The ratio of the areas of two squares is $\frac{32}{63}$. 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 $a+b+c$?

MathzSolver111 Nov 18, 2020

#1**+2 **

I'm not Chris, but I can help you with this

First, we need to find the square roots of the 2 squares, because the squares of them will give us 32 and 63. So, the square root of 32 is \(\sqrt{32}\), which simplifies to: \(\sqrt{16} \cdot \sqrt2 \implies 4\sqrt2\).

We can use this same thing to find the square root of 63.

\(\sqrt{63} \implies \sqrt9 \cdot \sqrt7 \implies 3\sqrt7\)

So when finding the square roots, our fraction becomes \(\frac{\sqrt{32}}{\sqrt{63}}\), which using our steps makes the fraction \(\frac{4\sqrt2}{3\sqrt7}\)

To rationalize this, we need to multiply \(\sqrt7\) on both sides, so this gives us:

\(\frac{4\sqrt{2} \cdot \sqrt7}{3\sqrt7 \cdot \sqrt7} \implies \frac{4\sqrt{14}}{21}\)

That makes a=4, b=14, and c=21. Adding 4+14+21 give us \(\boxed{39}\)

mobro Nov 18, 2020