dimensions of a rectangle whose length is 50 inches more than its width and whose perimeter is 500 inches.
2007-07-29 06:40:03 · 5 answers · asked by Patricia F 1 in Science & Mathematics ➔ Mathematics
say x=width then x+50=length
perimeter=2*width + 2 *length
500 = 2 *(x) + 2*(x+50)
500= 2x+2x+100
400=4x
x=100
so width=100 and length=150
2007-07-29 06:44:56 · answer #1 · answered by golffan137 3 · 0⤊ 0⤋
Width = w. Length= w+50. Perimeter = 2w+2(w+50) = 500
500=4w+100. 4w=400. Width =100, length= 150.
2007-07-29 13:47:11 · answer #2 · answered by Del Piero 10 7 · 0⤊ 0⤋
l = w + 50
2l + 2w = 500 ---> l + w = 250
substituting for l in the second equation:
(w+50) + w = 250
2 w + 50 = 250
2 w = 200
w = 100
using the first equation
l = (100) + 50
l = 150
verify:
Length (150) is 50 more than it's width (100), and
perimeter is 100 + 150 + 100 + 150 = 500.
2007-07-29 13:49:23 · answer #3 · answered by Skeptic 7 · 0⤊ 0⤋
P = L + L + W + W
500 = 2L + 2W
500 = 2(W+50) + 2W
500 = 2W + 100 + 2W
500 = 4W + 100
400 = 4W
100 = W
Proof:
500 = 2(L+ W) + 2W
500 = 2(50+W) + 2W
500 = 2(50 + 100) + 2(100)
2007-07-29 14:14:56 · answer #4 · answered by mcbetha 2 · 0⤊ 0⤋
100x150
2007-07-29 13:44:07 · answer #5 · answered by Kendra 3 · 0⤊ 0⤋