6. Let $A B C$ be an isosceles triangle with $A B = A C$. The altitude from $A$ measures 15, while the altitude from $B$ measures 24. What is the area of $A B C$?
(A) 180
(B) 300
(C) $240 \sqrt{2}$
(D) $200 \sqrt{3}$
(E) 320
6. The answer is (B). Let $M$ be the midpoint of side $B C$, which is also the foot of the altitude from $A$ (since $A B C$ is isosceles with base $B C$), and let $P$ be the foot of the altitude from $B$. Since the altitude from $A$ is shorter than the altitude from $B$, the angle at $A$ is obtuse, so the point $P$ falls outside the segment $A C$ (and is on the opposite side of $C$ relative to $A$).

![](https://cdn.mathpix.com/cropped/2024_04_17_9f6e2f6c8f4b083321c1g-08.jpg?height=257&width=834&top_left_y=1822&top_left_x=657)

The triangles $M C A$ and $P C B$ are similar, as they have two equal angles: indeed, the angles at $M$ and $P$ are right angles, and the angle at $C$ is common. Let $x = P C$ and $y = M B = M C$. From the similarity already observed, we get $\frac{M C}{A M} = \frac{P C}{B P}$, or $\frac{y}{15} = \frac{x}{24}$. The triangle $B P C$ is right-angled at $P$, so the Pythagorean theorem gives $P B^2 + P C^2 = B C^2$, which is

$$
24^2 + x^2 = (2 y)^2 \text{.}
$$

Substituting $x = \frac{24}{15} y = \frac{8}{5} y$ we then get

$$
24^2 + \frac{64}{25} y^2 = 4 y^2 \Rightarrow 24^2 = \frac{36}{25} y^2 \Rightarrow y = \sqrt{\frac{24^2 \cdot 5^2}{6^2}} = \frac{24 \cdot 5}{6} = 20
$$

The area of $A B C$ is therefore equal to $\frac{1}{2} \cdot B C \cdot A M = \frac{1}{2} \cdot 2 z \cdot 15 = 20 \cdot 15 = 300$.