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Question

Two circles each of which passes through the centre of the other intersect points M and N. A line from M intersects the circle at K and L as shown in the figure. If KL = 6 compute the area of ∆KLN.

I was attempting a Practice Paper and stumbled on this Question.

I think the tangent secant theorem must used somewhere which states that.

$Tangent^2=Outer Secant×Whole Secant$

But I don't know where to apply it so I am clueless.

I don't know where to start from.

Any help will be appreciated

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1 Answer 1

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enter image description here

Realize that $\angle NKM= 120^o$ and that $\angle KLN= 60^o$, which leaves $\Delta KLN$ an equilateral triangle. The answer is therefore $9\sqrt{3}$.

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  • $\begingroup$ Here is link for the basic angle theorems in a circle: mathbitsnotebook.com/Geometry/Circles/CRAngles.html $\endgroup$
    – Larry
    Commented Oct 16, 2019 at 18:27
  • $\begingroup$ How did you find the aswer to be 9√3? $\endgroup$
    – Crocogator
    Commented Oct 17, 2019 at 2:52
  • $\begingroup$ How did you find angle NKM to be 120°? I've gone through the links but none of the theorems help me find the angle. $\endgroup$
    – Crocogator
    Commented Oct 17, 2019 at 3:01
  • $\begingroup$ I added another image as a hint. You already know that $\angle KLN= 60^o$, so according to the theorem I just added, $\angle KLN = 180^o - \angle KLN= 180^o- 60^o=120^o$. And the area of an equilateral triangle can be calculated if you know one side of the triangle. $\endgroup$
    – Larry
    Commented Oct 17, 2019 at 11:27
  • $\begingroup$ $\angle NOM=\angle NKM$ since these two angles are on the same intercepted arc. $\endgroup$
    – Larry
    Commented Oct 17, 2019 at 11:31

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