The perimeter of rectangle P is 12 feet. The perimeter of rectangle Q is 18 feet. Both rectangles have the same area.Find the area and dimensoin of each rectangle.
1 by 8 is the dimensions I drew them both and divided the perimeters and then substitued starting from 1 and it worked the first time, but yeah the answer is 8
P= L+W+L+W
P = 8+1+8+1
P = 9 + 9 = 18
A = L * W
A = 8 * 1
A = 8
P = 4 + 2 +4 + 2
P = 6 + 6 = 12
A = 4 * 2
A = 8
Both perimeters Area is 8
Both perimeters Area is 8 square feet
A=8
8(2)+1(2)=16+2=18
A=8*1=8
4(2)+2(2)=8+4=12
A=4*2=8
Perimeters 18 and 12
I don't think there is an answer to this. ( I got it for homework!)
To find the area and dimensions of each rectangle, let's start with the perimeter.
Perimeter of a rectangle is given by the formula: 2(L + W), where L is the length and W is the width.
For rectangle P, the perimeter is given as 12 feet. So we can write the equation: 2(Lp + Wp) = 12.
Similarly, for rectangle Q, the perimeter is given as 18 feet. So we can write the equation: 2(Lq + Wq) = 18.
Now, we are given that both rectangles have the same area. The formula for the area of a rectangle is A = L * W.
Let's find the area of rectangle P:
A = Lp * Wp.
Let's also find the area of rectangle Q:
A = Lq * Wq.
Since the areas are the same for both rectangles, we can set up an equation:
Lp * Wp = Lq * Wq.
Now we have a system of equations:
1. 2(Lp + Wp) = 12
2. 2(Lq + Wq) = 18
3. Lp * Wp = Lq * Wq
To solve this system of equations, we can use substitution.
From equation 1, we can solve for Lp + Wp:
Lp + Wp = 6.
Similarly, from equation 2, we can solve for Lq + Wq:
Lq + Wq = 9.
Now, let's express Lp in terms of Wp using equation 3:
Lp = (Lq * Wq) / Wp.
Substitute this value of Lp in equation 1:
((Lq * Wq) / Wp) + Wp = 6.
Simplify equation 4:
Lq + (Wq^2) / (Wp) = 6.
Using equation 2, we know that Lq + Wq = 9, so:
9 + (Wq^2) / (Wp) = 6.
Rearranging this equation, we get:
(Wq^2) / (Wp) = 6 - 9 = -3.
Now, let's consider the case where Wp is not equal to 0. In that case, we can divide equation 5 by Wp:
(Wq^2) / (Wp^2) = -3 / Wp.
Now, cross multiply the equation:
Wq^2 * Wp = -3.
Simplify this equation:
Wq^2 = -3/Wp.
Since we know that both length and width cannot be negative, we can conclude that this equation is not possible.
This means that the case where Wp is not equal to 0 is not possible. Hence, Wp must be equal to 0.
From equation 6, we know that (Wq^2) / (Wp) = -3, substituting Wp = 0, we get:
(Wq^2) / 0 = -3.
This is not possible, and since there is no solution, it means that there is an error in the problem statement or the given information.
Therefore, we cannot determine the area and dimensions of rectangles P and Q based on the given information.
1x8 and 5x1 don't have the same area. Wait for someone else to answer, but I am trying to figure out, logically, how they could possibly have the same area.
Eight is not a logical answer. I hope you have figured it out by now... 8+8=16. Even dividing 18 would not come to an answer of 8, it would come to an answer of 9.
Neither triangle has the same area. That answer is wrong.