Question

find the area of the minor segment of a circle of radius 42 cm .if the length of the corresponding arc is 44cm

Answer 1

Answer:length of arc= x/360×2×22/7×42

Silving this u get that teta =60

Atea of corresponding sector=60/360×22/7×42×42

=924cm^2

Area of the minor segment =924-1/2×42×42×sin60

=924-441root3 cm^2

Step-by-step explanation:

Answer 2

Solution:

Radius of circle = 42 cm

Length of corresponding arc = 44 cm

If circle of radius r , has perimeter = 2 π r

Then, if circle of radius = 42 cm has perimeter

 $=2\times \frac{22}{7}\times 42\\\\ =264$  cm

Major segment has arc length = 264 - 44= 220 cm

Minor segment has arc length = 44 cm

Now coming to area part,

Suppose, area of minor segment= x cm²

Area of circle having radius 42 cm

$=\frac{22}{7}\times 42 \times 42\\\\= 132 \times 42\\\\=5544$

So, area of major segment = (5544 - x )cm²

The ratio of circumference to area of both the parts should be equal.

$\frac{44}{220}=\frac{x}{5544-x}\\\\ \frac{1}{5}=\frac{x}{5544-x}\\\\ 5 x= 5544-x\\\\ 6 x= 5544\\\\ x= 924$

So, Area of minor segment= 924 cm²