Practice Set 15.5 8th Std Maths Answers Chapter 15 Area
Question 1
Maharashtra Board Solution
Find the areas of given plots. (All measures are in meters.)
Solution & Step-by-Step Answer:
i. Here, ∆QAP, ∆RCS are right angled triangles and ☐QACR is a trapezium. In ∆QAP, l(AP) = 30 m, l(QA) = 50 m A(∆QAP) = x product of sides forming the right angle = x l(AP) x l(QA) = x 30 x 50 = 750 sq. m In ☐QACR, l(QA) = 50 m, l(RC) = 25 m, l(AC) = l(AB) + l(BC) = 30 + 30 = 60 m A(☐QACR) = x sum of lengths of parallel sides x height = x [l(QA) + l(RC)] x l(AC) = x (50 + 25) x 60 = x 75 x 60 = 2250 sq.m In ∆RCS, l(CS) = 60 m, l(RC) = 25 m A(∆RCS) = x product of sides forming the right angle = x l(CS) x l(RC) = x 60 x 25 = 750 sq. m In ∆PTS, l(TB) = 30 m, l(PS) = l(PA) + l(AB) + l(BC) + l(CS) = 30 + 30 + 30 + 60 = 150m A(∆PTS) = x base x height = x l(PS) x l(TB) = x 150 x 30 = 2250 sq. m ∴ Area of plot QPTSR = A(∆QAP) + A(☐QACR) + A(∆RCS) + A(∆PTS) = 750 + 2250 + 750 + 2250 = 6000 sq. m ∴ The area of the given plot is 6000 sq.m.
ii. In ∆ABE, m∠BAE = 90°, l(AB) = 24 m, l(BE) = 30 m
∴ [l(BE)]² = [l(AB)]² + [l(AE)]²
…[Pythagoras theorem]
∴ (30)² = (24)² + [l(AE)]²
∴ 900 = 576 + [l(AE)]²
∴ [l(AE)]² = 900 – 576
∴ [l(AE)]² = 324
∴ l(AE) = √324 = 18 m
…[Taking square root of both sides]
A(∆ABE)
= x product of sides forming the right angle
= x l(AE) x l(AB)
= x 18 x 24
= 216 sq. m
In ∆BCE, a = 30m, b = 28m, c = 26m
∴ Area of plot ABCDE
= A(∆ABE) + A(∆BCE) + A(∆EDC)
= 216 + 336 + 224
= 776 sq. m
∴ The area of the given plot is 776 sq.m.
[Note: In the given figure, we have taken l(DF) = 16 m]