1.In figure 7.43, A is the centre of thecircle. ABC = 45Â° and AC =

1.	1.In figure 7.43, A is the centre of thecircle. ABC = 45Â° and AC = 712 cmFind the area of segment BXC.45Fig. 7.43
Answer» Area of Shaded Portion:area of sector- area of triangle AB = AC (Radius)therefore ang ABC = ang ACB THEREFORE ang BAC = 90 Area of sector= pi x (r)^2 x (Theta)/360= (22/7)(7√2)^2 x(90/360)= (22/7)(98)(1/4)by solvingArea = 77cm^2 area of triangle right angle trianglePythagoras theoremAB^2 + AC^2 = BC^2(7√2)^2 + (7√2)^2 = BC^22x 98 = BC^2BC^2 = 196BC = 14 area = (1/2) Base x Height= (1/2) (7√2)(7√2)= 49 Area of Shaded = 77-49= 28cm^2 Hit like if you find it useful!

1.In figure 7.43, A is the centre of thecircle. ABC = 45Â° and AC = 712 cmFind the area of segment BXC.45Fig. 7.43

Answer»

Area of Shaded Portion:area of sector- area of triangle

AB = AC (Radius)therefore ang ABC = ang ACB

THEREFORE ang BAC = 90

Area of sector= pi x (r)^2 x (Theta)/360= (22/7)(7√2)^2 x(90/360)= (22/7)(98)(1/4)by solvingArea = 77cm^2

area of triangle

right angle trianglePythagoras theoremAB^2 + AC^2 = BC^2(7√2)^2 + (7√2)^2 = BC^22x 98 = BC^2BC^2 = 196BC = 14

area = (1/2) Base x Height= (1/2) (7√2)(7√2)= 49

Area of Shaded = 77-49= 28cm^2

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