# The ratio of the areas of two squares is $\frac{192}{80}$. What is the value of the sum $a+b+c$?

The ratio of the areas of two squares is $\frac{192}{80}$. After rationalizing the denominator, the ratio of their side lengths can be expressed in the simplified form $\frac{a\sqrt{b}}{c}$ where $a$, $b$, and $c$ are integers. What is the value of the sum $a+b+c$?

I have no idea how to approach this problem. Ant hints would be really appreciated.

Call $x$ and $y$ the sides of the two squares

If

$\dfrac{x^2}{y^2}=\dfrac{192}{80}=\dfrac{12}{5}$

then

$\dfrac{x}{y}=\dfrac{\sqrt{12}}{\sqrt{5}}=\dfrac{\sqrt{12}\sqrt{5}}{\sqrt{5}\sqrt{5}}=\dfrac{\sqrt{60}}{5}=\dfrac{\sqrt{4\cdot 15}}{5}=\dfrac{2\sqrt{15}}{5}$

Therefore $a+b+c=22$

Hope it is useful

• $\dfrac{2\sqrt{15}}{5} = \dfrac{4\sqrt{15}}{10}$. There must be another condition, such as $\gcd(a,c)=1$ etc. – Math Lover Oct 21 '17 at 18:15
• @MathLover I thought the same but the OP text says "simplified form"... – Raffaele Oct 21 '17 at 18:46