Question

The sum of the areas of two squares is $400m^2$. If the difference between their perimeters is $16m$, find the smallest side of two squares.

• A. $3m$
• B. $8m$
• C. $12m$
• D. $21m$

Answer 1

Answer: (C)

$12m$

Let the sides of the squares be $s_1$ and $s_2$
Area of square $={\text{side}}^2$

Then,
${s_1}^2+{s_2}^2=400$.....$\left(i\right)$
and $4s_2-4s_1=16$
$s_2-s_1=4$
$s_2=4+s_1$
Put this value in equation $\left(i\right)$
${4+s_1}^2+{s_1}^2=400$
$16+{s_1}^2+8s_1+{s_1}^2=400$
$2{s_1}^2+8s_1-384=0$
${s_1}^2+4s_1-192=0$
$(s_1+16)+(s_1-12)=0$
$s_1=12$ or $s_1=-16$

Since length cannot be negative, $s_1=12$, and $s_2=s_1+4=16$
Hence, the smaller side of the two squares is $12m$