GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
What is the area of a rectangle whose length is twice its wi
[#permalink]
31 Jan 2019, 15:53
31 Jan 2019, 15:53
Expert Reply
2
Bookmarks
00:00
Question Stats:
77% (01:30) correct
22% (01:47) wrong based on 75 sessions
Hide Show timer Statistics
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
[#permalink]
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
[#permalink]
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
