Question

The centre of a circle is at O. AB and CD are two chords
of length d and respectively. If P is the mid point of CD,
then the length 《OP》 is:

●●●Hey please give a step by step explanation!!!●●●​

Answer 1

• CP = $\bf{\dfrac{l}{2}}$

$\huge\mathcal{\boxed{\fcolorbox{lime}{orange}{SOLUTION}}}$

$\sf{\implies\:(OP)^2\:=\:\Big(\dfrac{d}{2}\Big)^2\:-\:\Big(\dfrac{1}{2}\Big)^2\:}$

$\sf{\implies\:(OP)^2\:=\:\dfrac{d^2}{4}\:-\:\dfrac{l^2}{4}\:}$

$\sf{\implies\:(OP)^2\:=\:\dfrac{d^2\:-\:l^2}{4}\:}$

$\sf\green{\implies\:OP\:=\:\dfrac{\sqrt{d^2\:-\:l^2}}{2}\:}$

✔️ Plz mark as Brainliest Answer....

Answer 2

• CP = $\bf{\dfrac{l}{2}}$

$\huge\mathcal{\boxed{\fcolorbox{lime}{orange}{SOLUTION}}}$

$\sf{\implies\:(OP)^2\:=\:\Big(\dfrac{d}{2}\Big)^2\:-\:\Big(\dfrac{1}{2}\Big)^2\:}$

$\sf{\implies\:(OP)^2\:=\:\dfrac{d^2}{4}\:-\:\dfrac{l^2}{4}\:}$

$\sf{\implies\:(OP)^2\:=\:\dfrac{d^2\:-\:l^2}{4}\:}$

$\sf\green{\implies\:OP\:=\:\dfrac{\sqrt{d^2\:-\:l^2}}{2}\:}$

✔️ Plz mark as Brainliest Answer....