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A rectangle is inscribed in a circle of diameter 2. Which of

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A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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New post 04 Jan 2008, 00:23
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A rectangle is inscribed in a circle of diameter 2. Which of the following can be the area of the rectangle?

I. 0.01
II. 2.00
III. 3.20

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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Re: PS: Area of Rectangle  [#permalink]

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New post 03 Aug 2012, 07:17
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voodoochild wrote:
walker wrote:
Area of a rectangle ranges from 0 to area of square=2. S0, I and II is possible.


Bunuel / Walker,
I have a question. If the diameter = 2 => diagonal of square =2 => 2(side of a square)^2 = 2 => (side of a square)^2 = 1.

Hence, the area of square is 1!

Thoughts?


If the length of a diagonal of a square is 2, then \(side^2+side^2= diagonal^2\) --> \(2*side^2=2^2\) --> \(side^2=area=2\).

You could get the side in another way: since the angle between a diagonal and a side in a square is 45 degrees, then \(side=\frac{diagonal}{\sqrt{2}}\) (from the properties of 45-45-90 triangle).

You could also get the area directly: \(area_{square}=\frac{diagonal^2}{2}=2\).

Complete solution:
A rectangle is inscribed in a circle of diameter 2. Which of the following can be the area of the rectangle?

I. 0.01
II. 2.00
III. 3.20

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Look at the diagram below:
Attachment:
m24-05.png
m24-05.png [ 10.63 KiB | Viewed 8324 times ]

If the width of blue rectangle is small enough then its area could be 0.01.

Generally, the are of the inscribed rectangle is more than 0 and less than or equal to the area of the inscribed square (inscribed square has the largest area from all rectangles that can be inscribed in a given circle).

Now, since the area of the inscribed square in a circle with the diameter of 2 is 2, then the area of the inscribed rectangle is \(0<area\leq{2}\). So, I and II are possible values of the area. (Else you can notice that the area of the circle is \(\pi{r^2}=\pi\approx{3.14}\) and the area of the inscribed rectangle cannot be greater than that, so III is not possible)

Answer: D.
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Re: PS: Area of Rectangle  [#permalink]

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New post 04 Jan 2008, 01:54
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Area of a rectangle ranges from 0 to area of square=2. S0, I and II is possible.
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Re: PS: Area of Rectangle  [#permalink]

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New post 04 Jan 2008, 03:27
walker wrote:
Area of a rectangle ranges from 0 to area of square=2. S0, I and II is possible.



Simple enough now! Thanks a bunch Walker.
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Re: PS: Area of Rectangle  [#permalink]

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New post 04 Jan 2008, 06:38
GK_Gmat wrote:
walker wrote:
Area of a rectangle ranges from 0 to area of square=2. S0, I and II is possible.



Simple enough now! Thanks a bunch Walker.


moreover, the area of the circle is 3,14...thus 3,20 is the area of a figure which is not inscribed....
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Re: PS: Area of Rectangle  [#permalink]

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New post 03 Aug 2012, 07:02
walker wrote:
Area of a rectangle ranges from 0 to area of square=2. S0, I and II is possible.


Bunuel / Walker,
I have a question. If the diameter = 2 => diagonal of square =2 => 2(side of a square)^2 = 2 => (side of a square)^2 = 1.

Hence, the area of square is 1!

Thoughts?
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Re: A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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New post 03 Aug 2012, 07:13
diagonal of Square = a* root 2=2, Hence a (side of sq)= root 2.

Area of Square = root 2 X root 2 = 2
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Re: PS: Area of Rectangle  [#permalink]

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New post 03 Aug 2012, 07:15
voodoochild wrote:
walker wrote:
Area of a rectangle ranges from 0 to area of square=2. S0, I and II is possible.


Bunuel / Walker,
I have a question. If the diameter = 2 => diagonal of square =2 => 2(side of a square)^2 = 2 => (side of a square)^2 = 1.

Hence, the area of square is 1!

Thoughts?


Area of a square is also diagonal * diagonal / 2, so in our case is 2 * 2 / 2 =2.

If you want to find the side of the square, then don't forget to square: 2(side of a square)^2 = 2^2=4...

