Question

Find an expression for $y$ in terms of $x$ for the triangle $\Delta FGH$ shown in the above figure.

• A. $x+4$
• B. $\sqrt{x^2+4}$
• C. $\sqrt{x^2+8}$
• D. $\sqrt{x^2-16}$
• E. $\sqrt{x^2+16}$

Answer 1

The correct option is E $\sqrt{x^2+16}$

In triangle $FGH$, $\angle F={90}^{\circ }$

$\therefore (HG{)}^2=(FG{)}^2+(HF{)}^2$

$\Rightarrow y^2=x^2+4^2$

$\Rightarrow y=\sqrt{x^2+16}$