1.A rectangle is twice as long as it is wide. If both its dimensions are increased by 4 m, its area is increased by 88 m2. Make a sketch as in Oral Exercise 1. Find the dimensions of the original rectangle.
A.
6 m wide, 12 m long
B.
8 m wide, 16 m long
C.
10 m wide, 20 m long
D.
12 m wide, 24 m long
2006-11-15 06:40:12 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x+4)(2x+4)-2x^2=88
2x^2+12x+16-2x^2=88
12x+16=88
adding -16
12x=72
dividing by 12
x=6
so the dimensions of the original rectangle are 6m*12m
2006-11-15 06:51:03 · answer #1 · answered by raj 7 · 0⤊ 0⤋
A is the correct answer. 6 x 12 =72 72+ 88 =160
6 + 4 = 10
12 + 4 = 16
16 x 10 = 160
2006-11-15 06:58:25 · answer #2 · answered by READER 1 5 · 0⤊ 0⤋
Let x = original short side, 2x = original long side.
new short side = x +4, new long side = 2x + 4
Original Area = x * 2x = 2x^2
New Area = (2x+4)(x+4) = 2x^2 + 12x + 16
Original Area + 88 = New Area
2x^2 + 88 = 2x^2 + 12x +16
12x = 72
x = 6
A) 6m wide, 12 m long
2006-11-15 06:50:00 · answer #3 · answered by T 5 · 0⤊ 0⤋
x = original width
2x = original length
2x^2 = original area
x+4 = new width
2x+4 = new length
(x+4)(2x+4) = 2x^2 + 12x +16 = new area
Thus 2x^2 +12x +16 - 2x^2 =88
12x +16 = 88
12x = 72
x = 6 = original width
2x = 12 =original length
Sorry, the answer is A
2006-11-15 06:55:02 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
Original dimentions: l = 2*w
Second area is 88 more than first area:
(l + 4) * (w + 4) = l * w + 88
Mulitply it out:
l*w + 4w + 4l + 16 = l*w + 88
Substitute 2w for every l and solve:
2w² + 12w + 16 = 2w² + 88
12w = 72
w = 6
l = 12
The answer is "a". Sorry.
2006-11-15 06:51:09 · answer #5 · answered by Dave 6 · 0⤊ 0⤋