2020 AMC 10A Problem 20

Triangle $ACD$ is right-angled at $C$, so $[ACD] = \frac12 \cdot 20 \cdot 30 = 300$. It remains to find $[ABC] = \frac12 \cdot 20 \cdot h$, where $h$ is the distance from $B$ to line $AC$.

Let $F$ be the foot of the perpendicular from $B$ to $AC$. Since $BF \parallel DC$ (both perpendicular to $AC$), triangles $BFE$ and $DCE$ are similar with ratio
$$\frac{FE}{BF} = \frac{CE}{DC} = \frac{15}{30} = \frac12,$$
using $CE = 20 - 5 = 15$. So $FE = \frac{h}{2}$, and since $B$ and $D$ lie on opposite sides of $AC$, $F$ is on the $A$ side of $E$: $AF = 5 - \frac h2$ and $FC = 15 + \frac h2$.

Because $\angle ABC = 90^\circ$, the altitude to the hypotenuse satisfies $BF^2 = AF \cdot FC$:
$$h^2 = \left(5 - \tfrac h2\right)\left(15 + \tfrac h2\right) = 75 - 5h - \tfrac{h^2}{4}.$$
Multiplying by $4$: $5h^2 + 20h - 300 = 0$, so $h^2 + 4h - 60 = 0$ and $h = 6$.

Then $[ABC] = \frac12 \cdot 20 \cdot 6 = 60$ and
$$[ABCD] = 300 + 60 = 360.$$

The answer is $\boxed{\textbf{(D)}\ 360}$.

The diagonal $AC$ splits the quadrilateral into two triangles sharing a base, so only the heights matter. The intersection point $E$ links the two heights through similar triangles, and the right angle at $B$ supplies the second equation via the geometric-mean relation. Two conditions, one unknown height.

Coordinates: $A = (0,0)$, $C = (20,0)$, $D = (20,30)$, $E = (5,0)$. Line $DE$ has slope $2$, so $B = (5 - t, -2t)$ for some $t > 0$. The right angle at $B$ puts $B$ on the circle with diameter $AC$: $(t + 5)^2 + 4t^2 = 100$, giving $t = 3$ and $B = (2, -6)$. Height $6$, area $300 + 60 = 360$.

• Assuming BD is a diameter or that ABCD is cyclic in a way that fixes B without using AE = 5, or computing [ACD] = 300 and guessing.
• Placing $F$ on the wrong side of $E$ (using $AF = 5 + \frac h2$), which leads to a quadratic with no sensible root.