A chord $PQ = x$ cm of a circle with centre $O$ has $OR \perp PQ$. If $OR = (x-2)$ cm and the radius equals the diameter

Question

A chord $PQ = x$ cm of a circle with centre $O$ has $OR \perp PQ$. If $OR = (x-2)$ cm and the radius equals the diameter logic gives $x$ cm such that $x^2 = (x-2)^2 + \left(\dfrac{x}{2}\right)^2$—the value of $x$ is—

Accepted Answer

$\dfrac{16}{3}$

Answer

$\dfrac{8}{3}$

Answer

$\dfrac{4}{3}$

Answer

$\dfrac{2}{3}$

A chord PQ=xPQ = xPQ=x cm of a circle with centre OOO has OR⊥PQOR \perp PQOR⊥PQ. If OR=(x−2)OR = (x-2)OR=(x−2) cm and the radius equals the diameter logic gives xxx cm such that x2=(x−2)2+(x2)2x^2 = (x-2)^2 + \left(\dfrac{x}{2}\right)^2x2=(x−2)2+(—the value of xxx is—

A chord $PQ = x$ cm of a circle with centre $O$ has $OR \perp PQ$ . If $OR = (x-2)$ cm and the radius equals the diameter logic gives $x$ cm such that $x^2 = (x-2)^2 + \left(\dfrac{x}{2}\right)^2$ —the value of $x$ is—