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Two circles of equal radius intersect at points C and D.The centres of the two circles are points A and B respectively.If their radius is 10 units,the area of triangle ABC is 40,then how do we find the distance between A and B?

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Draw a figure.

You have to find the length of the third side of an isosceles triangle with two sides length = 10 and area= 40.

This is equivalent to finding the length of twice one of the short sides of a right angled triangle with hypotenuse 10 and area 20.

So there are two solutions.

Introduce notation.

If $p, q$ are the lengths of the short sides in the right-angled triangle (so you are looking for AB = $2p$ or $2q$) then you have to solve $$ p^2 + q^2 = 10^2, \mbox{ (Pythagoras)}$$ $$ \frac{pq}{2} = 20. \mbox{ (area triangle)}$$

If you solve this quadratic equation you get e.g. $$p = \sqrt{80} \mbox{ or } p = \sqrt{20},$$ $$q = \frac{20}{p}.$$

Edit: corrected formula for area of triangle and solution for $p$.

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