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Area Problems

Area Problems
The radius of a circular field is 25 m. Find the area of the field.

25 ? sq m
135 ? sq m
125 ? sq m
625 ? sq m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Required area = ?r2 = ? x 25 x 25 = 625? sq m.

Area Problems
The side of a square is 5 cm which is 13 cm less than the diameter of a circle. What is the approximate area of the circle?

255 sq cm
265 sq cm
235 sq cm
245 sq cm

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Diameter of the circle = 13 + 5 = 18 cm? Radius = Diameter/2 =18/2 = 9 cm Area of the circle = ?r2 = (22/7) x 92 = (22 x 81)/9 = 1782/7 = 254.57 sq cm= 255 sq cm

Area Problems
The length of minute hand of a wall clock is 7 cms. The area swept by minute hand in 30 minutes is?

154 sq. cm
147 sq. cm
210 sq. cm
77 sq. cm

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Angle swept in 30 min= 180° Area swept = [(22/7) x 7 x 7] x [180°/360°] cm2 = 77 cm2

Area Problems
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

88
34
40
68

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft

Area Problems
If the circumference of a circle is increased by 50%, then its area will be increased by?

225%
100%
50%
125%

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Original circumference = 2?r New circumference = (150 /100) x 2 ?r = 3?r 2?R = 3?r? R = 3r/2 Original area = ?r2New area = ?R2= ?9r2 / 4 = 9?r2/4Increase in area = (9?r2/4 ) - (?r2)= (5/4) ?r2Req. increase per cent = [{(5/4) ?r2} / {?r2}] x 100 = 125 %

Area Problems
The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

25% increase
75% decrease
50% increase
50% decrease

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let original length = x and original breadth = y. Original area = xy. New length = x . 2 New breadth = 3y. New area = ❨ x x 3y ❩ = 3 xy. 2 2 ∴ Increase % = ❨ 1 xy x 1 x 100 ❩% = 50%

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