Difference between revisions of "2017 AMC 10B Problems/Problem 15"
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}</math> | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
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Let <math>F</math> be the foot of the perpendicular from <math>E</math> to <math>AB</math>. Since <math>EF</math> and <math>BC</math> are parallel, <math>\Delta AFE</math> is similar to <math>\Delta ABC</math>. Therefore, we have <math>\frac{AF}{AB}=\frac{AE}{AC}=\frac{9}{25}</math>. Since <math>AB=3</math>, <math>AF=\frac{27}{25}</math>. Note that <math>AF</math> is an altitude of <math>\Delta AED</math> from <math>AD</math>, which has length <math>4</math>. Therefore, the area of <math>\Delta AED</math> is <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | Let <math>F</math> be the foot of the perpendicular from <math>E</math> to <math>AB</math>. Since <math>EF</math> and <math>BC</math> are parallel, <math>\Delta AFE</math> is similar to <math>\Delta ABC</math>. Therefore, we have <math>\frac{AF}{AB}=\frac{AE}{AC}=\frac{9}{25}</math>. Since <math>AB=3</math>, <math>AF=\frac{27}{25}</math>. Note that <math>AF</math> is an altitude of <math>\Delta AED</math> from <math>AD</math>, which has length <math>4</math>. Therefore, the area of <math>\Delta AED</math> is <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | ||
− | ===Solution 2=== | + | ==Solution 2== |
+ | From similar triangles, we know that <math>AE=\frac{9}{5}</math> (see Solution 1). Furthermore, we also know that <math>AD=4</math> from the rectangle. Using the sine formula for area, we have | ||
+ | <cmath>\frac{1}{2}(AE)(AD)(\sin(m\angle EAD)).</cmath> | ||
+ | But, note that <math>\sin(m\angle EAD)=\sin(m\angle CAD)=\frac{O}{H}=\frac{3}{5}</math>. Thus, we see that | ||
+ | <cmath>[AED]=\frac{1}{2}\cdot \frac{9}{5}\cdot 4\cdot\frac{3}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.</cmath> | ||
+ | ~coolwiz | ||
+ | |||
+ | ==Solution 3== | ||
Alternatively, we can use coordinates. Denote <math>D</math> as the origin. We find the equation for <math>AC</math> as <math>y=-\frac{4}{3}x+4</math>, and <math>BE</math> as <math>y=\frac{3}{4}x+\frac{7}{4}</math>. Solving for <math>x</math> yields <math>\frac{27}{25}</math>. Our final answer then becomes <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | Alternatively, we can use coordinates. Denote <math>D</math> as the origin. We find the equation for <math>AC</math> as <math>y=-\frac{4}{3}x+4</math>, and <math>BE</math> as <math>y=\frac{3}{4}x+\frac{7}{4}</math>. Solving for <math>x</math> yields <math>\frac{27}{25}</math>. Our final answer then becomes <math>\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | ||
− | + | ==Solution 4== | |
− | We note that the area of <math>ABE</math> must equal area of <math>AED</math> because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of <math>ABE</math> to be <math>\frac{1}{2}*\frac{9}{5}*\frac{12}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.</math> | + | We note that the area of <math>ABE</math> must equal the area of <math>AED</math> because they share the base <math>AE</math> and the height of both is the altitude of congruent triangles. Therefore, we find the area of <math>ABE</math> to be <math>\frac{1}{2}*\frac{9}{5}*\frac{12}{5}=\boxed{\textbf{(E)}\frac{54}{25}}.</math> |
− | + | ||
− | We know all right triangles are 5-4-3, so the areas are proportional to the square | + | ==Solution 5== |
+ | We know all right triangles are 5-4-3, so the areas are proportional to the square of corresponding sides. Area of <math>ABE</math> is <math> (\dfrac{3}{5})^2</math> of <math>ABC = \frac{54}{25}</math>. Using similar logic in Solution 4, Area of <math>AED</math> is the same as <math>ABE</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | Drop an altitude from <math>E</math> to <math>BC</math> and call its foot <math>X</math>. We have that <math>EX \cdot BC=BE \cdot EC</math> since both are equal to two times the area of <math>BEC</math>. Since <math>BC=4</math>, <math>BE=\frac{12}{5}</math>, and <math>EC=\frac{16}{5}</math>, we can calculate that <math>EX=\frac{48}{25}</math>. If <math>EX</math> is extended to meet <math>AD</math> at point <math>Y</math>, <math>EY=3-\frac{48}{25}=\frac{27}{25}</math>. Therefore, <math>[AED]=\frac{EY \cdot AD}{2}=\frac{\frac{27}{25} \cdot 4}{2}=\boxed{\textbf{(E)}\frac{54}{25}}</math>. | ||
+ | |||
+ | - Fasolinka | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/GrCtzL0S-Uo?t=369 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by Math4All999== | ||
+ | https://www.youtube.com/watch?v=hSrgGKxduho | ||
+ | |||
+ | ~Math4All999 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2017|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:20, 4 December 2023
Contents
Problem
Rectangle has and . Point is the foot of the perpendicular from to diagonal . What is the area of ?
Solution 1
First, note that because is a right triangle. In addition, we have , so . Using similar triangles within , we get that and .
Let be the foot of the perpendicular from to . Since and are parallel, is similar to . Therefore, we have . Since , . Note that is an altitude of from , which has length . Therefore, the area of is
Solution 2
From similar triangles, we know that (see Solution 1). Furthermore, we also know that from the rectangle. Using the sine formula for area, we have But, note that . Thus, we see that ~coolwiz
Solution 3
Alternatively, we can use coordinates. Denote as the origin. We find the equation for as , and as . Solving for yields . Our final answer then becomes
Solution 4
We note that the area of must equal the area of because they share the base and the height of both is the altitude of congruent triangles. Therefore, we find the area of to be
Solution 5
We know all right triangles are 5-4-3, so the areas are proportional to the square of corresponding sides. Area of is of . Using similar logic in Solution 4, Area of is the same as .
Solution 6
Drop an altitude from to and call its foot . We have that since both are equal to two times the area of . Since , , and , we can calculate that . If is extended to meet at point , . Therefore, .
- Fasolinka
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=369
~ pi_is_3.14
Video Solution by Math4All999
https://www.youtube.com/watch?v=hSrgGKxduho
~Math4All999
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AMC 10 Problems and Solutions |
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