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Carcass
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x > 0 [#permalink] New post   17 Dec 2017, 12:52
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\(x > 0\)

Quantity A
Quantity B
The area of a square region with diagonal of length \(\sqrt{2} x\)
The area of a circular region with diameter of length x


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.
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A
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Popo
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Re: x > 0 [#permalink] New post   Updated on: 18 Dec 2017, 18:41
The diagonal of a square can be found by the following formula : Squroot 2 * side
Since we already know the length = Squroot 2 * side = Squroo 2 * x
We can divide the square root and we have side = x. So the area is x^2

For Quantity B we know that the diameter is x, so the radii must be x/2 and for the area it is multiplied by pi.

Lets plug some values.
For x = 2.
QA = Area is equal to 4
QB= Area is 3,14
QA is greater than B.

Lets try x =36
QA = 36
QB = 9,42

But what is bizarre it that when I plug 1 B is greater

Originally posted by Popo on 18 Dec 2017, 01:59.
Last edited by Popo on 18 Dec 2017, 18:41, edited 1 time in total.
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Re: x > 0 [#permalink] New post   18 Dec 2017, 11:48
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The answer is A
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Re: x > 0 [#permalink] New post   18 Dec 2017, 18:41
Carcass wrote:
The answer is A


Can I ask you what is the right reasoning?
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Re: x > 0 [#permalink] New post   19 Dec 2017, 13:35
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Popo wrote:
Carcass wrote:
The answer is A


Can I ask you what is the right reasoning?


When you plug 1. The Quanity A is \(x^2=1\)
and Quantiy B is \(\frac{3.14}{4}x^2 = 0.785\).

Hence quanity A is always greater.
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Re: x > 0 [#permalink] New post   19 Dec 2017, 21:30
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Carcass wrote:
\(x > 0\)

Quantity A
Quantity B
The area of a square region with diagonal of length \(\sqrt{2} x\)
The area of a circular region with diameter of length x


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.



Here let us put statement 1

Area of the square = \(\frac{{diagonal^2}}{2}\)

or \((\sqrt{2}x)^2\) * \(\frac{1}{2}\)
or \(\frac{2x^2}{2}\)

From statement 2

Area of circle = \(\pi * r^2\)

or \(\frac{22}{7}\) * \((\frac{Diameter}{2})^2\)

or \(\frac{22x}{28}\)


From the two results statement 1 > statement 2 for any positive value of x
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Re: x > 0 [#permalink] New post   04 Jun 2018, 19:58
sandy wrote:
Popo wrote:
Carcass wrote:
The answer is A


Can I ask you what is the right reasoning?


When you plug 1. The Quanity A is \(x^2=1\)
and Quantiy B is \(\frac{3.14}{4}x^2 = 0.785\).

Hence quanity A is always greater.

For quantity B, why are we dividing by 4? When a diameter is given, we divide by 2 to get the radius, we should divide x by 2 to get the radius, square it and then multiply by the value of pi
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Re: x > 0 [#permalink] New post   05 Jun 2018, 12:51
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@Emike56

You are right if \(x\) is the diameter then radius of circle is \(\frac{x}{2}\).

Area of the circle is \(\pi \times radius^2= \pi \times (\frac{x}{2})^2=\pi \times \frac{x^2}{2^2}= \pi \times \frac{x^2}{4}\).

Hope this explains the 4.
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Re: x > 0 [#permalink] New post   19 Jan 2019, 14:30
Ans A
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Re: x > 0 [#permalink] New post   19 Jan 2019, 18:53
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Carcass wrote:
\(x > 0\)

Quantity A
Quantity B
The area of a square region with diagonal of length \(\sqrt{2} x\)
The area of a circular region with diameter of length x


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.


The area of a square region with diagonal of length \(\sqrt{2} x\).
If the diagonal is \(x\sqrt{2}\), the sides are x as \(side^2+side^2=(x\sqrt{2})^2=2x^2=2side^2....side=x\)
so area = \(side^2=x^2=A\)

The area of a circular region with diameter of length x
so radius = x/2.
area = B = \(\pi*(\frac{x}{2})^2=\pi*\frac{x^2}{4}=\frac{22}{28}*x^2=\frac{11}{14}*A\).

Thus A is greater
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Re: x > 0 [#permalink] New post   30 Aug 2021, 03:06
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Re: x > 0 [#permalink]
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