Difference between revisions of "2015 AMC 8 Problems/Problem 6"
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In <math>\bigtriangleup ABC</math>, <math>AB=BC=29</math>, and <math>AC=42</math>. What is the area of <math>\bigtriangleup ABC</math>? | In <math>\bigtriangleup ABC</math>, <math>AB=BC=29</math>, and <math>AC=42</math>. What is the area of <math>\bigtriangleup ABC</math>? | ||
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+ | <math>\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701</math> | ||
==Solutions== | ==Solutions== |
Latest revision as of 14:55, 21 January 2024
Contents
[hide]Problem
In , , and . What is the area of ?
Solutions
Solution 1
We know the semi-perimeter of is . Next, we use Heron's Formula to find that the area of the triangle is just .
Solution 2 (easier)
Splitting the isosceles triangle in half, we get a right triangle with hypotenuse and leg . Using the Pythagorean Theorem , we know the height is . Now that we know the height, the area is .
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution 1
https://www.youtube.com/watch?v=Bl3_W2i5zwc ~David
Video Solution 2
~savannahsolver
Note
20-21-29 is a Pythagorean Triple (only for right triangles!)
~SaxStreak
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.