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

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 !!!!

You did this

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

why find QM then Subtract it from PQ

we can directly find PQ = √[(AP)² - (AM)²]

JUST POINTING OUT !!!!

THANKS AGAIN