Question

Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.

[2 Marks]

Answer 1

By pythagoras theorem,

OA${}^2$ = OC${}^2$ + AC${}^2$

5${}^2$ = 3${}^2$ + AC${}^2$

⇒ AC = √(25-9) = √16

⇒ AC = 4 cm [0.5 Marks]

By using the theorem which states that “The perpendicular drawn from the centre of a circle to a chord bisects the chord” [0.5 Marks]

Since OC ⟂ AB. It divides AC into two equal segments as per the above mentioned theorem. [0.5 Marks]

⇒ AB = AC + CB

⇒ AB = AC + AC [AC = CB]

⇒ AB = 4 + 4

⇒ AB = 8 cm [0.5 Marks]