The area of the ice surface of a skating rink is about 221 yd2. The rink is about the shape of a rectangle where the ice-surface width is 4 yd longer than its length. Find the dimensions of the surface
Let x----------> length of rectangle surface y---------->width of rectangle surface we know that x*y=221------------> equation 1 and y=x+4--------------> equation 2
I substitute 2 in 1 x*[x+4]=221--------> x²+4x-221=0 using a graph tool---------> to calculate the quadratic equation
see the attached figure the solution is x=13 y=x+4-------> y=13+7-----> y=17
the answer is the length of rectangle surface is 13 yd the width of rectangle surface is 17 yd
