Difference between revisions of "2013 AMC 10A Problems/Problem 3"

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==Problem==
+
#redirect [[2013 AMC 12A Problems/Problem 1]]
 
 
Square <math> ABCD </math> has side length <math>10</math>. Point <math>E</math> is on <math>\overline{BC}</math>, and the area of <math> \triangle ABE </math>
 
is <math>40</math>. What is <math> BE </math>?
 
<asy>
 
pair A,B,C,D,E;
 
A=(0,0);
 
B=(0,50);
 
C=(50,50);
 
D=(50,0);
 
E = (30,50);
 
  draw(A--B);
 
  draw(B--E);
 
  draw(E--C);
 
draw(C--D);
 
draw(D--A);
 
draw(A--E);
 
dot(A);
 
dot(B);
 
dot(C);
 
dot(D);
 
dot(E);
 
label("A",A,SW);
 
label("B",B,NW);
 
label("C",C,NE);
 
label("D",D,SE);
 
label("E",E,N);
 
 
 
</asy>
 
 
 
 
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
 
[[Category: Introductory Geometry Problems]]
 
 
 
==Solution==
 
 
 
We know that the area of <math>\triangle ABE</math> is equal to <math>\frac{AB(BE)}{2}</math>.  Plugging in <math>AB=10</math>, we get <math>80 = 10BE</math>.  Dividing, we find that <math>BE=\boxed{\textbf{(E) }8}</math>
 
 
 
==See Also==
 
 
 
{{AMC10 box|year=2013|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2013|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 

Latest revision as of 22:18, 23 November 2020

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