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>? | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <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>. | ||
− | The area of a triangle: | + | The area of a triangle is: |
<math>A = \frac{bh}{2}</math> | <math>A = \frac{bh}{2}</math> | ||
Line 9: | Line 40: | ||
<math>40 = \frac{10b}{2}</math> | <math>40 = \frac{10b}{2}</math> | ||
− | and solving for b, | + | and solving for <math>b</math>, |
+ | |||
+ | <math>b = 8</math>, which is <math>\boxed{\textbf{(E)}}</math> | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/wW_GidbHnnM | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=2vf843cvVzo?t=0 | ||
+ | ~sugar_rush | ||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2013|ab=A|num-b=2|num-a=4}} | ||
+ | {{AMC12 box|year=2013|ab=A|before=First Question|num-a=2}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | [[Category:Area Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:10, 1 July 2023
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 is:
Using as the height of
,
and solving for ,
, which is
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=0 ~sugar_rush
See also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 10 Problems and Solutions |
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.