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**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 ABC and if , then

area of ABC =

Since ABC is given as equilateral of side = a

Therefore here a = b = c (= a) and

Now ,

and

Therefore

... (1)

To find the area when its perimeter = 180 cm

Here a + a + a = 180 cm

3a = 180 cm

a = 60 cm

Therefore Required area cm^{2
} cm^{2}

[On putting a = 60 in (1)]

**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 . A company hired one of its walls for 3 months. How much rent did it pay?**

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

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 [By Heron's formula]

Rent charges = Rs 5000 per per year

Therefore Rent charged from a company for 3 month

**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.**

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

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

Therefore Area painted in blue colour = Area of side wall

**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

c = (42 – 18 – 10) cm = 14 cm

Now,

and

Therefore Area of the triangle

**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

a = 12 k, b = 17 k and c = 25 k

Also, perimeter = 540 cm

a + b + c = 540

12k + 17k + 25 k = 540

54 k = 540

i.e., k = 10

Thus a = 12 × 10 = 120 cm, b = 17 × 10 cm = 170 and c = 25 × 10 = 250 cm

Now,

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

**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

c = (30 – a – b) cm

c = (30 –12 – 12) cm

=(30 – 24) cm

= 6 cm

Now ,

and

Therefore, Area of the triangle

Hence, the area of triangle is = cm^{2}.

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