Circle Chord Of Contact Lenth + Area (How to Prove ??)

L/sinB = AB/sinP (triangle APB)
AB/sin(180-P) = R/sin(90-B) (triangle QAB)
tan(P/2) = R/L (if you join PQ then triangle PAQ)

using these three u can eliminate B and P and hence get what you want

Too easy....

Join PQ which meets AB at M

clearly, triangle BPM and APM are congruent by SAS criterion...

=> LPMA = 90°
Now, In triangle APQ we have,

AP = L and AQ = R, LQAP = 90°, => PQ = √(L² + R²)
Thus, 1/2 x AQ x AP = 1/2 x PQ x AM [both equal to area of triangle APQ]

=> AQ x AP = PQ x AM
=> AM = AQ X AP / PQ = R x L / √(L² + R²)

or, AM = R x L / √(L² + R²)

But, AB = 2 AM
=> AB = 2RL /√(L² + R²)

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Now comes the area of triangle ABP
So do this,

since, AM = RL / (L² + R²) and AQ = R and LAMQ = 90°

So by pythagoras' theorem we find QM and then subtract it from PQ to find PM

i.e., PM = [√(R² + L²) - R²/√(R² + L²)] = L²/√(R² + L²)

Thus, Area of triangle ABP = 1/2 x AB x AM = 1/2 x 2RL/√(R² + L²) x L²/√(R² + L²)

or, Area of triangle ABP = RL³/(R² + L²)

Hope this helps u...!!

Thanks

I am feeling very Stupid . I couldn't even solve this !!!!