Thursday, 11 September 2014

ncert solution for class 9 maths Chapter 12 – Heron's Formula Exercise 12.1

Exercise 12.1

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?

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

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 m2 per year. A company hired one of its walls for  3 months. How much rent did it pay?

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 m2 area per year = Rs 5000
Rent of 1 m2 area per month = Rs 5000/12
Rent of 1320 m2 area for 3 months =

= Rs (5000 × 330) = Rs 1650000
Therefore, the company had to pay Rs 1650000.

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 m2 is Rs.15000, find the height of the hall.

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

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
Perimeter of the floor of hall = 2(l + b)
= 250 m
Area of four walls = 2(l + b) h = 250h m2 Cost of painting per m2 area = Rs 10
Cost of painting 250h m2 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.

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

Let the third side of the triangle be x.
Perimeter of the given triangle = 42 cm
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.

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
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 cm2

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

Let the third side of this triangle be x.
Perimeter of triangle = 30 cm
12 cm + 12 cm + x = 30 cm
x = 6 cm
By Heron’s formula,

Click here for Exercise 12.2

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