Question

Find the length of a chord which is at a distance of $5cm$ from the center of a circle of radius $10cm.$

Answer 1

Let $AB$ be a chord of a circle with radius $10cm.$

$OC$ $\perp AB$

$\therefore OA=10cm$

$OC=5cm$

$\because$ $OC$ divides $AB$ into two equal parts. [Perpendicular from the center bisects the chord]

i.e., $AC=CB$

Now in right $\Delta OAC$, we have

$O{A}^2=O{C}^2+A{C}^2$ (Pythagoras Theorem)

$\Rightarrow (10{)}^2=(5{)}^2+A{C}^2$

$\Rightarrow 100=25+A{C}^2$

$\Rightarrow A{C}^2=100-25=75$

$\therefore AC=\sqrt{75}=\sqrt{25\times 3}=5\times 1.732$ [$\because \sqrt{3}=1.732$]

$\therefore AB=2\times AC=2\times 5\times 1.732$

$=10\times 1.732=17.32cm$

Hence, the length of the chord is $17.32cm$