Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.5 8th Std Maths Answers Solutions Chapter 15 Area.
Practice Set 15.5 8th Std Maths Answers Chapter 15 Area
Question 1.
Find the areas of given plots. (All measures are in meters.)
Solution:
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)
= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(AP) x l(QA)
= \(\frac { 1 }{ 2 }\) 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)
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
= \(\frac { 1 }{ 2 }\) x [l(QA) + l(RC)] x l(AC)
= \(\frac { 1 }{ 2 }\) x (50 + 25) x 60
= \(\frac { 1 }{ 2 }\) x 75 x 60
= 2250 sq.m
In ∆RCS, l(CS) = 60 m, l(RC) = 25 m A(∆RCS)
= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(CS) x l(RC)
= \(\frac { 1 }{ 2 }\) 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) = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x l(PS) x l(TB)
= \(\frac { 1 }{ 2 }\) 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)
= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(AE) x l(AB)
= \(\frac { 1 }{ 2 }\) 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]