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O is the center of the circle and OS=SQ.
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15 Jul 2021, 05:59
15 Jul 2021, 05:59
Expert Reply
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Question Stats:
42% (01:02) correct
57% (01:34) wrong based on 14 sessions
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GRE O is the center of the circle and OS=SQ..jpg [ 10.26 KiB | Viewed 1310 times ]
O is the center of the circle and OS=SQ.
Quantity A |
Quantity B |
PQ |
OR |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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O is the center of the circle and OS=SQ.
[#permalink]
15 Jul 2021, 08:15
15 Jul 2021, 08:15
1
Let's Join O and P as shown in the image below
image.jpg [ 9.84 KiB | Viewed 1264 times ]
In \(\triangle\)POR
-> OP = OR = Radius
=> \(\triangle\)POR is an isosceles triangle.
\(\angle\)OSP = \(90^{\circ}\) [ As \(\angle\)PSQ= \(90^{\circ}\) ]
And we know that, in an Isosceles Triangle the perpendicular drawn from the vertex tot he non equal side, bisects the side
=> OS will bisect PR
=> PS = SR
Now, consider \(\triangle\)PSQ and \(\triangle\)RSO
PS = SR (proved above)
OS = OQ (given)
And the included angle, \(\angle\)PSQ = \(\angle\)RSO = \(90^{\circ}\)
=> \(\triangle\)PSQ ≅ \(\triangle\)RSO [Congruent triangles, all corresponding sides and all corresponding angles will be equal]
=> PQ = OR
=> Quantity A = Quantity B
So, answer will C
Hope it helps!
Watch the following video to Learn Basics of Circles
Attachment:
image.jpg [ 9.84 KiB | Viewed 1264 times ]
In \(\triangle\)POR
-> OP = OR = Radius
=> \(\triangle\)POR is an isosceles triangle.
\(\angle\)OSP = \(90^{\circ}\) [ As \(\angle\)PSQ= \(90^{\circ}\) ]
And we know that, in an Isosceles Triangle the perpendicular drawn from the vertex tot he non equal side, bisects the side
=> OS will bisect PR
=> PS = SR
Now, consider \(\triangle\)PSQ and \(\triangle\)RSO
PS = SR (proved above)
OS = OQ (given)
And the included angle, \(\angle\)PSQ = \(\angle\)RSO = \(90^{\circ}\)
=> \(\triangle\)PSQ ≅ \(\triangle\)RSO [Congruent triangles, all corresponding sides and all corresponding angles will be equal]
=> PQ = OR
=> Quantity A = Quantity B
So, answer will C
Hope it helps!
Watch the following video to Learn Basics of Circles
gmatclubot
O is the center of the circle and OS=SQ. [#permalink]
15 Jul 2021, 08:15
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