Difference between revisions of "2014 AMC 12B Problems/Problem 9"
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Note that by the pythagorean theorem, <math>AC=5</math>. Also note that <math>\angle CAD</math> is a right angle because <math>\triangle CAD</math> is a right triangle. The area of the quadrilateral is the sum of the areas of <math>\triangle ABC</math> and <math>\triangle CAD</math> which is equal to | Note that by the pythagorean theorem, <math>AC=5</math>. Also note that <math>\angle CAD</math> is a right angle because <math>\triangle CAD</math> is a right triangle. The area of the quadrilateral is the sum of the areas of <math>\triangle ABC</math> and <math>\triangle CAD</math> which is equal to | ||
<cmath>\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}</cmath> | <cmath>\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}</cmath> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/4_x1sgcQCp4?t=923 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2014|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:34, 24 January 2021
Contents
Problem
Convex quadrilateral has , , , , and , as shown. What is the area of the quadrilateral?
Solution
Note that by the pythagorean theorem, . Also note that is a right angle because is a right triangle. The area of the quadrilateral is the sum of the areas of and which is equal to
Video Solution
https://youtu.be/4_x1sgcQCp4?t=923
~ pi_is_3.14
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 |
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