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Area Problems

Area Problems
The ratio of the area of a square to that of the square drawn on its diagonal is

1:2
4:1
3:1
2;3

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

atio =  a 2 2 a 2

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

225%
50%
125%
100%

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 ratio between the length and breadth of a rectangular field is 5:4. If the breadth is 20 meters less than the length, the perimeters of the field is?

260 m
280 m
none of these
360 m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let length of rectangular field = 5y,so width = 4y.From question5y - 4y = 20m? x = 20m ? Length = (5 x 20)m = 100 mBreadth = (4 x 20) m = 80 m ? Perimeter = 2 (100 + 80 ) m = 360 m

Area Problems
A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

30
20
24
33

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres. AC = √2x = (1.41x) m. Saving on 2x metres = (0.59x) m. Saving % = ❨ 0.59x x 100 ❩% = 30%

Area Problems
The length and breadth of a playground are 36 m and 21 m respectively. Flagstaffs are required to be fixed on all along the boundary at a distance of 3 m apart. The number of flagstaffs will be?

38
40
37
39

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Perimeter = 2 x (36 + 21 ) m = 144 m ? Number of flagstaffs = 144 / 3 = 38

Area Problems
A cost of cultivating a square field at a rate of ? 135 per hectare is ? 1215. The cost of putting a fence around it at the rate of 75 paise per meter wouldbe

? 1800
? 810
? 900
? 360

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Area of the field =1215/135 = 9 hec= 90000 m2 [1 hec =10000 m2]? Side of the field = ?90000 = 300 mPerimeter of the field = 4 x 300 = 1200 mNow, cost of putting a fence around field = (1200 x 75)/100 = ? 900

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