Question

The length of the minor arc AB of a circle ,centre o is 2 pi cm and the length of the major arc is 22 pi cm. A) find the radius B) the acute angle AOB​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that radius of circle be 'r' cm.

According to statement,

It is given that,

↝ Length of minor arc of circle = 2 pi cm

↝ Length of major arc of circle = 22 pi cm

We know,

↝ Length of major arc + Length of minor arc = Circumference of circle.

We know,

$\red{ \boxed{ \sf{ \:Circumference_{(circle)} = 2\pi \: r }}}$

So,

$\rm :\longmapsto\:2\pi \: r = \: 2\pi \: + \: 22 \: \pi$

$\rm :\longmapsto\:2\pi \: r = \: 24 \: \pi$

$\bf\implies \:r \: = \: 12 \: cm$

Let we further assume that minor arc subtends an angle p at the centre.

We know,

$\red{ \boxed{ \sf{l \: = \: 2 \: \pi \: r \: \dfrac{p}{360 \degree \: } }}}$

where

• l is the length of minor arc

• r is the radius of circle

• p is the central angle subtended at the centre by minor arc

Now,

We have

• l = 2 pi cm

• r = 12 cm

↝ On Substituting all the values, we get

$\rm :\longmapsto\:2 \: \pi \: = \: 2 \: \pi \: \times \: 12 \times \dfrac{p}{360\degree \: }$

$\rm :\longmapsto\:1 = \dfrac{p}{30\degree \: }$

$\bf :\implies\:p \: = \: 30\degree \:$

Additional Information :-

$\rm :\longmapsto\:Area_{(sector)} \: = \: \pi \: {r}^{2} \: \dfrac{ \theta \: }{360\degree \: }$

$\rm :\longmapsto\:Area_{(major \: sector)} \: = \: \pi \: {r}^{2} \: \dfrac{(360\degree \: - \theta \: )}{360\degree}$

$\rm :\longmapsto\:Area_{(minor \: segment)} = \dfrac{\pi \: {r}^{2} \theta }{360\degree \: } - \dfrac{1}{2} {r}^{2} \: sin \theta$

$\rm :\longmapsto\:Area_{(major \: segment)} = \pi {r}^{2} - \dfrac{\pi \: {r}^{2} \theta }{360\degree \: } + \dfrac{1}{2} {r}^{2} \: sin \theta$