Geometry Problem

[bretania]

Can someone please shed light on this problem? Please see the attached file for illustration. Basically, the illustration is so simple you may not have a need for it (but you'll probably need to draw one yourself when solving the problem).

There is a tunnel that takes the shape of 3/4 of a circle for it's entrance. If the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance (i.e., what's the perimeter of the 3/4 of the circle).

A.9 PI
B.12 PI
C. 9 PI * SQRT(2)
D.18 PI
E. (9 PI)/SQRT(2)

[Crack_Mat]

answer is C.

area of the tunnel = 1/2 * r^2 * theta = 3/4 * area of circle = 3/4 *PI *r^2
sp theta = 3*PI/2 -----------(1)

and

(12/2) / r = sin (theta/2) = sin (3*PI/4) = sin(PI/4) = 1/root2
6 / r = 1/root2
r = 6*root2

now desired perimeter s = r * angle made at the center
= (6*root2) * 3*PI/2
= 9*PI*root2


but I think it is wrong!!!!!!!!!!!!!

area made by an arc at the center is 1/2*r^2*theta. It does not include the area of the triangle made by the triangle made at center by the two end points.

So if it is saying that 3/4 of the circle that should mean:
area of the arc + area of the triangle = 3/4 of area of circle.

Please comment if you are agree.

[Dr_GMAT]

I agree

[bretania]

Crack Mat, first off, thanks AGAIN!
Yes, the OA here is C, you're right.

Sorry I can't comment however on your solution because you're using trig. I have totally forgotten trig and am not learning it for GMAT because I have no more time.

The prob though because still be solved without using trig.

I'm not sure but this is my take.

because we are told that the tunnel is 3/4 of a circle, we can form a right triangle
with a 90 deg angle on the center of the circle. This tirangle is not only right but also an isosceles (with 2 sides having the same length as the radius of the circle).

Hence, 12 = r*SQRT(2)
--> r = 6*SQRT(2)

Hence 3/4 of the perim of the circle becomes
=3/4*PI* r^2
Substituting r,
we get 9PI * SQRT(2).

[wanderwanderer]

answer C.
3/4 of the circle taken to form the arch. so 1/4 th ie 90deg is the vacant part to which the bases of the arch are attached to and they are 12ft apart. So you have an isosceles triangle whose base is 12ft and the two arms are each (radius). so R^2 + R^2 = 12^2 => R = 6(sqrt2)
now for 3/4 of the circle :
circumference = (3/4) * 2*Pi * 6(sqrt2) = 9 Pi (sqrt2)

Thanks

[sonikamadala]

Originally Posted by wanderwanderer

answer C.
3/4 of the circle taken to form the arch. so 1/4 th ie 90deg is the vacant part to which the bases of the arch are attached to and they are 12ft apart. So you have an isosceles triangle whose base is 12ft and the two arms are each (radius). so R^2 + R^2 = 12^2 => R = 6(sqrt2)
now for 3/4 of the circle :
circumference = (3/4) * 2*Pi * 6(sqrt2) = 9 Pi (sqrt2)

Thanks

Very good explanation. Thank u.

[shailender.bisht]

C. 9 Pi * Sqrt(2)

[grad.mad]

it is C