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The figure above shows a drop-leaf table. With all four leaves down th

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Bunuel
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The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   26 Apr 2019, 07:05
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The figure above shows a drop-leaf table. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?

A. 1/2

B. \(\frac{\sqrt{2}}{2}\)

C. 1

D. \(\sqrt{2}\)

E. 2


PS26602.01
Quantitative Review 2020 NEW QUESTION


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B
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   23 May 2019, 06:41
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Solution


Given
In this question, we are given
    • The figure of a drop-leaf table.
    • With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle.

To Find
We need to determine
    • The radius, in meters, of the tabletop when all four leaves are up.

Approach & Working
When all four leaves are up, the table will look like the following:


In this case, the diameter of the circle is equal to the diagonal of the table.
    • As the square table has a side length of 1m, its diagonal will be √2 m.
    • Hence, diameter of the circular table = √2 m.
    • Therefore, radius of the table = (√2)/2 m

Hence, the correct answer is option B.

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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   26 Apr 2019, 12:38
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Bunuel wrote:

The figure above shows a drop-leaf table. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?

A. 1/2

B. \(\frac{\sqrt{2}}{2}\)

C. 1

D. \(\sqrt{2}\)

E. 2



PS26602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
2019-04-26_1803.png


We can use simple geometric rules to solve this problem, a Precise approach.
The radius of the circle is half its diameter. Since the diameter of the circle is just the diagonal of the (square) tabletop, and since the diagonal of a square with a side of 1 is sqrt(2) (remember your 45-45-90 triangle!), then our answer is 1/2 * sqrt(2).

(B) is our answer.
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   28 Apr 2019, 04:49
diagonal of square= √2
and diamater = √2
so when sides are up radius √2/2
IMO B


Bunuel wrote:

The figure above shows a drop-leaf table. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?

A. 1/2

B. \(\frac{\sqrt{2}}{2}\)

C. 1

D. \(\sqrt{2}\)

E. 2


PS26602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
2019-04-26_1803.png
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   22 May 2019, 22:44
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DavidTutorexamPAL wrote:
Bunuel wrote:

The figure above shows a drop-leaf table. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?


A. 1/2

B. \(\frac{\sqrt{2}}{2}\)

C. 1

D. \(\sqrt{2}\)

E. 2



PS26602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
2019-04-26_1803.png


We can use simple geometric rules to solve this problem, a Precise approach.
The radius of the circle is half its diameter. Since the diameter of the circle is just the diagonal of the (square) tabletop, and since the diagonal of a square with a side of 1 is sqrt(2) (remember your 45-45-90 triangle!), then our answer is 1/2 * sqrt(2).

(B) is our answer.


I can't visualize the question.
Please explain with drawing. Thanks

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DavidTutorexamPAL
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   23 May 2019, 10:30
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shaonkarim wrote:
DavidTutorexamPAL wrote:
We can use simple geometric rules to solve this problem, a Precise approach.
The radius of the circle is half its diameter. Since the diameter of the circle is just the diagonal of the (square) tabletop, and since the diagonal of a square with a side of 1 is sqrt(2) (remember your 45-45-90 triangle!), then our answer is 1/2 * sqrt(2).

(B) is our answer.


I can't visualize the question.
Please explain with drawing. Thanks

Posted from my mobile device

Hey shaonkarim

See the attached drawing.
BD is the diameter of the circle and also the diagonal of the square.
Since the side of the square is 1, then BD is sqrt(2) so the radius is 1/2 * sqrt(2)

Hope this helps!
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   27 May 2019, 18:58
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Bunuel wrote:

The figure above shows a drop-leaf table. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?

A. 1/2

B. \(\frac{\sqrt{2}}{2}\)

C. 1

D. \(\sqrt{2}\)

E. 2


PS26602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
2019-04-26_1803.png


We see that when the leaves are fully extended, the diameter of the circle is the diagonal of the square. Since the side of the square is 1, then its diagonal is √2. Therefore, the diameter is also √2, and hence the radius is √2/2.

Answer: B
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   12 Sep 2019, 23:55
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The sooner you realise the shape is just a square inscribed in a circle, the easier your life becomes.
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The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   20 Nov 2020, 00:41
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Bunuel wrote:

The figure above shows a drop-leaf table. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?

A. 1/2

B. \(\frac{\sqrt{2}}{2}\)

C. 1

D. \(\sqrt{2}\)

E. 2


PS26602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
The attachment 2019-04-26_1803.png is no longer available


Actually, this is a square inscribed in a circle.

We need to figure out the diagonal of the square and half the diagonal will be the radius of the circle.

Side of the Square is \(1\) so, the diagonal will be \(1\sqrt{2}\) \(= \) \(\sqrt{2}\)

Thus the radius is \(\frac{\sqrt{2}}{2}\)

The answer is B.
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The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   13 May 2022, 11:52
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Bunuel wrote:

The figure above shows a drop-leaf table. With all four leaves down the tabletop is a square, and with all four leaves up the tabletop is a circle. What is the radius, in meters, of the tabletop when all four leaves are up?

A. 1/2

B. \(\frac{\sqrt{2}}{2}\)

C. 1

D. \(\sqrt{2}\)

E. 2


PS26602.01
Quantitative Review 2020 NEW QUESTION


Show Spoiler
Attachment:
2019-04-26_1803.png



Ballparking! Yay!

The radius is half of the diagonal of the square.

Is that longer than the side, the same as the side, or shorter than the side? Shorter. Cool, eliminate C, D, and E.
Is it half as long as the side (if it were, then the side and the diagonal would be equal, which they aren't)? Nahhhh. Eliminate A.

Answer choice B.

No math.

ThatDudeKnowsBallparking
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink] New post   09 Mar 2023, 21:20
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Diagonal of the square is equal to the diameter of the circle. Since each side of square =1, diagonal = root2*side.
Now a circle inscribing a square share a relationship wherein, diagonal of the square is equal to diameter of the circle.
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Re: The figure above shows a drop-leaf table. With all four leaves down th [#permalink]
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