Problem 10.6. Given trapezoid $ABCD$. Segments $AB$ and $CD$ - its lateral sides - are equal to 24 and 10, respectively. Points $X$ and $Y$ are marked on side $AB$ such that $AX=6, XY=8, YB=10$. It is known that the distances from points $X$ and $Y$ to line $CD$ are 23 and 27, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4c5d21501884c712855g-6.jpg?height=297&width=493&top_left_y=1418&top_left_x=480)

(a) (1 point) Find the area of triangle $ACD$.

(b) (3 points) Find the area of trapezoid $ABCD$.
# Answer:

(a) (1 point) 100.

(b) (3 points) 260.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4c5d21501884c712855g-7.jpg?height=362&width=503&top_left_y=255&top_left_x=467)

Fig. 7: to the solution of problem 10.6

Solution. We will prove that the distances from points $A$ and $B$ to the line $C D$ are 20 and 32, respectively. Let $A_{1}, X_{1}, Y_{1}, B_{1}$ be the projections of points $A, X, Y, B$ onto the line $C D$ respectively (Fig. 7). Draw a line through $X$ parallel to $C D$, and let it intersect the lines $A A_{1}$, $Y Y_{1}$, $B B_{1}$ at points $P, Q, R$ respectively. In the similar right triangles $A P X$, $Y Q X$, $B R X$, the hypotenuses are in the ratio $A X: X Y: Y B=6: 8: 18=3: 4: 9$, so $A P: Y Q: B R=3: 4: 9$. Considering that $Y Q=Y Y_{1}-Q Y_{1}=Y Y_{1}-X X_{1}=27-23=4$, we get that $A P=3$ and $B R=9$. Then $A A_{1}=X X_{1}-A P=23-3=20$ and $B B_{1}=X X_{1}+B R=23+9=32$. Therefore, $S_{A C D}=\frac{1}{2} \cdot A A_{1} \cdot C D=\frac{1}{2} \cdot 20 \cdot 10=100$ and $S_{B C D}=\frac{1}{2} \cdot B B_{1} \cdot C D=\frac{1}{2} \cdot 32 \cdot 10=160$.

In the trapezoid $A B C D$, the areas of triangles $A B C$ and $B C D$ are equal, since side $B C$ is common to both, and the heights dropped from points $A$ and $D$ to $B C$ are equal. Therefore,

$$
S_{A B C D}=S_{A C D}+S_{A B C}=S_{A C D}+S_{B C D}=100+160=260
$$