Exercise 12.1 Complete Solutions
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?
For an equilateral triangle with side a, all three sides are equal.
Sides of the triangle are: a, a, a.
First, find the semi-perimeter (s):
Using Heron’s formula, Area = √[s(s − a)(s − b)(s − c)]
Area = √[ (3a/2) × (3a/2 − a) × (3a/2 − a) × (3a/2 − a) ]
Area = √[ (3a/2) × (a/2) × (a/2) × (a/2) ]
Area = √(3a4 / 16)
Area = (√3 / 4)a2
Now, it is given that the perimeter of the board is 180 cm.
Substitute a = 60 cm into the area formula:
Area = (√3 / 4) × (60)2
Area = (√3 / 4) × 3600 = 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. 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?
Let the sides of the triangular wall be a = 122 m, b = 120 m, and c = 22 m.
Semi-perimeter (s):
Using Heron's formula, Area = √[s(s − a)(s − b)(s − c)]
Area = √[ 132(132 − 122)(132 − 120)(132 − 22) ]
Area = √[ 132 × 10 × 12 × 110 ]
Area = √[ (12 × 11) × 10 × 12 × (11 × 10) ]
Area = √[ 12² × 11² × 10² ]
Area = 12 × 11 × 10 = 1320 m²
Now, let's calculate the rent to be paid:
- Rent for 1 m² for 1 year (12 months) = ₹5000
- Rent for 1320 m² for 12 months = 1320 × 5000
- Rent for 1320 m² for 3 months = 1320 × 5000 × (3/12)
- Total Rent = 1320 × 5000 × (1/4) = 330 × 5000 = 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". If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Let the sides of the triangular wall be a = 15 m, b = 11 m, and c = 6 m.
Semi-perimeter (s):
Using Heron's formula, Area = √[s(s − a)(s − b)(s − c)]
Area = √[ 16(16 − 15)(16 − 11)(16 − 6) ]
Area = √[ 16 × 1 × 5 × 10 ]
Area = √[ 16 × 50 ]
Area = 4 × √(25 × 2) = 4 × 5√2 = 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.
Let the sides be a = 18 cm and b = 10 cm. Let the third side be c.
Given, Perimeter = 42 cm
a + b + c = 42
18 + 10 + c = 42
28 + c = 42 ⇒ c = 42 − 28 = 14 cm
Semi-perimeter (s) = Perimeter / 2 = 42 / 2 = 21 cm
Using Heron's formula, Area = √[s(s − a)(s − b)(s − c)]
Area = √[ 21(21 − 18)(21 − 10)(21 − 14) ]
Area = √[ 21 × 3 × 11 × 7 ]
Area = √[ (7 × 3) × 3 × 11 × 7 ]
Area = √[ 7² × 3² × 11 ]
Area = 7 × 3√11 = 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.
Let the sides of the triangle be 12x, 17x, and 25x.
Given, Perimeter = 540 cm
12x + 17x + 25x = 540
54x = 540 ⇒ x = 540 / 54 = 10
So, the actual lengths of the sides are:
- a = 12 × 10 = 120 cm
- b = 17 × 10 = 170 cm
- c = 25 × 10 = 250 cm
Semi-perimeter (s) = 540 / 2 = 270 cm
Using Heron's formula, Area = √[s(s − a)(s − b)(s − c)]
Area = √[ 270(270 − 120)(270 − 170)(270 − 250) ]
Area = √[ 270 × 150 × 100 × 20 ]
Area = √[ (9 × 30) × (5 × 30) × 100 × (4 × 5) ]
Area = √[ 9 × 30² × 5² × 100 × 4 ]
Area = 3 × 30 × 5 × 10 × 2 = 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 equal sides of the isosceles triangle be a = 12 cm and b = 12 cm. Let the third side be c.
Given, Perimeter = 30 cm
a + b + c = 30
12 + 12 + c = 30
24 + c = 30 ⇒ c = 30 − 24 = 6 cm
Semi-perimeter (s) = 30 / 2 = 15 cm
Using Heron's formula, Area = √[s(s − a)(s − b)(s − c)]
Area = √[ 15(15 − 12)(15 − 12)(15 − 6) ]
Area = √[ 15 × 3 × 3 × 9 ]
Area = √[ 15 × 3² × 3² ]
Area = 3 × 3√15 = 9√15 cm²
