A circular playground has 2 parellel roads running through it. Their lengths are 112m and 50m. The perpendicular distance between these roads is 92m.

Find the diameter of the playground.

Any line that intersects in two points a circle, is always perpendicular to the circle's center. In this case, there are two parallel lines crossing the circle, so they both have to be perpendicular to the center, this is why I decided to represent the problem using half of each length of roads (25m and 56m) aligned horizontally or parallel to the x-axis with the origin in the 25m road. The two roads in this arrangement give out a difference of 31m (56m-25m=31m). The hypotenusa would be √9425. This hypotenusa is also perpendicular to the center of the circle, so we find the inclination respect to the x-axis of the perpendicular line that pases trough the middle of the hypotenusa and touches the x-axis. It results that the angle of aperture between the described perpendicular line respect to the x-axis is 18.621579793665 degrees. We use this angle to find the distance between the vertex where the 25m road part intersects with the circle and the point where the perpendicular line calculated previously touches the x-axis. This distance would be √9425/(2*sin(18.621579793665))=152.016129032m. We add 25m to 152.016129032m and we get 177.016129032m from the point of intersection of the perpendicular line from the origin. We find the intersection of perpendiculars were the center should be and is at a height of 177.016129032*tan(18.621579793665)=59.6467391303m from the origin. Finally we make a right triangle with sides of 59.6467391303m, 25m and an hypotenusa equal to the radius of such circular playground. The radius would be 64.6740557633m thus the diameter is 129.348111527m

Posted by Antonio on 2003-09-03 17:05:31