Geometry Prob.

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Geometry Prob.

by dkumar.83 » Tue May 11, 2010 12:27 pm
Image

Help with the solution.
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by debmalya_dutta » Tue May 11, 2010 12:46 pm
7.5
The width of the traffic lane is 12 ft. So , the road is 6 ft on either side of the centre of the circle. The top of the tunnel is lowest near edges of the traffic lane. height of the roof of the tunnel from the edge of the lane = square roof of (100-36) or 8 . But it is mentioned that a clearance of 0.5 ft is required.
hence the answer = 8-0.5=7.5

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by kstv » Tue May 11, 2010 10:28 pm
debmalya_dutta wrote:7.5
The width of the traffic lane is 12 ft. So , the road is 6 ft on either side of the centre of the circle. The top of the tunnel is lowest near edges of the traffic lane. height of the roof of the tunnel from the edge of the lane = square roof of (100-36) or 8 . But it is mentioned that a clearance of 0.5 ft is required.
hence the answer = 8-0.5=7.5
The edge is 6 ft from the centre of the road and 4 ft from the egde of the tunnel.
The radius of the semi circle is 10 ft.
Height from the tunnel to the edge on the road is
10²-6² = √64

@ d_dutta ---------Very elegant sloution

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by KKGMATTER » Mon Jun 07, 2010 6:43 am
10²-6² = √64 - dumb question ..but wat formula is this ?

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by jeffedwards » Mon Jun 07, 2010 5:08 pm
KKGMATTER wrote:10²-6² = √64 - dumb question ..but wat formula is this ?

Yeah, that took me a bit too. What they are using is the Pythagorean Theorem - a^2 + b^2 = c^2.

Hopefully you can picture what I'm explaining.
We know that the radios is 10 right (the diameter is stated as 20; the radios is the length from the center of the street to the top of the wall on the edge)
We also know that the base is 6 (the center to the edge of the lane)
Now we just need to find the height

a^2 + b^2 = c^2
6^2 + b^2 = 10^2
36 + b^2 = 100
b^2 + 64
b = sqrt64
b=8

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by KKGMATTER » Mon Jun 07, 2010 5:57 pm
That really helps