Chapter 12 Heron’s Formula Ex 12.1 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?
Answer - 1 : - We know that, an equilateral triangle has equal sides. So, all sides are equal to a.
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 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 - 2 : -
Leta = 122m,
b = 22m
c = 120m
We have, b2 + c2 = (22)2 + (120)2 = 484 + 14400 = 14884= (122)2 = a2
Hence, the side walls are in right triangled shape.
Question - 3 : - There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6m, find the area painted in colour.
Answer - 3 : -
The given figure formed a triangle whose sides are
a = 15m
b = 11m,
c = 6m
Question - 4 : - Find the area of a triangle two sides of which are18 cm and 10 cm and the perimeter is 42 cm.
Answer - 4 : - Solution Let the sides of a triangle, a = 18cm b = 10cm and c
We have, perimeter = 42 cm
Question - 5 : - Sides of a triangle are in the ratio of 12 : 17 : 25and its perimeter is 540 cm. Find its area.
Answer - 5 : - Suppose that the sides in cm, are 12x, 17x and 25x.
Then, we know that 12x + 17x + 25x = 540 (Perimeter of triangle)
⇒ 54x = 540
⇒ x=10
So, the sides of the triangle are 12 x 10 cm, 17 x 10 cm 25 x 10 cm
i.e., 120 cm, 170 cm, 250 cm
Question - 6 : - An isosceles triangle has perimeter 30 cm and eachof the equal sides is 12 cm. Find the area of the triangle.
Answer - 6 : -
Let in isosceles ∆ ABC,
AB = AC = 12 cm (Given)
Question - 7 : - A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. Howmuch area does it occupy?
Answer - 7 : -
Let us joinBD.
In ΔBCD, applying Pythagoras theorem,
BD2 = BC2 +CD2
= (12)2 +(5)2
= 144 + 25
BD2 = 169
BD = 13 m
Area of ΔBCD
For ΔABD,
By Heron’sformula,
Area of triangle
Area of ΔABD 
Area of thepark = Area of ΔABD + Area of ΔBCD
= 35.496 + 30 m2 = 65.496 m2 =65.5 m2 (approximately)
Find the area of a quadrilateral ABCD in which AB = 3 cm,BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Answer - 8 : - 
For ΔABC,
AC2 = AB2 +BC2
(5)2 = (3)2 +(4)2
Therefore, ΔABC is a right-angled triangle, right-angledat point B.
Area of ΔABC
For ΔADC,
Perimeter = 2s =AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 7 cm
By Heron’s formula,
Area of triangle.
Area ofABCD = Area of ΔABC + Area of ΔACD
= (6 + 9.166) cm2 =15.166 cm2 = 15.2 cm2 (approximately)
Radha madea picture of an aeroplane with coloured papers as shown in the given figure.Find the total area of the paper used.
Answer - 9 : - 
For triangle I
This triangle is an isosceles triangle.
Perimeter = 2s =(5 + 5 + 1) cm = 11cm
Area of the triangleFor quadrilateralII
This quadrilateral is a rectangle.
Area = l× b = (6.5 × 1) cm2 = 6.5 cm2
For quadrilateral III
This quadrilateral is a trapezium.
Perpendicular height of parallelogram 
Area = Areaof parallelogram + Area of equilateral triangle
= 0.866 + 0.433 = 1.299 cm2
Area oftriangle (IV) = Area of triangle in (V)
Total areaof the paper used = 2.488 + 6.5 + 1.299 + 4.5 × 2
= 19.287 cm2