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Subject: Math , asked on 31/10/14

# Find the area of a quadrant of a circle whose circumference is 22 cm. Take PIE= 22/7

hope it helps..

• 1

circumference of circle= 2 XpieXr

22= 2X(22/7)Xr

(22X7)/(22X2)=r

7/2=r

area of quadrant= 1/4X pie X r^2

=(1/4)X(22/7)X(7/2)^2

=77/8

• 0

as u knw circumference of a circle = 2 * pie * radius, 22 = 2 * 22/7 * r , r = 7/2

area of circle = 1/4 * pie * radius * radius = 1/4 * 22/7 * 7/2 * 7/2 = 77/8 ans

• 0

circumference = 2*pi*r

22 = 2 * 22/7 * r

r = 22 * 7/22 * 1/2

r = 7/2 cm

area of quadrant = 1/4 * pi * r * r

= 1/4 * 22/7 * 7/2 * 7/2

= 77/8 cm sq.

• 0

2 π r =22

R= 22 *7/22 *1/2

So r= 7/2

Now area of quadrant = π ϴ /360

22/7*49/4*90/360

• 0

circumference of circle = 22 cm

therefore  2 pi r = 22

r = 22 x 7 / 2 x 22

on cancellation

r = 7 / 2 = 3.5 cm

therfore area of quadrant = pi r square / 4

= 22 / 7 x 3.5 x 3.5 x 1 / 4

= 38.5 / 4 cm square = 9.625 cm square

• 0

Circumference of a quadrant is 1/4 the circumference so the circumference of the circle is 4x22=88cm, Also circumference of a circle is 2(pi)R.

=>2x22/7xR=88cm

>R=88x7/22x2

>R=88x7/44

>R=2x7

>R=14cm