Difference between revisions of "2009 AMC 10B Problems/Problem 18"
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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 AMC \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>: <math>(8 - x)^2 + 6^2 = x^2.</math> 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 <math>5^2 + y^2 = \left(\frac{25}{4} \right)^2.</math> 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 == | ||
{{AMC10 box|year=2009|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2009|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:43, 9 October 2021
Contents
Problem
Rectangle has and . Point is the midpoint of diagonal , and is on with . What is the area of ?
Solution 1 (Coordinate Geo)
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
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 $\frac{\frac{15}{4} \cdot 5}{2} = \frac{75}{8} \implies \boxed{D}}.$ (Error compiling LaTeX. ! Extra }, or forgotten $.)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.