![2.9 cm^2](/tpl/images/1024/9050/23703.png)
Step-by-step explanation:
Given,
Angle of sector = 30°
Radius of circle, r = 8 cm
Area of the shaded region, A =?
The diagram is attached below.
Now,
Area of shaded region = Area of sector - Area of triangle
Area of triangle = ![\frac{1}{2}\times base \times height](/tpl/images/1024/9050/976f6.png)
We know that,
![\sin \theta = \dfrac{P}{H}](/tpl/images/1024/9050/53867.png)
![\sin 30^\circ= \dfrac{P}{8}](/tpl/images/1024/9050/bc2d4.png)
![P = \dfrac{8}{2}](/tpl/images/1024/9050/ff04f.png)
![P = 4\ cm](/tpl/images/1024/9050/d8f86.png)
And
![\cos \theta =\dfrac{B}{H}](/tpl/images/1024/9050/e35ba.png)
![\cos 30^\circ = \dfrac{B}{8}](/tpl/images/1024/9050/1a5d3.png)
![B = \dfrac{8\times \sqrt{3}}{2}](/tpl/images/1024/9050/dc01f.png)
![B = 4\sqrt 3](/tpl/images/1024/9050/e0e7b.png)
Area of sector = ![\dfrac{\theta}{360^\circ}\times \pi r^2](/tpl/images/1024/9050/1577c.png)
Area of sector = ![\dfrac{30^\circ}{360^\circ}\times \pi \times 8^2](/tpl/images/1024/9050/c85b7.png)
= 16.75 cm^2
Area of triangle = ![\dfrac{1}{2} \times 4\sqrt 3 \times 4](/tpl/images/1024/9050/25773.png)
= 13.85 cm^2
Area of shaded region = 16.75 - 13.85
= ![2.9 cm^2](/tpl/images/1024/9050/23703.png)
![GCSE 995381
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Past Papers
8 cm
30°
A
OAB is the sector of a circle, centre 0](/tpl/images/1024/9050/ba386.jpg)