Perimeter of V

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akshatgupta87: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Wed Sep 01, 2010 11:11 am; Thanked: 1 times; Followed by:2 members

Perimeter of V

by akshatgupta87 » Fri Apr 08, 2011 4:24 am

In the figure to the right, 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?

a)(5/3)pi+5(sqrt3)
b)(5/3)pi+10(sqrt3)
c)(10/3)pi+10(sqrt3)
d)(10/3)pi+5(sqrt3)
e)(10/3)pi+20(sqrt3)

sqrt=Square Root

Please explain...

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Source: Beat The GMAT — Problem Solving | Fri Apr 08, 2011 4:24 am

HSPA: Legendary Member; Posts: 1101; Joined: Fri Jan 28, 2011 7:26 am; Thanked: 47 times; Followed by:13 members; GMAT Score:640

by HSPA » Fri Apr 08, 2011 4:54 am

Is it only I or no one can see the image,,kindly help

First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.

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GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

by Anurag@Gurome » Fri Apr 08, 2011 4:59 am

akshatgupta87 wrote:In the figure to the right, 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?

a)(5/3)pi+5(sqrt3)
b)(5/3)pi+10(sqrt3)
c)(10/3)pi+10(sqrt3)
d)(10/3)pi+5(sqrt3)
e)(10/3)pi+20(sqrt3)

sqrt=Square Root

Please explain...

Given: CD is || to diameter AB, so x = 30Âº, which implies angle CBE = 2*30Âº = 60Âº
So, angle COE = 120Âº (angle subtended by arc CE at the center of circle is two times the angle subtended at the circumference.
Hence, length of arc CE = 2(pi)(5)(120)/(360) = 10(pi)/3
Also, angles ACB and AEB will be 90Âº (angle subtended at the circumference by the diameter is always 90Âº).
Since, angle ACB = 90Âº and angle ABC = 30Âº, so angle BAC = 60Âº, which means triangle ABC is a 30-60-90 triangle. Hence, ratio of the sides are 1:2:âˆš3
AB:AC:BC = 2a

aâˆš3, and a = 5 (as 2a = 10)
BC = 5âˆš3
Perimeter of the shaded region = EB + BC + arc CE = 10âˆš3 + 10(pi)/3

The correct answer is C.

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akshatgupta87: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Wed Sep 01, 2010 11:11 am; Thanked: 1 times; Followed by:2 members

by akshatgupta87 » Fri Apr 08, 2011 5:28 am

thanks..

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GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770