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

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

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

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..

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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