The new playground at the elementary school is 15 and 4/5 longer than it is wide. if the perimeter that requires fencing is 177 1/5 ft, what are the dimensions of the playground?

if i could get help asap that would be great! :)

Question

The new playground at the elementary school is 15 and 4/5 longer than it is wide. if the perimeter that requires fencing is 177 1/5 ft, what are the dimensions of the playground?

if i could get help asap that would be great! :)

jackofalltradez · Answer

I'm here for ya :)

jackofalltradez · Answer

OK, use this equation: 177.2 = 15.8(2w) + 2w

jackofalltradez · Answer

You know the length is 15.8 times longer than the width, so that's where 15.8(2w) comes from

anonymous · Answer

omg thank you soo much!!!! i will work on this right now. thanks jarod i appreciate it:)

jackofalltradez · Answer

No problem :)
I had trouble with these problems too, but my excellent math teacher helped me out

mathstudent55 · Answer

@happybird7
@JackofallTradez 
I think you misunderstood the problem.
I think the problem is stating that the length is 15 4/5 ___ft___ longer than the width, not that the length is 15 4/5 ___times___ longer than the width.
The statement of the problem is ambiguous because there is something missing there. There is no "ft" or "times" between "15 and 4/5" and "longer", but I still think the correct interpretation is that the length is 15 4/5 feet longer than the width.

mathstudent55 · Answer

If the length is 15 4/5 ft longer than the width, then this is how you solve it:
Let the width = w
Then the length = w + 15.8
The perimeter is 2(l + w) = 2(w + 15.8 + w) = 2(2w + 15.8)
Since the perimeter is 177.2 ft, we let the perimeter expression equal 177.2

2(2w + 15.8) = 177.2

Now you need to solve for w. That is the width.
Then add 15.8 ft to the width to find the length.

anonymous · Answer

thank you both @JackofallTradez @mathstudent55

jackofalltradez · Answer

Thank you for clarifying that, @mathstudent55

mathstudent55 · Answer

You're welcome.