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.
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.
To solve this problem, we'll use algebraic equations to represent the given information and find the dimensions of the first rectangle.
Let's assume the length of the first rectangle is 4x (since the length and width are in a ratio of 4:3). Therefore, the width of the first rectangle would be 3x.
We know that the area of a rectangle is given by the formula: Area = Length x Width.
So, the area of the first rectangle would be: Area1 = (4x) x (3x) = 12x^2.
According to the problem, the second rectangle is 4 units longer and 2 units wider than the first one. Hence, the length of the second rectangle would be 4x + 4, and the width would be 3x + 2.
The area of the second rectangle is said to be twice as great as the first one. Therefore, we can write the following equation:
Area2 = 2 x Area1
(4x + 4) x (3x + 2) = 2 x 12x^2
Simplifying the equation, we get:
(4x + 4)(3x + 2) = 24x^2
Expanding the terms on the left side of the equation, we get:
12x^2 + 8x + 8x + 8 = 24x^2
Combining like terms, we have:
24x^2 - 12x^2 + 16x - 8 = 0
Simplifying the equation further, we get:
12x^2 + 16x - 8 = 0
Now, we can solve this quadratic equation for x using factoring, completing the square, or the quadratic formula.
By factoring, we can rewrite the equation as:
4(3x^2 + 4x - 2) = 0
Setting each factor equal to zero, we get two possible solutions for x:
1) 3x^2 + 4x - 2 = 0
2) 3x^2 + 4x - 2 = 0
Now, we can solve each equation separately to find the value(s) of x.
By using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), we can find the roots of the quadratic equation.
For the first equation, a = 3, b = 4, and c = -2:
x = (-4 ± √(4^2 - 4(3)(-2)) / (2(3))
Simplifying the equation, we get:
x = (-4 ± √(16 + 24)) / 6
x = (-4 ± √40) / 6
x ≈ -0.47 or x ≈ 0.8
Since the dimensions cannot be negative, we discard -0.47 as a solution. Therefore, x ≈ 0.8.
Now, we can find the dimensions of the first rectangle:
Length = 4x = 4(0.8) = 3.2 units
Width = 3x = 3(0.8) = 2.4 units
Hence, the dimensions of the first rectangle are approximately 3.2 units (length) and 2.4 units (width).