Find area of the semicircle | Geometry Problem | Important Geometry and Algebra Skills Explained

[@Ramkabharosa]

Let O be the center of the circle, |OD| = a, & |OB| = |OA| = r. Then r²= |OB|²= |OD|²+|DB|²= a²+2² & r²= |OA|²= |OC|²+|CA|² = (6-a)²+4². So a²+2²= (6-a)²+4². ∴ a²+4= a²+36+16-12a. So 12a= 48 ∴ a= 4. Thus r²= a²+2²= 4²+2²= 20. So area of semi-circle = πr²/2 = 10π.

[@isg9792003]

It’s simple. Distance from centre Xand 6-X. By Pythagorean theorem 4^2+X^2=(6-X)^2+2^2 X=2 R^2= 4^2+2^2=20 Area=pyex20/2 = 10 pye

[@vestelshirley8887]

Yours is the best one. I initially just wrote off the presenters solution because PQ>AE and got out and watched a presentation on the golden ratio. But I decided to get back in to the problem. Sense I'm no geometer and I'm trying to stave off dementia, I took to the comments instead of slugging it out myself - cheap fix for sure but your's is so elegant.

[@Diph64]

My solution is exactly the same

[@vestelshirley8887]

2 right triangles of heights 4 and 2 joined at the center of the semicircle with hypoteneuses(?) Of the same length. The Pythagorean Theorem yields triangle bases 4 and 2 with heights of 2 and 4 respectfully. The resultant radius is 20^(1/2) and area of 10*pi.

[@miguelgnievesl6882]

If you join the center of the semicircle with A and then with B, you get 2 right triangles, which generates a system of 2 equations with 2 unknowns.

[@glenmcneill1675]

Yes much much simpler!

[@dileepmv7438]

Yes, why need so much hazzle

[@robertbrianhart818]

I solved this in my head by finding where the perpendicular bisector of the chord (AB) intersected the base of the semi circle, calculated the radius to be sqrt(4^2+2^2)=sqrt(20). Thus pi*r^2/2 = 10pi.

[@honestadministrator]

Let O be the centre of this semi circle OA^2 = AC^2 + CO^2 OB^2 = BD^2 + OD^2 AC^2 - BD^2 = (CO + OD) ( CO - OD). Hereby CO - OD = (AC^2 - BD^2)/(CO + OD) = (16 - 4)/6 = 2 CO = ( 6 +2)/2 = 4 AO^2 = AC^2 + CO^2 = 32 sq unit

[@じーちゃんねる-v4n]

Radius r Circle center O ∠AOC=α ∠BOD=β then rsinα=4 rsinβ=2 rcosα+rcosβ=6=rsinα+rsinβ ∴cosα+cosβ=sinα+sinβ ∴(cosα-sinα)^2=(sinβ-cosβ)^2 ∴sinαcosα=sinβcosβ ∴sin2α=sin2β ∴(2α+2β)/2=90° ∴β=90°-α ∴rsinβ=rsin(90°-α)=rcosα=2 ∴(4/r)^2+(2/r)^2=1 ∴r^2=20 ∴S=20π/2=10π

[@murdock5537]

Nice! CD = 6 = CM + DM = 2 + 4 → r = √(16 + 4) = 2√5 → area = 10π or: AB = √(4 + 36) = 2√10 = r√2 → r = 2√5 → area = 10π 🙂

[@WahranRai]

you can use the height theorem in a right triangle. AC^2 = PC*CQ = 16 BD^2 = PD*DQ = 4 CQ = CD + DQ = 6+DQ PD = PC+ CD = 6 + PC Solve with PC and DQ unknown ---> diameter of circle

[@d-8664]

This is explained in the video, right? Your solution method is the same.

[@WahranRai]

@@d-8664 Nooo . Show me where is explained Review your theorems on the right triangle

[@echandler]

@@d-8664 A triangle inscribed inside a semicircle is a right triangle. Thus its altitude is the mean proportional between the segments of the diameter that it cuts.

[@Grizzly01]

@@WahranRai Your first 2 lines are basically just stating the intersecting chords theorem, as per the video.

[@WahranRai]

@@Grizzly01 No. I used this theorem (translated from french) :In a right-angled triangle, the height resulting from the right angle (h) is proportional average between the 2 segments which it determines on the hypotenuse (m and n). not ) . Right triangles PAQ and PBQ

[@robertbourke7935]

Got It 🙂🙂

[@tellerhwang364]

結論:R^2=4^2+2^2 →R=2sqrt5😊

[@vestelshirley8887]

PQ>AE because PQ goes through the center of the circle and AE does not.

[@MathBooster]

Yes, if you calculate PQ then it will be greater than AE that is 8.

[@glenmcneill1675]

Again I do not like you relying on unproven expressions, I did this with two simple triangles and Pythagorean using about 10% of the paper, no quadratic etc.

[@Grizzly01]

What is it you consider to be 'unproven expressions' here? I used broadly the same method as the video, although I went the route of calculating x - y, then y.

[@glenmcneill1675]

@@Grizzly01 I did not like the use of the intersecting line theorem in the application of a proof. It is not required, and honestly I was not familiar with it.

[@charlesbromberick4247]

nifty