Difference between revisions of "2014 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | In rectangle <math>ABCD</math>, <math>DC = | + | In rectangle <math>ABCD</math>, <math>DC = 2 \cdot CB</math> and points <math>E</math> and <math>F</math> lie on <math>\overline{AB}</math> so that <math>\overline{ED}</math> and <math>\overline{FD}</math> trisect <math>\angle ADC</math> as shown. What is the ratio of the area of <math>\triangle DEF</math> to the area of rectangle <math>ABCD</math>? |
<asy> | <asy> | ||
Line 14: | Line 14: | ||
</asy> | </asy> | ||
− | <math> \textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D) | + | <math> \textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math> |
+ | [[Category: Introductory Geometry Problems]] | ||
+ | |||
+ | ==Solution 1== | ||
− | |||
Let the length of <math>AD</math> be <math>x</math>, so that the length of <math>AB</math> is <math>2x</math> and <math>\text{[}ABCD\text{]}=2x^2</math>. | Let the length of <math>AD</math> be <math>x</math>, so that the length of <math>AB</math> is <math>2x</math> and <math>\text{[}ABCD\text{]}=2x^2</math>. | ||
− | Because <math>ABCD</math> is a rectangle, <math>\angle ADC=90^{\circ}</math>, and so <math>\angle | + | Because <math>ABCD</math> is a rectangle, <math>\angle ADC=90^{\circ}</math>, and so <math>\angle ADE=\angle EDF=\angle FDC=30^{\circ}</math>. Thus <math>\triangle DAE</math> is a <math>30-60-90</math> right triangle; this implies that <math>\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}</math>, so <math>\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}</math>. Now drop the altitude from <math>E</math> of <math>\triangle DEF</math>, forming two <math>30-60-90</math> triangles. |
+ | |||
+ | Because the length of <math>AD</math> is <math>x</math>, from the properties of a <math>30-60-90</math> triangle the length of <math>AE</math> is <math>\frac{x\sqrt{3}}{3}</math> and the length of <math>DE</math> is thus <math>\frac{2x\sqrt{3}}{3}</math>. Thus the altitude of <math>\triangle DEF</math> is <math>\frac{x\sqrt{3}}{3}</math>, and its base is <math>2x</math>, so its area is <math>\frac{1}{2}(2x)\left(\frac{x\sqrt{3}}{3}\right)=\frac{x^2\sqrt{3}}{3}</math>. | ||
+ | |||
+ | To finish, <math>\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>. | ||
− | + | ==Solution 2== | |
+ | WLOG, let <math>AD = 1</math> and <math>DC = 2</math>. Furthermore, drop an an altitude from <math>F</math> to <math>CD</math>, which meets <math>CD</math> at <math>X</math>. Since <math>\angle ADC</math> is right and has been trisected, it follows that <math>\triangle ADE</math> and <math>\triangle DXF</math> are both <math>30-60-90</math> triangles. Therefore, <math>AE = \frac{\sqrt{3}}{3}</math>, and <math>DX = AF = \sqrt{3}</math>. Hence, it follows that <math>EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}</math>. Since <math>\angle ADE</math> is right, the height and base of <math>\triangle DEF</math> are <math>1</math> and <math> \frac{2\sqrt{3}}{3}</math>, respectively. Thus, the area of <math>\triangle DEF</math> is <math>\frac{\sqrt{3}}{3}</math>, and the area of rectengle <math>ABCD</math> is <math>2</math>, so the ratio beween the area of <math>\triangle DEF</math> and <math>ABCD</math> is <math>\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>. Note that we are able to assume that <math>AD=1</math> and <math>DC = 2</math> because we were asked to find the ratio between two areas. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:02, 31 October 2022
Contents
Problem
In rectangle , and points and lie on so that and trisect as shown. What is the ratio of the area of to the area of rectangle ?
Solution 1
Let the length of be , so that the length of is and .
Because is a rectangle, , and so . Thus is a right triangle; this implies that , so . Now drop the altitude from of , forming two triangles.
Because the length of is , from the properties of a triangle the length of is and the length of is thus . Thus the altitude of is , and its base is , so its area is .
To finish, .
Solution 2
WLOG, let and . Furthermore, drop an an altitude from to , which meets at . Since is right and has been trisected, it follows that and are both triangles. Therefore, , and . Hence, it follows that . Since is right, the height and base of are and , respectively. Thus, the area of is , and the area of rectengle is , so the ratio beween the area of and is . Note that we are able to assume that and because we were asked to find the ratio between two areas.
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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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