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What is the area of a rectangle whose length is twice its wi
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31 Jan 2019, 15:53
31 Jan 2019, 15:53
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What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?
A) \(1\)
B) \(6\)
C) \(\frac{2}{3}\)
D) \(\frac{4}{3}\)
E) \(\frac{8}{9}\)
A) \(1\)
B) \(6\)
C) \(\frac{2}{3}\)
D) \(\frac{4}{3}\)
E) \(\frac{8}{9}\)
Re: What is the area of a rectangle whose length is twice its wi
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23 Aug 2019, 01:09
23 Aug 2019, 01:09
Expert Reply
The key information here from which is good to start is that
whose perimeter is equal to that of a square whose area is 1
PR (perimeter rectangle) = PS (perimeter square)
The area of the square is 1. Now to have a square area = 1, one side of the square must be 1 as well. Area square is \(1^2=1\)
So its perimeter is \(1 \times 4 = 4\); \(PS = 4\)
The area of a rectangle is \(PR = \ell \times w\)
In our case, the length is twice its width; \(PR = \ell + 2w + 2we + \ell = 2\ell + 4w\)
\ell = w
we do have that PR = 6w
\(PR=PS >>>>>>> 4 = 6w >>> w=\frac{4}{6} = \frac{2}{3}\)
\(2w = \frac{4}{3}\)
Area rectangle \(\ell \times w = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}\)
whose perimeter is equal to that of a square whose area is 1
PR (perimeter rectangle) = PS (perimeter square)
The area of the square is 1. Now to have a square area = 1, one side of the square must be 1 as well. Area square is \(1^2=1\)
So its perimeter is \(1 \times 4 = 4\); \(PS = 4\)
The area of a rectangle is \(PR = \ell \times w\)
In our case, the length is twice its width; \(PR = \ell + 2w + 2we + \ell = 2\ell + 4w\)
\ell = w
we do have that PR = 6w
\(PR=PS >>>>>>> 4 = 6w >>> w=\frac{4}{6} = \frac{2}{3}\)
\(2w = \frac{4}{3}\)
Area rectangle \(\ell \times w = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}\)
Re: What is the area of a rectangle whose length is twice its wi
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05 Sep 2019, 08:38
05 Sep 2019, 08:38
1
Carcass wrote:
What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?
A) \(1\)
B) \(6\)
C) \(\frac{2}{3}\)
D) \(\frac{4}{3}\)
E) \(\frac{8}{9}\)
A) \(1\)
B) \(6\)
C) \(\frac{2}{3}\)
D) \(\frac{4}{3}\)
E) \(\frac{8}{9}\)
Let length be l and breadth be b
Given that L=2B
For Square; side =1
so, its perimeter will be 4
Thus, 2(L+B)=4; which simplifies to B=2/3 (After plugging in L=2B)
Which implies; => L=4/3
Thus Area of rectangle=LxB => 8/9
