Question

solve kro

kaha ho ab aao sab​

Answer 1

The right angled triangle, ABC with right angle at B, BC = 15 cm and AB = 8 cm, will have AC, the hypotenuse as [15^2+8^2]^0.5 = [225+64]^0.5 289^0.5 = 17 cm.

Let the radius of the inscribed circle = r.

Let the circle touch BC at D, AC at E and AB at F.

AB = 8 = r+AF. Or AF = 8-r

BC = 15 = r+CD. Or CD = 15-r.

AC = 17 = AE+EC = AF+CD = (8-r)+(15-r) = 23–2r, or

2r = 23–17 = 6, or

r the inradius = 6/2 = 3 cm.