A school gym has a perimeter of 346 feet and an area of 7,326 square feet. What are the dimensions of the gym?

Question

Help this math is due tommorrow

Answer 1

P=2l+2w

346=2l+2w

173=l+w

therefore l=173-w

A=lw

7326=(173-w)w

7326=173w-w^2

w^2-173w+7326=0

now you have to factor this equation

this is the algebra involved in solving the problem

in reality it is just a matter of factoring 7326 and grouping the factors such that the two factors sum to 173

so,....

7326/2=3663

3663/3=1221

1221/3=407

407/11=37

7326=2*3*3*11*37

now you have to group the factors accordingly such that the two factors sum to 173

2*3*3*11*37=(3*3*11)*(2*37)=99*74 and 99+74=173

w^2-173w+7326=0

(w-99)(w-74)=0

w-99=0

w=99

w-74=0

w=74

the dimensions of the gym are 74 feet by 99 feet

Answer 2

I'm Happy to help with the setup.

 

Equations needed: A = lw                 P = 2l +2w

 

7326 = lw                                        346 = 2l +2w

 

 

Solving this system of equations should lead you to the answer. (find one variable using one of these and plug into the other equation)