Two circle of radii 5 cm and 3 vm intersect at two points and the distance between their centre is 4 cm find the length of common chord
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Answer:
In ∆ACO,
5^2=x^2+AC^2 -(1)st
In ∆ACO',
5^2=(4-x)^2+AC^2 -(2)nd
adding(1)+(2)
5^2-x^2=3^2-(4-x)^2
by using a^2+2ab+b^2 identity we can expand(4-x)^2
so,we get 16-8x+x^2
25-x^2 = 9-16-8x+x^2
canceling -x^2 and+x^2
we get,
25=9-16-8x
25=-7-8x
8x=25+7
8x=32
x=32÷8
x=4
AC^2=√5^2-√4^2
AC^2=√25-√16
AC^2=√9
AC=3
so,
AB=Ac×2
AB=3×2
AB=6cm
so.the length of the common chord =6cm
Answered by
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so the length of comman chord is 6 cm.
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