Mensuration Made Great : With Area Word Problems For Class 8th,9th,10th[11]

WORD PROBLEMS ON MENSURATION
FOR CLASS 8TH,9TH,10TH

MENSURATION :

Q 1]. ABCD and WXYB are two identical squares as shown in the figure given , with side of length 8cm . If P is the midpoint of AD and WX . Find the area of the hexagon BCDPXYB.

Solution :

So , ∠A = ∠W = 90°
as each of these figures is a square . Join P to B .

Q2]. There are two parks shown below . Rani goes round Park A and Ria goes round Park B .

(a) Who covers more distance ?
(b) Which park encloses more area and by how much ?

SOLUTION:

(a) Distance covered by Rani : 2l + b + π r
= 2 x 70 + 35 + 22/7 x 35/2
= 140 + 35 + 11 x 5
= 175 + 55
= 230 m

Distance covered by Ria = 2l + b + π r
= 2 x 70 + 35 + 22/7 x 35/2
= 140 +35 + 11 x 5
=175 + 55
= 230 m

Thus , both of them cover the same distance .


(b) Area of Park A

= l x b – 1/2 (π r )²
= 70 x 35 -1/2 x 22/7 x 35/2 x 35/2
= 2450 – 481.25
= 1968.75 m²

Area of Park B = l x b + 1/2(π r )²
= 70 x 35 + 1/2 x 22/7 x 35/2 x 35/2
= 2450 + 481.25
= 2931.25 m²

Ans :- Park B encloses more area by 2931.25 – 1968.75 = 962.5m² .

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Q3]. Find the area of the following figure by dividing it into parts . Also verify its area using formula for the area of a trapezium.

Solution : Draw CE ⊥ AB . Since AB ∥ DC and CE ⊥ AB, AECD is a rectangle with
DC = AE = 5cm

AND

BE =AB – AE = 8 -5 = 3 cm
Now, consider the right – angled triangle BEC .

BC² = BE² + CE² (By Pythagoras Theorem )

5² = 3² + CE²

CE² = 25 -9 = 16

So, CE = 4 cm

Thus, the area of trapezium ABCD = Area of rectangle AECD + Area of △ BEC

= 5 x 4 + 1/2 x 3 x 4 = 20 + 6 = 26 cm²

Using formula ,
area of trapezium ABCD = 1/2 (AB + CD) x CE
= 1/2 (8 + 5 ) x 4
= 13 x 2 = 26 cm²

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Q 4]. In the given figure , PQTS is a parallelogram in which PQ = TS = 12 cm and the area of △QRT is 75 cm² . If TR = 15 cm , find the area of trapezium PQRS.

Solution :

Draw QU ⊥ RT and let QU = h
Area of △QRT = 1/2 x b x h = 1/2 x 15 x h

75 = 1/2 x 15 x h
h = (75 x 2 )/ 15

h = 10 cm

Area of trapezium PQRS = 1/2 h (PQ + RS ) = 1/2 x 10 x (12 + 27 ) = 5 x 39 = 195 cm²

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Q 5]. Find the area of the following quadrilaterals .

Solution :

(a) Area of quadrilateral ABCD = 1/2 x AC x( DE + BF )
= 1/2 x 7.5 x ( 2+4)
= 1/2 x 7.5 x (6)
= 3 x 7.5 = 22.5 cm²

(b) Area of quadrilateral WXYZ = 2 x 1/2 x WY x OZ
= 2 x 1/2 x 16 x 6
= 16 x 6
= 96 cm²

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Q 6]. Find the area of the quadrilateral PQRS shown below .

Solution :

Area of quadrilateral PQRS = AREA 1 + AREA 2

= 1/2 x PS x QS + 1/2 x QR x QS
= 1/2 x 7 x 12 + 1/2 x 15 x 12
= 7 x 6 + 15 x 6
=42 + 90
= 132 cm²

So , answer is 132 cm²

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Q 7] . Find the area of a rhombus with side 26 cm, one of whose diagonals is 20 cm .

Solution : Let ABCD be the rhombus , with diagonals meeting O .The diagonals of a rhombus bisect each other at 90° . So , OA = OC = 10 cm .

Also, let OB = OD = x .

And ,
In △OCD , CD² = OC² + OD²
(By Pythagoras Theorem )

or 26² = 10² + x
or x² = 26²– 10²
or x² = 676 – 100
or x² = 576
or x = 24 cm

So , BD = 2x = 2 x 24 = 48 cm

Thus , area of rhombus = 1/2 AC x BD
= 1/2 x 20 x 48
= 10 x 48
= 480 cm²

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