Difference between revisions of "2015 AMC 8 Problems/Problem 19"
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+ | == Problem == | ||
A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle? | A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle? | ||
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<asy> | <asy> | ||
− | |||
draw((1,0)--(1,5),linewidth(.5)); | draw((1,0)--(1,5),linewidth(.5)); | ||
draw((2,0)--(2,5),linewidth(.5)); | draw((2,0)--(2,5),linewidth(.5)); | ||
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</asy> | </asy> | ||
− | ==Solution 1== | + | ==Solutions== |
+ | |||
+ | ===Solution 1=== | ||
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 <math>2</math> 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 <math>2</math> 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== | + | ===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== | + | ===Solution 3=== |
− | By the Shoelace | + | 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>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math> | This means the fraction of the total area is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math> | ||
− | ==Solution 4== | + | ===Solution 4=== |
The smallest rectangle that follows the grid lines and completely encloses <math>\triangle ABC</math> has an area of <math>12</math>, where <math>\triangle ABC</math> splits the rectangle into four triangles. The area of <math>\triangle ABC</math> is therefore <math>12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5</math>. That means that <math>\triangle ABC</math> takes up <math>\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid. | The smallest rectangle that follows the grid lines and completely encloses <math>\triangle ABC</math> has an area of <math>12</math>, where <math>\triangle ABC</math> splits the rectangle into four triangles. The area of <math>\triangle ABC</math> is therefore <math>12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5</math>. That means that <math>\triangle ABC</math> takes up <math>\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid. | ||
− | ==Solution 5== | + | ===Solution 5=== |
− | Using Pick's Theorem, the area of the triangle is <math>4 + \dfrac{4}{2} - 1=5</math>. Therefore, the triangle takes up <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid. | + | Using [[Pick's Theorem]], the area of the triangle is <math>4 + \dfrac{4}{2} - 1=5</math>. Therefore, the triangle takes up <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid. |
− | ==Solution 6 (Heron's Formula, Not Recommended)== | + | ===Solution 6 (Heron's Formula, Not Recommended)=== |
− | We can find the lengths of the sides by using the Pythagorean | + | We can find the lengths of the sides by using the [[Pythagorean Theorem]]. Then, we apply [[Heron's Formula]] to find the area. |
<cmath> \sqrt{\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-2\sqrt{5}\right)}. </cmath> | <cmath> \sqrt{\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-2\sqrt{5}\right)}. </cmath> | ||
This simplifies to | This simplifies to | ||
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<cmath> \sqrt{5\cdot 5}=5. </cmath> | <cmath> \sqrt{5\cdot 5}=5. </cmath> | ||
The area of the grid is <math> 6\cdot 5=30. </math> Thus, the answer is <math> \frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}} </math>. -BorealBear | The area of the grid is <math> 6\cdot 5=30. </math> Thus, the answer is <math> \frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}} </math>. -BorealBear | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/EyDGtLc6xGE | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 13:05, 14 April 2022
Contents
Problem
A triangle with vertices as , , and is plotted on a grid. What fraction of the grid is covered by the triangle?
Solutions
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 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
Solution 4
The smallest rectangle that follows the grid lines and completely encloses has an area of , where splits the rectangle into four triangles. The area of is therefore . That means that takes up of the grid.
Solution 5
Using Pick's Theorem, the area of the triangle is . Therefore, the triangle takes up of the grid.
Solution 6 (Heron's Formula, Not Recommended)
We can find the lengths of the sides by using the Pythagorean Theorem. Then, we apply Heron's Formula to find the area. This simplifies to Again, we simplify to get The middle two terms inside the square root multiply to , and the first and last terms inside the square root multiply to This means that the area of the triangle is The area of the grid is Thus, the answer is . -BorealBear
Video Solution
~savannahsolver
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.