Question

OPQ is the sector of the circle having centre at O and radius 15 cm .If measure angle POQ ia equal to 30 degree find the area enclosed by arc PQ and Chord PQ​

Answer 1

Answer:

The required area is 2.68 square cm.

Step-by-step explanation:

Area of sector POQ

$=\frac{\theta}{360}*\pi\:r^2$

$=\frac{30}{360}*\frac{22}{7}*15*15$

$=\frac{1}{12}*\frac{22}{7}*225$

$=\frac{1}{6}*\frac{11}{7}*225$

$=\frac{2475}{42}\:cm^2$

Area of triangle POQ

$=\frac{1}{2}*a*b*sinC$

$=\frac{1}{2}*r*r*sin30$

$=\frac{1}{2}*15*15*\frac{1}{2}$

$=\frac{225}{4}$

The area enclosed by arc PQ and Chord PQ

=​Area of sector POQ-Area of triangle POQ

$=\frac{2475}{42}-\frac{225}{4}$

$=\frac{4950}{84}-\frac{4725}{84}$

$=\frac{4950-4725}{84}$

$=\frac{225}{84}$

$=2.68\:cm^2$

Answer 2

Step-by-step explanation:

The required area is 2.68 square cm.

Step-by-step explanation:

Area of sector POQ

=\frac{\theta}{360}*\pi\:r^2=

360

θ

∗πr

2

=\frac{30}{360}*\frac{22}{7}*15*15=

360

30

∗

7

22

∗15∗15

=\frac{1}{12}*\frac{22}{7}*225=

12

1

∗

7

22

∗225

=\frac{1}{6}*\frac{11}{7}*225=

6

1

∗

7

11

∗225

=\frac{2475}{42}\:cm^2=

42

2475

cm

2

Area of triangle POQ

=\frac{1}{2}*a*b*sinC=

2

1

∗a∗b∗sinC

=\frac{1}{2}*r*r*sin30=

2

1

∗r∗r∗sin30

=\frac{1}{2}*15*15*\frac{1}{2}=

2

1

∗15∗15∗

2

1

=\frac{225}{4}=

4

225

The area enclosed by arc PQ and Chord PQ

=Area of sector POQ-Area of triangle POQ

=\frac{2475}{42}-\frac{225}{4}=

42

2475

−

4

225

=\frac{4950}{84}-\frac{4725}{84}=

84

4950

−

84

4725

=\frac{4950-4725}{84}=

84

4950−4725

=\frac{225}{84}=

84

225

=2.68\:cm^2=2.68cm

2