What is the area of a 60° sector of a circle with area #42pim^2#?

2 Answers
Dec 3, 2016

#7pim^2#

Explanation:

A full circle is #360^@#

Let area of the #60^@# sector = #A_S# and area of the circle = #A_C#
#A_S= 60^@/360^@A_C=1/6A_C#
Given that #A_C=42pim^2#,
#=> A_S=(1/6)*42pim^2=7pim^2#

Dec 3, 2016

#7pi# #m^2#

Explanation:

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We need to find the area of the sector. For that we use the formula

#color(blue)("Area of a sector"=x/360*pir^2#

Where #x# is the angle on the vertex of the sector(#60^circ#)

(Note: #pir^2# is the area of the whole circle which is #42pi#)

Let's put everything into the formula

#rarrx/360*pir^2#

#rarr60/360*42pi#

#rarr1/6*42pi#

#rarr1/cancel6^1*cancel42^7pi#

#color(green)(rArr7pi# #color(green)(m^2#

Hopefully this helps!!! :)