Difference between revisions of "2014 AMC 10B Problems/Problem 15"
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==Solution== | ==Solution== | ||
+ | ===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>. | ||
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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>. | 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> | + | 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>BC = 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>. | ||
==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 17:05, 1 January 2018
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
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 .
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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