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PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR = OS. Then, the ratio of area of the circle to the area of the square is :
A. Pi/3
B. 11/7
C. 3/Pi
D. 4/Pi
E. 7/11
Attachment:
Screenshot_20191205-134829~2.png
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05 Dec 2019, 03:11
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ab965
PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR = OS. Then, the ratio of area of the circle to the area of the square is :
A. Pi/3
B. 11/7
C. 3/Pi
D. 4/Pi
E. 7/11
Attachment:
The attachment Screenshot_20191205-134829~2.png is no longer available
Step-by-step explanation:
We are given that PQRS is a square and a circle with centre O
We are given that SR is a tangent to the circle and TR=OR
Let OR =x
OR is a radius of a given circle
Then TR=x
OR=OT+TR=x+x=2x
All radius of a circle are equal
We know that a radius is perpendicular to the tangent line
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Ansh777
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Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
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05 Dec 2019, 04:12
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Given TR = OS= r
Also OS= OT since both are radius of the circle
Consider Traingle OSR
OR=2r OS= r and traingle is right angled traingle.
by pythagoras theorem
RS= Side of the square = \(\sqrt{3 }\)r
area of circle = Pie* r^2
Area of square= (\(\sqrt{3 }\)r)^2
ratio = pie/ 3
Answer : A
Also OS= OT since both are radius of the circle
Consider Traingle OSR
OR=2r OS= r and traingle is right angled traingle.
by pythagoras theorem
RS= Side of the square = \(\sqrt{3 }\)r
area of circle = Pie* r^2
Area of square= (\(\sqrt{3 }\)r)^2
ratio = pie/ 3
Answer : A
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
