A field in the shape of a right triangle (with the two shortest sides measuring 60 feet and 80 feet) has roads on all three sides that don't drain properly. As a result, muddy water puddles collect and when the cars pass through, they splash all the snow that is within 5 feet of any one of the three roads.
If all other snow not splashed by cars is kept clean, what percent is clean?
(In reply to different answer, four years on
"By the way, the quadrilaterals in the corners are referred to as kites in most of the comments, but they're not. They're parallelograms. I wonder if that's what the issue is? "
That might be the case.
For sake of picturing this, consider the 80-ft side as the base and the 60-ft side upright on the right hand side.
The 5-ft band, on the right of course shortens the base, for the innter triangle by 5-feet, as that corner is a square 5 ft x 5 ft.
Consider the inner triangle's vertex on the left (the smallest angle of the triangle). That point is at the larger angle of a quite small right triangle at the bottom left. The vertical (shorter) leg of this tiny right triangle is 5 ft, and the base is 5*80/60 = 20/3 = 6.666..., the amount by which this also decreases the base of the inner triangle as compared to the outer. So the new triangle has base 80 - 5 - 6.6666.... = 68.33333....
So the pieces at the acute angles are in fact kites, the one at the smallest side having lengths in sequence of 6.666..., 6.6666..., 5,5. (the two length 5's are adjacent, not opposite), as they are hypotenuse-to-hypotenuse congruent right triangles that are mirror images of each other.
An interesting way to get the ratio is to take (68.333.../80)^2, as the inner and outer triangle are similar, as are each of the tiny triangles at the acute corners.
Posted by Charlie
on 2009-04-30 18:46:35