Q11.    Sum of the areas of two squares is 468 m2 . If the difference of their perimeters is 24 m, find the sides of the two squares.

Answers (1)
D Divya Prakash Singh

Let the sides of the squares be 'x'\ and\ 'y'.              (NOTE: length are in meters)

And the perimeters will be: 4x\ and\ 4y respectively.

Areas x^2\ and\ y^2 respectively.

It is given that,

x^2 + y^2 = 468\ m^2                    .................................(1)

4x-4y = 24\ m                       .................................(2)

Solving both equations:

x-y = 6  or  x= y+6  putting in equation (1), we obtain

(y+6)^2 +y^2 = 468

\Rightarrow 2y^2+36+12y = 468

\Rightarrow y^2+6y - 216 = 0

Solving by the factorizing method:

\Rightarrow y^2+18y -12y-216 = 0

\Rightarrow y(y+18) -12(y+18) = 0

\Rightarrow (y+18)(y-12)= 0

Here the roots are: \Rightarrow y = -18,\ 12

As the sides of a square cannot be negative.

Therefore, the sides of the squares are 12m and (12\ m+6\ m) = 18\ m.

 

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