Difference between revisions of "2015 AMC 8 Problems/Problem 19"
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===Solution 2=== | ===Solution 2=== | ||
− | Note angle <math>\angle ACB</math> is right | + | Note angle <math>\angle ACB</math> is right, thus the area is <math>\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=(10)\times \dfrac{1}{2}=5</math> thus the fraction of the total is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math> |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=18|num-a=20}} | {{AMC8 box|year=2015|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:26, 25 November 2015
A triangle with vertices as , , and is plotted on a grid. What fraction of the grid is covered by the triangle?
Solution 1
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 .
Solution 2
Note angle is right, thus the area is thus the fraction of the total is
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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