Question

AD is a diameter of a circle and AB is a chord. If AD=34cm ,AB=30cm,find the distance of AB from the center of the circle

Answer 1

distance of AB from center of circle is 8 cm by Pythagoras theorem

Answer 2

$\huge{\red{\underline{\underline{\mathfrak{\blue{\bigstar Given:: \bigstar}}}}}}$

• Chord AB = 30 cm
• Diameter AD = 34 cm
• Radius = 34 / 2 = 17 cm

$\huge{\red{\underline{\underline{\mathfrak{\blue{\bigstar Sol^{n}:: \bigstar}}}}}}$

➣ Dram OM ⊥ AB as shown in the figure.

➣ 2AM = AB [Perpendicular drawn from the center bisects the chord]

➣ AM = 30 / 2 = 15 cm

➣ By Pythagoras Theorem in Δ AOM:

➣ OM² + AM² = OA²

➣ OM² + 15² = 17²

➣ OM² = 289 - 225

➣ OM² = 64 cm

➣ OM = √64

➣ OM = 8 cm

Hope it helps..♪