GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2019, 01:01

### GMAT 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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58427
In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

### Show Tags

26 Feb 2019, 01:20
2
2
00:00

Difficulty:

65% (hard)

Question Stats:

34% (02:16) correct 66% (02:21) wrong based on 27 sessions

### HideShow timer Statistics

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 623 times ]

_________________
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5003
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

### Show Tags

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

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.
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 269
In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

### Show Tags

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

### Show Tags

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)

Answer: E
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 269
Re: In the figure above, a circle with center O is inscribed in square WXY  [#permalink]

### Show Tags

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

oh my bad!

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

E can be the answer

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

### Show Tags

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
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne