joylive wrote:

If the vertices of a quadrilateral PQRS lie on the circumference of the circle, is PQRS a square?

1. Side PS is equal in length to the radius of the circle

2. The degree measure of the minor arc QR is 90 degrees

Can you explain this?

Would like to know what is the source of this problem and how OA is D? It should be solved as below:

First consider this,

if a square has all of its veritces on circumference of the circle and if radius of circle is r and side of square is x. then \(x= r*\sqrt{2}\)

This can be proved using pyth theorem.

Second diagonals are perpendicular to each other.

Check the image screenshot 1

Now lets go to statements 1 by 1.

Statement 1 says:

PS = radius (where PS is one of the sides)

Or basically x=r

however we know for a square \(x= r*\sqrt{2}\)

another way to look at same is - if side is equal to radius - all angles are 60 in the triangle POS. However, for a square angle POS should be 90.

Hence, we know that PQRS is not a square, as it doesnt meet even the basic requirement.

Hence statement is Sufficient

Statement 2 says:

QOR =90 Where O is center of circle.

Hence, using pyth theorem in triangle QOR,

QR^2 = OR^2 + OQ^2

or

x^2 = r^2 + r^2

or \(x= r*\sqrt{2}\)

that means, side length is fine. side QS meets minimum condition, but not sufficient to answer anything else since we dont know anything about rest of the sides or angles, it could be a quad as shown in image 2.

Hence statement 2 is not sufficient.

Ans is A.

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