Question

Practice set 6.2
Radius of circle is 10 cm. There are two chords of length 16 cm ea
ch. what will be the
distance of these chords from the centre of the circle?
inere are two chords of length 16 cm each. What will be the
distance of these chords from the centre of this circle

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Answer 1

Answer:

both 6cm

Step-by-step explanation:

draw the diagram and join the centre to the chord normally to find the distance

by using Pythagoras theorem on the triangle formed,

we get

distance from the centre - ?

hypotenuse = 10 cm (radius)

other side = 8 cm ( since perpendicular from the center bisects the chord)

10^2 = 8^2 + x^2

=> 100 - 64 = x^2

=> x^2 = 36

x = 6 cm

=> distance of one chord from the center is 6 cm

but since both chords are equal = 16 cm they both are at a distance of 6 cm from the centre since equal chords are equidistant from the center

Answer 2

Hii There!!!

Given :- Length of the chord AB = 16cm and radius ( r ) = 10cm .

To find :- OM

Construction :- Draw a perpendicularOM from centre O such that, it bisects the chord AB.

Proof :- In right angled ∆OMB,

OB^2 = OM^2 + BM^2

=> 10^2 = OM^2 + 8^2

=> 100 = OM^2 + 64

=> OM^2 = 36

=> OM = √36

=> OM = 6 cm

Hence, the distance of the chord from the centre of the circle is 6cm.

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Hope it helps you.

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