Difference between revisions of "2009 AMC 10B Problems/Problem 18"
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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== | ||
+ | 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>. | ||
== See Also == | == See Also == |
Revision as of 22:05, 19 August 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 .
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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