Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In the figure, circle O has center O, diameter AB and a [#permalink]
11 Feb 2012, 14:21

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

53% (03:02) correct
46% (01:54) wrong based on 120 sessions

Attachment:

Perimeter.PNG [ 7.32 KiB | Viewed 6924 times ]

In the figure, 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√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3

I can think of few pointers such as

The value of x will be 30 degrees because CD is parallel to diameter. Angle COA and AOE will each be 60 degrees as the central angle is double the inscribed angle. But I am still struggling to get through the question.

Re: Perimeter of the shaded region. [#permalink]
11 Feb 2012, 14:49

2

This post received KUDOS

Expert's post

enigma123 wrote:

In the figure, 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√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3

Attachment:

Perimeter.PNG [ 7.32 KiB | Viewed 6542 times ]

Since CD is parallel to AB then <CBA=<BCD=30 --> <CBE=60 --> <COE=2*60=120 (according to the central angle theorem) --> length of minor arc CE is \frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi;

Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus \frac{BC}{AB}=\frac{\sqrt{3}}{2}, (BC is opposite to 60 degrees so corresponds to \sqrt{3}) --> \frac{BC}{10}=\frac{\sqrt{3}}{2} (AB=diameter=2r=10) --> BC=5\sqrt{3};

Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}.

Re: In the figure, circle O has center O, diameter AB and a [#permalink]
01 Mar 2012, 19:40

I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.

Re: In the figure, circle O has center O, diameter AB and a [#permalink]
01 Mar 2012, 20:18

Expert's post

priyalr wrote:

I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.

Pls explain.

The red parts are not correct.

The point is that AB=diameter=2*radius=2*5=10, not 5,. You've done everything right after this step but got different numerical values because of this one mistake. It should be: AC=\frac{10}{2}=5 --> CB=5\sqrt{3} --> CB+BE=5\sqrt{3}+5\sqrt{3}=10\sqrt{3}.

Re: In the figure, circle O has center O, diameter AB and a [#permalink]
01 Mar 2012, 23:29

priyalr wrote:

I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.

Pls explain.

thnx a ton. My bad !

I want to give my exam by APril end. I have studied the quant notes for this specific topic and leter solve the PS q, for the given topic. The problem is I get stuck at the start or in middle of my solving. I seem to not apply the concepts learned on to the q. What shld i do ?

Re: Perimeter of the shaded region. [#permalink]
01 Mar 2012, 23:41

Bunuel wrote:

Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus \frac{BC}{AB}=\frac{\sqrt{3}}{2}, (BC is opposite to 60 degrees so corresponds to \sqrt{3}) --> \frac{BC}{10}=\frac{\sqrt{3}}{2} (AB=diameter=2r=10) --> BC=5\sqrt{3};

I didnt get this part. could u please explain sqroot3/2, or give some links where I can read theory?

btw, Bunuel, is there another way to solve this question?
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: Perimeter of the shaded region. [#permalink]
02 Mar 2012, 02:42

Expert's post

LalaB wrote:

Bunuel wrote:

Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus \frac{BC}{AB}=\frac{\sqrt{3}}{2}, (BC is opposite to 60 degrees so corresponds to \sqrt{3}) --> \frac{BC}{10}=\frac{\sqrt{3}}{2} (AB=diameter=2r=10) --> BC=5\sqrt{3};

I didnt get this part. could u please explain sqroot3/2, or give some links where I can read theory?

btw, Bunuel, is there another way to solve this question?

Triangle ACB is a 30°-60°-90° right triangle. AC is opposite the smallest angle 30°, BC is opposite 60° angle, and hypotenuse AB is opposite the largest angle 90°.

• A right triangle where the angles are 30°, 60°, and 90°.

This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio 1 : \sqrt{3}: 2. Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).

So, for our question BC (the side opposite 60° angle) and AB (the side opposite the largest angle 90°) must be in the ratio \sqrt{3}:2 --> \frac{BC}{AB}=\frac{\sqrt{3}}{2}, and since AB=diameter=2r=10 then \frac{BC}{10}=\frac{\sqrt{3}}{2} --> BC=5\sqrt{3}.

Re: Perimeter of the shaded region. [#permalink]
01 Jul 2012, 21:22

Bunuel

This is the only part that alvvays troubles me

right triangle (AB=diameter means that <C=90)

Hovv I knovv angle at center subtended by diameter is 90 but cant form it here please explain

Bunuel wrote:

enigma123 wrote:

In the figure, 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√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3

Attachment:

Perimeter.PNG

Since CD is parallel to AB then <CBA=<BCD=30 --> <CBE=60 --> <COE=2*60=120 (according to the central angle theorem) --> length of minor arc CE is \frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi;

Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus \frac{BC}{AB}=\frac{\sqrt{3}}{2}, (BC is opposite to 60 degrees so corresponds to \sqrt{3}) --> \frac{BC}{10}=\frac{\sqrt{3}}{2} (AB=diameter=2r=10) --> BC=5\sqrt{3};

Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}.

Re: Perimeter of the shaded region. [#permalink]
02 Jul 2012, 00:28

Expert's post

venmic wrote:

Bunuel

This is the only part that alvvays troubles me

right triangle (AB=diameter means that <C=90)

Hovv I knovv angle at center subtended by diameter is 90 but cant form it here please explain

Bunuel wrote:

enigma123 wrote:

In the figure, 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√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3

Attachment:

The attachment Perimeter.PNG is no longer available

Since CD is parallel to AB then <CBA=<BCD=30 --> <CBE=60 --> <COE=2*60=120 (according to the central angle theorem) --> length of minor arc CE is \frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi;

Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus \frac{BC}{AB}=\frac{\sqrt{3}}{2}, (BC is opposite to 60 degrees so corresponds to \sqrt{3}) --> \frac{BC}{10}=\frac{\sqrt{3}}{2} (AB=diameter=2r=10) --> BC=5\sqrt{3};

Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}.