**
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?
**

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We know that an equilateral triangle has equal sides.

So, all sides are equal to a.

If Perimeter of triangle = 180 cm (Given)

\(\Rightarrow \) a + a + a = 180

\(\Rightarrow \) 3a = 180

\(\Rightarrow \) a = 60cm

\(\therefore \) \(s = \frac{a + a + a}{2}\)

(Since, 2s = a + b + c)

\(\Rightarrow \) \(s = \frac{3a}{2} = \frac{180}{2}\)

\(\Rightarrow \) s = 90cm

We also, know that, Area of an equilateral triangle

= \(\sqrt{s(s - a)(s - a)(s - a)}\)

(Since, Heron's formula = \(\sqrt{s(s - a)(s - b)(s - c)}\))

\(\Rightarrow \) Area

= \(\sqrt{90(90 - 60)(90 - 60)(90 - 60)}\)

= \(\sqrt{90 × 30 × 30 × 30}\)

= \(30 × 30\sqrt{3}\)

\(\Rightarrow \)Area of an equilateral triangle = \(900\sqrt{3} {cm}^2\))

**
The triangular side walls of a flyover have been used for advertisements. The sides of
the wails are 122 m, 22 m and 120 m (see figure). The advertisements yield earnings of Rs. 5000 per \({m}^2\) per year. A company hired one of its walls for 3 months. How much rent did it pay?
**

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

Also, we have,

\(b^2 + c^2 = (22)^2 + (120)^2 = 484 + 14400 = 14884 = (122)^2 = a^2\)

Thus, we observe that the side walls are in right triangular shape.

Thus, the area of the triangular side walls

= \(\frac{1}{2} × a × c\)

\(\Rightarrow \)= \(\frac{1}{2} × 22 × 120\)

\(\Rightarrow \) = \(11 × 120 = 1320 m^2\)

We know that, yearly rent = Rs.5000 per \(m^2\)

Therefore, yearly rent

= Rs.5000 × \(\frac{1}{2} \) per \( m^2\)

Now, the company has hired one of its walls for 3 months.

Thus, rent paid by the company for 3 months

= 1320 × \(\frac{5000}{12}\) × 3

= 110 × 5000 × 3

= Rs. 1650000

Therefore, rent paid by the company for 3 months = Rs. 1650000.

**
There is a slide in a park. One of its side walls has been painted in same 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.
**

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The given figure formed a triangle whose sides are:

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

\(s = \frac{15 + 11 + 6}{2}\)

(Since, 2s = a + b + c)

\(\Rightarrow \) \(s = \frac{32}{2}\)

\(\Rightarrow \) s = 16m

Therefore, area painted in colour

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

(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

\(\therefore \) Area

= \(\sqrt{16 × 1 × 5 × 10}\)

= \(\sqrt{2 × 2 × 2 × 2 × 5 × 5 × 2}\)

= \(\sqrt{400 × 2} {m}^2\)

\(\therefore \) the area painted in colour is \(20\sqrt{2} m^2\))

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

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Let the sides of a triangle a = 18 cm, b = 10 cm and c

We have, perimeter = 42 cm

So, a + b + c = 42

By substituting the values,

\(\Rightarrow \)
18 + 10 + c = 42

\(\Rightarrow \)
28 + c = 42cm

\(\Rightarrow \)
c = 42 - 28 cm

\(\Rightarrow \) c = 14cm

Now, we know that,

\(s = \frac{a + b + c}{2}\)

\(\therefore \) \(s = \frac{18 + 10 + 14}{2} = \frac{42}{2} = 21cm\)

Now, Area of triangle

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

(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{21 × 3 × 11 × 7}\)

= \(\sqrt{7 × 3 × 3 × 11 × 7}\)

= \(21\sqrt{11} {cm}^2\)

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

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Suppose that the sides in cm, are 12x, 17x and 25x.

Then, we know that

12x + 17x + 25x = 540

(Given, Perimeter of the triangle)

54x = 540

\(\Rightarrow \) x = 10

So, the sides of the triangle are 12 × 10cm, 17 × 10cm, 25 × 10cm \(\Rightarrow \) 120cm, 170cm and 250cm.

Now, we know that,

\(s = \frac{a + b + c}{2}\)

\(\therefore \) \(s = \frac{120 + 170 + 250}{2} = \frac{540}{2} = 270cm\)

Now, Area of triangle

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

(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{270 × 150 × 100 × 20}\)

= \(\sqrt{27 × 10 × 3 × 15 × 10 × 100 × 20}\)

= \(100\sqrt{27 × 15 × 10 × 2} {cm}^2\)

= \(100\sqrt{9 × 3 × 3 × 5 × 10 × 2} {cm}^2\)

= \(100\sqrt{9 × 3 × 3 × 10 × 10} {cm}^2\)

= 100 × 10 × 9

Therefore, area of the given triangle is 9000\({cm}^2\)

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

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Let \(\triangle{ABC}\) be an isosceles triangle, in which,

We have,

AB = AC = 12cm ...(Given)

Now, AB + AC + BC = 30

(Given, perimeter = 30cm)

\(\Rightarrow \)
12 + 12 + BC = 30

\(\Rightarrow \) BC = 30 - 24

\(\Rightarrow \) BC = 6cm

Now, we know that,

\(s = \frac{a + b + c}{2}\)

\(\therefore \) \(s = \frac{30}{2} = 15cm\)

Now, Area of triangle

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

(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{15 × 3 × 9 × 3}\)

= \(\sqrt{3 × 5 × 9 × 9}\)

= \(9\sqrt{15} {cm}^2\)