If a square ABCD is inscribed in a circle and AB = 4 cm, the radius of the circle is:

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Solution

$\begin{matrix} W e h a v e A B = 4 c m w e h a v e t o f i n d r . N o w, ∠ A B C i s 90^{\circ} . (a s a l l t h e a n g l e s o f a s q u a r e a r e 90^{\circ} .) b y p y t h a g o r a s t h e o r e m, A C^{2} = A B^{2} + B C^{2} b u t, A C i s t h e d i a m e t e r = 2 r \Rightarrow {(2 r)}^{2} = 2 {(4)}^{2} . . . . . . {a s A B = B C = 4 c m} \Rightarrow 4 r^{2} = 32 \Rightarrow r^{2} = 8 ∴ r = 2 \sqrt{2} H e n c e, t h e r a d i u s o f t h e c i r c l e i s 2 \sqrt{2} c m . \end{matrix}$

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