Radius of circle is 10 cm. there are two chords of length 16 cm each. what will be the distance of these chords from the centre of the circle?
Answers
Answer:
Length of the chord AB=16cm and radius(r)=10cm.
To find:- OM
Construction: Draw a perpendicular OM from centre O such that it bisects the chord AB
Proof: In right angled ΔOMB
OB
2
=OM
2
+BM
2
⇒(10)
2
=OM
2
+(8)
2
⇒100=OM
2
+64
⇒OM
2
=100−64
⇒OM
2
=36
⇒OM=
36
⇒OM=6cm
∴ Hence, the distance of the chord from the centre of the circle is 6cm.
Answer:
Given :
Radius of Circle = 10cm
Length of Chord = 16cm
To Find :
Find the distance of these chords from the centre of the circle ?
Solution :
OR = OP = 10cm [Radius]
PQ = RS = 16cm [Chord]
Perpendicular drawn from the centre of the circle to the chord bisects the chord,
➣ PU = ½ × PQ
➣ PU = ½ × 16
➣ PU = 8cm
Applying Pythagoras Theorem in ∆OUP :
➣ (OP)² = (OU)² + (PU)²
➣ (10)² = (OU)² + (8)²
➣ 100 = (OU)² + 64
➣ 100 - 64 = (OU)²
➣ 36 = (OU)²
➣ √36 = OU
➣ 6cm = OU
Therefore,
Congruent chords of the circle are equidistant from the circle are :
➣ OU = OT = 6cm
Hence,
The distance of the chord from the centre is 6cm.