Maharashtra Board 8th Class Maths Practice Set 15.4 Solutions Chapter 15 Area

Maharashtra Board 8th Class Maths Practice Set 15.4 Solutions Chapter 15 Area

Practice Set 15.4 8th Std Maths Answers Chapter 15 Area

Question 1.
Sides of a triangle are 45 cm, 39 cm and 42 cm, find its area.

Question 2.
Look at the measures shown in the given figure and find the area of ☐PQRS.

Solution:
A (☐PQRS) = A(∆PSR) + A(∆PQR)
In ∆PSR, l(PS) = 36 m, l(SR) = 15 m
A(∆PSR)
= 1/2 x product of sides forming the right angle
= 1/2 x l(SR) x l(PS)
= 1/2 x 15 x 36
= 270 sq.m
In ∆PSR, m∠PSR = 90°
[l(PR)]² = [l(PS)]² + [l(SR)]²
…[Pythagoras theorem]
= (36)² + (15)²
= 1296 + 225
∴ l(PR)² = 1521
∴ l(PR) = 39m
…[Taking square root of both sides]
In ∆PQR, a = 56m, b = 25m, c = 39m

A(☐PQRS) = A(∆PSR) + A(∆PQR)
= 270 + 420
= 690 sq. m
∴ The area of ☐PQRS is 690 sq.m

Question 3.
Some measures are given in the figure, find the area of ☐ABCD.

Solution:
A(☐ABCD) = A(∆BAD) + A(∆BDC)
In ∆BAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m
A(∆BAD) = 1/2 x product of sides forming the right angle
= 1/2 x l(AB) x l(AD)
= 1/2 x 40 x 9
= 180 sq. m
In ∆BDC, l(BT) = 13m, l(CD) = 60m
A(∆BDC) = 1/2 x base x height
= 1/2 x l(CD) x l(BT)
= 1/2 x 60 x 13
= 390 sq. m
A (☐ABCD) = A(∆BAD) + A(∆BDC)
= 180 + 390
= 570 sq. m
∴ The area of ☐ABCD is 570 sq.m.