In a circle of radius 3 cm, two chords of length 2 cm and 3 cm lie on the different side of a diameter. What is the perpendicular distance between the two chords?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 09 Dec 2022 Shift 2)

Answer 1

Given:

the radius of circle = 3 cm

length of chord AB = 2 cm

length of chord CD = 3 cm

Formula used:

Hypotenuse² = Perpendicular² + Base²

Calculation:
F1 Vinanti SSC 26.06.23 D1
Now from △OFB,

⇒ OF² = 3² − 1²

⇒ OF2 = 9 - 1

⇒ OF2 = 8

⇒ OF = \(2\sqrt2\)

From △OEC,

⇒ OE² = 3² − \(({3\over 2})^2\)

⇒ OE2 = 9 - \(({9\over 4})\)

⇒ OE2 = \({27\over 4}\)

⇒ OE = \(3\sqrt3\over2\)

So, the distance between AB and CD,

⇒ FE = OF + OE

⇒ FE = \(2\sqrt2\) + \(3\sqrt3\over2\)

⇒ FE = \(\bf \frac{{4\sqrt 2 + 3\sqrt 3 }}{2} \) cm

∴ The perpendicular distance between the two chords is \(\bf \frac{{4\sqrt 2 + 3\sqrt 3 }}{2} \) cm.