In the figure, circle O has center O, diameter AB and a radius of 5.

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: \(AC=\frac{10}{2}=5\) --> \(CB=5\sqrt{3}\) --> \(CB+BE=5\sqrt{3}+5\sqrt{3}=10\sqrt{3}\).

Hope it's clear.

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 ?

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 \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) --> \(BC=5\sqrt{3}\).

For more on this subject check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.

Oh, I forgot that ABS ia a right triangle. thats why I didnt get whats sqroot3/2. my fault, sorry.

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

Look at the diagram below:Hope it helps.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Hi Bunuel, 1 doubt.. when you calculated length of the arc CE, why you used angle as 120 degree, why not 60 degree...??

Thanks

Arc CE is a part of the circumference --> central angle COE is 120 degrees --> a central angle in a circle determines an arc.

For more check here: math-circles-87957.html

Hope this helps.

Yes definitely, I was thinking why 60* s not considered , Now I am clear that central angle is considered while calculating the length of the arc.

Thank You Bunuel..

TO calculate the length of the arc CAE...I considered B as the center of a larger circle with radius 5\sqrt{3}..and the arc substending an angle of 60 at the center..But I got a different answer...The legth of arc comes out to be 5\sqrt{3}/3

Dunno what did I do wrong here?

Ok Got it..it will not be a circle but an ellipse..My bad..

Hi Bunuel,
How did you find that BC and BE are equal?
Thanks!

The correct Answer is D.
The perimeter of the shaded region is the sum of lengths of CB + BE + arc EC.
Now let us find out the individual lengths.
1. arc EC
Length of an arc in a circle making an angle X with center O and radius 'r' is given by
[X][/360]*2*pi*r
In the given question we need to find out angle COE which is X in the above formula.
Since CD||AB, angle DCB = angle CBA = 30
Now consider the triangle BOC. This is an Isosceles triangle with OC = OB = 5 and thus the angles in the triangle are as follows:
angle OBC = 30 (given), angle OCB = 30 (property of an Isosceles triangle) and angle BOC = 120 (180-(30 +30))
Now consider the parallel lines CD and AB with CO as the transverse. angle DCO = 60 which is equal to angle COA i.e., 60.
Now the total angle X = angle COE = angle COA + angle AOE = 120
Hence length of arc CE is [120][/360]*2*pi*5 = [10][/3]*pi

2. CB = BE
Consider triangle ACB which is 30-60-90 triangle where angle ACB = 90, angle ABC = 30 and angle CAB = 60.
Now, Sine (angle CAB) = [opposite side][/Hypotenuse] i.e., Sine 60 = [BC][/AB] i.e., sqrt(3)/2 = BC/10 which gives BC as [10*sqrt(3)]/2

Now Perimeter of the shaded region = arc CE + (2 * CB) = [10][/3]*pi + 10*sqrt(3)

Hello Dienekes,
Triangles ACB and AEB are Similar Triangles with a common side AB. As per the property of Similar Triangles, ratios of sides are equal, hence BC = BE. Hope this clarifies your doubt.

Regards,
Bharat.

BC and BE are deviated by the same degree from the diameter AB, thus they are mirror images of each other around AB, therefore they must be equal.

Does this make sense?