In the figure to the right [edit: above], circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

Apologies, here are the answer choices:
A (5/3)pi + 5sqrt(3)
B (5/3)pi + 10sqrt(3)
C (10/3)pi + 5sqrt(3)
D (10/3)pi + 10sqrt(3)
E (10/3)pi + 20sqrt(3)

I chose B: Perimeter = 10sqrt(3) + 5pi/3.

Can you describe what you difficulty is with this question? As all of our CAT questions include a detailed explanation, I'd rather just discuss the particular part you're having a hard time understanding. Thanks.

can you please explinn how is the ans B

I won't take you all the way through it here (go for the challenge!), but I'll get you started...

First, notice that we have the angle 30 degrees. By knowing that the lines are parallel, we can see that we have the two parallel lines cut by a transversal. Thus, the interior angle CBA must also be 30 degrees. From this, we can determine that the central angle COA is 60 degrees. This allows us to find the portion of the shaded area that is the sector formed by the circle and angle COE. We now just need to find the remaining portion, which is made of two triangles of equal size, the top being the triangle CBO. But how do we find the measurements here? Well, we know the base, BO, but what about the height? Here you have to get creative. Try looking to form another triangle with an angle that you already know (that original 30 degrees) to find the height of the triangle CBO...

I think the answer should be D

How do we can get the height of the triangle?

Thanks.

I think we can get the length of arc AC = 2*pi*5/6.

Now let us consider triangle COB. We know angle COA = 60. So angle COB = 120 and hence angle OCB = 30. Since length of side OB = 5, it follows that OC = 5 ( Triangle OCB is isosceles)

Let us draw a perpendicular line to AB from C. Lets say the line meets AB at point X.

Lets consider the triangle CXO. Its right angled with angle CXO = 90. We know that angle COX = 60 as angle COX = inscribed angle at the center COA. Since this a right triangle with angles 30,60 and 90 and the hypotneuse of length 5. The sides are in the ratio 2:sqrt(3):1, So length of CX = 5*sqrt(3)/2

All this effort is to get the length of CB and we are getting there. So let us consider triangle CXB. We know angle CBX = 30 and angle CXB = 90. So this is another right triangle with sides of ratio 2:sqrt(3):1. We know CX = 5*sqrt(3)/2. Hence the length of hypotneuse CB is 5*sqrt(3).

So CB = EB and length of arc AC = AE.
Perimeter = 2*2*pi*5/6+10*sqrt(3) = 10*pi/3 + 10*sqrt(3) - Choice D

Let me know if there is an easier way to do this

To the last poster--I drew slightly different triangles, and I think the solution is a little easier as a result.

Like you, I drew line OC to create triangle CBO. But from there, you drew other lines outside of that triangle. I think it makes sense to focus just on triangle CBO.

1) We want the length of CB, one side of CBO.
2) We already know the lengths of the other two sides, as they are radii of the circle. BO = OC = 5.
3) As you noted, angles OCB and CBO are 30 degrees. Not only does this mean that triangle CBO is an isosceles, but also that it can be split into two identical 30-60-90 triangles. We'll follow your lead and use the property that the side ratios for 30-60-90 triangles are 1:sqrt3:2.

I drew my line from point O perpendicular to line segment CB, calling the intersection point X. Look at triangle XBO, which is 30-60-90.

OX : XB : BO = 1y : (sqrt3)y : 2y, where y is just an unknown multiplier.

BO = 5 = 2y, so y = 5/2.
XB = (sqrt3)y = (5sqrt3)/2.

CB = 2 * XB = 5sqrt3
BE = CB = 5sqrt3

Hope this helps. For me, the toughest thing about geometry questions is knowing which lines to draw and which parts of the picture to focus on. There's more than one way to do so here, and any method that gets you to an answer in reasonable time is OK in my book.

_________________
Emily Sledge
Instructor
ManhattanGMAT

Ans : D
as CD || AB = > x = 30
thus angle CBE =60
now angle COB = 2 (angle CBE) =120 ; as angle subtended by ARC CE on center is twice the angle subtended on circumference.
thus length of arc CE = (120 /360) * 2* pi * 5 = 10/ 3 pi-------------------------(1)

For length BC and BE :
Take the triangle ACB : angle ACB = 90 ; as angle subtended by diameter is always 90.
angle ABC = x= 30
thus =>angle BAC = 60
now as we know if the angle are 30 , 60 , 90 then opposite sides are in the ratio of k : k sqrt(3) : 2k
= > AC : BC : AB = k : k sqrt(3) : 2k and AB = 2k = 10 diameter.
thus k = 5
=> BC = 5*sqrt(3) = BE ;as triangle ABC and triangle ABE are similar---------------------(2)

thus perimeter of shaded region is BC + BE + arc CE
=> 10/ 3 pi + 5*sqrt(3) + 5*sqrt(3) = 10/ 3 pi + 10*sqrt(3)
thus Answer is D[/img]

Thanks all for the effort! OA is D

Nice work. Notice that this problem neatly disguises the transveral that cuts the two parallel lines.

Can you elaborate more on the COE and COA? I thought the perimeter of arc CAE is (60/360)*10pi=5pi/3.

[Degree/360]*[(2pi)radius]

Arc CAE is 120 degrees, not 60 degrees; this makes your calculation off by a factor of 2. Once we know that x=30 and it represents an inscribed angle, the arc it intercepts (CA or AE) is twice that angle, or 60 degrees. thus CAE is 60 + 60 = 120 degrees..

_________________
Tim Sanders
Manhattan GMAT Instructor