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Question

The sum of the areas of two squares is 400m2. If the difference between their perimeters is 16m, find the smallest side of two squares.

A
3m
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B
8m
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C
12m
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D
21m
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Solution

The correct option is B 12m
Let the sides of the squares be s1 and s2
Area of square =side2
Then,
s12+s22=400.....(i)
and 4s24s1=16
s2s1=4
s2=4+s1
Put this value in equation (i)
4+s12+s12=400
16+s12+8s1+s12=400
2s12+8s1384=0
s12+4s1192=0
(s1+16)+(s112)=0
s1=12 or s1=16
Since length cannot be negative, s1=12, and s2=s1+4=16
Hence, the smaller side of the two squares is 12m

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