Question

Pe cerc de centru O si raza =24 cm se considera A,B,C in aceasta ordine. Astfel incat AB=24√2 cm si AC=24√3 cm. Aflati aria sectorului circular determinat de arcul BC

Answer 1

calculam inaltimile triunghiurilor AOB si AOC
, h1 respecriv h2
h1=√(R^2-(AB/2)^2)=√(24^2-2 x 12^2)
h1=12√2
se observa ca h1=AB/2 ⇒ ∡AOB=45° ⇒ ∡AOB=90°

h2=√(R^2-(AC/2)^2)=√(24^2-3 x 12^2)
h2=12, h2=R/2 ⇒ ∡OCA=∡OAC =30° (teorema ∡ de 30°)
prin urmare ∡AOC=180-30-30=120°
∡BOC=360-∡AOB-∡AOC=360-90-120=150°
la 360°....................πR^2
150°.........................x
x=150 x 24^2 x π/360
x=240π= aria sector BOC
sa-mi spui ce rezultat e in culegere