A chord ab of a circle of radius 10cm

Asked by admin @ 16/08/2021 in Math viewed by 279 People

A chord AB of a radius 10cm makes a right angle at the centre of the circle.Find the area of the minor and major segments.

Answered by admin @ 16/08/2021

Answer:

Area of Minor Segment = 28.57 $cm^2$ and Area of Major Segment = 285.72 $cm^2$

Step-by-step explanation:

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 $cm^2$ and Area of Major Segment = 285.72 $cm^2$

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