Question

Find the areas of given plots. (All measures are in meters.)

i. 

Answer 1

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) = (1/2) x product of sides forming the right angle 

= (1/2) x l(AP) x l(QA) 

= (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) = (1/2) x sum of lengths of parallel sides x height 

= (1/2) x [l(QA) + l(RC)] x l(AC) 

= (1/2) x (50 + 25) x 60 

= (1/2) x 75 x 60

= 2250 sq.m 

In ∆RCS, l(CS) = 60 m, l(RC) = 25 m

A(∆RCS) = (1/2) x product of sides forming the right angle 

= (1/2) x l(CS) x l(RC) 

= (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) = (1/2) x base x height 

= (1/2) x l(PS) x l(TB) 

= (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) = (1/2) x product of sides forming the right angle

= (1/2) x l(AE) x l(AB) 

= (1/2) x 18 x 24 = 216 sq. m 

In ∆BCE, a = 30m, b = 28m, c = 26m

= (1/2) x 28 x 16

= 224 sq. m.

∴ 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]