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perimeter of a tunnel

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perimeter of a tunnel

by vivekjaiswal » Fri Sep 25, 2009 2:40 am
The figure above (in the attachment) shows the shape of a tunnel entrance. If the curved portion is 3/4 of a 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) 9Ï€
(B) 12Ï€
(C) 9Ï€*sqrRt(2)
(D) 18Ï€
(E) 9Ï€/sqrRt(2)


OA is C
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Re: perimeter of a tunnel

by mohitsharda » Fri Sep 25, 2009 3:00 am
vivekjaiswal wrote:The figure above (in the attachment) shows the shape of a tunnel entrance. If the curved portion is 3/4 of a 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) 9Ï€
(B) 12Ï€
(C) 9Ï€*sqrRt(2)
(D) 18Ï€
(E) 9Ï€/sqrRt(2)


OA is C
Join the end points of the base with the centre. Curved portion represents 3/4th of the circle
=> the angle covered is 360*3/4 = 270 degrees.

The angle subtended by the base on the centre is 360-270 = 90 degrees.

Also, note that the base is a chord of the circle.
From the centre of the circle drop a perpendicular on the base. It will bisect the base as well as the angle subtended at the centre.
now, using trigonometry, we will get
(radius r)
r sin 45 = (12/2)
=> r = 6 sqrt 2

=> perimeter of curved portion = (3/4) * 2 pi *r
= 9*pi*sqrt 2

=> C
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by sanju09 » Fri Sep 25, 2009 3:03 am
The base of the entrance, which is 12 feet across, is actually the hypotenuse of the right angled isosceles triangle with each of the equal side r feet (let's say r be the radius); such that r 2 = 12, or r = 6 2 feet. Now ¾ of the circumference = ¾ (2 × π × 6 2) = 9 π 2.

Go with [spoiler]C[/spoiler]

Note: Read '' as 'square root of' and '×' as 'multiplied by'.
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by vivekjaiswal » Fri Sep 25, 2009 3:17 am
thanks a lot guys...it makes sense to me now :)
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by sanju09 » Wed Feb 17, 2010 5:07 am
vivekjaiswal wrote:thanks a lot guys...it makes sense to me now :)
The only difference between a rut and a grave is the depth. :wink:
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