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# In the figure above, a circle with center O is inscribed in square WXY

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In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

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26 Feb 2019, 01:20
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Difficulty:

55% (hard)

Question Stats:

29% (01:39) correct 71% (02:12) wrong based on 22 sessions

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In the figure above, a circle with center O is inscribed in square WXYZ. If the circle has radius 3, then PZ =

A. 6

B. $$3\sqrt{2}$$

C. $$6 + \sqrt{2}$$

D. $$3 +\sqrt{3}$$

E. $$3\sqrt{3}$$

Attachment:

2019-02-26_1209_001.png [ 32.21 KiB | Viewed 381 times ]

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In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

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Updated on: 26 Feb 2019, 08:12
1
square XYWZ has each side 6
and diagonal length = 6√2
so OZ = 6√2/2 = 3√2
for triangle OPZ we can say
pz^2= oz^2+op^2
pz^2 = (3√2)^2+ (3)^2
pz^2 = 18+9
pz=3√3
IMO E

Bunuel wrote:

In the figure above, a circle with center O is inscribed in square WXYZ. If the circle has radius 3, then PZ =

A. 6

B. $$3\sqrt{2}$$

C. $$6 + \sqrt{2}$$

D. $$3 +\sqrt{3}$$

E. $$3\sqrt{3}$$

Attachment:
2019-02-26_1209_001.png

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Originally posted by Archit3110 on 26 Feb 2019, 02:44.
Last edited by Archit3110 on 26 Feb 2019, 08:12, edited 1 time in total.
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Posts: 278
In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

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26 Feb 2019, 07:10
1
Bunuel wrote:

In the figure above, a circle with center O is inscribed in square WXYZ. If the circle has radius 3, then PZ =

A. 6

B. $$3\sqrt{2}$$

C. $$6 + \sqrt{2}$$

D. $$3 +\sqrt{3}$$

E. $$3\sqrt{3}$$

Attachment:
2019-02-26_1209_001.png

okay

I will go with option "A"

since the circle has radius 3, each side of the square measures 6, so far so good

so the diagonal of the square is 6 root over 3

now OZ = (6 root over 3) / 2, and

(PZ)^2 = (3 root over 3)^ 2 + 3^2

thus PZ = 6

So I think the answer can be A

thanks
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Posts: 2
Re: In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

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26 Feb 2019, 07:49
1
By doing the calculations, I believe the answers that have been given above are incorrect, I will therefore post my answer to this problem.

As mentioned above, knowing radius is 3, we can quickly realise that the square has length= 3x2 = 6.

Therefore, being a square and knowing the COMMON RATIOS of a 90-45-45 triangle we can derive the length of the diagonal of the square.

Diagonal = length of side *sqrt(2), therefore Diagonal = 6*sqrt(2)

To find PZ we would need to know half of that distance, hence (6*sqrt(2) )/ 2 = 3 sqrt(2)

Then, we realise PO and OZ are perpendicular, meaning that angle ZOP is a right angle (90°), we can therefore apply the pythagorean theorem.

PZ^2 = OZ^2 + PO^2

PZ^2= (3sqrt(2))^2 + 3^2

PZ^2= 9*2 + 9

PZ^2= 27

PZ = sqrt(27)

27 can be expressed as 3 times 3, which is 3^3, or, alternatively, 3^2 * 3. We therefore take out the 3^2 from the square root, and PZ will result in:

PZ= 3 sqrt(3)

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Re: In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

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26 Feb 2019, 08:02
testcracker wrote:
Bunuel wrote:

In the figure above, a circle with center O is inscribed in square WXYZ. If the circle has radius 3, then PZ =

A. 6

B. $$3\sqrt{2}$$

C. $$6 + \sqrt{2}$$

D. $$3 +\sqrt{3}$$

E. $$3\sqrt{3}$$

Attachment:
2019-02-26_1209_001.png

okay

I will go with option "A"

since the circle has radius 3, each side of the square measures 6, so far so good

so the diagonal of the square is 6 root over 3

now OZ = (6 root over 3) / 2, and

(PZ)^2 = (3 root over 3)^ 2 + 3^2

thus PZ = 6

So I think the answer can be A

thanks

the diagonal of the square is 6 root over 2

and OZ = (6 root over 2) /2, and

(PZ)^2 = (3 root over 2)^2 + 3^2

so PZ = 3 root over 3

thanks
Intern
Joined: 12 Nov 2018
Posts: 2
Re: In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

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26 Feb 2019, 08:07
I believe it is misleading to write 'over', as in common english, 'over' means 'divided by'.

Instead, what you mean is 'times', which means 'multiplied'.

hence, 6 over root 2 would mean 6 / sqrt(2), which is not the case.

6 TIMES sqrt(2) is instead equals to = 6 *sqrt(2) which is the case in questions.

It was just a matter of math language I guess..
Re: In the figure above, a circle with center O is inscribed in square WXY   [#permalink] 26 Feb 2019, 08:07
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# In the figure above, a circle with center O is inscribed in square WXY

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