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

Hello! mark me brainliest
answer:
congruent segments
step-by-step explanation:
congruent segments. take a line, cut it in half, superipose one half over the other half; you'll have congruence.