## MGMAT Geometry

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### MGMAT Geometry

by rommysingh » Thu Sep 03, 2015 11:30 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?

(5/3) + 5

(5/3) + 10

(10/3) + 5

(10/3) + 10

(10/3) + 20

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by gmatdriller » Fri Sep 04, 2015 3:43 am
Where is the diagram?

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by umasarath52 » Fri Sep 04, 2015 4:44 am
Since CD is parallel to AB. Angle DCB = Angle CBA = 30

Now angle CBE = 30 + 30 = 60 degree. Angle COE = 120 which is n in the length of arc formulae.

Length of an arc = n/360 x 2 x pi x radius = 120/360 x 2 x pi x 5 = 10pi/3

Angle CBA is 30 and ACB is 90. We get Right angle triangle. (Property of triangle inscribed in Circle with one of its side as diameter, opposite angle to diameter is 90 degree) and the triangle is 30-60-90 and sides will be in the ratio of 1: root 3: 2

Length of side BC is 5 root 3 = length of the side BE

Perimeter = BC+BE+arc EC

=5root3 + 5root 3 + 10pi/3

= 10 root 3 + 10pi/3

I got this answer where i don't see this in any of the options

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### MGMAT Geometry

by Brent@GMATPrepNow » Fri Sep 04, 2015 6:15 am
As umasarath52 noted, the answer choices in rommysingh's original post are incorrect.

Below are the original answer choices.
rommysingh 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 + 5âˆš3
B. (5/3)pi + 10âˆš3
C. (10/3)pi + 5âˆš3
D. (10/3)pi + 10âˆš3
E. (10/3)pi + 20âˆš3

IMPORTANT: unless stated otherwise, the diagrams in Problem Solving geometry questions are DRAWN TO SCALE.
For more on this, see our free video: https://www.gmatprepnow.com/module/gmat- ... /video/863

I'll solve this question using estimation.
Since the diameter AB = 10, we can ESTIMATE the length of CB.
It looks like CB is just a little bit shorter than AB.
So, I'll say that the length of side CB is approximately 9.
This means the length of side EB is approximately 9 as well.
Finally, arc EC looks a little bit shorter than sides CB and EB, so I'll estimate it to be length 8

So, the TOTAL perimeter = 9 + 9 + 8 = 26

ASIDE: On test day, everyone should know the following apprximations:
âˆš2 â‰ˆ 1.4
âˆš3 â‰ˆ 1.7
âˆš5 â‰ˆ 2.2
Also, we'll say that pi â‰ˆ 3

A. (5/3)pi + 5âˆš3 â‰ˆ 5 + 8.5 â‰ˆ 13.5
B. (5/3)pi + 10âˆš3 â‰ˆ 5 + 17 â‰ˆ 22
C. (10/3)pi + 5âˆš3 â‰ˆ 10 + 8.5 â‰ˆ 18.5
D. (10/3)pi + 10âˆš3 â‰ˆ 10 + 17 â‰ˆ 27
E. (10/3)pi + 20âˆš3 â‰ˆ 10 + 34 â‰ˆ 44

Of these, it appears that D is the closest.

Aside: We can see that answer choice B is pretty close too. At this point, you have a time-management decision. You can either stick with D, and use your extra time elsewhere, or your can spend time trying to be more certain of the answer. Your choice.

That said, D is the correct answer.

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Sat Apr 28, 2018 9:04 am, edited 1 time in total.
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by GMATGuruNY » Fri Sep 04, 2015 7:18 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)Ï€ + 5âˆš3
B (5/3)Ï€ + 10âˆš3
C (10/3)Ï€ + 5âˆš3
D (10/3)Ï€+ 10âˆš3
E (10/3)Ï€ + 20âˆš3

Perimeter of the shaded region = arc CAE + CB + EB.

Since AB || CD, âˆ DCB and âˆ CBA are congruent.
Thus, x=30, and âˆ CBE = 60.

An inscribed angle is formed by two chords.
Thus, âˆ CBE is an inscribed angle.
The degree measurement of an inscribed angle = 1/2 the degree measurement of the intercepted arc.
âˆ CBE intercepts arc CAE.
The, arc CAE = 120 degrees.
Since the entire circle = 360 degrees, the length of arc CAE = 120/360 = 1/3 of the entire circumference.
C = 2Ï€r = 10Ï€.
Thus, the length of arc CAE = (1/3)(10Ï€) = (10/3)Ï€.

The correct answer must include (10/3)Ï€.
Eliminate A and B.

Since âˆ CBA = âˆ ABE, the shaded sector above the diameter is congruent to the shaded sector below the diameter.
Thus, CB = EB.
Since OC=5 and AB=10, the length of CB must be between 5 and 10.
Thus, CB and EB are each between 5 and 10, implying that CB + EB is between 10 and 20.

Of the remaining answer choices, only D works:
(10/3)Ï€ + 10âˆš3 â‰ˆ (10/3)Ï€ + 10(1.7) â‰ˆ (10/3)Ï€ + 17.

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by rommysingh » Fri Sep 04, 2015 12:07 pm
Hi,
thanks for the correction and help. Do you think such question should be attempted or skipped in real time..

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by bschool3 » Fri Sep 04, 2015 4:56 pm
Another way to get to D

please forgive my lack of knowledge on typing math symbols, still new here.

First find length of arc CAE
We know that x = 30 deg; then 2x = 60 deg. Therefore arc CAE= (120/360)th or 1/3 of the circumference = 1/3 * 2pi*r = 1/3*2pi* 5 = 10pi/3

Secondly, we find either CB or EB and multiply by 2.
Let's pick EB. Recognize that EOB is isosceles with EO = OB, and EOB = 120 deg (given that OBE = OEB = 30 deg).
Now, a bisector of EOB would cut EB in half; let's call that half-way point Y, such that EY = YB. Focus on the triangle OYB. Recognize that the triangle OYB is a 30-60-90, with OB = 5. At this point it's just trig. We know that sin 60 = (root 3)/2 = YB/5. Therefore, YB = 5(root 3)/2. Then YB+EY =EB =5(root 3). Then EB+BC = 2* EB = 10(root3).

Then perimeter = 10(root 3)+ 10pi/3
Last edited by bschool3 on Fri Sep 04, 2015 5:08 pm, edited 1 time in total.

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by Matt@VeritasPrep » Fri Sep 04, 2015 5:07 pm
Here's a visual:

• Page 1 of 1