PROBLEMS ON AREA OF SECTOR AND SEGMENT OF CIRCLE

Find the area of the shaded region. Leave your answer in terms of π and in simplest radical form.

Problem 1 :

segmentofcircle

Solution:

Area of sector=𝜃360×𝜋r2=300360×𝜋×(8)2=1603𝜋 Area of equilateral triangle=34a2=34×(8)2=163Area of shaded region=1603𝜋 +163

Problem 2 :

segmentofcircleq1

Solution:

Area of sector=𝜃360×𝜋r2=270360×𝜋×(3)2=274𝜋 Area of isosceles triangle=12a2=12×(3)2=92Area of shaded region=274𝜋 +92

Problem 3 :

segmentofcircleq2

Solution:

areaofsectorandsegment

sin 60 = x/7

√3/2 = x/7

x = 7√3/2

AB = 7√3

cos 60 = h/7

1/2 = h/7

h = 7/2

Base of the triangle is  7√3 and height is 3.5

Area of triangle which is having central angle as 120 degree is

= (1/2) x  (7√3) x 3.5

= 12.25√3 ---(2)

(1) + (2)

= 98π/3 + 12.25√3

Find the area of each shaded segment. Round your answer to the nearest tenth.

Problem 4 :

segmentofcircleq3

Solution:

The diameter will divide the circle into two semicircles, from the picture it is clear, we have three sectors with same measure of central angle.

areaofsectorandsegmentq1

sin 30 = x/4

1/2 = x/4

x = 2

AB = 4

cos 30 = h/4

√3/2 = h/4

h = 2√3

Area of triangle = (1/2) x 4 x2√3

= 6.928 --(2)

Required area = 8.37 - 6.928

= 1.442

= 1.4 mm2

Problem 5 :

segmentofcircleq4

Solution:

Diameter = 10 in

radius = 10/2 = 5 in

triangle2
sin 36°=x55×sin 36°=xx=5×0.588x=2.9base=2×2.9=5.8cos 36°=y50.809=y5y=4.04Area of right angle triangle=12ab=12×5.8×4.04=11.716Area of shaded segment=Area of sector-area of right angle triangle=72360×3.14×52-11.716=15.7-11.716=3.98 in2

Problem 6 :

segmentofcircleq5

Solution:

60+150+150 ==> 360

The central angle of two equal arcs will be 150 degree.

Diameter = 12 cm

radius = 12/2 = 6 cm

triangle3
sin 15°=x66×sin 15°=xx=6×0.258x=1.553cos 15°=y60.965=y6y=5.79Area of right angle triangle=12ab=12×1.55×5.79=4.487 cm2Area of segment=2×area of triangle=2×4.487=8.974 cm2Area of shaded segment=Area of sector-area of right angle triangle=47.1-8.974=38.1 cm2

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