2020 AMC 10A Problems/Problem 20

Problem

Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$

$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$

Solution

It's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$ , we get $[ACD]=300$ . Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$ . Let $FE=x$ . Since $AE=5$ , then $AF=5-x$ . By dropping this altitude, we can also see two similar triangles, $BFE$ and $DCE$ . Since $EC$ is $20-5=15$ , and $DC=30$ , we get that $BF=2x$ . Now, if we redraw another diagram just of $ABC$ , we get that $(2x)^2=(5-x)(15+x)$ . Now expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$ . This factors to $(x+5)(x-3)$ . Since lengths cannot be negative, $x=3$ . Since $x=3$ , $[ABC]=60$ . So $[ABCD]=[ACD]+[ABC]=300+60=\boxed {D)360}$

(I'm very sorry if you're a visual learner)

~Ultraman

Video Solution

https://youtu.be/RKlG6oZq9so

~IceMatrix

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2020 AMC 10A Problems/Problem 20

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