Question

A school gym has an area of 4,641 square feet and a perimeter of 284 feet. What are the dimensions of the gym?

Answer 1

91 ft × 51 ft

Explanation:

while the other answer is basically correct, it makes a to at the beginning of the calculation (but the continues correctly).

and it assumes you can find the right factoring of the expression.

so, I will show you here how to generally solve this, even if you cannot find the right factorization :

l×w = 4,641 ft²

w = 4641/l

2l + 2w = 284

2l + 2(4641/l) = 284

2l + 9282/l = 284

2l² + 9282 = 284l

2l² - 284l + 9282 = 0

to make things easier let's divide everything by 2

l² - 142l + 4641 = 0

the general solution to a squared equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = l

a = 1

b = -142

c = 4641

l = (142 ± sqrt((-142)² - 4×1×4641))/(2×1) =

= (142 ± sqrt(20,164 - 18,564))/2 =

= (142 ± sqrt(1600))/2 = (142 ± 40)/2 = 71 ± 20

l1 = 71 + 20 = 91

l2 = 71 - 20 = 51

w1 = 4641/91 = 51

w2 = 4641/51 = 91

so, they are interchangeable (it does not matter which we call length and which width).

the dimensions are 91 ft × 51 ft

Answer 2

dimensions are 51x91

Explanation:

Assuming that the gym is rectangular in shape, then...

l*w=4641

and 2l+2w=284

We have two variables and two equations so to solve this we plug one equation into another.

Solve for one variable:

l*w=4641

w=4641/l

Plug into the other equation...

2l+2(4641/l)=284

Solve

l*(2l+9282/l)=284

2l^2+9282=284l

2l^2-284l+9282=0

Factor the trinomial:

2(l−51)(l−91)=0

l−51=0 or l−91=0(Set factors equal to 0)

l=51 or l=91

now, lets plug it into one of the equations to find w

51*w=4641

w=91, l=51

or...

91*w=4641

w=51, l=91