2013 AMC 12A Problems/Problem 1: Difference between revisions
| 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
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
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 | |
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