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Area Problems

Area Problems
If the sides of a rectangle are increased by 5%, find the percentage increase in its diagonals.

4%
9%
6%
5%

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

According to the formula,Percentage increase in diagonals = 5%

Area Problems
The two parallel sides of a trapezium are 1 meters and 2 meters respectively . The perpendicular distance between them is 6 meters. The area of the trapezium is?

6 sq. meters .
9 sq. meters .
18 sq. meters .
12 sq. meters .

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Area of trapezium = 1/2 x sum of parallel sides x perpendicular distance between them= 1/2 (1 + 2) x 6 m2= 9 m2

Area Problems
The inner circumference of a circular race track, 14 m wide is 440 m . Then the radius of the outer circle is?

77 m
84 m
56 m
70 m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Circumference of a circular = 2?r? 2 x (22 / 7) x r = 440? r = [440 x (7/22) x (1/2)] = 70 m? Radius of outer circle = (70 + 14) m = 84 m

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

? 360
? 1800
? 810
? 900

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

Area Problems
The area of sector of a circle of radius 5 cm, formed by an arc of length 3.5 cm is?

8. 75 sq.cms
17.5 sq.cms
35 sq.cms
55 sq.cms

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Area of sector = ( arc length x radius ) / 2 cm2= (3.5 x 5 ) / 2= 8.75 cm2

Area Problems
The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

1520 m2
2480 m2
2420 m2
2520 m2

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. ∴ Area = (l x b) = (63 x 40) m2 = 2520 m2

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