Question

In a circle of radius 13 𝑐𝑚, two chords MN and JK are at a distance of 5 𝑐𝑚 from the centre. Find the value of
(MN + 2 × JK). [4 Marks]

Answer 1

Let OP be the perpendicular drawn from the centre of the circle to the chord MN.
The perpendicular drawn from the centre of the circle to the chord bisects the chord.
∴ MP = PN
We know that, equal chords are equidistant from the centre of the circle, and vice versa.
Here, OP = OQ = 5 cm (given)
According to the theorem, we get
MN = JK [1 Mark]
OM = 13 cm (radius of the circle)
Consider triangle OPM.
Applying Pythagoras theorem,
OM${}^2$ = OP${}^2$ + MP${}^2$
13${}^2$ = 5${}^2$ + MP${}^2$
MP${}^2$ = 169 − 25 = 144
MP = 12 [1 Mark]
∴ MP = PN = 12 cm
MN = 24 cm
∴ MN = JK = 24 cm [1 Mark]
MN + 2×JK = (24 + 48) cm = 72 cm
Answer: MN + 2×JK = 72 cm [1 Mark]