Considering the following problem, how far should the end of the ramp be from the base of the building?
Problem: You are in charge of building a wheelchair ramp for a doctor's office. Federal regulations require that the ramp must extend 12 inches for every 1 inch of rise. The ramp needs to rise to a height of 18 inches.
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You can answer this question by considering the ratio of similar triangles:
In 12 inches, the ramp rises 1 inch-- > 12/1
In x inches, the ramp must rise 18 inches --> x /18
Both of these triangles have concruent angles. Therefore their ratios are equal:
12/1 = x / 18
x = 12*18 = 216 inches = 18 feet
Assuming that you are going to go with a single straight ramp with a constant slope (which you probably wouldn't, but we'll assume), all you have to do to find this is to multiply 12 (which is of course 1 foot) by 18. The reason for this is that you need 12 inches of "run" for every inch of "rise."
So the answer to this is that you need to have the end of the ramp be 18 feet from the base of the building.
If you need the answer to be in inches, you multiply 18 by 12 and you get 216 inches.
For every 1 inch rise the ramp must extend 12 inches.
For a rise of n inches the ramp must extend by n*12 inches.
The required rise = n = 18 inches.
The required length of ramp = 18*12 = 216 inches
This is equivalent to 216/12 = 18 feet.
Distance of end of ramp from building = 18 feet.
An alternative, quicker way to solve this problem is as follows:
As the ramp must extend 1 foot (12 inches) for every 1 inch rise, the ramp must extend by 18 feet for 18 inches rise.
We draw ( or here imagine) the geometrical figure of an inclined ramp AC inclined against a vertical wall . The ends of the ramp on the ground and the vertical wall ar A and C repectively. Let G be the feet of perpendicular of C on the ground.
Now the question is how far the distance of AG is. We know the one end of the ramp is vetically above by a distance of GC = 18 inches.
We consider the right angled triangle AGC in a vertical plane.
AC^2 = AG^2+GC^2.
AG^2 = AC^2 - GC^2
By given data, GC = 18 inches. AC = 12*18 = 216 inches by Law.
Therefore AG^2 = 216^2- 18^2 = 46332
Therefore AG = sqrt(46332) = 215.2487 inches nearly.
So the distance of the other end of the ramp from the base of the building is AG = 215.2487 inch approximately.
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