Difference between revisions of "2015 AMC 8 Problems/Problem 19"
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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>. | 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>. | ||
− | + | ==Solution 2== | |
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> | 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> | ||
− | + | ==Solution 3== | |
By the shoelace theorem the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math> | By the shoelace theorem the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math> | ||
This means the fraction of the total area is <math>\boxed{\textbf{(A)}~\dfrac{1}{6}}</math> | This means the fraction of the total area is <math>\boxed{\textbf{(A)}~\dfrac{1}{6}}</math> |
Revision as of 20:48, 25 November 2015
A triangle with vertices as , , and is plotted on a grid. What fraction of the grid is covered by the triangle?
Contents
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
Solution 3
By the shoelace theorem the area of This means the fraction of the total area 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.