# Thread: The diameter of the circle...

1. ## Appreciable as well as pivotal Argument Originally Posted by zymeth02 Hi, KBTA, i see that u used the angle and area relationship to solve question 1. I have a question though.

Let's say angle E is the the third angle of the triangle with angles C and D.

The equation should be E/360 = area of circular region covered by E / area of circle, right?

The area of the circular region covered by E includes the small unshaded triangle, and a small shaded region. However, in your equation, you seem to have ignored the area of the shaded part, and only considered the area of the small triangle (20pi*1/2). Could you explain why?
Hi zymeth02,
Yes, you are right. I have wrongfully neglected the shaded area behind the tringle. We should wait to get the better solution.
Regards,
Jakir  Reply With Quote

2. Originally Posted by Sandeep Bansal There are two ways to get it
easy by approximation as angle subtended at the center has nothing to do with the area of traingle , but with the sector....
1)
See it is told that area of shaded region is 80pi
and area of circle is pi*(R^2)=pi*100
so the area of traingle is 100pi - 80 pi= 20pie
now the two chords are diameters if they intersect, they at the centre and bisect equally and engendering two equal trianlges....
approx(area of one trinagle will be ) 10pi

in circle total area is find out by sector ;;;;
if we can fine the sector we can get the answer...
ratio of the areas=20pi/100pi(100pi is the total area , full circle)
that is ratio of angles should be for two triangles (or so called sector) --->360*(2/10)=72
for 1 sector angle should be 360*(1/10)=36
A+B+C+D+2x(say)=180+180
A+B+C+D= 360-2*36
A+B+C+D=288
Hence proved without interring further into the formula.....

Other probable solution is
Let E be the thrid angle in the traingle

area of unshaded cirle comes out to be=100pie-80pie=20pie
so,
E=180-(c+d),180-(a+b)

1/2*100*SinE + 1/2*100*SinE =20pie
E comes out to be 39 near about
so
360 -(39+39) = 282 (approx) = 288  now the two chords are diameters if they intersect,?
pl explain how u concluded the chords are diameters  Reply With Quote

3. ## Hi ,

It is totally based on assumption there is no way out , for this problem ......
otherwise , they should have provided with another option.......
f)Information given is incomplete or if we can prove these two traingles equal........the only way...

Regards
Sandeep  Reply With Quote

4. ## Soln of No. 1

Hi Everybody,
The solution of no. 1 problem I have given before is totally correct. Thebasic principle:

The angle subtended at the center is same for arc as well as the triangle formed with radius and chord connecting the arc end points.
Kapil leo, a newcomer of our GMAT Testmagic forum has helped me to clarify the mystery that was us before. Cheers Kapil leo.

For further clarification we can solve the following problem that I have borrowed from TM GMAT Problem Solving
p></p>
4.The shape of a tunnel entrance. is of a circle which is 3/4th of the full circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?

(A) 9π
(B) 12π
(C)
(D) 18π
(E) 9π/sqrt2

Rgds,
Jakir  Reply With Quote

5. ## Re: The diameter of the circle... Originally Posted by rohit_oberoi  The diameter of the circle is 20 and the area of the shaded region is 80 pi. What is the value of a + b + c + d

A) 144
B) 216
C) 240
D) 270
E) 288 How do you solve these fast?
Could you please insert the figure of this problem  Reply With Quote