A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #22 #. 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/8 #, vertex B has an angle of #( pi)/6  #, and the triangle's area is #22 #.  What is the area of the triangle's incircle?

Savannah Aguilar · Accepted Answer

The area of incircle of triangle is #7.47# sq.unit.

#/_A = pi/8= 180/8=22.5^0 , /_B = pi/6=180/6= 30^0# # /_C= 180-(22.5+30)=127.5^0: A_t=22 # We know Area ,# A_t= 1/2*b*c*sinA# or #b*c=(2*22)/sin22.5 ~~114.98 #, similarly ,#a*c=(2*22)/sin30# #:.a*c = 88.0; a*b=(2*22)/sin127.5 ~~ 55.46# #(a*b)*(b*c)*(c.a)=(abc)^2= (55.46*114.98*88.0)# # =561165.55 :. abc =sqrt(561165.55)=749.11 # #a= (abc)/(bc)=749.11/114.98~~6.52# #b= (abc)/(ac)=749.11/88.00~~8.51# #c= (abc)/(ab)=749.11/55.46~~13.51# Semi perimeter : #S/2=(6.52+8.51+13.51)/2 ~~14.27# Incircle radius is #r_i= A_t/(S/2) = 22.0/14.27 ~~1.54# Incircle Area = #A_i= pi* r_i^2= pi*1.54^2 ~~ 7.47# sq.unit. The area of incircle of triangle is #7.47# sq.unit. [Ans]

Evelyn Arboleda · Answer

undefined To find the area of the triangle's incircle, use the formula $A = rs$, where $A$ is the area of the triangle, $r$ is the radius of the incircle, and $s$ is the semiperimeter of the triangle. Calculate the semiperimeter $s$ using the formula $s = \frac{a + b + c}{2}$, where $a$, $b$, and $c$ are the lengths of the sides of the triangle. Then, use the given angles to find the lengths of the sides using trigonometric ratios. Once you have $s$, use it along with the given area $A$ to find $r$. Finally, use the formula for the area of a circle $A = \pi r^2$ to find the area of the incircle.