Difference between revisions of "2013 AMC 12A Problems/Problem 1"
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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>. | ||
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<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 18:25, 22 February 2013
Problem
Square has side length . Point is on , and the area of is . What is ?
Solution
We are given that the area of is , and that .
The area of a triangle:
Using as the height of ,
and solving for b,
, which is
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 |