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**area of a sector**= the number**of**degrees * the**area of**the circle square the radius, multiply...**of**degrees and you get that the number**of**degrees**of**the**sector**is 26.6666667 degrees.2 Answers · Education & Reference · 06/05/2013

... you the fraction

**of**the semicircle that the**sector**takes up. Then multiply by... pi*(radius^2), since that's the**area of**the whole semicircle. That is, half**of a**circle's**area**. The other option...1 Answers · Education & Reference · 28/06/2009

**Area of a sector**is 1/2r²Θ and theta has to be in radians...24 2/3)*π/180 = =.4305 radians now you can use the first formula A =1/2 (38)² (.4305) A =310.8 m²2 Answers · Science & Mathematics · 06/12/2009

**A**= pi * 150^2 * (175/360) Watch out for rounding on pi and the fraction:**A**= 3.1416 * 22500 * 0.4861... = 34,360.4646... which rounds down to 34,360**A**= (pi * 22500 * 175) / 360 = 34,361.16964 which rounds down to 34,3613 Answers · Science & Mathematics · 11/03/2011

Area of a circular sector , A ( sector ) = ½θ...2π – [ ⅙π + 33/60ths of ⅙π] simplifying...θ = (101/120)π radians ∴ A ( sector ) = ½θr² = ...

1 Answers · Science & Mathematics · 14/10/2011

...=t Then OP=r showing that r= a *cos(t) is a circle on OB as diameter. Its area is πa²/4. You can also do it by integrating (1...

2 Answers · Science & Mathematics · 01/05/2013

...: radius x radius x pie circumference ??? cant remember soz to find the

**area of a sector**circle do the**area**as if it was**a**full circle (see above) then divide by 2 if...4 Answers · Education & Reference · 19/03/2008

**A**presumed answer would be to find the**area of**each section and add the areas up to find the total**area of**the shape. But the question is very vague.2 Answers · Science & Mathematics · 03/05/2008

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**of**all, "theata" is spelled "theta" anyway...and radius R so what you are saying is that**A**= 1 / TR, this is not the correct formula in actuality,**A**...1 Answers · Science & Mathematics · 18/06/2008

(45/360) * pi * 6^2 = 4.5pi cm^2

4 Answers · Science & Mathematics · 18/11/2012