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# AREA MCQS SET-13

1. The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .
A) 22 cm
B) 24 cm
C) 26 cm
D) 28 cm
Ans. B

We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm

Area of third Square = 10*10 – 8*8
= 36 cm

So side of third square = 6 [because area of square = side*side]Perimeter = 4*Side = 4*6 = 24 cm
2. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.
A) 4:25
B) 4:15
C) 3:25
D) 3:15
Ans. A
Let the diagonals of the squares be 2x and 5x.
Then ratio of their areas will be

Area of square=12∗Diagonal2
12∗2×2 : 12∗5×2 4×2 : 25×2 = 4 : 25 Area of square = 12 ∗ Diagonal2 12 ∗ 2×2: 12 ∗ 5×2 4x 2: 25×2 = 4 : 25 Area of square
= 12∗Diagonal2 12 ∗ 2×2 : 12∗ 5×2 4×2 : 25×2 = 4:25

3. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.
A) 10 m * 5 m
B) 15 m * 5 m
C) 20 m * 5 m
D) 25 m * 5 m
Ans. C
From the question, 2b+l = 30
=> l = 30-2b

Area=100m2
=l×b=100
=b(30−2b) =100b2−15b+50=0
=(b−10)(b−5)
=0Area=100m2
=l×b=100
=b(30−2b)
=100b2−15b+50=0
=(b−10)(b−5)
=0Area=100m2
=l×b=100
=b(30−2b)
=100b2−15b+50=0
=(b−10)(b−5)=0
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m

4. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is ?
A) 25%
B) 26%
C) 27%
D) 28%
Ans. D
Let original length = x
and original width = y
Decrease in area will be

=xy−(80 x 100 × 90y 100)
=(xy−1825xy) = 725xy Decrease
= (7xy 25xy × 100) %
=28% = xy−(80 x 100 × 90y 100)
=(xy−1825xy)
=725 xy Decrease
= (7xy 25xy × 100) %
=28%
=xy − (80x 100 × 90y 100)
=(xy−1825xy)
=725 xy Decrease
= (7xy25xy×100)%=28%

5. What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?
A) Rs. 3430
B) Rs. 3440
C) Rs. 3450
D) Rs. 3460

Ans. B
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344

So required cost will be = 344 * 10 = 3440