In the figure, AB and CD are two parallel chords of the circle with centre O. If AB = 16cm, CD = 30cm and OA = 17cm, find the distance between AB and CD.
Answers
Answer:
It is given that AB = 30cm and CD = 16cm
Join the lines OA and OC
We know that AO = 17cm and CO = 17cm
Construct OM ⊥ CD and OL ⊥ AB
The perpendicular from the centre of a circle to a chord bisects the chord
We know that
AL = ½ × AB
By substituting the values
AL = ½ × 30
So we get
AL = 15cm
We know that
CM = ½ × CD
By substituting the values
CM = ½ × 16
So we get
CM = 8cm
Consider △ ALO
Using the Pythagoras theorem it can be written as
AO^2 = OL^2 + AL^2
By substituting the values
17^2 = OL^2 + 15^2
So we get
OL^2 = 17^2 - 15^2
On further calculation
OL^2 = 289 – 225
By subtraction
OL^2 = 64
By taking the square root
OL = √64
OL = 8cm
Consider △ CMO
Using the Pythagoras theorem it can be written as
CO^2 = CM^2 + OM^2
By substituting the values
17^2 = 82 + OM^2
So we get
OM^2 = 17^2 - 8^2
On further calculation
OM^2 = 289 – 64
By subtraction
OM^2 = 225
By taking the square root
OM = √225
OM = 15cm
So the distance between the chords = OM + OL
By substituting the values
Distance between the chords = 8 + 15 = 23cm
Therefore, the distance between the chords is 23 cm.
In the figure, AB and CD are two parallel chords of the circle with centre O. If AB = 16cm, CD = 30cm and OA = 17cm, find the distance between AB and CD.
