Difference between revisions of "2009 AMC 10B Problems/Problem 18"
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</math> | </math> | ||
− | == Solution 1 (Coordinate | + | == Solution 1 (Coordinate Geometry)== |
− | Set <math>A</math> to <math>(0,0)</math>. Since <math>M</math> is the midpoint of the diagonal, it would be <math>(4, | + | Set <math>A</math> to <math>(0,0)</math>. Since <math>M</math> is the midpoint of the diagonal, it would be <math>(4,3)</math>. The diagonal <math>AC</math> would be the line <math>y = \frac{3x}{4}</math>. Since <math>ME</math> is perpendicular to <math>AC</math>, its line would be in the form <math>y = -\frac{4x}{3} + b</math>. Plugging in <math>4</math> and <math>3</math> for <math>x</math> and <math>y</math> would give <math>b = \frac{25}{3}</math>. To find the x-intercept of <math>y = -\frac{4x}{3} + \frac{25}{3}</math> we plug in <math>0</math> for <math>y</math> and get <math>x = \frac{25}{4}</math>. Then, using the Shoelace Formula for <math>(0,0)</math> , <math>(4,3)</math>, and <math>(\frac{25}{4}, 0)</math>, we find the area is <math>\frac{75}{8}</math>. |
== Solution 2 == | == Solution 2 == | ||
Line 70: | Line 70: | ||
Draw <math>EC</math> as shown from the diagram. Since <math>AC</math> is of length <math>10</math>, we have that <math>AM</math> is of length <math>5</math>, because of the midpoint <math>M</math>. Through the Pythagorean theorem, we know that <math>AE^2 = AM^2 + ME^2 \implies 25 + ME^2</math>, which means <math>AE = \sqrt{25 + ME^2}</math>. Define <math>ME</math> to be <math>x</math> for the sake of clarity. We know that <math>EB = 8 - \sqrt{25 + x^2}</math>. From here, we know that <math>CE^2 = CB^2 + BE^2 = ME^2 + MC^2</math>. From here, we can write the expression <math>6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}</math>. Now, remember <math>CE \neq \frac{15}{4}</math>. <math>x = \frac{15}{4} = ME</math>, since we set <math>x = ME</math> in the start of the solution. Now to find the area <math>\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE</math> | Draw <math>EC</math> as shown from the diagram. Since <math>AC</math> is of length <math>10</math>, we have that <math>AM</math> is of length <math>5</math>, because of the midpoint <math>M</math>. Through the Pythagorean theorem, we know that <math>AE^2 = AM^2 + ME^2 \implies 25 + ME^2</math>, which means <math>AE = \sqrt{25 + ME^2}</math>. Define <math>ME</math> to be <math>x</math> for the sake of clarity. We know that <math>EB = 8 - \sqrt{25 + x^2}</math>. From here, we know that <math>CE^2 = CB^2 + BE^2 = ME^2 + MC^2</math>. From here, we can write the expression <math>6^2 + (8 - \sqrt{25 + x^2})^2 = 5^2 + x^2 \implies 36 + (64 - 16\sqrt{25 + x^2} + 25 + x^2) = 25 + x^2 \implies 100 - 16\sqrt{25 + x^2} \implies \sqrt{25 + x^2} = \frac{25}{4} \implies 25 + x^2 = \frac{625}{16} \implies 400 + 16x^2 = 625 \implies 16x^2 = 225 \implies x = \frac{15}{4}</math>. Now, remember <math>CE \neq \frac{15}{4}</math>. <math>x = \frac{15}{4} = ME</math>, since we set <math>x = ME</math> in the start of the solution. Now to find the area <math>\frac{15}{4} \cdot 5 \cdot \frac{1}{2} = \frac{75}{8} = CE</math> | ||
− | ==Solution 4== | + | ==Solution 4 (Similarity) == |
We know <math>AM = \frac{10}{2} = 5</math> by the Pythagorean theorem, and furthermore, <math>\triangle AME</math> is similar to <math>\triangle ABC</math>. Therefore, <math>ME = 5 \cdot \frac{6}{8} = \frac{15}{4}</math>, and the area of the triangle is <math>5 \cdot \frac{15}{4} \cdot \frac{1}{2} = \boxed{\frac{75}{8}}</math>. | We know <math>AM = \frac{10}{2} = 5</math> by the Pythagorean theorem, and furthermore, <math>\triangle AME</math> is similar to <math>\triangle ABC</math>. Therefore, <math>ME = 5 \cdot \frac{6}{8} = \frac{15}{4}</math>, and the area of the triangle is <math>5 \cdot \frac{15}{4} \cdot \frac{1}{2} = \boxed{\frac{75}{8}}</math>. | ||
