This Article consist of most important ” Area Questions and Answers ” that are mostly asked in all competitive exams. We collected these questions from the students who appeared in exams. Now try to solve these questions.
Question : 1 The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm.The area of the triangle is :
A. 70 sq.cm
B. 140 sq.cm
C. 210 sq.cm
D. 420 sq.cm
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Correct Answer : C
Explanation :
Let the radius of incircle be r cm. Then ,
2*Pi*r=88 , r=(88*7*1)/(22*2), r=14 cm
Semi Perimeter,s=(30/2) cm=15 cm
So, Area = r*s=(14*15)=220 sq.cm
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Question : 2 A typist uses a sheet measuring 20 cm by 30 cm lengthwise.If a margin of 2 cm is left on each side and a 3 cm margin on top and bottom, then percent of the page used for typing is :
A. 40
B. 60
C. 64
D. 72
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Correct Answer : C
Explanation :
Area of the sheet = (20*30) sq.cm
Area used for typing=[(20-4)*(30-6)] sq.cm=384 sq.cm
so, Required percentage=(384/600)*100=64%
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Question : 3 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?
A. 34
B. 40
C. 68
D. 88
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Correct Answer : D
Explanation :
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft
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Question : 4 What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
A. 814
B. 820
C. 840
D. 844
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Correct Answer : A
Explanation :
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm2.
Required number of tiles = (1517 x 902)/(41 x 41)
= 814
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Question : 5 The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
A. 1520 m2
B. 2420 m2
C. 2480 m2
D. 2520 m2
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Correct Answer : D
Explanation :
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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