Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?

yes, it is true for minor segment since ar.minor segment

=

ar.sector-triangle made by the radii and the chord of the segment.

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Yes, it is only true for major segment since ar.minor segment

= -

ar.sector-triangle made by the radii and the chord of the segment

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No, it is only true for major segment since ar.minor segment
=ar.sector-triangle made by the radii and the chord of the segment.

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No, it is only true for minor segment since ar.minor segment
=ar.sector-triangle made by the radii and the chord of the segment.

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Solution

The correct option is D No, it is only true for minor segment since ar.minor segment
=ar.sector-triangle made by the radii and the chord of the segment.
$G i v e n - A B i s t h e c h o r d o f t h e c i r c l e w i t h c e n t r e O . T h e m i n o r s e g m e n t i s A B P & t h e m i n o r s e c t o r i s A O B P . A l s o t h e m a j o r s e g m e n t i s A B Q & t h e m a j o r s e c t o r i s A B Q . T o f i n d o u t - t h e v a l i d i t y o f t h e s t a t e m e n t t h a t a r . s e g m e n t < a r . s e c t o r . S o l u t i o n - I n t h e f i g u r e a r . m i n o r s e g m e n t A B P = a r . m i n o r s e c t o r A O B P - a r . Δ O A B ∴ a r . m i n o r s e g m e n t A B P < a r . m i n o r s e c t o r A O B P . A g a i n, i n t h e f i g u r e, a r . m a j o r s e g m e n t A B Q = a r . m a j o r s e c t o r A B Q + a r . Δ O A B . a r . m a j o r s e g m e n t > a r . m a j o r s e c t o r . ∴ T h e s t a t e m e n i s v a l i d o n l y f o r m i n o r s e g m e n t . A n s - O p t i o n D .$

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