Question : If the area of a circle is $616 \mathrm{~cm}^2$ and a chord $XY=10 \mathrm{~cm}$, then find the perpendicular distance from the centre of the circle to the chord $XY$.
Option 1: $\sqrt{171} \mathrm{~cm}$
Option 2: $\sqrt{161 } \mathrm{~cm}$
Option 3: $\sqrt{117 } \mathrm{~cm}$
Option 4: $\sqrt{ 181} \mathrm{~cm}$
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Correct Answer: $\sqrt{171} \mathrm{~cm}$
Solution :
Given: The area of a circle is $616 \mathrm{~cm}^2$.
A chord $XY=10 \mathrm{~cm}$
Let $O$ be the centre, $XY$ be the chord, $OX$ be the radius and $OZ \perp XY$.
$XY=10$ cm therefore, $XZ=5$ cm
Area of circle = $\pi r^{2}$
⇒ $616=\pi r^{2}$
⇒ $r^{2}=616×\frac{7}{22}=196$
⇒ $r=14=OX$
$\triangle OXZ$ is a right-angled triangle, by using Pythagoras' theorem,
$OX^{2} = XZ^{2}+OZ^{2}$
⇒ $14^{2} = 5^{2}+OZ^{2}$
⇒ $OZ^{2}=196–25$
⇒ $OZ=\sqrt{171}$ cm
Hence, the correct answer is $\sqrt{171} \mathrm{~cm}$.
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