Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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+ | ===Solution 5 === | ||
+ | Think of <math>\triangle ABC</math> and <math>\triangle MCD</math> being independent. Now to find area's we just solve for ratios between the triangles that we can plug in the value of 2 (for a side of ABC) for. Looking at the information, we see that <math>C</math> is the midpoint of <math>\overline{BD}</math>, and this means that it bisects BD which results in <math>\BC=CD</math>. Now for the height, we can see that <math>M</math> is the midpoint of <math>\overline{AC}</math> which means that <math>\AM=CM, and in turn means that the height of </math>\MCD<math> is half of that of </math>\ABC<math>, and now plugging the ratios of the bases being the same while the height is half of the other triangle, we end up with the area of </math>\MCD<math> being half of that of </math>\ABC<math>. Now all that's left is to find the area of </math>\ABC<math>, and for that, we plug in 2 which leads us to the answer of </math>\3<math>, but since we need to divide by two, our final answer is </math>\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}.$ | ||
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== See Also == | == See Also == |
Revision as of 21:10, 30 April 2024
Contents
[hide]Problem
Equilateral has side length , is the midpoint of , and is the midpoint of . What is the area of ?
Solutions
Solution 1 (trig)
The area of a triangle can be given by . because it is the midpoint of a side, and because it is the same length as . Each angle of an equilateral triangle is so . The area is . Note: Even if you don't know the value of , you can use the fact that , so . You can easily calculate to be using equilateral triangles.
~Minor Edits by doulai1
Solution 2
In order to calculate the area of , we can use the formula , where is the base. We already know that , so the formula becomes . We can drop verticals down from and to points and , respectively. We can see that . Now, we establish the relationship that . We are given that , and is the midpoint of , so . Because is a triangle and the ratio of the sides opposite the angles are is . Plugging those numbers in, we have . Cross-multiplying, we see that Since is the height , the area is .
Solution 3
Draw a line from to the midpoint of . Call the midpoint of . This is an equilateral triangle, since the two segments and are identical, and is . Using the Pythagorean Theorem, point to is . Also, the length of is 2, since is the midpoint of . So, our final equation is , which just leaves us with .
Solution 4
Drop a vertical down from to . Let us call the point of intersection and the midpoint of , . We can observe that and are similar. By the Pythagorean theorem, is .
Since we find Because is the midpoint of and Using the area formula,
~ sdk652
Solution 5
Think of and being independent. Now to find area's we just solve for ratios between the triangles that we can plug in the value of 2 (for a side of ABC) for. Looking at the information, we see that is the midpoint of , and this means that it bisects BD which results in $\BC=CD$ (Error compiling LaTeX. Unknown error_msg). Now for the height, we can see that is the midpoint of which means that $\AM=CM, and in turn means that the height of$ (Error compiling LaTeX. Unknown error_msg)\MCD\ABC\MCD\ABC\ABC\3\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}.$
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.