Difference between revisions of "2013 AMC 12A Problems/Problem 1"

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We are given that the area of ∆ABE is 40, and that side AB is of length 10.
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== Problem ==
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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>?
  
Side AB can be used as the height of ∆ABE.
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<asy>
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pair A,B,C,D,E;
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A=(0,0);
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B=(0,50);
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C=(50,50);
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D=(50,0);
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E = (40,50);
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draw(A--B);
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draw(B--E);
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draw(E--C);
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draw(C--D);
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draw(D--A);
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draw(A--E);
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dot(A);
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dot(B);
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dot(C);
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dot(D);
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dot(E);
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label("A",A,SW);
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label("B",B,NW);
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label("C",C,NE);
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label("D",D,SE);
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label("E",E,N);
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</asy>
  
Since A= (bh)/2,
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<math>\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad </math>
  
40=(10h)/2
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== Solution ==
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We are given that the area of <math>\triangle ABE</math> is <math>40</math>, and that <math>AB = 10</math>.
  
and solving for h will give us 8, or E.
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The area of a triangle:
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<math>A = \frac{bh}{2}</math>
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Using <math>AB</math> as the height of <math>\triangle ABE</math>,
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<math>40 = \frac{10b}{2}</math>
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and solving for b,
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<math>b = 8</math>, which is <math>E</math>
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== Video Solution ==
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https://www.youtube.com/watch?v=2vf843cvVzo?t=0
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~sugar_rush
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== See also ==
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{{AMC12 box|year=2013|ab=A|before=First Question|num-a=2}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Area Problems]]
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{{MAA Notice}}

Latest revision as of 14:14, 19 January 2021

Problem

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

[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (40,50); 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]

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

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 $E$

Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=0 ~sugar_rush

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
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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