Difference between revisions of "2020 AMC 10A Problems/Problem 12"
(→Solution 5 (Medians)) |
(Solution where the figure is drawn out and area is approximated deleted due to inability to directly construct figure with given information) |
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<math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math> | <math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math> | ||
− | == Solution == | + | == Solution 1 == |
Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that <math>\triangle AUV</math> has <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus, | Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that <math>\triangle AUV</math> has <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus, | ||
<cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath> | <cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath> | ||
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<cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath> | <cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath> | ||
− | + | ==Solution 2 (Trapezoid)== | |
− | ==Solution 2 | ||
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<asy> | <asy> | ||
draw((-4,0)--(4,0)--(0,12)--cycle); | draw((-4,0)--(4,0)--(0,12)--cycle); | ||
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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, diagram by programjames1 | 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, diagram by programjames1 | ||
− | ==Solution | + | ==Solution 3 (Medians)== |
Draw median <math>\overline{AB}</math>. | Draw median <math>\overline{AB}</math>. | ||
<asy> | <asy> |
Revision as of 23:25, 31 January 2020
Contents
Problem
Triangle is isoceles with . Medians and are perpendicular to each other, and . What is the area of
Solution 1
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 (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, diagram by programjames1
Solution 3 (Medians)
Draw median .
Since we know that all medians of a triangle intersect at the incenter, we know that passes through point . We also know that medians of a triangle divide each other into segments of ratio . Knowing this, we can see that , and since the two segments sum to , and are and , respectively.
Finally knowing that the medians divide the triangle into sections of equal area, finding the area of is enough. .
The area of . Multiplying this by gives us ~quacker88
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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