)(15-12)(15-12)}$

$= \sqrt{15(9)(3)(3)}$

$= 9\sqrt{15}$ cm2

Hence, the area of the triangle is $9\sqrt{15}$ cm2.