Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4 (Trignometry))
(→‎Solution 4 (Trignometry))
Line 31: Line 31:
 
Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, we find <math>BE=3\sqrt5</math>. Since <math>EC=15</math> and <math>DC=30</math>, <math>DE=15\sqrt3</math>, which is five times <math>BE</math>,  <math>[ABCD]=[ACD]+\frac{1}/{5}[ACD]=300+60=\boxed {\textbf{D) }360}</math>
 
Using the law of cosines, express <math>AB^2</math> and <math>BC^2</math> in terms of <math>\angle{AEB}</math>. The sum of these two equations is <math>AC^2</math> by the Pythagorean Theorem. Solving for <math>BE</math>, we find <math>BE=3\sqrt5</math>. Since <math>EC=15</math> and <math>DC=30</math>, <math>DE=15\sqrt3</math>, which is five times <math>BE</math>,  <math>[ABCD]=[ACD]+\frac{1}/{5}[ACD]=300+60=\boxed {\textbf{D) }360}</math>
  
−
(This solution is incomplete, can someone complete it please)
+
(This solution is incomplete, can someone complete it please-Lingjun)
−
-Lingjun
 
 
Latex edited by kc5170
 
Latex edited by kc5170
  

Revision as of 15:46, 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 (Just Drop An Altitude)

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 $[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{D) }360}$

(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 $\sqrt{20a}$ and $\sqrt{20b}$, 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 $\boxed {\textbf{D) }360}$.

~tigershark22 ~(edited by HappyHuman)

Solution 3 (coordinates)

[asy] draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); draw(circle((0,0),10)); label("D",(10,30),N); label("C",(10,0),E); label("B",(-8,-6),S); label("A",(-10,0),W); [/asy] 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 $(0,10)$ and $(-8,-6)$. We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}$.

Solution 4 (Trignometry)

Using the law of cosines, express $AB^2$ and $BC^2$ in terms of $\angle{AEB}$. The sum of these two equations is $AC^2$ by the Pythagorean Theorem. Solving for $BE$, we find $BE=3\sqrt5$. Since $EC=15$ and $DC=30$, $DE=15\sqrt3$, which is five times $BE$, $[ABCD]=[ACD]+\frac{1}/{5}[ACD]=300+60=\boxed {\textbf{D) }360}$

(This solution is incomplete, can someone complete it please-Lingjun) Latex edited by kc5170

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
2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&oldid=116334"