Difference between revisions of "2013 AMC 12A Problems/Problem 1"
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− | ==Problem== | + | == 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>? | 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>? | ||
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<asy> | <asy> | ||
Line 12: | Line 9: | ||
D=(50,0); | D=(50,0); | ||
E = (40,50); | E = (40,50); | ||
− | + | draw(A--B); | |
− | + | draw(B--E); | |
− | + | draw(E--C); | |
draw(C--D); | draw(C--D); | ||
draw(D--A); | draw(D--A); | ||
Line 28: | Line 25: | ||
label("D",D,SE); | label("D",D,SE); | ||
label("E",E,N); | label("E",E,N); | ||
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</asy> | </asy> | ||
− | + | <math>\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad </math> | |
+ | == 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 47: | Line 44: | ||
<math>b = 8</math>, which is <math>E</math> | <math>b = 8</math>, which is <math>E</math> | ||
− | ==Video Solution== | + | == Video Solution == |
− | https://www.youtube.com/watch?v=2vf843cvVzo?t= | + | https://www.youtube.com/watch?v=2vf843cvVzo?t=0 |
~sugar_rush | ~sugar_rush | ||
Latest revision as of 13:14, 19 January 2021
Contents
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
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=0 ~sugar_rush
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.