A square ABCD is inscribed in a semicircle of radius unity. Find the length of the side of the square.

\frac{2}{\sqrt{3}}

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\frac{\sqrt{3}}{2}

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

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

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Solution

The correct option is D $\frac{2}{\sqrt{5}}$
According to figure $r = 1$
$(O B)^{2} = (O C)^{2} + (B C)^{2} O B = {(\frac{l}{2})}^{2} + l^{2} = \frac{l^{2}}{4} + l^{2} 1 = (\frac{5}{4}) l^{2} l = \frac{2}{\sqrt{5}}$

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A square ABCD is inscribed in a semicircle of radius unity. Find the length of the side of the square.

The correct option is D 2√5According to figure r=1(OB)2=(OC)2+(BC)2OB=(l2)2+l2=l24+l21=(54)l2l=2√5

The correct option is D $\frac{2}{\sqrt{5}}$
According to figure $r = 1$
$(O B)^{2} = (O C)^{2} + (B C)^{2} O B = {(\frac{l}{2})}^{2} + l^{2} = \frac{l^{2}}{4} + l^{2} 1 = (\frac{5}{4}) l^{2} l = \frac{2}{\sqrt{5}}$