Suppose that the width of a rectangle is 5 inches shorter than the length and that then perimeter of the rectange is 50.
A. Set up an equation for the perimeter involving only L, the length of the rectangle.
2007-07-18 01:32:22 · 11 answers · asked by yuleydis77 1 in Science & Mathematics ➔ Mathematics
According to you the length of the rectangle is L inches. Then the width of the rectangle is L-5 inches.
The perimeter, let it be P, therefore, comes out to be
P = 2{L+(L-5)} inches.
or P=4L - 10 inches
If the perimeter is 50 inches then
50 = 4L - 10
or 4L = 50 + 10 = 60
and L = 60/4 = 15 inches
and width is 15 - 5 = 10 inches
Similarly you can have equation of area of the rectangle involving L only.
A (area) = L*(L-5)
or A = L^2 - 5L square inches
If area is given, length and width of the rectangle can be obtained.
2007-07-18 01:51:20 · answer #1 · answered by Indian Primrose 6 · 0⤊ 0⤋
If the length is L, then the width is L-5. with the perimeter 50, then 50 = 2*L + 2*(L-5)
2007-07-18 08:37:52 · answer #2 · answered by Mark S, JPAA 7 · 0⤊ 0⤋
The length is L. The width is 5 inches shorter, making it L - 5. The perimeter is twice the length plus twice the width, or 2L + 2(L - 5). Set this equal to 50 and solve.
2007-07-18 08:37:19 · answer #3 · answered by DavidK93 7 · 0⤊ 0⤋
Let x = length of the rectangle, x - 5 = the width.
Equation:
(x + x - 5) * 2 = 50
(2x - 5) * 2 = 50
4x - 10 = 50
4x = 50 + 10
4x = 60
x = 60 / 4
x = 15 answer as to the length
Proof:
2(15 + 10) = 50
2 * 25 = 50
2007-07-22 06:02:03 · answer #4 · answered by Jun Agruda 7 · 2⤊ 0⤋
Let the length of the rectangle be Linches, then the width is L-5
the Perimeter of the rectangle=2[]L+L-5] =4L-10=50
4L=60
L=15''.ANS.
2007-07-18 08:39:29 · answer #5 · answered by Anonymous · 0⤊ 0⤋
width = L-5
length = L
perimeter = 2(L+W)
eqn: P=2(L+L-5)
If we solve it, 50=2(2L-5)
50 = 4L-10
4L=60
L=15inches
W=10 inches
2007-07-18 08:56:10 · answer #6 · answered by tbaz4us 2 · 0⤊ 0⤋
2L + 2(L - 5) = 50
4L - 10 = 50
4L = 60
L = 15
Length is 15 inches.
Width is 15 - 5 = 10 inches.
2 sides of 15 inches + 2 sides of 10 inches = 30 + 20 = 50 inches (Hence Proved)
2007-07-18 08:37:12 · answer #7 · answered by Doctor Q 6 · 1⤊ 0⤋
length (L)
width (W) = L - 5
perimeter (P) = 2L + 2(L-5)
P = 2L + 2L - 10
P = 4L - 10 <=== equation for perimeter
solving for L
50 = 4L - 10
4L = 50 + 10
4L = 60
L = 60/4
L = 15
solving for W
W = L - 5
W = 15 - 5
W = 10
2007-07-18 08:46:32 · answer #8 · answered by ronjo 2 · 0⤊ 0⤋
w = L - 5
2w + 2L = 50
2(L-5) + 2L = 50
2L - 10 + 2L = 50
4 L = 40
L = 10
w = 5
2007-07-18 08:39:23 · answer #9 · answered by CPUcate 6 · 0⤊ 0⤋
equation is:
2(L+L-5)=50
->4L-10=50
->4L=60
_.L=15
2007-07-18 08:37:18 · answer #10 · answered by aviral17 3 · 0⤊ 0⤋