A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{12}$ $pi/12$ , vertex B has an angle of $\frac{π}{2}$ $(pi)/2$ , and the triangle's area is $3$ $3$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-12-02T11:11:41

Area of triangle's incircle is $0.95$ $0.95$ sq.unit.

$∠ A = \frac{π}{12} = \frac{180}{12} = 15^{0}, ∠ B = \frac{π}{2} = \frac{180}{2} = 90^{0}$ $/_A = pi/12= 180/12=15^0 , /_B = pi/2=180/2= 90^0$

$:. /_C= 180-(90+15)=75^0 ; A_t=3$

We know Area , $A_t= 1/2*b*c*sinA or b*c=(2*3)/sin15 ~~23.18$ .

Similarly $a*c=(2*3)/sin90 = 6.0$ , and $a*b=(2*3)/sin75~~ 6.21$

$(a*b)*(b*c)*(c.a)=(abc)^2= (6.21*23.18*6.00) ~~863.69$

$:.abc=sqrt(863.69) ~~29.39 ; a=(abc)/(bc)=29.39/23.18~~1.27$

$b= (abc)/(ac)=29.39/6.0~~4.90$ and

$c= (abc)/(ab)=29.39/6.21~~4.73$

Semi perimeter : $S/2=(1.27+4.90+4.73)/2~~5.45$

Incircle radius is $r_i= A_t/(S/2) = 3/5.45~~0.55$

Incircle Area = $A_i= pi* r_i^2= pi*0.55^2 ~~ 0.95$ sq.unit

Area of triangle's incircle is $0.95$ sq.unit [Ans]

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 π12pi/12 , vertex B has an angle of (pi)/2 π2(pi)/2 , and the triangle's area is 3 33 . What is the area of the triangle's incircle?