PSEB 9th Class Maths Solutions Chapter 12 Heron’s Formula Ex 12.1

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 12 Heron’s Formula Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 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?
Answer:
In equilateral ∆ ABC, the length of each side is a.
∴ a = a, b = a, c = a
and semiperimeter s = \(\frac{a+b+c}{2}\) = \(\frac{a+a+a}{2}\) = \(\frac{3}{2}\)a
Now,
s – a = \(\frac{3}{2}\)a – a = \(\frac{1}{2}\)a,
s – b = \(\frac{3}{2}\)a – a = \(\frac{1}{2}\)a, and
s – c = \(\frac{3}{2}\)a – a = \(\frac{1}{2}\)a,

Now, the perimeter of equilateral ∆ ABC is 180 cm.
∴ Length of each side = \(\frac{180}{3}\) = 60 cm and
semiperimeter s = \(\frac{180}{2}\) = 90 cm.
Here, a = b = c = 60 cm and s = 90 cm
∴ s – a = 90 – 60 = 30 cm,
s – b = 90 – 60 = 30 cm and
s – c = 90 – 60 = 30 cm.
Area of ∆ ABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{90 \times 30 \times 30 \times 30}\) cm²
= \(\sqrt{3 \times 900 \times 900}\) cm²
= 30 × 30 × √3 cm²
= 900 √3 cm²

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 (see the given figure). The advertisements yield an earning of ₹ 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer:
For triangular side wall of the flyover,
a = 122 m, b = 120 m and c = 22 m.
Semiperimeter s = \(\frac{a+b+c}{2}\)
= \(\frac{122+120+22}{2}\)
= \(\frac{264}{2}\)
= 132 m
Area of triangular side wall
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{132(132-122)(132-120)(132-22)}\) m²
= \(\sqrt{132 \times 10 \times 12 \times 110}\) m²
= \(\sqrt{12 \times 11 \times 10 \times 12 \times 11 \times 10}\) m²
= 12 × 11 × 10 m²
= 1320 m²
∴ Annual rent of one wall = ₹ (1320 × 5000)
∴ Rent of one wall for 3 months
= ₹ (1320 × 5000 × \(\frac{3}{12}\))
= ₹ 16,50,000

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” (see the given figure). If the sides of the wall are 15 m, 11m and 6 m, find the area painted in colour.

Answer:
The lengths of the triangular side wall are
15 m, 11m and 6 m.
∴ a = 15 m, b = 11m, c = 6m and
semiperimeter s = \(\frac{a+b+c}{2}\) = \(\frac{15+11+6}{2}\) = \(\frac{32}{2}\) = 16 cm
Then, s – a = 16 – 15 = 1 m,
s – b = 16 – 11 = 5 m, and
s – c = 16 – 6 = 10m.
Area of the triangular region painted in colour
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{16 \times 1 \times 5 \times 10}\) m²
= \(\sqrt{16 \times 5 \times 5 \times 2}\) m²
= 4 × 5 × √2 m²
= 20√2 m²

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:
Here, perimeter of the triangle = 42 cm
∴ Semiperimeter s = \(\frac{42}{2}\) = 21 cm.
Now, a = 18 cm and b = 10 cm.
s = \(\frac{a+b+c}{2}\)
∴ 21 = \(\frac{18+10+c}{2}\)
∴ 42 = 28 + c
∴ c = 14 cm
Now,
s – a = 21 – 18 = 3 cm,
s – b = 21 – 10 = 11 cm and
s – c = 21 – 14 = 7 cm.
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{21 \times 3 \times 11 \times 7}\) cm²
= \(\sqrt{21 \times 21 \times 11}\) cm²
= 21√11 cm²

Question 5.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Answer:
Suppose the sides of the triangle measure 12x cm, 17x cm and 25x cm.
Perimeter of a triangle = Stun of three sides
∴ 540 = 12x + 17x + 25x
∴ 540 = 54x
∴ x = 10
Then, the measures of the sides of the triangle are,
a = 12 × 10 = 120 cm,
b = 17 × 10 = 170 cm and
c = 25 × 10 = 250 cm.
Now, s – a = 270 – 120 = 150 cm,
s – b = 270 – 170 = 100 cm and
s – c = 270 – 250 = 20 cm.
Area of a triangle
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= 7\(\sqrt{270(150)(100)(20)}\) cm²
= \(\sqrt{270 \times 30 \times 5 \times 100 \times 5 \times 4}\) cm²
= \(\sqrt{8100 \times 25 \times 400}\) cm²
= \(\sqrt{(90)^{2} \times(5)^{2} \times(20)^{2}}\) cm²
= 90 × 5 × 20 cm²
= 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.
Answer:
Let, the sides of the isosceles triangle be a = 12 cm, b = 12 cm and c cm.
Perimeter of triangle = Sum of three sides
∴ 30 = 12 + 12 + c
∴ 30 = 24 + c
∴ c = 6 cm
Now, semiperimeter s = \(\frac{\text { Perimeter }}{2}\) = \(\frac{30}{2}\) = 15 cm
Then, s – a = 15 – 12 = 3 cm,
s – b = 15 – 12 = 3 cm and
s – c = 15 – 6 = 9 cm.
Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{15 \times 3 \times 3 \times 9}\) cm²
= \(\sqrt{15 \times 9 \times 9}\) cm²
= 9 √15 cm²