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 $(16+4)\ cm=20\ cm$. 

Tutorialspoint

e

Updated on: 10-Oct-2022

23 Views

Kickstart Your Career

Get certified by completing the course

Get Started