Difference between revisions of "2017 AMC 10B Problems/Problem 15"

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==Solution==
 
==Solution==
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(E) <math>\frac{54}{25}</math>
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}}
 
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{{MAA Notice}}

Revision as of 16:11, 16 February 2017

Problem

Rectangle $ABCD$ has $AB=3$ and $BC=4$. Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}$. What is the area of $\triangle ABC$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}$

Solution

(E) $\frac{54}{25}$

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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 10 Problems and Solutions

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