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In the diagram, points A, B, and C are on the diameter of the circle with center B. Additionally, all arcs pictured are semicircles. Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded region above the line YB to the area of the shaded region below the line YB? (Note: Diagram is not drawn to scale and angles drawn are not accurate.)
(A) ¾
(B) 5/6
(C) 1
(D) 7/5
(E) 9/7
Attachment:
Circle.png [ 4.27 KiB | Viewed 55450 times ]
08 Feb 2012, 18:59
Kudos
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In the diagram, points A, B, and C are on the diameter of the circle with center B. Additionally, all arcs pictured are semicircles. Suppose angle YXA = 105 degrees. What is the ratio of the area of the shaded region above the line YB to the area of the shaded region below the line YB? (Note: Diagram is not drawn to scale and angles drawn are not accurate.)
(A) ¾
(B) 5/6
(C) 1
(D) 7/5
(E) 9/7
According to the central angle theorem <ABY=2*(180-105)=150 (for more on this check Circles chapter of Math Book: math-circles-87957.html). Hence <CBY=180-150=30.
The area of sector \(ABY=\frac{150}{360}*\pi{r^2}=\frac{5}{12}\pi{r^2}\);
The area of sector \(CBY=\frac{30}{360}*\pi{r^2}=\frac{1}{12}\pi{r^2}\);
The area of each of two small semicircles is \(\frac{\pi{(\frac{r}{2})^2}}{2}=\pi{\frac{r^2}{8}}\) (as its radius is half of the radius of the big circle);
The are of the shaded region above BY is \(\frac{5}{12}\pi{r^2}-\pi{\frac{r^2}{8}=\frac{7}{24}\pi{r^2}\);
The are of the shaded region below BY is \(\frac{1}{12}\pi{r^2}+\pi{\frac{r^2}{8}=\frac{5}{24}\pi{r^2}\);
Ratio of the areas of the shaded regions is \(\frac{7}{5}\).
Answer: D.
Attachment:
untitled.PNG [ 4.59 KiB | Viewed 56081 times ]
12 Jun 2014, 02:36
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Bunuel
It can also be solved the following way:
Three new triangles can be formed which will be isosceles triangles (please refer the attachment)
Hence,
x+y = 105 (given)
2(x+y+z) = 360 since ACYX is a quadrilateral
x+y+z = 180
-> z = 75
in Triangle, BYC, <YBC will be 180 - 2*z = 180 - 2*75 = 30 degrees.
(1) Area of the total shaded portion is half the area of the circle pi*(r^2)/2
(2) Area below the red line = area of segment BYC of circle + area of shaded semicircle BC
= (30/360)*(pi*r^2) + pi ((r/2)^2)/2
=pi*(r^2)/12+pi*(r^2)/8
=5 * pi * (r^2) / 24
(3) Area above the red line is (1)-(2) above
= [ pi * (r^2)/2 ] - [ 5 * pi* (r^2) / 24 ]
= 7 * pi * (r^2) / 24
Answer is (3) / (2) which is 7/5
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Attachments
Circle 1.png [ 6.08 KiB | Viewed 46791 times ]
