Question

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.

Answer 1

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 $cm^2$

Answer 2

Step-by-step explanation:

Here's the required answer!!!