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2013 AMC 12A Problems/Problem 1: Difference between revisions

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==Problem==
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> \bigtriangleup ABE </math> is <math> 40 </math>. What is <math> BE </math>?
<math>\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad </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>
==Solution==
We are given that the area of <math>\triangle ABE</math> is <math>40</math>, and that <math>AB = 10</math>.
We are given that the area of <math>\triangle ABE</math> is <math>40</math>, and that <math>AB = 10</math>.


Line 12: Line 46:


<math>b = 8</math>, which is <math>E</math>
<math>b = 8</math>, which is <math>E</math>
== See also ==
{{AMC12 box|year=2013|ab=A|before=First Question|num-a=2}}

Revision as of 17:25, 22 February 2013

Problem

Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\bigtriangleup ABE$ is $40$. What is $BE$?

$\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad$

[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]

Solution

We are given that the area of $\triangle ABE$ is $40$, and that $AB = 10$.

The area of a triangle:

$A = \frac{bh}{2}$

Using $AB$ as the height of $\triangle ABE$,

$40 = \frac{10b}{2}$

and solving for b,

$b = 8$, which is $E$

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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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