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Maharashtra State BoardSSC (English Medium) 8th Standard

Find the areas of the given plot. (All measures are in metres.)

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Question

Find the areas of the given plot. (All measures are in metres.)

Sum
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Solution

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) = `sqrt324`

= 18 m …[Taking square root of both sides]

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

= `(1/2) xx l("AE") xx l("AB")`

= `(1/2) xx 18 xx 24`

= 216 sq. m

In ∆BCE, a = 30m, b = 28m, c = 26m

Semiperimeter of ∆BCE,

S =  `1/2` (a + b + c) 

= `(30 + 28 + 26)/2`

= `84/2`

= 42 m

A (∆BCE) = `sqrt(s(s - a)(s - b)(s - c))`

= `sqrt(42 (42 - 30)(42 - 28)(42 - 26))`

= `sqrt(42 xx 12xx 14 xx 16)`

= `sqrt(6 xx 7 xx 6 xx 2 xx 2 xx 7 xx 8 xx 2)`

= `sqrt(6^2 xx 7^2 xx 8^2)`

= `6 xx 7 xx 8`

= 336 sq. m

In ∆EDC, l(CE) = 28 m, l(DF) = 16 m

A(∆EDC) = `1/2 xx "base" xx "height"`

`1/2 xx l("CE") xx l("DF")`

= `(1/2) xx 28 xx 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.

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Chapter 15: Area - Practice Set 15.5 [Page 102]

APPEARS IN

Balbharati Mathematics [English] Standard 8 Maharashtra State Board
Chapter 15 Area
Practice Set 15.5 | Q 1.2 | Page 102
Balbharati Mathematics Integrated [English] Standard 8 Maharashtra State Board
Chapter 15 Area
Practice Set 15.5 | Q 1. (2) | Page 57
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