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

(Problem)
(Problem)
Line 2: Line 2:
 
Rectangle <math>ABCD</math> has <math>AB=3</math> and <math>BC=4</math>. Point <math>E</math> is the foot of the perpendicular from <math>B</math> to diagonal <math>\overline{AC}</math>. What is the area of <math>\triangle ABC</math>?
 
Rectangle <math>ABCD</math> has <math>AB=3</math> and <math>BC=4</math>. Point <math>E</math> is the foot of the perpendicular from <math>B</math> to diagonal <math>\overline{AC}</math>. What is the area of <math>\triangle ABC</math>?
  
<math>\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}</math>
+
<math>\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}</math>
  
 
==Solution==
 
==Solution==

Revision as of 12:33, 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

Placeholder

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png