Difference between revisions of "2021 Fall AMC 12A Problems/Problem 21"
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== Solution 1 == | == 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^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 | ||
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==Video Solution by Mathematical Dexterity== | ==Video Solution by Mathematical Dexterity== | ||
https://www.youtube.com/watch?v=TCeYkekkjrU | 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 03:40, 23 January 2023
Contents
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)
Solution 3
Since both and
are perpendicular to
, we have
.
Since the trapezoid is equilateral, . Hence,
.
Notice that both triangles and
share a common base
.
Therefore,
~RunyangWang
Video Solution by The Power of Logic
~math2718281828459
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=TCeYkekkjrU
Video Solution by TheBeautyofMath
~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 |