In triangle $A B C$, if $A B=A C=x+1$ and $B C=2 x-2$, where $x>1$, then the area of the triangle is always equal to
(A) $(x-1) \sqrt{2 x^{2}+2}$
(B) $2(x-1)$
(C) $\frac{1}{2}(x+1)^{2}$
(D) $(x+1)(x-1)$
(E) $2(x-1) \sqrt{x}$

## Part C: Each correct answer is worth 8.
Draw the altitude from $A$ to $P$ on $B C$.

Since $\triangle A B C$ is isosceles, then $P$ is the midpoint of $B C$, so $B P=P C=x-1$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-222.jpg?height=331&width=480&top_left_y=301&top_left_x=882)

By the Pythagorean Theorem,

$$
A P=\sqrt{A B^{2}-B P^{2}}=\sqrt{(x+1)^{2}-(x-1)^{2}}=\sqrt{\left(x^{2}+2 x+1\right)-\left(x^{2}-2 x+1\right)}=\sqrt{4 x}=2 \sqrt{x}
$$

Therefore, the area of $\triangle A B C$ is equal to

$$
\frac{1}{2}(B C)(A P)=\frac{1}{2}(2 x-2)(2 \sqrt{x})=2(x-1) \sqrt{x}
$$

ANswer: (E)