Difference between revisions of "2013 AMC 12A Problems/Problem 1"
(→See also) |
(→Video Solution) |
||
(2 intermediate revisions by 2 users not shown) | |||
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>\boxed{\textbf{(E)}}</math> | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/wW_GidbHnnM | ||
+ | |||
+ | ~Education, the Study of Everything | ||
− | |||
== Video Solution == | == Video Solution == |
Latest revision as of 12:10, 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.