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3. 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

4. O A B 8 cm O A B 100 o 12.2 cm 50 o 6.1 cm h M h 2 = 8 2 - 6.1 2 (Pythagoras) h =  (8 2 - 6.1 2 ) = 5.18 cm Area of triangle AOB = ½ (12.2 x 5.18) = 31.60 cm 2 Segment area = 55.85 – 31.60 = 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?

5. A B 6.3 cm O O A B 8.4 cm h 2 = 6.3 2 - 4.2 2 (Pythagoras) h =  (6.3 2 - 4.2 2 ) = 4.70 cm Area of triangle AOB = ½ (8.4 x 4.70) = 19.74 cm 2 Segment area = 28.96 - 19.74 = 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. 83.6 o 4.2 cm h M  Not to SCALE Alternatively use ½ absinC for area of triangle. ½ x 6.3 2 x sin 83.6 o = 19.72 cm 2

6. h 2 = 2 2 - 1.95 2 (Pythagoras) h =  (2 2 - 1.95 2 ) = 0.44 m Area of triangle AOB = ½ (3.9 x 0.44) = 0.86 m 2 Segment area = 5.38 - 0.86 = 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

7. A B 4.7 cm O O A B 6.2 cm h 2 = 4.7 2 - 3.1 2 (Pythagoras) h =  (4.7 2 - 3.1 2 ) = 3.53 cm Area of triangle AOB = ½ (6.2 x 3.53) = 10.94 cm 2 Segment area = 15.92 - 10.94 = 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. 82.6 o 3.1 cm h M  Not to SCALE Alternatively use ½ absinC for area of triangle. ½ x 4.7 2 x sin 82.6 o = 10.95 cm 2

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

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