in a circle of radius 10 cm find the length of a chord which is distance of 5cm drom centre (take√3=1.732)
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We will draw a figure like this...
OB is 10 cm because it is a radius.
Angle OCD will be 90.
We can apply PGT in ∆OCB:
CB^2 = OC^2 + OB^2
CB^2 = (5×5) + (10×10)
CB= √25+100
CB= √125
So, AB will be 2√125 = 10√5 cm
OB is 10 cm because it is a radius.
Angle OCD will be 90.
We can apply PGT in ∆OCB:
CB^2 = OC^2 + OB^2
CB^2 = (5×5) + (10×10)
CB= √25+100
CB= √125
So, AB will be 2√125 = 10√5 cm
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