Difference between revisions of "2020 AMC 10A Problems/Problem 12"
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==Solution 3 (Trapezoid)== | ==Solution 3 (Trapezoid)== | ||
+ | <asy> | ||
+ | draw((-4,0)--(4,0)--(0,12)--cycle); | ||
+ | draw((-2,6)--(4,0)); | ||
+ | draw((2,6)--(-4,0)); | ||
+ | draw((-2,6)--(2,6)); | ||
+ | label("M", (-4,0), W); | ||
+ | label("C", (4,0), E); | ||
+ | label("A", (0, 12), N); | ||
+ | label("V", (2, 6), NE); | ||
+ | label("U", (-2, 6), NW); | ||
+ | label("P", (0, 5.65685425), NW); | ||
+ | </asy> | ||
+ | |||
We know that <math>\triangle AUV \sim \triangle AMC</math>, and since the ratios of its sides are <math>\frac{1}{2}</math>, the ratio of of their areas is <math>(\frac{1}{2})^2=\frac{1}{4}</math>. | We know that <math>\triangle AUV \sim \triangle AMC</math>, and since the ratios of its sides are <math>\frac{1}{2}</math>, the ratio of of their areas is <math>(\frac{1}{2})^2=\frac{1}{4}</math>. | ||
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This is <math>\frac{3}{4}</math> of the triangle, so the area of the triangle is <math>\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}</math> ~quacker88 | This is <math>\frac{3}{4}</math> of the triangle, so the area of the triangle is <math>\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}</math> ~quacker88 | ||
+ | |||
+ | ==Solution 4== | ||
+ | Again, call the intersection of the medians <math>P</math>. There are many ways to find the lengths of | ||
+ | |||
+ | <math>\triangle UPV \sim \triangle CPM</math>, and since <math>\frac{\overline{UV}}{\overline{MC}}=\frac{1}{2}</math>, <math>\frac{\overline{UP}}{\overline{PC}}=\frac{1}{2}</math>, and since <math>\overline{UP} + \overline{PC} = 12</math>, <math>\overline{UP}</math> and <math>\overline{PC}</math> are <math>4</math> and <math>8</math>, respectively. Proceed with the Pythagorean Theorem like the other solutions. | ||
==Video Solution== | ==Video Solution== |
Revision as of 23:27, 31 January 2020
Contents
Problem
Triangle is isoceles with . Medians and are perpendicular to each other, and . What is the area of
Solution
Since quadrilateral has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that has the area of triangle by similarity, so Thus,
Solution 2 (CHEATING)
Draw a to-scale diagram with your graph paper and straightedge. Measure the height and approximate the area.
Solution 3 (Trapezoid)
We know that , and since the ratios of its sides are , the ratio of of their areas is .
If is the area of , then trapezoid is the area of .
Let's call the intersection of and . Let . Then . Since , and are heights of triangles and , respectively. Both of these triangles have base .
Area of
Area of
Adding these two gives us the area of trapezoid , which is .
This is of the triangle, so the area of the triangle is ~quacker88
Solution 4
Again, call the intersection of the medians . There are many ways to find the lengths of
, and since , , and since , and are and , respectively. Proceed with the Pythagorean Theorem like the other solutions.
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 AMC 10 Problems and Solutions |
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