The area of a certain rectangle is 288yd^2.the perimeter is 68yd.if you double the lengths and widths,what will be the new area and perimeter of the new rectangle be?
p doubles
a grows by 2^2 = 4x
Whatever the original surface may be, if you double the length of a rectangle the surface will also be doubled. Just the same for the width.
So doing both the operations, will multiply the surface by 2x2 = 4
Jake ate 1/6 of the cookies. If he ate 5 cookies, How many cookies were there in all?
To solve this question, let's start by denoting the length of the rectangle as "l" and the width as "w". We are given two pieces of information:
1. The area of the rectangle is 288yd²: Area = l * w = 288yd².
2. The perimeter of the rectangle is 68yd: Perimeter = 2(l + w) = 68yd.
From the information given, we have two equations:
1. lw = 288 (Equation 1)
2. 2(l + w) = 68 (Equation 2)
Now, let's solve these equations to find the values of l and w.
From Equation 2, we can simplify it as follows:
2(l + w) = 68
l + w = 34 (Divide both sides by 2)
We can rewrite Equation 1 as:
lw = 288
l = 288/w (Divide both sides by w)
Substitute the value of l from Equation 1 into Equation 2:
(288/w) + w = 34
Now, let's solve this quadratic equation to find the value of "w".
Rearranging the equation:
w² - 34w + 288 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Factoring gives us:
(w - 18)(w - 16) = 0
So we get two possible values for "w":
w = 18 or w = 16
If w = 18, we can substitute it into Equation 2 to find the value of l:
l + 18 = 34
l = 16
If w = 16, we get:
l + 16 = 34
l = 18
So, we have two sets of dimensions for the original rectangle:
Case 1: Length = 16yd, Width = 18yd
Case 2: Length = 18yd, Width = 16yd
Now, let's move on to the second part of the question: doubling the lengths and widths.
Case 1:
For the rectangle with length = 16yd and width = 18yd, if we double both dimensions, we get:
New length = 2 * 16yd = 32yd
New width = 2 * 18yd = 36yd
New area = new length * new width = 32yd * 36yd = 1152yd²
New perimeter = 2 * (new length + new width) = 2 * (32yd + 36yd) = 2 * 68yd = 136yd
Case 2:
For the rectangle with length = 18yd and width = 16yd, if we double both dimensions, we get:
New length = 2 * 18yd = 36yd
New width = 2 * 16yd = 32yd
New area = new length * new width = 36yd * 32yd = 1152yd²
New perimeter = 2 * (new length + new width) = 2 * (36yd + 32yd) = 2 * 68yd = 136yd
Therefore, regardless of which case we consider, the new area of the rectangle will be 1152yd², and the new perimeter will be 136yd.