A triangle has vertices A, B, and C. Vertex A has an angle of $pi/6$ , vertex B has an angle of $(pi)/12$ , and the triangle's area is $24$ . What is the area of the triangle's incircle?

Question

A triangle has vertices A, B, and C. Vertex A has an angle of $pi/6$ , vertex B has an angle of $(pi)/12$ , and the triangle's area is $24$ . What is the area of the triangle's incircle?

1 Answer

Impact of this question

1357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-03T11:24:30

The area of triangle's incircle is $6.45$ sq.unit.

Explanation:

$/_A = pi/6= 180/6=30^0 , /_B = pi/12=180/12= 15^0 , /_C= 180-(30+15)=135^0 , A_t=24$
We now Area , $A_t= 1/2*b*c*sinA or b*c=(2*24)/sin30 = 96$ , similarly , $a*c=(2*24)/sin15 = 185.46$ , and $a*b=(2*24)/sin135 = 67.88$

$(a*b)*(b*c)*(c.a)=(abc)^2= (96*185.46*67.88)$ or
$abc=sqrt((96*185.46*67.88)) = 1099.35$

$a= (abc)/(bc)=1099.35/96~~11.45$
$b= (abc)/(ac)=1099.35/185.46~~5.93$
$c= (abc)/(ab)=1099.35/67.88~~16.19$

Semi perimeter : $S/2=(11.45+5.93+16.20)/2=16.79$
Incircle radius is $r_i= A_t/(S/2) = 24/16.79=1.43$
Incircla Area = $A_i= pi* r_i^2= pi*1.43^2 =6.45$ sq.unit.

The area of triangle's incircle is $6.45$ sq.unit. [Ans]

Science

Math

Humanities

... and beyond

A triangle has vertices A, B, and C. Vertex A has an angle of $pi/6$ , vertex B has an angle of $(pi)/12$ , and the triangle's area is $24$ . What is the area of the triangle's incircle?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A triangle has vertices A, B, and C. Vertex A has an angle of pi/6 pi/6 , vertex B has an angle of (pi)/12 (pi)/12 , and the triangle's area is 24 24 . What is the area of the triangle's incircle?

1 Answer

Explanation:

Related questions

Impact of this question

A triangle has vertices A, B, and C. Vertex A has an angle of $pi/6$ , vertex B has an angle of $(pi)/12$ , and the triangle's area is $24$ . What is the area of the triangle's incircle?