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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 21"

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<math>\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}</math>
 
<math>\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}</math>
  
−
== Solution ==
+
== Solution 1 ==
−
First realize that <math>\triangle BCY \sim \triangle DAX.</math> Thus, because <math>CY: XA = 2:3,</math> we can say that <math>BY = 2s</math> and <math>DX = 3s.</math> From the Pythagorean Theorem, we have <math>AB =(2s)^2 + 4^2 = 4s^2 + 16</math> and <math>CD = (3s)^2 + 3^2 = 9s^2 + 9.</math> Because <math>AB = CD,</math> from the problem statement, we have that <cmath>4s^2 + 16 = 9s^2 + 9.</cmath> Solving, gives <math>s = \frac{\sqrt{7}}{\sqrt{5}}.</math> To find the area of the trapezoid, we can compute the area of <math>\triangle ABC</math> and add it to the area of <math>\triangle ACD.</math> Thus, the area of the trapezoid is <math>\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \left(\frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 \right) = \frac{15\sqrt{7}}{{5}} = 3\sqrt{35}.</math> Thus, the answer is <math>\boxed{\textbf {(C)} \: 3\sqrt{35}}.</math>
+
First realize that <math>\triangle BCY \sim \triangle DAX.</math> Thus, because <math>CY: XA = 2:3,</math> we can say that <math>BY = 2s</math> and <math>DX = 3s.</math> From the Pythagorean Theorem, we have <math>AB^2 =(2s)^2 + 4^2 = 4s^2 + 16</math> and <math>CD^2 = (3s)^2 + 3^2 = 9s^2 + 9.</math> Because <math>AB = CD,</math> from the problem statement, we have that <cmath>4s^2 + 16 = 9s^2 + 9.</cmath> Solving, gives <math>s = \frac{\sqrt{7}}{\sqrt{5}}.</math> To find the area of the trapezoid, we can compute the area of <math>\triangle ABC</math> and add it to the area of <math>\triangle ACD.</math> Thus, the area of the trapezoid is <math>\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \left(\frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 \right) = \frac{15\sqrt{7}}{\sqrt{5}} = 3\sqrt{35}.</math> Thus, the answer is <math>\boxed{\textbf{(C)} \: 3\sqrt{35}}.</math>
  
 
~NH14
 
~NH14
 +
 +
== Solution 2 ==
 +
We put the graph to a coordinate plane. We put <math>X</math> to the origin, <math>AC</math> to the <math>x</math>-axis, and <math>DX</math> to the <math>y</math>-axis.
 +
 +
We have the coordinates of the following points: <math>X = \left( 0 , 0 \right)</math>, <math>A = \left( 3 , 0 \right)</math>, <math>Y = \left( - 1 , 0 \right)</math>, <math>C = \left( - 3 , 0 \right)</math>.
 +
 +
Denote <math>BY = u</math>, <math>DX = v</math>. Hence, <math>B = \left( - 1 , u \right)</math>, <math>D = \left( 0 , - v \right)</math>.
 +
 +
Hence, the slope of <math>BC</math> is <math>m_{BC} = \frac{u}{2}</math>.
 +
The slope of <math>AD</math> is <math>m_{AD} = \frac{v}{3}</math>.
 +
 +
Because <math>BC \parallel AD</math>, <math>m_{BC} = m_{AD}</math>. Hence,
 +
<cmath>
 +
\[
 +
\frac{u}{2} = \frac{v}{3} . \hspace{1cm} (1)
 +
\]
 +
</cmath>
 +
 +
In <math>\triangle AYB</math>, following from the Pythagorean theorem, we have <math>AB^2 = AY^2 + BY^2 = 4^2 + u^2</math>.
 +
 +
In <math>\triangle CXD</math>, following from the Pythagorean theorem, we have <math>CD^2 = CX^2 + XD^2 = 3^2 + v^2</math>.
 +
 +
Because <math>AB = CD</math>,
 +
<cmath>
 +
\[
 +
4^2 + u^2 = 3^2 + v^2 . \hspace{1cm} (2)
 +
\]
 +
</cmath>
 +
 +
Solving Equations (1) and (2), we get <math>u = \frac{2 \sqrt{35}}{5}</math>, <math>V = \frac{3 \sqrt{35}}{5}</math>.
 +
 +
Therefore,
 +
<cmath>
 +
\begin{align*}
 +
{\rm Area} \ ABCD
 +
& = {\rm Area} \ \triangle ABC + {\rm Area} \ \triangle  ADC \\
 +
& = \frac{1}{2} AC \cdot BY + \frac{1}{2} AC \cdot DX \\
 +
& = \frac{1}{2} AC \left( BY + DY \right) \\
 +
& = 3 \sqrt{35} .
 +
\end{align*}
 +
</cmath>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(C)} \: 3\sqrt{35}}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
 +
==Solution 3==
 +
 +
Since both <math>\overline{BY}</math> and <math>\overline{DX}</math> are perpendicular to <math>\overline{AC}</math>, we have <math>BD^2=XY^2+(BY+DX)^2</math>.
 +
 +
Since the trapezoid is equilateral, <math>BD=AC=6</math>. Hence, <math>BY+DX=\sqrt{6^2-1^2}=\sqrt{35}</math>.
 +
 +
Notice that both triangles <math>ABC</math> and <math>ADC</math> share a common base <math>\overline{AC}</math>.
 +
 +
Therefore, <cmath>[ABCD]=[ABC]+[ADC]=\frac{1}{2}AC\cdot(BY+DX)=\frac{1}{2}\cdot6\cdot\sqrt{35}=\boxed{\textbf{(C)} \: 3\sqrt{35}}.</cmath>
 +
~RunyangWang
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/R1cyesL2t8A
 +
 +
~math2718281828459
 +
 +
==Video Solution by Mathematical Dexterity==
 +
https://www.youtube.com/watch?v=TCeYkekkjrU
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/9dyA0hSqfXE
 +
 +
~IceMatrix
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/FWmrHV1dWPM?t=950
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=20|num-a=22}}

