Find the radius of the quarter circle | A Nice Geometry Problem | 2 Different Methods to Solve

[@valisahranavard]

با سپاس،دو روش زیبا

[@giuseppemalaguti435]

r^2=10^2+(r-5)^2

[@Okkk517]

r=25/2

[@olumuyiwaasaolu7674]

This is so obvious and is the fastest approach.

[@fadvis]

Excelent. If we generalize 5=a, then: R=2.5a.

[@honestadministrator]

∆ B D E is similar to ∆ BOA DB / OB =DE / AO = BE /AB or DE = BD Again intersecting chord Theorem tells us BE . AE = CE . (3 DE) (√2 AO - BE) BE = 3 DE^2 Again BE /DE = √2 Hereby 2 ( AO/DE) = 3 + 2 AO = 5 * DE /2

[@じーちゃんねる-v4n]

Think in terms of a unit circle with radius 1 C(cosθ, sinθ) E(cosθ, (sinθ)/2) since E on AB cosθ+(sinθ)/2=1 ∴sinθ=4/5 DE=2/5 Return to actual size R=5/(2/5)=25/2

[@hkgupta1954]

Excellent 👍👍

[@MathBooster]

Thank you 🙂

[@piman9280]

The intersecting chords theorem can be visualised fairly simply, thus 5(2r - 5) =(10)(10)=> 2r - 5 = 20 => 2r = 25 => r = 12.5. It's a twenty seconds problem, so why bother with any other method?

[@rohitmadashri7250]

A computer can solve this in a trillionth of a second. So why bother with Intersecting Chords theorem? This is not a contest and there are no prizes for quick solutions. The point is not just the quickest solution, but also to explore other possibilities. That is the beauty of Mathematics.

[@piman9280]

@@rohitmadashri7250 The real beauty of mathematics is finding the most elegant solution. There is little to be gained by "beating about the bush" and spending ten minutes over this particular problem - both methods shown needed no more than five minutes of explanation in total (maybe less).