SOLUTION: The length of a rectangle is 8cm more then 3 times the width. The perimeter of the rectangle is 64cm. What are the dimensions of the rectangle?

Click here to see ALL problems on Miscellaneous Word Problems

Question 105824: The length of a rectangle is 8cm more then 3 times the width. The perimeter of the rectangle is 64cm. What are the dimensions of the rectangle?
Found 2 solutions by checkley75, fmo:
Answer by checkley75(3666) (Show Source):
You can put this solution on YOUR website!
2L+2W=PERIMETER.
L=3W+8
2(3W+8)+2W=64
6W+16+2W=64
8W=64-16
8W=48
W=48/8
W=6CM FOR THE WIDTH.
L=3*6+8
L=18+8
L=26 CM FOR THE LENGTH.
PROOF
2*26+2*6=64
52+12=64
64=64

Answer by fmo(8) (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangle is 2w + 2l where w is the width and l is the length. We know that:
2w + 2l = 64.
We also know from the problem statement that the length is 3 times the width plus another 8 cm. Or:
l = 3w + 8
Plugging that into the above formula we now have:
2w + 2(3w + 8) = 64 which we can solve as follows:
2w + 6w + 16 = 64 (multiplied 2 and (3w + 8)
8w + 16 = 64 (added like terms on the left side)
8w = 48 (subtracted 16 from both sides)
w = 6 (divided both sides by 8)
To find out the length given the width equals 6, we simply substitute this value into our equation, l = 3w + 8. This gives us:
l = 3(6) + 8
l = 18 + 8
l = 26
So, the dimensions of the rectangle are 6 cm wide by 26 cm long.