Two squares have sides $x$ cm and $(x + 4)$ cm. The sum of their areas is $656\ cm^2$. Find the sides of the squares.
Given:
Two squares have sides $x$ cm and $(x + 4)$ cm. The sum of their areas is $656\ cm^2$.
To do:
We have to find the sides of the squares.
Solution:
We know that,
Area of a square of side 's' is $s^2$.
This implies,
Area of the square of side $x = x^2$
Area of the square of side $(x+4)\ cm = (x+4)^2\ cm^2$
According to the question,
$x^2+(x+4)^2=656$
$x^2+x^2+2(4)(x)+4^2=656$ (Since $(a+b)^2=a^2+2ab+b^2$)
$2x^2+8x+16=656$
$2x^2+8x+16-656=0$
$2x^2+8x-640=0$
$2(x^2+4x-320)=0$
$x^2+4x-320=0$
Solving for $x$ by factorization method, we get,
$x^2+20x-16x-320=0$
$x(x+20)-16(x+20)=0$
$(x+20)(x-16)=0$
$x+20=0$ or $x-16=0$
$x=-20$ or $x=16$
Length cannot be negative. Therefore, $x=16$.
The sides of the squares are $16$ cm and $(16+4)\ cm=20\ cm$.
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