Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/6`, and the triangle's area is `98`. What is the area of the triangle's incircle?

Answer 1

`color(chocolate)("Area of incircle " = pi r^2 = pi * (3.89)^2 = 47.59`

Explanation:


`hat A = pi/2, hat B pi/6, hat C = pi/3, A_t = 98`

`A_t = (1/2) ab sin C = (1/2(bc sin A = (1/2) ac sin B`

`ab =( 2A_t) /sin C = (2 * 98) / sin (pi/3) = 226.32`

`bc = (2 * 98) / sin (pi/2) = 196`

`ca = (2 * 98) / sin (pi/6) = 392`
4169.97
`a = sqrt(abc)^2 / (bc) =sqrt (226.32 * 196 * 392) / 196 =21.28`

`b = sqrt(abc)^2 / (ac) =sqrt (226.32 * 196 * 392) / 392 =10.64`

`c = sqrt(abc)^2 / (ab) =sqrt (226.32 * 196 * 392) / 226.32 =18.43`

`"Semiperimeter " = s = (a = b + c) / 2 = (21.28 + 10.64 + 18.43) / 2 = 25.18`

`color(chocolate)("Incircle radius " = r = A_t / s = 98 / 25.18 = 3.89`

`color(chocolate)("Area of incircle " = pi r^2 = pi * (3.89)^2 = 47.59`