Mensuration Made Great : With Word Problems For Class 8th ,9th,10th[12]

Q.1] Find the area of the polygon given below .

Solution :

Area of polygon RSTUV = Area of △RUV + Area of △RTU + Area of △RST

= 1/2 RU x VA + 1/2RU x UT + 1/2 RT x SB

Also , △RTU is a right – angled triangle .

So, RT² = RU² + UT²

or , RT² = 12² + 5²

or, RT² = 144 + 25

or , RT² = 169

or , RT = 13 cm

So, area of polygon RSTUV

= 1/2 x 12 x 4.5 x + 1/2 x 12 x5 + 1/2 x 13 x 8

= 27 + 30 + 52 = 109 cm²

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Q2] Sonia and Shreya are involved in a paper – cutting activity . Each of them is given an A4 size sheet . Both of them cut out the following polygons . Find out which polygon has the larger area .

Solution :

(a) = Sonia ‘s polygon
(b) = Shreya’s polygon


Area of polygon ABCDEFG = Area of parallelogram ABCG + Area of trapezium GCDF + Area of △DEF

= b x h + 1/2 h (a + b) + 1/2 b x h

= 6 x 4 + 1/2 x 5 x (10 + 6 ) + 1/2 x 10 x 4.8

= 24 + 40 + 24

= 88 cm²


Area of polygon PQRSTUVW = Area of △PWQ + Area of trapezium WQRV + Area of right – angled triangle UVR + Area of rectangle UTSR

Also, UVR is a right – angled triangle , so
VR² = UV² + UR² = 6² + 8² =100

VR = 10 cm

= 1/2 x 3 x 2 + 1/2 x (10 +36 ) x 6 + 1/2 x 6 x8 + 8 x 4.5

= 3 + 13 x 3 + 8 x 3 + 8 x 4.5

= 3 + 39 + 24 + 36

= 102 cm²

Thus , Shreya’s polygon has a larger area .

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Q 3] . The given figure shows the side view of a house . Find the area of this side wall . Also, find the cost of repairing it at the rate of rupees 150 per m² .

Solution :

Area of side wall = Area of trapezium DCFE + Area of square ABCF

= 1/2( DE + FC ) x DX + AB²

=1/2 (1.6 + 2.8) x 2.2 + 2.8 x 2.8

= 1.1 x 4.4 + 7.84

= 4.84 + 7.84

= 12.68 m²

So, the cost of repairing the wall

= 12.68 x 150 = rupees 1902 .

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