A rectangular poster has an area of 190in(squared). The height of the poster is 1in. less than twice its width. Find the dimensions of the poster.
A = Area= 190 in^2
H = Hight
W = Width
H = 2W-1
A = W*H
A = W*(2W-1)
190 = 2W*W-1*W
190 = 2W^2-W
2W^2-W-190 = 0
The exact solutions of this equation are:
W=10
and
W= -19/2
Width can't be negative so:
W = 10 in
H = 2*W-1
H = 2*10-1
H = 20-1
H = 19 in
W = 10 in
H = 19 in
A = W*H
A = 10*19
A = 190 in^2
If you don't know solve quadratic equation, in google type:
quadratic equation online
When you see list of results click on:
Free Online Quadratic Equation Solver:Solve By Quadratic Formula
When page be open in rectangle type:
2W^2-W-190=0
and click option: solve it!
You will see solution step-by-step
Well, well, well, let's solve this puzzler together, shall we?
Let's call the width of the poster "W" (because I'm a big fan of short names, you know), and its height "H".
According to the information given, we have two clues:
1. The area of the poster is 190 square inches.
2. The height is 1 inch less than twice the width.
Alright, let's tackle the area first. We know that the area of a rectangle is the length multiplied by the width. In this case, since we only have the width and height, we can rewrite the formula as width multiplied by height = area.
That gives us the equation W * H = 190.
Moving on to the second clue, we're told that the height is 1 inch less than twice the width. So we can write this as H = (2W - 1).
(Why anyone would want to make things so complicated is beyond me!)
Now, let's substitute this expression for H into our area equation, which gives us:
W * (2W - 1) = 190.
Expanding this, we get:
2W^2 - W - 190 = 0.
Now, let's solve this quadratic equation. (Don't worry, I promise to keep it entertaining!)
Using the quadratic formula, we get:
W = (-b ± √(b^2 - 4ac)) / 2a.
Plugging in the values, we have:
W = (1 ± √(1 - 4 * 2 * -190)) / (2 * 2).
Calculating this, we end up with two possible values for W: approximately -10.61 and 9.11.
Since we're talking about dimensions here, negative values won't do us much good. So we'll go with the positive value, W ≈ 9.11 inches.
(That's a pretty specific approximation, isn't it?)
Now that we have the width, we can find the height by substituting it into our expression: H = (2W - 1).
H = 2 * 9.11 - 1
H ≈ 17.22 inches.
So, the dimensions of the poster are approximately 9.11 inches by 17.22 inches.
Voila! We've cracked the code and solved the puzzle.
Let's assume the width of the poster is "w" inches.
According to the given information, the height of the poster is 1 inch less than twice its width. So, the height can be represented as: (2w - 1) inches.
The area of a rectangle is given by the formula A = length × width. In this case, the length is the height of the poster (2w - 1) inches, and the width is w inches. Therefore, we can set up the equation:
A = (2w - 1) × w
Substituting the given area value of 190 square inches:
190 = (2w - 1) × w
Now, let's solve this equation step by step:
Expand the equation:
190 = 2w^2 - w
Rearrange the equation to move all terms to one side:
0 = 2w^2 - w - 190
At this point, we have a quadratic equation. To solve for "w," we need to factorize or use the quadratic formula. Factoring this equation might not be straightforward or possible, so let's use the quadratic formula:
w = [-b ± sqrt(b^2 - 4ac)] / 2a
In the given equation, a = 2, b = -1, and c = -190.
w = [-(-1) ± sqrt((-1)^2 - 4 × 2 × -190)] / (2 × 2)
Simplifying:
w = [1 ± sqrt(1 + 1520)] / 4
w = [1 ± sqrt(1521)] / 4
Now, let's calculate the values:
w = (1 + sqrt(1521)) / 4 ≈ 10.8 inches (rounded to one decimal place)
w = (1 - sqrt(1521)) / 4 ≈ -9.8 inches (rounded to one decimal place)
Since measurements cannot be negative, we disregard the negative value for the width.
Therefore, the width of the poster is approximately 10.8 inches.
To find the height, we substitute the value of "w" into the expression for the height:
Height = 2w - 1
Height = 2(10.8) - 1
Height ≈ 21.6 - 1
Height ≈ 20.6 inches (rounded to one decimal place)
So, the dimensions of the rectangular poster are approximately 10.8 inches (width) and 20.6 inches (height).
To find the dimensions of the rectangular poster, we can set up an equation based on the given information.
Let's assume the width of the poster is x inches.
According to the problem, the height of the poster is 1 inch less than twice its width, which can be written as (2x - 1) inches.
The area of a rectangle is given by the formula: Area = Length x Width.
In this case, the area is 190 square inches.
Therefore, we have the equation: x * (2x - 1) = 190.
Now we can solve this equation to find the value of x, which will give us the width of the poster.
Expanding the equation, we get: 2x^2 - x = 190.
Rearranging the equation, we have: 2x^2 - x - 190 = 0.
Solving this quadratic equation, we can either factorize it, complete the square, or use the quadratic formula. Let's use the quadratic formula to find the value of x:
x = (-b ± √(b^2 - 4ac)) / (2a).
In our case, a = 2, b = -1, and c = -190.
Substituting these values into the quadratic formula, we get:
x = (-(-1) ± √((-1)^2 - 4(2)(-190))) / (2(2)).
Simplifying further:
x = (1 ± √(1 + 1520)) / 4.
x = (1 ± √1521) / 4.
Since we're dealing with a measurement, we can ignore the negative result:
x = (1 + √1521) / 4.
Calculating this further:
x = (1 + 39) / 4.
x = 40 / 4.
x = 10.
So the width of the poster is 10 inches.
To find the height, we can substitute this value of x back into the expression for the height: (2x - 1).
Height = 2x - 1 = 2(10) - 1 = 20 - 1 = 19 inches.
Therefore, the dimensions of the rectangular poster are 10 inches by 19 inches.