2020 AMC 10A Problem 20

The diagonal $AC$ splits $ABCD$ into triangles $ACD$ and $ABC$, which share the base $AC = 20$. So we need the height of each triangle above $AC$.

1. Triangle $ACD$. It has a right angle at $C$, so its legs are $AC = 20$ and $CD = 30$. Area $= \frac12 \times 20 \times 30 = 300$.
2. Triangle $ABC$. Its area is $\frac12 \times 20 \times h$, where $h$ is the distance from $B$ to line $AC$. Drop a perpendicular from $B$ to $AC$ and call its foot $F$, so $BF = h$.
3. Use the point $E$. $E$ is on $AC$ with $AE = 5$, so $EC = 20 - 5 = 15$. Both $BF$ and $DC$ are perpendicular to $AC$, so they are parallel. That makes triangles $BFE$ and $DCE$ similar (same angles). In triangle $DCE$, the side along $AC$ is $EC = 15$ and the upright side is $DC = 30$: the first is half the second. So in triangle $BFE$ as well, $FE$ is half of $BF$: $FE = \frac{h}{2}$.
4. Which side of $E$ is $F$ on? $D$ is above the $C$ end of the diagonal. The line from $D$ through $E$ keeps going past $E$ toward the $A$ end, and $B$ is on that line on the other side of $AC$. So $F$ lies between $A$ and $E$:
$$AF = 5 - \tfrac{h}{2}, \qquad FC = 15 + \tfrac{h}{2}.$$
5. Use the right angle at $B$. In right triangle $ABC$, the height $BF$ splits it into two smaller triangles, $AFB$ and $BFC$, with the same angles. Matching their sides gives $\frac{AF}{BF} = \frac{BF}{FC}$, that is,
$$h^2 = AF \times FC = \left(5 - \tfrac{h}{2}\right)\left(15 + \tfrac{h}{2}\right).$$
6. Solve. Multiply out: $h^2 = 75 + \frac{5h}{2} - \frac{15h}{2} - \frac{h^2}{4} = 75 - 5h - \frac{h^2}{4}$. Multiply by $4$ and collect: $5h^2 + 20h - 300 = 0$, so $h^2 + 4h - 60 = 0$, which factors as $(h + 10)(h - 6) = 0$. The height is positive, so $h = 6$.
7. Check: $AF = 5 - 3 = 2$, $FC = 15 + 3 = 18$, and $2 \times 18 = 36 = 6^2$. Good.
8. Add up. Triangle $ABC$ has area $\frac12 \times 20 \times 6 = 60$, so the total is $300 + 60 = 360$.

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

Splitting a four-sided shape along a diagonal gives two triangles with a shared base, so only their heights matter. The crossing point $E$ ties the two heights together through similar triangles, and the right angle at $B$ gives a second relation. Two facts, one unknown height: that is enough to solve.

Use coordinates. Put $A = (0, 0)$, $C = (20, 0)$, $D = (20, 30)$ and $E = (5, 0)$. The line from $D$ to $E$ drops $30$ while moving $15$ to the left, so it drops $2$ for every $1$ to the left. Continuing past $E$, its points are $(5 - t, -2t)$ for $t > 0$.

The angle at $B$ is a right angle, so $B$ lies on the circle with diameter $AC$: center $(10, 0)$, radius $10$. So
$$(5 - t - 10)^2 + (2t)^2 = 100, \quad\text{that is}\quad (t + 5)^2 + 4t^2 = 100.$$
This gives $5t^2 + 10t - 75 = 0$, or $t^2 + 2t - 15 = 0$, so $t = 3$ and $B = (2, -6)$. The height of $B$ below $AC$ is $6$, and the area is $300 + 60 = 360$.

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 $\boxed{\textbf{(D)}\ 360}$.