Latest revision as of 02:40, 23 January 2023

Problem

Let $ABCD$ be an isosceles trapezoid with $\overline{BC}\parallel \overline{AD}$ and $AB=CD$. Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$, as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ$, $AX = 3$, $XY = 1$, and $YC = 2$. What is the area of $ABCD?$

[asy] size(10cm); usepackage("mathptmx"); import geometry; void perp(picture pic=currentpicture, pair O, pair M, pair B, real size=5, pen p=currentpen, filltype filltype = NoFill){ perpendicularmark(pic, M,unit(unit(O-M)+unit(B-M)),size,p,filltype); } pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),Y=(2,0),X=(3,0),A=(6,0),B=(2,sqrt(5.6)),D=(3,-sqrt(12.6)); draw(A--B--C--D--cycle,p); draw(A--C,p); draw(B--Y,p); draw(D--X,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(X,q); dot(Y,q); label("2",C--Y,S); label("1",Y--X,S); label("3",X--A,S); label("$A$",A,2*E); label("$B$",B,2*N); label("$C$",C,2*W); label("$D$",D,2*S); label("$Y$",Y,2*sqrt(2)*NE); label("$X$",X,2*N); perp(B,Y,C,8,p); perp(A,X,D,8,p); [/asy] $\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}$

Solution 1

First realize that $\triangle BCY \sim \triangle DAX.$ Thus, because $CY: XA = 2:3,$ we can say that $BY = 2s$ and $DX = 3s.$ From the Pythagorean Theorem, we have $AB^2 =(2s)^2 + 4^2 = 4s^2 + 16$ and $CD^2 = (3s)^2 + 3^2 = 9s^2 + 9.$ Because $AB = CD,$ from the problem statement, we have that \[4s^2 + 16 = 9s^2 + 9.\] Solving, gives $s = \frac{\sqrt{7}}{\sqrt{5}}.$ To find the area of the trapezoid, we can compute the area of $\triangle ABC$ and add it to the area of $\triangle ACD.$ Thus, the area of the trapezoid is $\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \left(\frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 \right) = \frac{15\sqrt{7}}{\sqrt{5}} = 3\sqrt{35}.$ Thus, the answer is $\boxed{\textbf{(C)} \: 3\sqrt{35}}.$

~NH14

Solution 2

We put the graph to a coordinate plane. We put $X$ to the origin, $AC$ to the $x$-axis, and $DX$ to the $y$-axis.

We have the coordinates of the following points: $X = \left( 0 , 0 \right)$, $A = \left( 3 , 0 \right)$, $Y = \left( - 1 , 0 \right)$, $C = \left( - 3 , 0 \right)$.

Denote $BY = u$, $DX = v$. Hence, $B = \left( - 1 , u \right)$, $D = \left( 0 , - v \right)$.

Hence, the slope of $BC$ is $m_{BC} = \frac{u}{2}$. The slope of $AD$ is $m_{AD} = \frac{v}{3}$.

Because $BC \parallel AD$, $m_{BC} = m_{AD}$. Hence, \[ \frac{u}{2} = \frac{v}{3} . \hspace{1cm} (1) \]

In $\triangle AYB$, following from the Pythagorean theorem, we have $AB^2 = AY^2 + BY^2 = 4^2 + u^2$.

In $\triangle CXD$, following from the Pythagorean theorem, we have $CD^2 = CX^2 + XD^2 = 3^2 + v^2$.

Because $AB = CD$, \[ 4^2 + u^2 = 3^2 + v^2 . \hspace{1cm} (2) \]

Solving Equations (1) and (2), we get $u = \frac{2 \sqrt{35}}{5}$, $V = \frac{3 \sqrt{35}}{5}$.

Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle ABC + {\rm Area} \ \triangle ADC \\ & = \frac{1}{2} AC \cdot BY + \frac{1}{2} AC \cdot DX \\ & = \frac{1}{2} AC \left( BY + DY \right) \\ & = 3 \sqrt{35} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(C)} \: 3\sqrt{35}}$.

~Steven Chen (www.professorchenedu.com)

Solution 3

Since both $\overline{BY}$ and $\overline{DX}$ are perpendicular to $\overline{AC}$, we have $BD^2=XY^2+(BY+DX)^2$.

Since the trapezoid is equilateral, $BD=AC=6$. Hence, $BY+DX=\sqrt{6^2-1^2}=\sqrt{35}$.

Notice that both triangles $ABC$ and $ADC$ share a common base $\overline{AC}$.

Therefore, \[[ABCD]=[ABC]+[ADC]=\frac{1}{2}AC\cdot(BY+DX)=\frac{1}{2}\cdot6\cdot\sqrt{35}=\boxed{\textbf{(C)} \: 3\sqrt{35}}.\] ~RunyangWang

Video Solution by The Power of Logic

https://youtu.be/R1cyesL2t8A

~math2718281828459

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=TCeYkekkjrU

Video Solution by TheBeautyofMath

https://youtu.be/9dyA0hSqfXE

~IceMatrix

Video Solution by OmegaLearn

https://youtu.be/FWmrHV1dWPM?t=950

~ pi_is_3.14

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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