Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/3`, vertex B has an angle of `(5 pi)/12`, and the triangle's area is `4`. What is the area of the triangle's incircle?

Answer 1

see a solution step below;

Explanation:

firstly, you have to look for the area of triangle..

`"Area of a triangle" = l^2`

Where, `l = 4`

`"Area of a triangle" = 4^2`

`"Area of a triangle" = 16`square units

Let;

`A = pi/3, B = (5pi)/12`

Since; `A + B + C = pi`

Therefore; `C = pi - A - B`

`C = pi - pi/3 - (5pi)/12`

`C = (12pi - 4pi - 5pi)/12`

`C = (3pi)/12`

`C = pi/4`

Finding their respective angles..

`a xx b = "Area"/(1/2 xx sin C) = 16/(0.5 xx sin (pi/3)) = 36.95`

similarly..

`b xx c = "Area"/(1/2 xx sin A) = 16/(0.5 xx sin ((5pi)/12)) = 33.13`

similarly..

`c xx a = "Area"/(1/2 xx sin B) = 16/(0.5 xx sin (pi/4)) = 45.25`

`(ab xx bc xx ca) = 36.95 xx 33.13 xx 45.25`

`(abc)^2 = 55,392.946`

squareroot both sides..

`sqrt((abc)^2) = sqrt(55,392.946)`

`abc = 235.357`

Now to get `a`

divide both sides by `bc` and its value..

`(abc)/(bc) = 235.357/33.13`

`a = 71`

smilarly..

`abc = 235.357`

divide both sides by `ca` and its value

`(abc)/(ca) = 235.357/45.25`

`b = 5.2`

smilarly..

`abc = 235.357`

divide both sides by `ab` and its value..

`(abc)/(ab) = 235.357/36.95`

`c = 6.37`

For half a perimeter; `(a + b + c)/2 = (7.1 + 5.2 + 6.37)/2 = 9.34`

`"radius" (r) = "Area"/"half of perimeter" = A/"semi perimeter" = 16/9.34 = 1.71`

`"Area of a circle" = pir^2`

`"Area of a circle" = 3.142 xx (1.71)^2`

`"Area of a circle" = 3.142 xx 2.9241`

`"Area of a circle" = 9.1875`

`"Area of a circle" = 9.19 "square units"`