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

(→‎Solution)
(→‎Solution)
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==Problem==
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Let <math>ABCD</math> be an isosceles trapezoid with <math>\overline{BC}\parallel \overline{AD}</math> and <math>AB=CD</math>. Points <math>X</math> and <math>Y</math> lie on diagonal <math>\overline{AC}</math> with <math>X</math> between <math>A</math> and <math>Y</math>, as shown in the figure. Suppose <math>\angle AXD = \angle BYC = 90^\circ</math>, <math>AX = 3</math>, <math>XY = 1</math>, and <math>YC = 2</math>. What is the area of <math>ABCD?</math>
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 +
<asy>
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size(10cm);
 +
usepackage("mathptmx");
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import geometry;
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void perp(picture pic=currentpicture,
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pair O, pair M, pair B, real size=5,
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pen p=currentpen, filltype filltype = NoFill){
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perpendicularmark(pic, M,unit(unit(O-M)+unit(B-M)),size,p,filltype);
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}
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pen p=black+linewidth(1),q=black+linewidth(5);
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pair C=(0,0),Y=(2,0),X=(3,0),A=(6,0),B=(2,sqrt(5.6)),D=(3,-sqrt(12.6));
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draw(A--B--C--D--cycle,p);
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draw(A--C,p);
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draw(B--Y,p);
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draw(D--X,p);
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dot(A,q);
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dot(B,q);
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dot(C,q);
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dot(D,q);
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dot(X,q);
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dot(Y,q);
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label("2",C--Y,S);
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label("1",Y--X,S);
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label("3",X--A,S);
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label("$A$",A,2*E);
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label("$B$",B,2*N);
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label("$C$",C,2*W);
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label("$D$",D,2*S);
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label("$Y$",Y,2*sqrt(2)*NE);
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label("$X$",X,2*N);
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perp(B,Y,C,8,p);
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perp(A,X,D,8,p);
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</asy>
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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>
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== Solution ==
 
== Solution ==
 
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 + \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 = \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 =(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 + \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 = \frac{15\sqrt{7}}{{5}} = 3\sqrt{35}.</math> Thus the answer is <math>\boxed{\textbf {(C)} \: 3\sqrt{35}}.</math>
  
 
~NH14
 
~NH14

Revision as of 20:37, 23 November 2021

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

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 $AB =(2s)^2 + 4^2 = 4s^2 + 16$ and $CD = (3s)^2 + 3^2 = 9s^2 + 9.$ 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 $\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 = \frac{15\sqrt{7}}{{5}} = 3\sqrt{35}.$ Thus the answer is $\boxed{\textbf {(C)} \: 3\sqrt{35}}.$

~NH14

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