In the given figure, $A B C D$ is a square of side $5 c m$ inscribed in a circle. The radius of the circle is
[Take $π = 3.14$ ]

5 \sqrt{2}

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\frac{5}{2} \sqrt{2}

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10 \sqrt{2}

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\frac{5}{4} \sqrt{2}

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Solution

The correct option is B $\frac{5}{2} \sqrt{2}$ cm
$G i v e n - A B C D i s a s q u a r e o f s i d e s 5 c m, i n s c r i b e d i n a c i r c l e$

$T o f i n d o u t - t h e r a d i u s o f t h e c i r c l e = ?$

$S o l u t i o n - W e j o i n A C & B D a n d t h e y i n t e r s e c t a t O .$

$S i n c e A B C D i s a s q u a r e, i t s d i a g o n a l s A C = B D a n d t h e y w i l l b i s e c t e a c h o t h e r .$

$∴ O A = O B = O C = O D = \frac{1}{2} A C (o r B D) .$

$S o A, B, C & D a r e e q u i d i s t a n t f r o m t h e c e n t r e O o f t h e c i r c l e .$

$∴ O i s t h e c e n t r e o f t h e c i r c l e a n d O A = O B = O C = O D = r a d i u s o f t h e c i r c l e .$

$N o w t h e d i a g o n a l o f a s q u a r e = \sqrt{2} \times l e n g t h o f t h e s i d e .$

$∴ A C = B D = \sqrt{2} \times 5 c m = 5 \sqrt{2} c m .$

$∴ O A = O B = O C = O D = \frac{1}{2} A C (o r B D) = \frac{1}{2} \times 5 \sqrt{2} c m = \frac{5 \sqrt{2}}{2} c m .$

$S o t h e r a d i u s o f t h e c i r c l e = c m$

$A n s - O p t i o n B .$

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In the given figure, ABCD is a square of side 5 cm inscribed in a circle. The radius of the circle is[Take π=3.14]

In the given figure, $A B C D$ is a square of side $5 c m$ inscribed in a circle. The radius of the circle is
[Take $π = 3.14$ ]