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In a circle of diameter 10 cm, the length of each of 2 equal and parallel chords is 8 cm, find the distance between these two chords.

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a

4 cm

b

5 cm

c

6 cm

d

7 cm

answer is C.

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Detailed Solution

Question Image

In the figure, let's consider AB and CD as two chords of length 8 cm respectively.
So, let us draw a perpendicular OM to AB and ON to CD and as we know that a perpendicular drawn from the centre to the chord bisects the chord.
Hence, CN = ND = 4 cm, and AM = MB = 4 cm,
OM = radius of the circle =

\frac{diameter}{2}

=

\frac{10}{2}

= 5cm
According to Pythagoras theorem in triangle AOM and CON
Thus, in triangle AOM, we get

O A^{2} = A M^{2} + O M^{2}

5^{2} = 4^{2} + O M^{2}

O M^{2} = \sqrt{25 - 16} = \sqrt{9} = 3

And, in triangle CON, we get

O C^{2} = N C^{2} + O N^{2}

5^{2} = 4^{2} + O N^{2}

O N^{2} = \sqrt{25 - 16} = \sqrt{9} = 3

But MN = OM + ON
MN = 3 + 3 = 6 cm
Hence, the perpendicular distance between the two chords is 6 cm.

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In a circle of diameter 10 cm, the length of each of 2 equal and parallel chords is 8 cm, find the distance between these two chords.