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 1

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

Solution 2 (Pro Guessing Strats)

We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to $a\sqrt{20}$ and $b\sqrt{20}$ , and because the hypotenuse is 20 we get $a+b=20$ . Testing small numbers, we get that when $a=2$ and $b=18$ , $ab$ is indeed a square. The area of the triangle is thus 60, so the answer is $\boxed {D)360}$ .

~tigershark22

Solution 3 (coordinates)

Let the points be $A(-10,0)$ , $\:B(x,y)$ , $\:C(10,0)$ , $\:D(10,30)$ ,and $\:E(-5,0)$ , respectively. Since $B$ lies on line $DE$ , we know that $y=2x+10$ . Furthermore, since $\angle{ABC}=90^\circ$ , $B$ lies on the circle with diameter $AC$ , so $x^2+y^2=100$ . Solving for $x$ and $y$ with these equations, we get the solutions $(0,0)$ and $(-8,-6)$ . We immediately discard the $(0,0)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}360}$ .

Video Solution

https://youtu.be/RKlG6oZq9so

~IceMatrix

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search