Question

Two circles of radii 5 cm and 3 cm intersect at two
points and the distance between their centres
4 cm. Find the length of the common chord.
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Answer 1

In ∆ACO,

5^2=x^2+AC^2 -(1)st

In ∆ACO',

3^2=(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 get16-8x+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

Answer 2

Answer:

Here is the answer

Step-by-step explanation:

Hope it helped u