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

Question

https://gmatclub.com/forum/download/file.php?id=46654 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} 2019-02-26_1209_001.png

Archit3110 · Answer

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

testcracker · Answer

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

stecudi · Answer

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

testcracker · Answer

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

stecudi · Answer

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..

Fdambro294 · Answer

Connect P to Z

Then connect Z to center O

This creates triangle POZ

Since a Circle Inscribed in a Square shares the geometric center with the Square, the Side ZO connecting the Vertex Z to the Center of the Circle/Square 0 will lie on top of the Diagonal of the Square.

As such, ZO will be (1/2) the Diagonal of the Square

Since the Radius of the Inscribed circle is 3, the Side of the Circumscribed Square will be = (2) (r) = (2) (3) = 6

ZO = (1/2) * (Diagonal of Square with side 6) = (1/2) * (6) * sqrt(2)

(3) * sqrt(2)

Furthermore, a property of squares is that the diagonals are perpendicular Bisectors of each other.

Since YW is the other Diagonal of the square that passes through point O, Side ZO will be perpendicular to Diagonal YW at center O

Thus, Triangle PZO is a right triangle.

Leg 1 = side ZO = (3) * sqrt(2)

Leg 2 = OP = radius of the circle = 3

hypotenuse = PZ = ?

You can then use the Pythagorean theorem to get the answer for Hypotenuse PZ

(E)

PZ = sqrt(3)

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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

Prep Toolkit

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