Difference between revisions of "2006 AMC 10B Problems/Problem 19"
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== Problem == | == Problem == | ||
− | == Solution == | + | A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <math>AB</math> and <math>CB</math> are extended past <math>B</math> to meet the circle at <math>D</math> and <math>E</math>, respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting <math>D</math> and <math>E</math>? |
+ | |||
+ | <asy> | ||
+ | defaultpen(linewidth(0.8)); | ||
+ | pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; | ||
+ | fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); | ||
+ | clip(B--Arc(O, 2, 30, 60)--cycle); | ||
+ | draw(Circle(origin, 2)); | ||
+ | draw((-2,0)--(2,0)^^(0,-2)--(0,2)); | ||
+ | draw(A--D^^C--E); | ||
+ | label("$A$", A, dir(point--A)); | ||
+ | label("$C$", C, dir(point--C)); | ||
+ | label("$O$", O, dir(point--O)); | ||
+ | label("$D$", D, dir(point--D)); | ||
+ | label("$E$", E, dir(point--E)); | ||
+ | label("$B$", B, SW); | ||
+ | </asy> | ||
+ | |||
+ | <math> \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3}) | ||
+ | \qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math> | ||
+ | |||
+ | == Solutions == | ||
+ | === Solution 1 === | ||
+ | The shaded area is equivalent to the area of sector <math>DOE</math>, minus the area of triangle <math>DOE</math> plus the area of triangle <math>DBE</math>. | ||
+ | |||
+ | Using the [[Pythagorean Theorem]], <math>(DA)^2=(CE)^2=2^2-1^2=3</math> so <math>DA=CE=\sqrt{3}</math>. | ||
+ | |||
+ | Clearly, <math>DOA</math> and <math>EOC</math> are <math>30-60-90</math> triangles with <math>\angle EOC = \angle DOA = 60^\circ </math>. Since <math>OABC</math> is a square, <math> \angle COA = 90^\circ </math>. | ||
+ | |||
+ | <math>\angle DOE</math> can be found by doing some subtraction of angles. | ||
+ | |||
+ | <math> \angle COA - \angle DOA = \angle EOA </math> | ||
+ | |||
+ | <math> 90^\circ - 60^\circ = \angle EOA = 30^\circ </math> | ||
+ | |||
+ | <math> \angle DOA - \angle EOA = \angle DOE </math> | ||
+ | |||
+ | <math> 60^\circ - 30^\circ = \angle DOE = 30^\circ </math> | ||
+ | |||
+ | So, the area of sector <math>DOE</math> is <math> \frac{30}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{3} </math>. | ||
+ | |||
+ | The area of triangle <math>DOE</math> is <math> \frac{1}{2}\cdot 2 \cdot 2 \cdot \sin 30^\circ = 1 </math>. | ||
+ | |||
+ | Since <math>AB=CB=1</math> , <math>DB=ED=(\sqrt{3}-1)</math>. So, the area of triangle <math>DBE</math> is <math>\frac{1}{2} \cdot (\sqrt{3}-1)^2 = 2-\sqrt{3}</math>. Therefore, the shaded area is <math> (\frac{\pi}{3}) - (1) + (2-\sqrt{3}) = \frac{\pi}{3}+1-\sqrt{3} \Longrightarrow \boxed{\mathrm{(A)}} </math> | ||
+ | |||
+ | ==== Non-Trig Approach ==== | ||
+ | <math>\triangle{ODA}</math> has the same height as <math>\triangle{OBD}</math> which is <math>1.</math> | ||
+ | |||
+ | We already know that <math>BD = \sqrt{3} - 1.</math> | ||
+ | |||
+ | Therefore the area is <math>(\sqrt{3}-1) \cdot 1 \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2}.</math> | ||
+ | |||
+ | Since <math>\triangle{ODB} = \triangle{OBE} = \frac{\sqrt{3}-1}{2}.</math> | ||
+ | |||
+ | Therefore the sum of the areas is <math>2 \cdot \frac{\sqrt{3}-1}{2} = \sqrt{3}-1.</math> | ||
+ | |||
+ | Then the area of the shaded area becomes <math>\frac{\pi}{3} - (\sqrt{3} - 1) = \boxed{\textbf{(A)}\frac{\pi}{3} - \sqrt{3} + 1}.</math> | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | === Solution 2 === | ||
+ | From the pythagorean theorem, we can see that <math>DA</math> is <math>\sqrt{3}</math>. Then, <math>DB = DA - BA = \sqrt{3} - 1</math>. The area of the shaded element is the area of sector <math>DOE</math> minus the areas of triangle <math>DBO</math> and triangle <math>EBO</math> combined. Below is an image to help. | ||
+ | |||
+ | <asy> | ||
+ | defaultpen(linewidth(0.8)); | ||
+ | pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; | ||
+ | fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); | ||
+ | clip(B--Arc(O, 2, 30, 60)--cycle); | ||
+ | draw(Circle(origin, 2)); | ||
+ | draw((-2,0)--(2,0)^^(0,-2)--(0,2)); | ||
+ | draw(A--D^^C--E^^D--O^^E--O^^B--O); | ||
+ | label("$A$", A, dir(point--A)); | ||
+ | label("$C$", C, dir(point--C)); | ||
+ | label("$O$", O, dir(point--O)); | ||
+ | label("$D$", D, dir(point--D)); | ||
+ | label("$E$", E, dir(point--E)); | ||
+ | label("$B$", (1.33,1.04), SW); | ||
+ | </asy> | ||
+ | |||
+ | Using the Base Altitude formula, where <math>DB</math> and <math>BE</math> are the bases and <math>OA</math> and <math>CO</math> are the altitudes, respectively, <math>[DBO] = [EBO] = \frac{\sqrt{3}-1}{2}</math>. The area of sector <math>DOE</math> is <math>\frac{1}{12}</math> of circle <math>O</math>. The area of circle <math>O</math> is <math>4\pi</math>, and therefore we have the area of sector <math>DBE</math> to be <math>\frac{\pi}{3} + 1 - \sqrt{3} \Longrightarrow \boxed{A}</math> | ||
+ | |||
+ | === Solution 3 (Using Answer Choices) === | ||
+ | Like the first solutions, you find that the area of sector <math>DOE</math> is <math>\frac{\pi}{3}</math>. We also know that the triangles will not be in terms of <math>{\pi}</math>. Looking at the answers, choices <math>\text{(A)}</math> and <math>\text{(E)}</math> both contain <math>\frac{\pi}{3}</math>. However, based on the diagram, we observe that the answer must be less than <math>\frac {\pi}{3}</math>. Only <math>\boxed{A}</math> consists of a value less than <math>\frac{\pi}{3}</math>. | ||
+ | |||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=18|num-a=20}} | |
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] | ||
+ | [[Category:Circle Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:20, 30 December 2020
Contents
Problem
A circle of radius is centered at . Square has side length . Sides and are extended past to meet the circle at and , respectively. What is the area of the shaded region in the figure, which is bounded by , , and the minor arc connecting and ?
Solutions
Solution 1
The shaded area is equivalent to the area of sector , minus the area of triangle plus the area of triangle .
Using the Pythagorean Theorem, so .
Clearly, and are triangles with . Since is a square, .
can be found by doing some subtraction of angles.
So, the area of sector is .
The area of triangle is .
Since , . So, the area of triangle is . Therefore, the shaded area is
Non-Trig Approach
has the same height as which is
We already know that
Therefore the area is
Since
Therefore the sum of the areas is
Then the area of the shaded area becomes
~mathboy282
Solution 2
From the pythagorean theorem, we can see that is . Then, . The area of the shaded element is the area of sector minus the areas of triangle and triangle combined. Below is an image to help.
Using the Base Altitude formula, where and are the bases and and are the altitudes, respectively, . The area of sector is of circle . The area of circle is , and therefore we have the area of sector to be
Solution 3 (Using Answer Choices)
Like the first solutions, you find that the area of sector is . We also know that the triangles will not be in terms of . Looking at the answers, choices and both contain . However, based on the diagram, we observe that the answer must be less than . Only consists of a value less than .
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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