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

Revision as of 18:32, 1 March 2020

The following problem is from both the 2020 AMC 12A #18 and 2020 AMC 10A #20, so both problems redirect to this page.

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)

$[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1,1.5), N); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((0,0)--(1,1.5), dashed); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (1,1.5)--(1.714,1.143), NE); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); draw(rightanglemark((0,0),(1,1.5),(0,2))); [/asy]$

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 but now you have a diagram by ciceronii)

~ Solution by Ultraman

~ Diagram by ciceronii

Solution 3 (coordinates)

$[asy] size(10cm,0); draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); draw((-20,0)--(20,0)); draw((0,-15)--(0,35)); draw((10,30)--(-8,-6)); draw(circle((0,0),10)); label("E",(-4.05,-.25),S); label("D",(10,30),NE); label("C",(10,0),NE); label("B",(-8,-6),SW); label("A",(-10,0),NW); label("5",(-10,0)--(-5,0), NE); label("15",(-5,0)--(10,0), N); label("30",(10,0)--(10,30), E); dot((-5,0)); dot((-10,0)); dot((-8,-6)); dot((10,0)); dot((10,30)); [/asy]$ 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 (Trigonometry)

Let $\angle C = \angle{ACB}$ and $\angle{B} = \angle{CBE}.$ Using Law of Sines on $\triangle{BCE}$ we get $\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}$ and LoS on $\triangle{ABE}$ yields $\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.$ Divide the two to get $\tan{B} = 3 \tan{C}.$ Now, $\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}$ and solve the quadratic, taking the positive solution (C is acute) to get $\tan{C} = \frac{1}{3}.$ So if $AB = a,$ then $BC = 3a$ and $[ABC] = \frac{3a^2}{2}.$ By Pythagorean Theorem, $10a^2 = 400 \iff \frac{3a^2}{2} = 60$ and the answer is $300 + 60 \iff \boxed{\textbf{(D)}}.$

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

We could use the famous m-n rule in trigonometry in triangle ABC with Point E [Unable to write it here.Could anybody write the expression] We will find that BD is angle bisector of triangle ABC(because we will get tan (x)=1) Therefore by converse of angle bisector theorem AB:BC = 1:3. By using phythagorean theorem we have values of AB and AC. AB.AC = 120. Adding area of ABC and ACD Answer••360

Video Solution

On The Spot STEM https://www.youtube.com/watch?v=hIdNde2Vln4

@@ Line 36: / Line 36: @@
 ~ Diagram by ciceronii
-==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 <math>\sqrt{20a}</math> and <math>\sqrt{20b}</math>, and because the hypotenuse is 20 we get <math>a+b=20</math>. Testing small numbers, we get that when <math>a=2</math> and <math>b=18</math>, <math>ab</math> is indeed a square. The area of the triangle is thus 60, so the answer is <math>\boxed {\textbf{(D) }360}</math>.
-~tigershark22
-~(edited by HappyHuman)
 ==Solution 3 (coordinates)==

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

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