a chord of a circle of radius 10 cm subtends a right angle at the centre find the area of the corresponding First main minor Arc second major are used by equals to 3.14
Answers
Answer:
Area of Minor Segment = 28.57 cm^2and Area of Major Segment = 285.72 cm^2
Given: Radius of circle, r = 10 cm
Angle made by chord at center = 90°
To find: Area of minor and major Segment
Area of Segment = Area of Sector - Area of triangle made by chord and radii.
Area\:of\:Sector\:=\:\frac{\theta}{360^{\circ}}\times\pi r^2
⇒ Area\:of\:minor\:Sector\:AOB=\:\frac{90^{\circ}}{360^{\circ}}\times\frac{22}{7}\times 10^2
= \frac{1}{4}\times\frac{22}{7}\times 100
= \frac{550}{7}\:cm^2
Area of ΔAOB = \frac{1}{2}\times base\times heigth
= \frac{1}{2}\times OA\times OB
= \frac{1}{2}\times 10\times 10
= 5\times 10
= 50\:cm^2
⇒ Area of Minor Segment = \frac{550}{7}-50
= \frac{550-350}{7}
= \frac{200}{7}
= 28.57 cm^2
Area of Major segment = Area of circle - Area of minor segment
= \pi r^2 - 28.57
= \frac{22}{7}\times 10^2 - 28.57
= \frac{22}{7}\times 100 - 28.57
= \frac{2200}{7} - 28.57
= 314.29 - 28.57
= 285.72 cm^2
Therefore, Area of Minor Segment = 28.57 cm^2and Area of Major Segment = 285.72 cm^2
HOPE THIS WILL BE HELPFUL.........................................................