Question

Given ABC find the values of x and y. In your final answer, include all of your calculations.

Answer 1

So with this, we will be using proportions. To solve for x, our proportion will be $\frac{x}{6} =\frac{6}{x-9}$

Firstly, cross-multiply: $x^2-9x=36$

Next, subtract 36 on both sides: $x^2-9x-36=0$

Next, replace -9x with 3x-12x $x^2+3x-12x-36=0$

Next, factor x^2+3x and -12x-36 separately: $x(x+3)-12(x+3)=0$ . After, rewrite it as $(x-12)(x+3)=0$

Next, solve for x separately in the two parentheses:

$x-12=0\\ x=12$

$x+3=0\\ x=-3$

Since we can't have a negative length, x = 12.

To solve for y, our proportion will be $\frac{y}{9} =\frac{12}{y}$

Firstly, cross-multiply: $y^2=108$

Next, just square root each side, and your answer will be (rounded to hundredths) $y=10.39$

In short, x = 12 and y = 10.39.