4. The diagonal $\overline{A C}$ of trapezoid $A B C D$ has a length of $13 \mathrm{~cm}$. For the measurement numbers of the other diagonal $(d)$ and the height of the trapezoid $(v)$, the equation $d^{2}+v^{2}-30 d-24 v+369=0$ holds. Calculate the area of trapezoid $A B C D$.
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_32e0003636127c6bb5b9g-05.jpg?height=505&width=831&top_left_y=353&top_left_x=318)

Sketch:

The given equation can be written in the form \(d^{2}-30 d+v^{2}-24 v+369=0\)

or by completing the square \((d-15)^{2}-225+(v-12)^{2}-144+369=0\)

and finally \((d-15)^{2}+(v-12)^{2}=0\).

2 POINTS

The left side of the equation is equal to 0 if \(d-15=0\) and \(v-12=0\).

1 POINT

From \(d-15=0\) it follows that \(d=15 \text{ cm}\), and

from \(v-12=0\) it follows that \(v=12 \text{ cm}\).

1 POINT

On the extension of side \(\overline{A B}\) beyond vertex \(B\), choose point \(E\) such that \(d(B, E)=d(C, D)\).

Then \(d(A, E)=a+c\).

Quadrilateral \(B E C D\) is a parallelogram, so \(d(C, E)=d=15 \text{ cm}\).

1 POINT

From the right triangle \(\triangle A F C\), we get \(d(A, F)=\sqrt{13^{2}-12^{2}}=5 \text{ cm}\), and

from the right triangle \(\triangle C F E\), \(d(F, E)=\sqrt{15^{2}-12^{2}}=9 \text{ cm}\).

2 POINTS

Since \(d(A, F)+d(F, E)=d(A, E)=d(A, B)+d(B, E)=a+c=5+9=14 \text{ cm}\),

the area of the trapezoid is \(P=\frac{a+c}{2} \cdot v=84 \text{ cm}^{2}\).