Students can Download Class 9 Maths Chapter 8 Heron’s Formula Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 8 Heron’s Formula Ex 8.1

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?

Solution:

side of traffic the signal board = a

perimeter of traffic signal board = 3a

2S = 3a

S = \(\frac{3}{2}\)a

By Heron’s formula

Perimeter of traffic signal board = 180 cm

Side of traffic signalboard (a) = \(\frac{180}{3}\) = 60 cm

Using equation (1)

area of traffic signal board = \(\frac{\sqrt{3}}{4}\) (60)^{2}

= \(\frac{3600}{4}\) √3 = 900√3 cm^{2}

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Question 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. 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?

Solution:

Sides of triangle, a = 122m, b = 22m, c = 120m.

Rent for each year, each sq.m. = Rs. 5,000

Rent for each month, each sq.m. = \(\frac{5000}{12}\)

Rent for each quarterly, each sq.m. = 3 × \(\frac{5000}{12}\)

= Rs. 1250

∴ Rent for 1320 sq.m.= 1250 × 1320 = Rs. 1650000

∴ The amount the company has to pay is Rs. 1650000.

Question 3.

There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15m, 11m, and 6 m, find the area painted in colour.

Solution:

Sides of the coloured triangular wall are 15 m, 11 m, and 6 m.

Perimeter of this triangular wall

2s = 15 + 11 + 6

2s = 32

s = \(\frac{32}{2}\) = 16 m

Using Heron’s formula,

area of the triangular wall

Hence Area painted = 20√2 m^{2}

Question 4.

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

Solution:

Let Three sides of the triangle be

Let a = 18 cm, b = 10 cm, c = x cm.

Perimeter = 42 cm.

a + b + c = 42

18 + 10 + x = 42

28 + x = 42

∴ x = 42 – 28

∴ x = 14 cm

∴ c = 14 cm

Question 5.

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

Solution:

Let sides of triangle be

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

∴ Perimeter = 12x + 17x + 25x = 540

54x = 540

∴ x = \(\frac{540}{54}\)

∴ x = 10 cm.

∴ a = 12x = 12 × 10 = 120 cm.

b = 17x = 17 × 10 = 170 cm.

c = 25x = 25 × 10 = 250 cm.

∴ Area of the triangle is 9000 cm^{2}

Question 6.

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

Solution:

Let the sides of the isosceles triangle be a and b.

∴ a = b = 12 cm.

Let the third side be ‘c’.

∴ Perimeter = a + b + c = 30

= 12 + 12 + c = 30

24 + c = 30

c = 30 – 24

c = 6 cm.