Be more careful with your computations. On the real test you won't get help from the forum's members :o)
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Re: PS: Area of Rectangle  [#permalink]

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New post 03 Aug 2012, 08:06
Bunuel wrote:
voodoochild wrote:
walker wrote:
Area of a rectangle ranges from 0 to area of square=2. S0, I and II is possible.


Bunuel / Walker,
I have a question. If the diameter = 2 => diagonal of square =2 => 2(side of a square)^2 = 2 => (side of a square)^2 = 1.

Hence, the area of square is 1!

Thoughts?


If the length of a diagonal of a square is 2, then \(side^2+side^2= diagonal^2\) --> \(2*side^2=2^2\) --> \(side^2=area=2\).

You could get the side in another way: since the angle between a diagonal and a side in a square is 45 degrees, then \(side=\frac{diagonal}{\sqrt{2}}\) (from the properties of 45-45-90 triangle).

You could also get the area directly: \(area_{square}=\frac{diagonal^2}{2}=2\).

Complete solution:
A rectangle is inscribed in a circle of diameter 2. Which of the following can be the area of the rectangle?

I. 0.01
II. 2.00
III. 3.20

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Look at the diagram below:
Image
If the width of blue rectangle is small enough then its area could be 0.01.

Generally, the are of the inscribed rectangle is more than 0 and less than or equal to the area of the inscribed square (inscribed square has the largest area from all rectangles that can be inscribed in a given circle).

Now, since the area of the inscribed square in a circle with the diameter of 2 is 2, then the area of the inscribed rectangle is \(0<area\leq{2}\). So, I and II are possible values of the area. (Else you can notice that the area of the circle is \(\pi{r^2}=\pi\approx{3.14}\) and the area of the inscribed rectangle cannot be greater than that, so III is not possible)

Answer: D.


The link to the figure is not working/missing...

Did you mean something like this? (See the attached figure)

We can make the area of the inscribed rectangle as close as possible to 0, when for example the length is getting closer, and closer to the diameter, while the width is getting closer and closer to 0.

The maximum area can be obtained when the height of the right triangle, which is half of the rectangle and it is inscribed in the half circle, is equal to the radius of the circle. In this case, the rectangle becomes a square.
Attachments

MAxRectangleAreaInCircle.jpg
MAxRectangleAreaInCircle.jpg [ 24.29 KiB | Viewed 8307 times ]


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Re: PS: Area of Rectangle  [#permalink]

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Re: A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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New post 03 Aug 2012, 08:45
Thanks Bunuel...I missed this problem because of a silly mistake. :( <sad> Thanks again.
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Re: A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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New post 12 Dec 2013, 12:45
2
A rectangle is inscribed in a circle of diameter 2. Which of the following can be the area of the rectangle?

If the circle has a diameter of 2, then it's radius is 1 so it's area is pi*1^2 or simply pi. Pi is equal to 3.1415. A a rectangle of .01 or 2.0 could fit inside this circle but not a rectangle with an area larger than the circle.

D

I. 0.01
II. 2.00
III. 3.20

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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Re: A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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New post 13 Dec 2013, 10:34
Rectangle could be a square in this problem? The correct answer is still D, but I'm just curious
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Re: A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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New post 14 Dec 2013, 05:46
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Re: A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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New post 03 Aug 2018, 12:14
GK_Gmat wrote:
A rectangle is inscribed in a circle of diameter 2. Which of the following can be the area of the rectangle?

I. 0.01
II. 2.00
III. 3.20

A. I only
B. II only
C. III only
D. I and II only
E. II and III only


Let \(x\) be the length & \(y\) be the width of the rectangle. Its diagonal will be the diameter of the circle, as the rectangle is inscribed in the circle.


Therefore length of diagonal = \(2\) & Area of the rectangle = \(xy\)

Hence we have \(x^2 + y^2 = 4\)

or \((x - y)^2 + 2xy = 4\)

or \((x - y)^2 = 4 - 2xy\)

LHS has to be positive or 0, hence \(4 - 2xy >=0\)

We can see that only I & II satisfy the above inequality.


Answer D.



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Re: A rectangle is inscribed in a circle of diameter 2. Which of  [#permalink]

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Re: A rectangle is inscribed in a circle of diameter 2. Which of   [#permalink] 25 Aug 2019, 11:52
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