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Area Problems

Area Problems
What is the diameter of the circle having 3 cm radius?

12 cm
6 cm
8 cm
1.5 cm

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Radius = Diameter/2 ? Diameter = 2 x Radius = 2 x 3 = 6 cm

Area Problems
The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field?

30 m
50 m
20 m
40 m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let breadth = x meters. Then, Length = 115x / 100 meters Given that,  X × (115X / 100) = 460 => X = 20

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

70 m
56 m
84 m
77 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
The ratio of length and breadth of a rectangle is 5 : 3 .If length is 8 m more than breadth, find the area of the rectangle.

240 sq m
250 sq m
300 sq m
185 sq m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Let length of rectangle = 5kand breadth of rectangle = 3kAccording to the quecation,5k - 3k = 8 ? 2k =8? k = 4? Lenght = 5k = 5 x 4 = 20 mBreadth = 3k = 3 x 4 = 12 m? Required area = Lenght x Breadth = 20 x 12 = 240 sq m

Area Problems
A rectangle has 20 cm as its length and 200 sq cm as its area. If the area is increased by 1 1/ 5 time the original area by increase its length only then the perimeter of the rectangle so formed (in cm) is

60
64
68
72

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

l1 = 20 cm, A1 = 200 sq cm? b1 = 200/20 = 10 cmNow, A2 = 200 x 6/5 = 240 sq cm b2 = 10 cm? l2 = 240/10 = 24 cm? Perimeter of new rectangle = 2(l2 + b2)= 2(24 + 10) = 2 x 34 = 68 cm

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

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

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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