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Area Problems

Area Problems
The ratios of areas of two squares, one having its diagonal double than the other is

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

ANSWER DOWNLOAD EXAMIANS APP

Area Problems
The breadth of a rectangle is 25 m. The total cost of putting a grass bed on this field was ? 12375, at the rate of ? 15 per sq m. What is the length of the rectangular field?

32 m
33 m
27 m
30 m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Area to the rectangular field = 12375/15 = 825 sq mAccording to the question, (L x B) = 825 [L = length and B = breadth]? L x 25 = 825 ? L = 825/25 = 33 m

Area Problems
The length of a rectangular field is 100 m and its breadth is 40 m. What will be the area of the field?

(4 x 10) sq m
(4 x 102) sq m
(4 x 104) sq m
(4 x 103) sq m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Required area = Length x Breadth= 100 x 40 = 4000 sq m= 4 x 103 sq m

Area Problems
The area of a circle is increased by 22 sq cm when its radius is increased by 1 cm. Find the original radius of the circle.

3 cm
6 cm
3.2 cm
3.5 cm

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let original radius be r.Then, according to the questions,? (r + 1)2 - ?r2 = 22? ? x [(r + 1)2 - r2] = 22? (22/7) x (r + 1 + r ) x (r + 1 - r) = 22? 2r + 1 = 7 ? 2r = 6 ? r = 6/2 = 3 cm

Area Problems
A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is

rs.58
rs.258
rs.358
rs.158

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m cost of graveling = 344 x  (75/100) = Rs. 258

Area Problems
If the base of a rectangular is increased by 10% and the area is unchanged , then the corresponding altitude must to be decreased by?

10%
111/9 %
91/11 %
11%

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let base = b and altitude = h Then, Area = b x h But New base = 110b / 100 = 11b / 10Let New altitude = HThen, Decrese = (h - 10h /11 )= h / 11? Required decrease per cent = (h/11) x (1 / h ) x 100 %= 91/11 %

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