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Maharashtra State BoardSSC (English Medium) 8th Standard

Find the areas of the given plot. (All measures are in metres.) - Marathi (Second Language) [मराठी (द्वितीय भाषा)]

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Question

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

Sum

Solution

In given Figure,

PA = 30 m, AB = 30 m BC = 30 m, CS = 60 m, QA = 50 m, RC = 25 m, TB = 30 m.

A(ΔPAQ) = 12 × PA × QA

= 12×30×50

∴ A(ΔPAQ) = 750 m2   ......(1)

A(ΔRCS) = 12 × CS × RC

= 12×60×25

∴ A(ΔRCS) = 750 m2   ......(2)

∵ P − A − B − C − S

∴ PS = PA + AB + BC + CS

= 30 + 30 + 30 + 60

∴ PS = 150 m

Also, AC = AB + BC = 30 + 30 = 60 m

Also, AC = AB + BC = 30 + 30 = 60 m

A(ΔPTS) = 12 × PS × TB

= 12×150×30=2250

∴ A(ΔPTS) = 2250 m2   .....(3)

QA || RC

QRCA is a Trapezium.

A(QRCA) = 12×(QA + RC)×AC

= 12×(50+25)×60=12×75×60

∴ A(QRCA) = 2250 m2   ......(4)

Area of given Plot = A(ΔPAQ) + A(QRCA) + A(ΔRCS) + A(ΔPTS)

= 750 + 2250 + 750 + 2250    .......[From (1), (2), (3) and (4)]

∴ Area of the given Plot = 6000 m2

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

APPEARS IN

Balbharati Mathematics [English] 8 Standard Maharashtra State Board
Chapter 15 Area
Practice Set 15.5 | Q 1.1 | Page 102
Balbharati Integrated 8 Standard Part 4 [English Medium] Maharashtra State Board
Chapter 3.2 Area
Practice Set 15.5 | Q 1. (1) | Page 57
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