Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See also)
m (→‎Solution)
Line 32: Line 32:
 
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 40: 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>E</math>
 
<math>b = 8</math>, which is <math>E</math>

Revision as of 16:24, 9 November 2022

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 is:

$A = \frac{bh}{2}$

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 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_1&oldid=180565"