Difference between revisions of "2014 AMC 10B Problems/Problem 15"
TheCrafter (talk | contribs) (Added solution) |
TheCrafter (talk | contribs) m (→Solution) |
||
Line 21: | Line 21: | ||
Because <math>ABCD</math> is a rectangle, <math>\angle ADC=90^{\circ}</math>, and so <math>\angle ADC=\angle ECF=\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)(\frac{x\sqrt{3}}{3})=\frac{x^2\sqrt{3}}{3}</math>. | Because <math>ABCD</math> is a rectangle, <math>\angle ADC=90^{\circ}</math>, and so <math>\angle ADC=\angle ECF=\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)(\frac{x\sqrt{3}}{3})=\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{(D) }}\sqrt{3}{6}}</math> | + | To finish, <math>\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(D) }}\frac{\sqrt{3}}{6}}</math> |
==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}} |
Revision as of 17:13, 20 February 2014
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 ?
$\textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ (Error compiling LaTeX. Unknown error_msg)
Solution
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, $\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(D) }}\frac{\sqrt{3}}{6}}$ (Error compiling LaTeX. Unknown error_msg)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.