Difference between revisions of "2015 AMC 8 Problems/Problem 6"
(Created page with "In <math>\bigtriangleup ABC</math>, <math>AB=BC=29</math>, and <math>AC=42</math>. What is the area of <math>\bigtriangleup ABC</math>? <math>\textbf{(A) }100\qquad\textbf{(B...") |
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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> | <math>\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701</math> | ||
| + | The area of <math>\triangle ABC</math> is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>. We multiply these and divide by 2 to find the of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>. Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>. | ||
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| + | ==See Also== | ||
| + | |||
| + | {{AMC8 box|year=2015|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:24, 25 November 2015
In 
, 
, and 
. What is the area of ?
The area of 
is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is 
, and its base is 
. We multiply these and divide by 2 to find the of the triangle is 
. Since the grid has an area of 
, the fraction of the grid covered by the triangle is 
.
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. 
