Question

Two square have sides $x$ cm and $(x+4)$ cm. The sum of their areas is $656c{m}^2$. Find the sides of the squares.

• A. $8$ cm, $12$ cm
• B. $12$ cm, $15$ cm
• C. $6$ cm, $10$ cm
• D. $16$ cm, $20$ cm

Answer 1

Answer: (D)

$16$ cm, $20$ cm


Area of 1st square $=x^2$
Area of 2nd square $=(x+4{)}^2=x^2+8x+16$
Sum of areas = $x^2+x^2+8x+16=656$
Simplifying and subtracting $656$ from both sides
$2x^2+8x-640=0$
$x^2+4x-320=0$

This is a standard quadratic equation of the form $a{x}^2+bx+c=0$ with $a=1,b=4$ and $c=-320.$
Solving it we get the values of x as $x=16$
or $x=-20$
Since x cannot be negative, $x=16$
So, side of 1st square = $16$ and side of second square = $20$