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

(Created page with "== 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 = 3...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
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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, <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 = 15\frac{\sqrt{7}}{{5}} = 3\sqrt{35}.</math> Thus the answer is <math>\boxed{\textbf {(C)} \: 3\sqrt{35}}.</math>
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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 21:36, 23 November 2021

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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