Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/3`, and the triangle's area is `21`. What is the area of the triangle's incircle?

Answer 1

The area of the triangle's incircle is `10.2` sq.unit.

Explanation:

`/_A = pi/2= 180/2=90^0 , /_B = pi/3=180/3= 60^0 , /_C= 180-(90+60)=30^0 ; Delta_a=21`

This is a right triangle of `(30,60,90)`

In right triangle `(30,60,90)`, Base is `c=x` , Hypotenuse is

`a=2x` and Perpendicular is `b=sqrt3*x`

Area of the triangle ,`A_t= 1/2*c*b=21 or 21= 1/2*x*sqrt3*x`

`:.x^2=42/sqrt3 or x =sqrt(14*sqrt3) ~~ 24.25 :. c ~~ 4.924`

`b= sqrt3*x ~~ 8.53 ; a=2x~~ 9.85`

Semi perimeter is `S/2=(4.924+8.53+9.85)/2 ~~ 11.65`

Incircle radius is `r_i= A_t/(S/2) = 21/11.65~~1.8`

Incircle Area = `A_i= pi* r_i^2= pi*1.8^2 ~~10.2` sq.unit

The area of the triangle's incircle is `10.2` sq.unit

Answer 2

see a step process below;

Explanation:

Given;

`A = pi/2, B = pi/3, C = ?`

`A + B + C = pi`

`pi/2 + pi/3 + C = pi`

`C = pi - pi/2 - pi/3`

`C = (6pi - 3pi - 2pi)/6`

`C = pi/6`

Area `= 21`

Getting the length..

`A rArr 21/(1/2 xx sin (pi/2)) ~~ 48.50`

`B rArr 21/(1/2 xx sin (pi/3)) ~~ 42`

`C rArr 21/(1/2 xx sin (pi/6)) ~~ 84`

`(ABC)^2 = 84 xx 42 xx 48.50`

`(ABC)^2 = 171108`

`ABC = sqrt171108`

`ABC = 413.65`

`A rArr 413.65/48.50 ~~ 8.53`

`B rArr 413.65/42 ~~ 9.85`

`C rArr 413.65/84 ~~ 4.12`

Getting semi-perimeter..

`(8.53 + 9.85 + 4.12)/2 = 23.3/2 = 11.65`

Getting the radius..

`r = 21/11.65 = 1.80`

Recall;

`A = pir^2`

`A = 3.142(1.80)^2`

`A = 3.142 xx 3.24`

`A = 10.18 ~~ 10.2"sq units"`