2002 AMC 10A Problem 25

Through $C$ draw the segment parallel to $DA$, meeting $AB$ at $E$. Then $AECD$ is a parallelogram, so $AE = CD = 39$ and $CE = DA = 5$. The leftover piece is triangle $EBC$ with
$$EB = 52 - 39 = 13,\qquad CE = 5,\qquad BC = 12.$$
Since $5^2 + 12^2 = 13^2$, this triangle is right-angled at $C$, and its area is $\frac{1}{2}\cdot 5 \cdot 12 = 30$.

The height of the trapezoid equals the altitude of this triangle onto its hypotenuse $EB$ (which lies along $AB$):
$$h = \frac{2 \cdot 30}{13} = \frac{60}{13}.$$

Now apply the trapezoid area formula:
$$[ABCD] = \frac{52 + 39}{2}\cdot\frac{60}{13} = \frac{91}{2}\cdot\frac{60}{13} = 7 \cdot 30 = 210.$$

The answer is $\boxed{\textbf{(C)}\ 210}$.

A trapezoid is a parallelogram plus a triangle: translating one leg across to the other end of the long base isolates a triangle whose sides are the two legs and the difference of the bases. Here that triangle happens to be $5$-$12$-$13$, so the height falls out without any trigonometry. The moment you see legs $5$, $12$ and base difference $13$, suspect a right triangle.

Extend the legs to meet at a point $P$ above $CD$. Triangles $PDC$ and $PAB$ are similar with ratio $\frac{39}{52} = \frac{3}{4}$, so $PD = 15$ and $PC = 36$, making $PDC$ a $15$-$36$-$39$ right triangle (a scaled $5$-$12$-$13$). Then $[ABCD] = [PAB] - [PDC] = \frac12(20)(48) - \frac12(15)(36) = 480 - 270 = 210$.

• Treating the trapezoid as if the leg 12 or 5 were the height, giving 546 or 227.5, instead of finding the true height 60/13.
• Computing $\frac{52 + 39}{2} = 45.5$ and multiplying by the triangle's area $30$ or by $13$, confusing the triangle's dimensions with the trapezoid's height.