Problem 9.2. On the sides $AB$ and $AD$ of rectangle $ABCD$, points $M$ and $N$ are marked, respectively. It is known that $AN=7$, $NC=39$, $AM=12$, and $MB=3$.

(a) (1 point) Find the area of rectangle $ABCD$.

(b) (3 points) Find the area of triangle $MNC$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1f4981519479532effc1g-01.jpg?height=417&width=599&top_left_y=1261&top_left_x=423)
Answer: (a) 645. (b) 268.5.

Solution. In a rectangle, opposite sides are equal: $AB = CD$ and $BC = AD$. Then $CD = 12 + 3 = 15$. From the Pythagorean theorem for triangle $CDN$, it follows that $DN = \sqrt{CN^2 - CD^2} = \sqrt{39^2 - 15^2} = 36$. Then $BC = AD = 36 + 7 = 43$ and $S_{ABCD} = 15 \cdot 43 = 645$.

As is known, the area of a right triangle is half the product of its legs. Then

$$
S_{MNC} = S_{ABCD} - S_{AMN} - S_{BCM} - S_{DNC} = 645 - \frac{12 \cdot 7}{2} - \frac{3 \cdot 43}{2} - \frac{15 \cdot 36}{2} = 268.5
$$