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Difference between revisions of "2020 AMC 10A Problems/Problem 20"

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==Solution 3 (coordinates)==
 
==Solution 3 (coordinates)==
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Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,0)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,0)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}360}</math>.
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Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,0)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,0)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:54, 1 February 2020

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


See Also

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

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