# algebra

posted by on .

give the demensions of rectangles with the perimeters of 70 feet and length-to-width ratios of 3 to 4, 4 to 5, and 1 to 1.

Ok, the perimeter is 70 and perimeter = 2(length + width)
The ratio is length:with=3:4 so length = (3/4)width
70 = 2(length + width) =2((3/4)width +
width)
So 35 = (7/4)width
Can you finish it off?
If the ratio is 4:5 then 35=((4/5)width + width)
If it's 1:1 then 35 = 2*length = 2*width
Can you finish it off now?

I'm sorry I havn't done this in 20 years my son is in 7th grade doing 8th grade work and I guess the teacher expects the student to know this stuff. He has given no direction to the kids. Thank you for all your help. I will have to get a teaching book on Algebra so I can teach my child.

Ok. This is introductory algebra so you should have no trouble finding sources to help. If you look at the Jump box above you'll find lessons that cover this material on Jiskha.

Thank You, Thank you

You're welcome
Feel free to have your son post questions too. There are plenty of people helping here.

Let the width-to-length ratio be r. I am calling "length" the longest side, so r is less than or equal to 1.

Let a be the longest side length. The shortest side length is then a*r.

The perimeter is
70 = 2a (1 + r)
35 = a (1+r)
a = 35/(1+r)

Solve for a, using r values of 3/4, 4/5 and 1
When r = 3/4, a = 35/(7/4) = 20
The short side will then be 3/4 of 20, or 15. You do the others