If the vertices of a quadrilateral PQRS lie on the circumference of

Question

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

Vips0000 · Answer

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.

EvaJager · Answer

(1) The side of a square inscribed in a given circle is  R\sqrt{2} , where  R  is the radius of the circle.
Since PS = R, the quadrilateral PQRS cannot be a square. 
Sufficient.

(2) The line segment QR subtends an arc of 90 degrees, therefore QR equals the side of the square which can be inscribed in the given circle.
In an inscribed square, each side subtends an arc of 360/4 = 90 degrees.
Since we have no information about the other sides of the quadrilateral, PQRS isn't necessarily a square.
Not sufficient.

Answer A.

ashishv045 · Answer

Thanks for the answer,

1) However, what if the statement 2 read - minor arc as 60 ?
In that case Is it sufficient to answer the question ?

2) Also, in case when the minor arc is 90,
Can't we use the vertical angles concept and prove that other angles too are 90 ? and subsequently 360 for a circular angle ?

jlgdr · Answer

OA says E. Is there any good reason for this?

Thanks
Cheers
J

Bunuel · Answer

The correct answer must be A, not E. Edited the OA. Thank you for reporting this.

mridupawandas · Answer

option A states that radius is equal to one side this itself confirms that it cant be a square as for square it has to be rt 2 times the radius.

Harsh9676 · Answer

Hi  Bunuel ,  chetan2u

For St 2..

We know that the diagonals are equal >> Can be a square, Rectangle, Isosceles trapezoid

Given the Angle of minor arc is 90. Meaning that the diagonals intersect perpendicular >> True for Rhombus, Square, Kite

So combined, we can infer that the quadrilateral is a square.

Isn't B sufficient?? What am i missing?

Rajeet123 · Answer

Quite some confusion, can an expert please solve this?

Binod826 · Answer

What if Statement 2 says the degree measure of Minor arc QR is 45 degree?

If the vertices of a quadrilateral PQRS lie on the circumference of

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