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Difference between revisions of "2014 AMC 12B Problems/Problem 9"

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<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5 </math>
 
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5 </math>
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[[Category: Introductory Geometry Problems]]
  
 
==Solution==
 
==Solution==
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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>
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 +
== Video Solution by OmegaLearn==
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https://youtu.be/4_x1sgcQCp4?t=923
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 +
~ 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 02:40, 23 January 2023

Problem

Convex quadrilateral $ABCD$ has $AB=3$, $BC=4$, $CD=13$, $AD=12$, and $\angle ABC=90^{\circ}$, as shown. What is the area of the quadrilateral?

[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A); label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,E); draw(rightanglemark(A,B,C,25)); [/asy]

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5$

Solution

Note that by the pythagorean theorem, $AC=5$. Also note that $\angle CAD$ is a right angle because $\triangle CAD$ is a right triangle. The area of the quadrilateral is the sum of the areas of $\triangle ABC$ and $\triangle CAD$ which is equal to \[\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}\]

Video Solution by OmegaLearn

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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