Question

Sum of the areas of two squares is 305 m2
. If the difference of their perimeters is 36 m then find the
sides of the two squares.

Answer 1

Answer:

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Step-by-step explanation:

so let the two squares be S1 and S2 respectively

so let the side of square S1 be x.m and that of square S2 be y.m

so we know that area of square=side²

thus then area of square S1=x²

area of square S2= y²

so according to the first condition

x²+y²=305 (1)

moreover we know that

perimter of square=4×side

so perimter if square square=4x

perimter of square S2=4y

so according to second condition

4x-4y=36

ie 4(x-y)=36

ie x-y=9

so x=y+9 (2)

substitute value of x in (1) from (2)

we get

x²+y²=305

ie (y+9)²+y²=305

ie y²+81+18y+y2-305=0

ie 2y²+18y-224=0

dividing by 2 on both sides

we get

y²+9y-112=0

so y²+16y-7y-112=0

ie y(y+16)-7(y+16)=0

ie (y+16)(y-7)=0

so y=-16 or y=7

but as dimensions length are never negative

y=-16 is absurd

hence y=7

substitute value of y in any of two equations

we get x=16

hence length of square S1 is 16 m and that of square S2 is 7 m