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The angle in a semi-circle is a right an...

The angle in a semi-circle is a right angle.

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To prove that the angle in a semicircle is a right angle, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Circle and Points**: Let’s consider a circle with center O. We will take points P and Q on the circumference of the circle such that the line segment PQ is a diameter of the circle. Let R be any point on the circumference of the circle. 2. **Understand the Angles**: ...
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RD SHARMA-CIRCLE -All Questions
  1. Any angle subtended by a minor arc in the alternate segment is acute ...

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  2. The arc of a circle subtending a right angle at any point of the circl...

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  3. The angle in a semi-circle is a right angle.

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  4. Angle in the same segment of a circle are equal.

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  5. Prove that the circle drawn on any one of the equal sides of an isos...

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  6. A chord of a circle is equal to the radius of the circle find the angl...

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  7. In Figure., A B C is a triangle in which /B A C=30^0dot Show that B C ...

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  8. Theorem:-7 If the line segment joining two points subtends equal angle...

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  9. In Figure, O A and O B ar respectively perpendiculars to chords C D an...

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  10. Draw the graph of the fuction 4x+3y=12,at what pt it cut the coordinat...

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  11. If two equal chords of a circle in intersect within the circle, prove ...

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  12. prove that the line joining the mid-point of two equal chords of a c...

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  13. Show that if two chords of a circle bisect one another they must be ...

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  14. In Figure, equal chords A B and C D of a circle with centre O , cut at...

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  15. Two equal circles intersect in P and Q . A straight line through P mee...

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  16. Prove that all the chords of a circle through a given point within it,...

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  17. Prove that the diameter is the greatest chord in a circle.

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  18. Bisector A D of /B A C of A B C passes through the centre O of the ci...

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  19. If two sides of a cyclic quadrilateral are parallel, prove that the ...

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  20. The quadrilateral formed by angle bisectors of a cyclic quadrilateral ...

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