8. Point $D$ is the midpoint of the base $\overline{A B}$ of isosceles triangle $A B C$. The perimeter of triangle $A B C$ is $50 \mathrm{~cm}$, and the perimeter of triangle $A C D$ is $40 \mathrm{~cm}$. What is the area of triangle $A B C$?

No use of a pocket calculator or any reference materials is allowed.
8. 

![](https://cdn.mathpix.com/cropped/2024_05_30_8d7aca671b2af1896715g-3.jpg?height=423&width=345&top_left_y=1362&top_left_x=267)

From the labels on the image, we have $2 x + 2 b = 50$, i.e., $x + b = 25$ (cm). 1 POINT

Furthermore, $x + b + v = 40$ (cm), from which we get $v = 15$ (cm).

1 POINT

By applying the Pythagorean theorem to the right triangle ACD, we get:

$b^2 = v^2 + x^2$, i.e., $(25 - x)^2 = 15^2 + x^2$

2 POINTS

Further, $625 - 50x + x^2 = 225 + x^2$, from which we get $50x = 400$, 2 POINTS

i.e., $x = 8$ (cm), $a = 2x = 16$ (cm) and $b = 17$ (cm).

2 POINTS

The area of triangle ABC is $P = \frac{1}{2} \cdot 16 \cdot 15 = 120$ (cm²).

1 POINT