Question

In the adjoining figure,line l touches the circle with
centre o at point P, Q is the midpoint of radius
OP. RS is a chord through Q such that chords
RS || line l. If RS=12. Find the radius of the circle.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

r = 4√3 units.

Step-by-step explanation:

See the diagram attached.

Here, OP is perpendicular to the line I and RS is parallel to line I.

So, OQ is perpendicular to RS.

Now, Q is the midpoint of OP and OP = radius (r)

Hence, OQ = $\frac{r}{2}$

Taking Δ OQS which is a right triangle,

OS² = OQ² + QS²

⇒ $r^{2} = (\frac{r}{2} )^{2} + 6^{2}$ {Since, RS = 12 and OQ is the perpendicular bisector of RS}

⇒ $\frac{3}{4} r^{2} = 36$

⇒ $r^{2} = 48$

⇒ r = 4√3 units. (Answer)