A rectangle has a perimeter of 23cm.
Its area is 33 cm*2. Determine the dimensions of the rectangle.
2nd answer:
P=2x +2y=23
x+y=11.5
y=11.5-x
x(11.5-x)= 33
11.5x -x^2=33
(Multiply by 2 to get rid of decimal?)
23x-2x^2=66
23x-2x^2+66=0
(x-6)(2x-11)=0
Sub 6 into y= 11.5-x=11.5-6=5.5
Sub 5.5, 5.5=11.5-x, x=11.5-5.5=6
Therefore the dimensions of the rectangle are 6cm by 5.5cm. CORRECT?
but isnt 66 supposed to turn into -66??
L=11.5-w
Well, if we imagine that the rectangle is a magic trick, it has decided to play hide and seek with its dimensions. But fear not, because I, the mighty Clown Bot, shall find them!
Let's say the length of the rectangle is L and the width is W. The perimeter is the sum of all four sides, so we have 2L + 2W = 23.
Now, the area of a rectangle is calculated by multiplying the length and width, so we have L * W = 33.
Since we want to determine the dimensions, we'll need to solve these two equations simultaneously. But fear not, math is no match for my clowning abilities!
Let's solve for L in terms of W using the perimeter equation: L = (23 - 2W) / 2.
Substituting this into the area equation, we get ((23 - 2W) / 2) * W = 33.
Now, let's simplify: (23 - 2W)W / 2 = 33.
Multiplying both sides by 2, we get (23 - 2W)W = 66.
Expanding the equation, we have 23W - 2W^2 = 66.
Rearranging terms, we get 2W^2 - 23W + 66 = 0.
Now, it's time to solve for W using the quadratic formula! But first, let me grab my clown calculator...
Calculating... Voilà! We have two possible values for W: W = 3 or W = 11/2.
Now let's substitute these values back into the length equation: L = (23 - 2W) / 2.
For W = 3, we get L = (23 - 2 * 3) / 2 = 17 / 2.
For W = 11/2, we get L = (23 - 2 * 11/2) / 2 = 1.
So, the possible dimensions of the rectangle are either 17/2 cm by 3 cm or 1 cm by 11/2 cm.
Remember, math problems may sometimes make us feel like we're juggling too many numbers, but with a little humor and perseverance, we can solve them!
To determine the dimensions of the rectangle, we need to use the information given about the perimeter and area of the rectangle. Let's call the length of the rectangle "l" and the width of the rectangle "w".
1. Perimeter of a rectangle: The formula for the perimeter of a rectangle is: P = 2l + 2w, where P represents the perimeter.
According to the information given, the perimeter (P) of the rectangle is 23 cm. Therefore, we can write the equation as:
23 cm = 2l + 2w
2. Area of a rectangle: The formula for the area of a rectangle is: A = l * w, where A represents the area.
According to the information given, the area (A) of the rectangle is 33 cm^2. Therefore, we can write the equation as:
33 cm^2 = l * w
Now we have two equations:
1) 23 cm = 2l + 2w
2) 33 cm^2 = l * w
We can solve these equations to find the dimensions of the rectangle.
Let's rearrange the first equation to solve for one variable, say l:
2l = 23 cm - 2w
l = (23 cm - 2w) / 2
Now substitute the value of l in the second equation:
33 cm^2 = [(23 cm - 2w) / 2] * w
Simplify and solve this quadratic equation for the width (w) using factoring or the quadratic formula. Once you find the value of w, substitute it back into the equation for l to find its value.
Note: Since you mentioned that the rectangle has dimensions, we can assume that the length and width are positive numbers.
2L + 2W = 23
L= 11.5 - w
Area=33=l(w)
=(11.5-w)(w)
=11.5w - w*2
h=-11.5/2(1)
=5.75
L=11.5-5.75=5.75
Dimensions are 5.75m by 5.75m
So this would make it a square?