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

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

Revision as of 23:59, 31 January 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

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


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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