Question

for △fgh, shown below, which of the following is an expression for y in terms of x? 4 meters meters a. x + 4 b. $sqrt{x^{2}+4}$ c. $sqrt{x^{2}+8}$ d. $sqrt{x^{2}-16}$ e. $sqrt{x^{2}+16}$

Explanation:

Step1: Apply Pythagorean theorem

In right - triangle $\triangle FGH$, by the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $a = 4$, $b=x$, and $c = y$.
So, $4^{2}+x^{2}=y^{2}$.

Step2: Solve for $y$

We have $y^{2}=x^{2}+16$. Taking the square root of both sides (since $y>0$ as it represents a length), we get $y=\sqrt{x^{2}+16}$.

Answer:

E. $\sqrt{x^{2}+16}$