Question

Р
5.
A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the
centre of the circle. Find the area of the minor as well as the major segment.
(r = 3.14, 13 = 1.73)
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Answer 1

Answer:

minor- 20.4375cm2

major-686.0625cm2

Step-by-step explanation:

radius= 15cm

angle subtended=60°=£

area of minor segment = area of sector-area of triangle

area of sector=£/360*πr**2=1/6*3.14*15**2=117.75cm2

area of triangle=1/2*base*height=1/2*15*3/√2*15

=97.3125 cm2

area of minor segment=20.4375cm2

area of circle=πr**2=706.5 cm2

area of major segment =area of circle-area of minor segment=706.5-20.4375=686.0625cm2

Answer 2

Given radius =3.5 cm

Given radius =3.5 cmθ=60

Given radius =3.5 cmθ=60 o

Given radius =3.5 cmθ=60 o

Given radius =3.5 cmθ=60 o Area of minor segment =

Given radius =3.5 cmθ=60 o Area of minor segment = 360

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 −

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ.

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. =

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 360

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5)

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 =

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 6

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2 ∴ Area of major segment =πr

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2 ∴ Area of major segment =πr 2

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2 ∴ Area of major segment =πr 2 ×0.28

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2 ∴ Area of major segment =πr 2 ×0.28 =π×3.5×3.5×0.28.

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2 ∴ Area of major segment =πr 2 ×0.28 =π×3.5×3.5×0.28. =10.77 cm

Given radius =3.5 cmθ=60 o Area of minor segment = 360θ ×πr 2 − 21 ×OP×OQ. = 36060 ×3.14(3.5) 2 −0.5×3.5×3.5 = 638.465 −6.125 =0.28 cm 2 ∴ Area of minor segment =0.28 cm 2 ∴ Area of major segment =πr 2 ×0.28 =π×3.5×3.5×0.28. =10.77 cm 2