in figure o is the centre of a circle such that diameter AB=13 and AC=12. bc is joines
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ANSWER
In ΔABC,
In ΔABC,Angle in a semicircle, ∠C=90o
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theorem
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132BC2=169−144=25
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132BC2=169−144=25BC=5
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132BC2=169−144=25BC=5Area of shaded region = Area of semicircle – Area of triangle
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132BC2=169−144=25BC=5Area of shaded region = Area of semicircle – Area of triangle= 21πr2−21bh
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132BC2=169−144=25BC=5Area of shaded region = Area of semicircle – Area of triangle= 21πr2−21bh= 21×3.14×(213)2−21×5×12
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132BC2=169−144=25BC=5Area of shaded region = Area of semicircle – Area of triangle= 21πr2−21bh= 21×3.14×(213)2−21×5×12= 2×43.14×13×13−30
In ΔABC,Angle in a semicircle, ∠C=90oTherefore, by Pythagoras theoremBC2+AC2=AB2BC2+122=132BC2=169−144=25BC=5Area of shaded region = Area of semicircle – Area of triangle= 21πr2−21bh= 21×3.14×(213)2−21×5×12= 2×43.14×13×13−30Area of shaded region =36.39 cm2.
