2014 AMC 8 Problems/Problem 14
Problem
Rectangle and right triangle
have the same area. They are joined to form a trapezoid, as shown. What is
?
Solution
The area of is
. The area of
is
, which also must be equal to the area of
, which, since
, must in turn equal
. Through transitivity, then,
, and
. Then, using the Pythagorean Theorem, you should be able to figure out that
is a
triangle, so
, or
.
Solution 2
The area of the rectangle is Since the parallel line pairs are identical,
. Let
be
.
is the area of the right triangle. Solving for
, we get
According to the Pythagorean Theorem, we have a
triangle. So, the hypotenuse
has to be
.
Solution 3
This problem can be solved with the Pythagorean Theorem (a^2 + b^2 = c^2. We know AB = DC, so DC = 5. CE is twice the length of AD, so CE = 12. 5^2 + 12^2 = c^2. 5^2 = 25. 12^2 = 144. 25 + 144 = 169. 169 has a square root of 13, so the hypotenuse or DE is 13. The answer is .
——MiracleMaths
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 AJHSME/AMC 8 Problems and Solutions |
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