Difference between revisions of "2013 AMC 12A Problems/Problem 1"
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| − | ==Problem== | + | == 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>? | 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> | <asy> | ||
| Line 11: | Line 8: | ||
C=(50,50); | C=(50,50); | ||
D=(50,0); | D=(50,0); | ||
| − | E = ( | + | 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); | ||
| − | |||
</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>. | ||
| − | The area of a triangle: | + | The area of a triangle is: |
<math>A = \frac{bh}{2}</math> | <math>A = \frac{bh}{2}</math> | ||
| Line 43: | 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 == | == 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}} | {{AMC12 box|year=2013|ab=A|before=First Question|num-a=2}} | ||
[[Category:Introductory Geometry Problems]] | [[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 
?
![[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]](http://latex.artofproblemsolving.com/5/d/e/5deb6040701b50303c4593a48dd3b09550139cd9.png)
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. 
