Is it true to say that area of a square inscribed in a circle of diameter $p c m$ is $p^{2} c m^{2}$ ? Why?

No,

a r e a o f s q u a r e = \frac{1}{2} \times {d i a g o n a l}^{2}

since the diagonal of the square is

p c m

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No,

a r e a o f s q u a r e = \frac{1}{4} \times {d i a g o n a l}^{2} .

since the diagonal of the square is

p c m

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No,

a r e a o f s q u a r e = 2 \times {d i a g o n a l}^{2} .

since the diagonal of the square is

p c m

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Yes,

a r e a o f s q u a r e = {d i a g o n a l}^{2}

since the diagonal of the square is

p c m

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Solution

The correct option is C No, $a r e a o f s q u a r e = \frac{1}{2} \times {d i a g o n a l}^{2}$ since the diagonal of the square is $p c m$ .
$G i v e n - A B C D i s a s q u a r e i n s c r i b e d i n a c i r c l e o f d i a m e t e r p c m . T o f i n d o u t - W h e t h e r a r . A B C D = p^{2} {c m}^{2} o r n o t . S o l u t i o n - A B C D h a s b e e n i n s c r i b e d i n t h e c i r c l e . ∴ T h e p o i n t s A, B, C & D w i l l b e o n t h e g i v e n c i r c l e . i . e T h e d i a g o n a l o f t h e s q u a r e i n s c r i b e d i n t h e c i r c l e = d i a m e t e r o f t h e c i r c l e . S o T h e d i a g o n a l B D o f A B C D = d i a m e t e r p o f t h e g i v e n c i r c l e . ∴ a r . A B C D = \frac{1}{2} \times {d i a g o n a l}^{2} = \frac{1}{2} p^{2} {c m}^{2} . S o a r . A B C D \neq p^{2} {c m}^{2} . ∴ T h e g i v e n a s s u m p t i o n i s f a l s e . A n s - O p t i o n A .$

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