Difference between revisions of "2021 Fall AMC 12A Problems/Problem 21"
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+ | ==Problem== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <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> | ||
+ | <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 == | ||
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 21:37, 23 November 2021
Problem
Let be an isosceles trapezoid with
and
. Points
and
lie on diagonal
with
between
and
, as shown in the figure. Suppose
,
,
, and
. What is the area of
Solution
First realize that Thus, because
we can say that
and
From the Pythagorean Theorem, we have
and
Because
from the problem statement, we have that
Solving, gives
To find the area of the trapezoid, we can compute the area of
and add it to the area of
Thus the area of the trapezoid is
Thus the answer is
~NH14