+ | |||
+ | ==Solution 5 (Pythagorean Theorem)== | ||
+ | <asy> | ||
+ | unitsize(0.75cm); | ||
+ | defaultpen(0.8); | ||
+ | pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; | ||
+ | path ortho = shift(M)*rotate(-90)*(A--C); | ||
+ | pair Ep = intersectionpoint(ortho, A--B); | ||
+ | draw( A--B--C--D--cycle ); | ||
+ | draw( A--C ); | ||
+ | draw( M--Ep ); | ||
+ | filldraw( A--M--Ep--cycle, lightgray, black ); | ||
+ | draw( rightanglemark(A,M,Ep) ); | ||
+ | draw( C--Ep ); | ||
+ | label("$A$",A,SW); | ||
+ | label("$B$",B,SE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,NW); | ||
+ | label("$E$",Ep,S); | ||
+ | label("$M$",M,NW); | ||
+ | </asy> | ||
+ | |||
+ | By the [[Pythagorean Theorem]], we claim that <math>AC = 10</math>. It then follows that <math>AM \cong MC = 5.</math> | ||
+ | |||
+ | Because we have <math>AM \cong MC, \angle AME \cong \angle CME,</math> and reflexive side <math>EM</math>, it follows that <math>\triangle AME \cong \triangle CME.</math> By CPCTC, we have <math>AE \cong EC.</math> For the sake of simplicity, we'll call those side lengths <math>x</math>. Also, since <math>AE = x,</math> we get <math>BE = 8 - x.</math> We can now set up the Pythagorean theorem on <math>\triangle EBC</math>: <cmath>(8 - x)^2 + 6^2 = x^2.</cmath> Combining like terms and simplifying gives <math>-16x + 100 = 0</math> so <math>x = \frac{25}{4}.</math> | ||
+ | |||
+ | It helps to think that in order to find <math>[AME],</math> we must have <math>\overline{MC}</math> and <math>\overline{EM}.</math> Let <math>EM = y.</math> Applying the Pythagorean Theorem to <math>\triangle CME</math> gives <cmath>5^2 + y^2 = \left(\frac{25}{4} \right)^2.</cmath> Solving for <math>y</math> (this is not that difficult) gives <math>y = \frac{15}{4}.</math> So, the area of <math>\triangle AME</math> is <math>\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}.</math> | ||
== See Also == | == See Also == |
Latest revision as of 22:14, 18 January 2024
Contents
Problem
Rectangle has
and
. Point
is the midpoint of diagonal
, and
is on
with
. What is the area of
?
Solution 1 (Coordinate Geometry)
Set to
. Since
is the midpoint of the diagonal, it would be
. The diagonal
would be the line
. Since
is perpendicular to
, its line would be in the form
. Plugging in
and
for
and
would give
. To find the x-intercept of
we plug in
for
and get
. Then, using the Shoelace Formula for
,
, and
, we find the area is
.
Solution 2
By the Pythagorean theorem we have , hence
.
The triangles and
have the same angle at
and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is , hence the ratio of their areas is
.
And as the area of triangle is
, the area of triangle
is
.
Solution 3 (Only Pythagorean Theorem)
Draw as shown from the diagram. Since
is of length
, we have that
is of length
, because of the midpoint
. Through the Pythagorean theorem, we know that
, which means
. Define
to be
for the sake of clarity. We know that
. From here, we know that
. From here, we can write the expression
. Now, remember
.
, since we set
in the start of the solution. Now to find the area
Solution 4 (Similarity)
We know by the Pythagorean theorem, and furthermore,
is similar to
. Therefore,
, and the area of the triangle is
.
Solution 5 (Pythagorean Theorem)
By the Pythagorean Theorem, we claim that . It then follows that
Because we have and reflexive side
, it follows that
By CPCTC, we have
For the sake of simplicity, we'll call those side lengths
. Also, since
we get
We can now set up the Pythagorean theorem on
:
Combining like terms and simplifying gives
so
It helps to think that in order to find we must have
and
Let
Applying the Pythagorean Theorem to
gives
Solving for
(this is not that difficult) gives
So, the area of
is
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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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