Asked by Lina
The length and width of a rectangle are to each other as 4 is to 3. A second rectangle are 4 units longer and 2 units wider than the first one, and has twice as great an area as the first. Find the dimensions of the first rectangle.
Answers
Answered by
Reiny
let the length and width be 4x and 3x
old area = (4x)(3x) = 12x^2
new length = 4x+4
new width = 3x + 2
new area = (4x+4)(3x+2) = 12x^2 + 20x + 8
12x^2 + 20x + 8 = 2(12x^2) = 24x^2
12x^2 - 20x - 8 = 0
3x^2 - 5x - 2 = 0
(x-2)(3x + 1) = 0
x = 2 or x = -1/3, which is not admissible
the first triangle was 8 by 6
check:
8 by 6 gives an area of 48
new dimensions: 12 by 8
new area = 12(8) = 96
which is twice 48
All is good.
old area = (4x)(3x) = 12x^2
new length = 4x+4
new width = 3x + 2
new area = (4x+4)(3x+2) = 12x^2 + 20x + 8
12x^2 + 20x + 8 = 2(12x^2) = 24x^2
12x^2 - 20x - 8 = 0
3x^2 - 5x - 2 = 0
(x-2)(3x + 1) = 0
x = 2 or x = -1/3, which is not admissible
the first triangle was 8 by 6
check:
8 by 6 gives an area of 48
new dimensions: 12 by 8
new area = 12(8) = 96
which is twice 48
All is good.
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