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}}{\sqrt{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 + \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) | ||
+ | |||
==Video Solution by The Power of Logic== | ==Video Solution by The Power of Logic== |
Revision as of 21:52, 25 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 1
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
Solution 2
We put the graph to a coordinate plane. We put to the origin,
to the
-axis, and
to the
-axis.
We have the coordinates of the following points: ,
,
,
.
Denote ,
. Hence,
,
.
Hence, the slope of is
.
The slope of
is
.
Because ,
. Hence,
In , following from the Pythagorean theorem, we have
.
In , following from the Pythagorean theorem, we have
.
Because ,
Solving Equations (1) and (2), we get ,
.
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by The Power of Logic
~math2718281828459
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 |