Q.Two chords of a circle have their lengths in the ratio 4:3. If their distances from the center are 9 cm and 12 cm respectively, what is the length of the chord closer to the center?

Let O be the center of the circle, and let AB and PQ be two chords such that their length ratio is AB:PQ = 4:3. Also, their perpendicular distances from the center are OM = 9 cm and ON = 12 cm respectively. Let AB = 4x cm and PQ = 3x cm. Now, in triangle △OPN: OP² = ON² + PN² ⇒ OP² = 12² + (3x/2)² ⇒ OP² = 144 + (9x² / 4) ..........(i) Similarly, in triangle △OAM: OA² = OM² + AM² ⇒ OA² = 9² + (2x)² ⇒ OA² = 81 + 4x² ..........(ii) Since both OA and OP are radii of the circle, equating (i) and (ii): 81 + 4x² = 144 + (9x² / 4) ⇒ 4x² − (9x² / 4) = 144 − 81 ⇒ (16x² − 9x²) / 4 = 63 ⇒ 7x² = 63 × 4 ⇒ x² = (9 × 4) ⇒ x = √36 = 6 Therefore, the length of the chord closer to the center = 4 × 6 cm = 24 cm
Similar Questions