# Difference between revisions of "2018 AMC 10A Problems/Problem 9"

(→Solution 2) |
(→Solution 2) |
||

Line 25: | Line 25: | ||

=Solution 2= | =Solution 2= | ||

− | Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math> | + | Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}</math>. Then triangle <math>ADE</math> has an are of 16. So the area is <math>40 - 16 = 24</math>. |

## Revision as of 17:02, 9 February 2018

All of the triangles in the diagram below are similar to iscoceles triangle , in which . Each of the 7 smallest triangles has area 1, and has area 40. What is the area of trapezoid ?

## Solution 1

Let be the area of . Note that is comprised of the small isosceles triangles and a triangle similar to with side length ratio (so an area ratio of ). Thus, we have This gives , so the area of .

# Solution 2

Let the base length of the small triangle be . Then, there is a triangle encompassing the 7 small triangles and sharing the top angle with a base length of . Because the area is proportional to the square of the side, let the base be . Then triangle has an are of 16. So the area is .