# Heron's Formula : Exercise 12.1 (Mathematics NCERT Class 9th)

Q.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?
Sol.

To find the area of an equilateral triangle using Heron's formula.|
If a, b, c be the lengths of sides BC, CA and AB of $\Delta$ ABC and if $s = {1 \over 2}\left( {a + b + c} \right)$, then

area of $\Delta$ ABC = $\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
Since $\Delta$ ABC is given as equilateral of side = a

Therefore here a = b = c (= a) and $s = {1 \over 2}\left( {a + a + a} \right) = {{3a} \over 2}$
Now , $s - a = {{3a} \over 2} - a = {a \over 2}$
$s - b = {{3a} \over 2} - a = {a \over 2}$
and $s - c = {{3a} \over 2} - a = {a \over 2}$
Therefore $Area = \sqrt {{{3a} \over 2} \times {a \over 2} \times {a \over 2} \times {a \over 2}}$
$= {a \over 2} \times {a \over 2}\sqrt 3 = {{\sqrt 3 \,{a^2}} \over 4}$ ... (1)
To find the area when its perimeter = 180 cm
Here a + a + a = 180 cm
$\Rightarrow$ 3a = 180 cm

$\Rightarrow$ a = 60 cm
Therefore Required area $= {{\sqrt 3 } \over 4} \times {\left( {60} \right)^2}$ cm2
$= {{\sqrt 3 } \over 4} \times 3600$ cm2

[On putting a = 60 in (1)]
$= 900\,\sqrt 3 \,c{m^2}$

Q.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 ${m^2}\,per\,year$. A company hired one of its walls for 3 months. How much rent did it pay?

Sol.

First we find the semi perimeter of triangular side measuring 122 m, 22m and 120 m.
Let a = 122 m, b = 22 m and c = 120 m
Therefore $s = {1 \over 2}\left( {a + b + c} \right) = {1 \over 2}\left( {122 + 22 + 120} \right)m$
$= \left( {{1 \over 2} \times 264} \right)m = 132\,m$
Now, s – a = (132 – 122) m = 10 m
s – b = (132 – 22) m = 110 m
and s – c = (132 – 120) m = 12 m
Therefore Area of triangular wall $= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$ [By Heron's formula]
$= \sqrt {132 \times 10 \times 110 \times 12} \,{m^2}$
$= \sqrt {11 \times 12 \times 10 \times 10 \times 11 \times 12} \,{m^2}$
$= \sqrt {10 \times 10 \times 11 \times 11 \times 12 \times 12} \,{m^2}$
$= \left( {10 \times 11 \times 12} \right)\,{m^2} = 1320\,{m^2}$
Rent charges = Rs 5000 per ${m^2}$ per year
Therefore Rent charged from a company for 3 month
$= Rs\left( {5000 \times 1320 \times {3 \over {12}}} \right)$
$= Rs\,\,16,50,000\,$

Q.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.

Sol.

First find the semi perimeter of side wall measuring 15 m, 11 m and 6 m.
Let a = 15 m, b = 11 m, c = 6 m
and $s = {1 \over 2}\left( {a + b + c} \right) = {1 \over 2}\left( {15 + 11 + 6} \right)m$
$= \left( {{1 \over 2} \times 32} \right)m = 16\,m$
Now, s – a = (16 – 15)m = 1 m
s – b = (16 – 11) m = 5 m
and s – c = (16 – 6) m = 10 m
Area of side wall $= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {16 \times 1 \times 5 \times 10} \,{m^2}$
$= \sqrt {4 \times 4 \times 5 \times 5 \times 2} \,{m^2}$
$= 4 \times 5\sqrt 2 \,{m^2}$
$= 20\sqrt 2 \,{m^2}$
Therefore Area painted in blue colour = Area of side wall
$= 20\sqrt 2 \,{m^2}$

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

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
Therefore c = 42 – a – b
$\Rightarrow$ c = (42 – 18 – 10) cm = 14 cm
Now, $s = {1 \over 2}\left( {a + b + c} \right) = {1 \over 2} \times 42\,cm = 21\,cm$
$s - a = \left( {21 - 18} \right)cm = 3\,cm$
$s - b = \left( {21 - 10} \right)cm = 11\,cm$
and $s - c = \left( {21 - 14} \right)cm = 7\,cm$
Therefore Area of the triangle $= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {21 \times 3 \times 11 \times 7} \,\,c{m^2}$
$= \sqrt {3 \times 7 \times 3 \times 11 \times 7} \,\,c{m^2}$
$= \sqrt {3 \times 3 \times 7 \times 7 \times 11} \,\,c{m^2}$
$= 3 \times 7\sqrt {11} \,c{m^2} = 21\sqrt {11} \,c{m^2}$

Q.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 a, b and c of the triangle.
Therefore a : b : c = 12 : 17 : 25
$\Rightarrow$ ${a \over {12}} = {b \over {17}} = {c \over {25}} = k\left( {say} \right)$
$\Rightarrow$ a = 12 k, b = 17 k and c = 25 k
Also, perimeter = 540 cm
$\Rightarrow$ a + b + c = 540
$\Rightarrow$ 12k + 17k + 25 k = 540
$\Rightarrow$ 54 k = 540
$k = {{540} \over {54}}$ i.e., k = 10

Thus a = 12 × 10 = 120 cm, b = 17 × 10 cm = 170 and c = 25 × 10 = 250 cm
Now, $s = {1 \over 2} \times 540\,cm = 270\,cm$
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
$= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {270 \times 150 \times 100 \times 20} \,\,c{m^2}$
$= 100\sqrt {27 \times 15 \times 10 \times 2} \,\,c{m^2}$
$= 100\sqrt {3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2} \,\,c{m^2}$
$= 100 \times 3 \times 3 \times 5 \times 2\,\,c{m^2} = 9000\,c{m^2}$

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

Here a = b = 12 cm and a + b + c = perimeter = 30 cm
$\Rightarrow$ c = (30 – a – b) cm
$\Rightarrow$ c = (30 –12 – 12) cm
=(30 – 24) cm

= 6 cm

Now , $s = {1 \over 2} \times 30\,cm = 15\,cm$
$s - a = \left( {15 - 12} \right)cm = 3\,cm$
$s - b = \left( {15 - 12} \right)cm = 3\,cm$
and $s - c = \left( {15 - 6} \right)cm = 9\,cm$
Therefore, Area of the triangle $= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {15 \times 3 \times 3 \times 9} \,c{m^2}$
$= \sqrt {5 \times 3 \times 3 \times 3 \times 3 \times 3} \,c{m^2}$
$= 3 \times 3\sqrt {5 \times 3} \,c{m^2} = 9\sqrt {15} \,c{m^2}$
Hence, the area of triangle is = $9\sqrt {15}$ cm2.

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