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Home » The sum of the areas of two squares is . If the difference in their perimeter be , find the sides of the two square.
The sum of the areas of two squares is 640 \mathrm{~m}^{2}. If the difference in their perimeter be 64 \mathrm{~m}, find the sides of the two square.
CBSE | Class 10 | Exercise 10e | Maths | Quadratic Equations | RS Aggarwal
The sum of the areas of two squares is 640 \mathrm{~m}^{2}. If the difference in their perimeter be 64 \mathrm{~m}, find the sides of the two square.

Let the length of the side of the first and the second square be x and y \cdot respectively. According to the question:

x^{2}+y^{2}=640

Also,

\begin{array}{l} 4 x-4 y=64 \\ \Rightarrow x-y=16 \\ \Rightarrow x=16+y \end{array}

Putting the value of x in (i), we get:

\begin{array}{l} x^{2}+y^{2}=640 \\ \Rightarrow(16+y)^{2}+y^{2}=640 \\ \Rightarrow 256+32 y+y^{2}+y^{2}=640 \\ \Rightarrow 2 y^{2}+32 y-384=0 \\ \Rightarrow y^{2}+16 y-192=0 \\ \Rightarrow y^{2}+(24-8) y-192=0 \\ \Rightarrow y^{2}+24 y-8 y-192=0 \\ \Rightarrow y(y+24)-8(y+24)=0 \end{array}
\begin{array}{l} \Rightarrow(y+24)(y-8)=0 \\ \Rightarrow y=-24 \text { or } y=8 \\ \therefore y=8 \quad(\because \text { Side cannot be negative) } \\ \therefore x=16+y=16+8=24 m \end{array}

Thus, the sides of the squares are 8 \mathrm{~m} and 24 \mathrm{~m}.

i

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