Difference between revisions of "2005 AMC 10B Problems/Problem 14"
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− | <math> \ | + | <math> \textbf{(A) }\ \frac {\sqrt {2}}{2}\qquad \textbf{(B) }\ \frac {3}{4}\qquad \textbf{(C) }\ \frac {\sqrt {3}}{2}\qquad \textbf{(D) }\ 1\qquad \textbf{(E) }\ \sqrt {2}</math> |
== Solutions == | == Solutions == | ||
− | ===Solution 1 ( | + | ===Solution 1 (trig) === |
The area of a triangle can be given by <math>\frac12 ab \sin C</math>. <math>MC=1</math> because it is the midpoint of a side, and <math>CD=2</math> because it is the same length as <math>BC</math>. Each angle of an equilateral triangle is <math>60^\circ</math> so <math>\angle MCD = 120^\circ</math>. The area is <math>\frac12 (1)(2) \sin | The area of a triangle can be given by <math>\frac12 ab \sin C</math>. <math>MC=1</math> because it is the midpoint of a side, and <math>CD=2</math> because it is the same length as <math>BC</math>. Each angle of an equilateral triangle is <math>60^\circ</math> so <math>\angle MCD = 120^\circ</math>. The area is <math>\frac12 (1)(2) \sin | ||
120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}</math>. | 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}</math>. | ||
Note: Even if you don't know the value of <math>\sin 120^\circ</math>, you can use the fact that <math>\sin x = \sin (180^\circ - x)</math>, so <math>\sin 120^\circ = \sin 60^\circ</math>. | Note: Even if you don't know the value of <math>\sin 120^\circ</math>, you can use the fact that <math>\sin x = \sin (180^\circ - x)</math>, so <math>\sin 120^\circ = \sin 60^\circ</math>. | ||
You can easily calculate <math>\sin 60^\circ</math> to be <math>\frac{\sqrt3}{2}</math> using equilateral triangles. | You can easily calculate <math>\sin 60^\circ</math> to be <math>\frac{\sqrt3}{2}</math> using equilateral triangles. | ||
+ | |||
+ | ~Minor Edits by doulai1 | ||
===Solution 2=== | ===Solution 2=== | ||
− | In order to calculate the area of <math>\triangle CDM</math>, we can use the formula <math>A=\dfrac{1}{2}bh</math>, where <math>\overline{CD}</math> is the base. We already know that <math>\overline{CD}=2</math>, so the formula | + | In order to calculate the area of <math>\triangle CDM</math>, we can use the formula <math>A=\dfrac{1}{2}bh</math>, where <math>\overline{CD}</math> is the base. We already know that <math>\overline{CD}=2</math>, so the formula becomes <math>A=h</math>. We can drop verticals down from <math>A</math> and <math>M</math> to points <math>E</math> and <math>F</math>, respectively. We can see that <math>\triangle AEC \sim \triangle MFC</math>. Now, we establish the relationship that <math>\dfrac{AE}{MF}=\dfrac{AC}{MC}</math>. We are given that <math>\overline{AC}=2</math>, and <math>M</math> is the midpoint of <math>\overline{AC}</math>, so <math>\overline{MC}=1</math>. Because <math>\triangle AEB</math> is a <math>30-60-90</math> triangle and the ratio of the sides opposite the angles are <math>1-\sqrt{3}-2</math> <math>\overline{AE}</math> is <math>\sqrt{3}</math>. Plugging those numbers in, we have <math>\dfrac{\sqrt{3}}{MF}=\dfrac{2}{1}</math>. Cross-multiplying, we see that <math>2\times\overline{MF}=\sqrt{3}\times1\implies \overline{MF}=\dfrac{\sqrt{3}}{2}</math> Since <math>\overline{MF}</math> is the height <math>\triangle CDM</math>, the area is <math>\boxed{\textbf{(C) }\frac{\sqrt{3}}{2}}</math>. |
===Solution 3=== | ===Solution 3=== | ||
− | Draw a line from <math>M</math> to the midpoint of <math>\overline{BC}</math>. Call the midpoint of <math>\overline{BC}</math> <math>P</math>. This is an equilateral triangle, since the two segments <math>\overline{PC}</math> and <math>\overline{MC}</math> are identical, and <math>\angle C</math> is | + | Draw a line from <math>M</math> to the midpoint of <math>\overline{BC}</math>. Call the midpoint of <math>\overline{BC}</math> <math>P</math>. This is an equilateral triangle, since the two segments <math>\overline{PC}</math> and <math>\overline{MC}</math> are identical, and <math>\angle C</math> is <math>60^{\circ}</math>. Using the [[Pythagorean Theorem]], point <math>M</math> to <math>\overline{BC}</math> is <math>\dfrac{\sqrt{3}}{2}</math>. Also, the length of <math>\overline{CD}</math> is 2, since <math>C</math> is the midpoint of <math>\overline{BD}</math>. So, our final equation is <math>\frac{\sqrt{3}}{2}\times2\over2</math>, which just leaves us with <math>\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}</math>. |
===Solution 4 === | ===Solution 4 === | ||
− | Drop a vertical down from <math>M</math> to <math>BC</math>. Let us call the point of intersection <math>X</math> and the midpoint of <math>BC</math>, <math>Y</math>. We can observe that <math>\triangle AYC</math> and <math>\ | + | Drop a vertical down from <math>M</math> to <math>BC</math>. Let us call the point of intersection <math>X</math> and the midpoint of <math>BC</math>, <math>Y</math>. We can observe that <math>\triangle AYC</math> and <math>\bigtriangleup MXC</math> are similar. By the [[Pythagorean theorem]], <math>AY</math> is <math>\sqrt3</math>. |
+ | |||
+ | Since <math>AC:MC=2:1,</math> we find <math>MX=\frac{\sqrt3}{2}.</math> Because <math>C</math> is the midpoint of <math>BD,</math> and <math>BC=2,</math> <math>CD=2.</math> Using the area formula, <math>\frac{CD*MX}{2}=\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}.</math> | ||
~ sdk652 | ~ sdk652 | ||
+ | |||
+ | ===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 <math>2</math> (for a side of <math>\triangle ABC</math>) 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 <math>\overline{BD}</math> 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</math>, and in turn means that the height of <math>\triangle MCD</math> is half of that of <math>\triangle 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>\triangle MCD</math> being half of that of <math>\triangle ABC</math>. Now all that's left is to find the area of <math>\triangle ABC</math>, and for that, we plug in <math>2</math> 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}}.</math> | ||
+ | |||
+ | ~<math>\LaTeX</math> help by vadava_lx | ||
== See Also == | == See Also == |
Latest revision as of 15:15, 23 June 2024
Contents
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
(for a side of
) for. Looking at the information, we see that
is the midpoint of
, and this means that it bisects
which results in
. Now for the height, we can see that
is the midpoint of
which means that
, and in turn means that the height of
is half of that of
, 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
being half of that of
. Now all that's left is to find the area of
, and for that, we plug in
which leads us to the answer of
, but since we need to divide by two, our final answer is
~ help by vadava_lx
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.