Question

in figure o is the centre of a circle such that diameter AB=13 and AC=12. bc is joines​

Answer 1

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.