**Exercise 12.1**

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**Question 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?**

Answer:

Side of traffic signal board = a

Perimeter of traffic signal board = 3 × a

By Heron’s formula,

Perimeter of traffic signal board = 180 cm

Side of traffic signal board

Using equation (1), area of traffic signal board

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**Question 2:****The triangular side walls of a flyover have been used for advertisements.
The sides ****of the walls are
122m, 22m, and 120m
(see the given figure). The a dvertisements 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?

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

Answer:

The sides of the triangle (i.e., a, b, c) are of 122 m, 22 m, and 120 m
respectively. Perimeter of triangle = (122 + 22 + 120) m

2s = 264 m

s = 132 m

By Heron’s formula,

Rent of 1 m

^{2}area per year = Rs 5000
Rent of 1 m

^{2}area per month = Rs 5000/12
Rent of 1320 m

^{2}area for 3 months == Rs (5000 × 330) = Rs 1650000

Therefore, the company had to pay Rs 1650000.

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**Question 3:****The floor of a
rectangular hall has a perimeter 250 m. If the cost of panting the four walls at the rate of Rs.10 per m**^{2} is Rs.15000, find the height of the hall.

^{2}is Rs.15000, find the height of the hall.

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**[Hint: Area of the four walls = Lateral surface
area.] **

Answer:

Let length, breadth,
and height of
the rectangular hall
be l m,
b m, and h m respectively.

Area of four walls = 2lh + 2bh

= 2(l + b) h

= 2(l + b) h

Perimeter of the floor of hall = 2(l + b)

= 250 m

= 250 m

∴ Area of four walls = 2(l + b) h = 250h m

^{2 }Cost of painting per m^{2}area = Rs 10
Cost of painting 250h m

^{2}area = Rs (250h × 10) = Rs 2500h
However, it is given that the
cost of paining the walls is Rs 15000. ∴ 15000 = 2500h

h = 6

Therefore, the height of the hall
is 6 m.

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

Answer:

Let the third side of the
triangle be x.

Perimeter of the given triangle =
42 cm

18 cm + 10 cm + x = 42

18 cm + 10 cm + x = 42

x = 14 cm

By Heron’s formula,

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

Answer:

Let the common ratio between the
sides of the given triangle be x. Therefore, the side of the triangle will be
12x, 17x, and 25x.

Perimeter of this triangle = 540 cm

Perimeter of this triangle = 540 cm

12x + 17x + 25x = 540 cm

54x = 540 cm

x = 10 cm

Sides of the triangle will be 120
cm, 170 cm, and 250 cm.

By Heron’s formula,

Therefore, the area of this
triangle is 9000 cm

^{2}.## Question 6:An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Answer:

Let the third side of this
triangle be x.

Perimeter of triangle = 30 cm

12 cm + 12 cm + x = 30 cm

x = 6 cm

Perimeter of triangle = 30 cm

12 cm + 12 cm + x = 30 cm

x = 6 cm

By Heron’s formula,

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