What-is-the-area-of-eqilateral-triangle-whose-inscribed-circle-has-a-radius-2-

Question Number 12743 by tawa last updated on 30/Apr/17

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{eqilateral}\:\mathrm{triangle}\:\mathrm{whose}\:\mathrm{inscribed}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{a}\:\mathrm{radius}\:\mathrm{2} \\ $$

Answered by mrW1 last updated on 30/Apr/17

OC=2 OB=4=DO CB=2(√3) DC=4+2=6 AB=2×2(√3)=4(√3) A_(ABC) =(1/2)×4(√3)×6=12(√3) or A_(ABC) =6×A_(OCB) =6×(1/2)×2(√3)×2=12(√3)

$${OC}=\mathrm{2} \\ $$$${OB}=\mathrm{4}={DO} \\ $$$${CB}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${DC}=\mathrm{4}+\mathrm{2}=\mathrm{6} \\ $$$${AB}=\mathrm{2}×\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\boldsymbol{{A}}_{\boldsymbol{{ABC}}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}\sqrt{\mathrm{3}}×\mathrm{6}=\mathrm{12}\sqrt{\mathrm{3}} \\ $$$${or} \\ $$$$\boldsymbol{{A}}_{\boldsymbol{{ABC}}} =\mathrm{6}×{A}_{{OCB}} =\mathrm{6}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\sqrt{\mathrm{3}}×\mathrm{2}=\mathrm{12}\sqrt{\mathrm{3}} \\ $$

Commented by mrW1 last updated on 30/Apr/17