# How to solve for the variables of the figures below?

Apr 10, 2018

For Figure A $x = \frac{11}{2}$ and $y = 11 \frac{\sqrt{3}}{2}$.

For Figure B $x = 16$, and $y = 8 \sqrt{2}$.

#### Explanation:

Figure A shows a 30-60-90 triangle with a hypotenuse of length 11. The short leg is 1/2 the length of the hypotenuse, so $x = \frac{11}{2}$. The long leg is the square root of 3 times the length of the short leg so $y = 11 \frac{\sqrt{3}}{2}$.

Figure B shows a 45-45-90 triangle with a leg length of $8 \sqrt{2}$. The legs of the triangle have equal length, so $y = 8 \sqrt{2}$. We can get $y$ from the Pythagorean Theorem.

${\left(8 \sqrt{2}\right)}^{2} + {\left(8 \sqrt{2}\right)}^{2} = {x}^{2}$

$256 = {x}^{2}$

$x = 16$