Practice Set 15.4 8th Std Maths Answers Chapter 15 Area
Question 1
Maharashtra Board Solution
Sides of a triangle are 45 cm, 39 cm and 42 cm, find its area.
Solution & Step-by-Step Answer:
Sides of a triangle are 45 cm, 39 cm and 42 cm. Here, a = 45cm, b = 39cm, c = 42cm Semi perimeter of triangle = s = = = = 63 Area of a triangle ∴ The area of the triangle is 756 sq.cm.
Question 2
Maharashtra Board Solution
Look at the measures shown in the given figure and find the area of ☐PQRS.
Solution & Step-by-Step Answer:
A (☐PQRS) = A(∆PSR) + A(∆PQR) In ∆PSR, l(PS) = 36 m, l(SR) = 15 m A(∆PSR) = x product of sides forming the right angle = x l(SR) x l(PS) = 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
Maharashtra Board Solution
Some measures are given in the figure, find the area of ☐ABCD.
Solution & Step-by-Step Answer:
A(☐ABCD) = A(∆BAD) + A(∆BDC) In ∆BAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m A(∆BAD) = x product of sides forming the right angle = x l(AB) x l(AD) = x 40 x 9 = 180 sq. m In ∆BDC, l(BT) = 13m, l(CD) = 60m A(∆BDC) = x base x height = x l(CD) x l(BT) = 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.