Whiteboardmaths.com © 2004 All rights reserved

Major Segment Area of a Segment A B Minor Segment Major Segment A B Minor Segment O Area of minor segment = Area of sector AOB - Area of triangle AOB

O A B 8 cm O A B 100 o 12.2 cm 50 o 6.1 cm h M h 2 = (Pythagoras) h =  ( ) = 5.18 cm Area of triangle AOB = ½ (12.2 x 5.18) = cm 2 Segment area = – = 24.3 cm 2 (1 dp) Example Question 1: Find the area of the minor segment in the circle below. Not to SCALE Alternatively use ½ absinC for area of triangle. ½ x 8 2 x sin 100 o = 31.5 cm 2 Why are the answers slightly different?

A B 6.3 cm O O A B 8.4 cm h 2 = (Pythagoras) h =  ( ) = 4.70 cm Area of triangle AOB = ½ (8.4 x 4.70) = cm 2 Segment area = = 9.2 cm 2 (1 dp) Example Question 2: Find the area of the minor segment in the circle below. Angle AOB is needed to calculate the sector area o 4.2 cm h M  Not to SCALE Alternatively use ½ absinC for area of triangle. ½ x x sin 83.6 o = cm 2

h 2 = (Pythagoras) h =  ( ) = 0.44 m Area of triangle AOB = ½ (3.9 x 0.44) = 0.86 m 2 Segment area = = 4.4 m 2 (1 dp) Question 1: Find the cross-sectional area of the water in the cylindrical pipe. O A B 2 m 154 o 12.2 cm O A B 2 m 12.2 cm M h 77 o Not to SCALE 3.9 m 1.95 m Alternatively use ½ absinC for area of triangle. ½ x 2 2 x sin 154 o = 0.88 cm 2

A B 4.7 cm O O A B 6.2 cm h 2 = (Pythagoras) h =  ( ) = 3.53 cm Area of triangle AOB = ½ (6.2 x 3.53) = cm 2 Segment area = = 5.0 cm 2 (1 dp) Question 2: Find the area of the minor segment in the circle below. Angle AOB is needed to calculate the sector area o 3.1 cm h M  Not to SCALE Alternatively use ½ absinC for area of triangle. ½ x x sin 82.6 o = cm 2

Worksheets O A B 8 cm 100 o 12.2 cm O A B 6.3 cm 8.4 cm 1 2

O A B 2 m 154 o 12.2 cm 3.9 m O A B 4.7 cm 6.2 cm 1 2