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

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

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Answer 1 · 2018-02-02T11:42:29

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

Explanation:

$∠ A = \frac{π}{8} = \frac{180}{8} = {22.5}^{0}, ∠ B = \frac{π}{6} = \frac{180}{6} = 30^{0}$ $/_A = pi/8= 180/8=22.5^0 , /_B = pi/6=180/6= 30^0$

$∠ C = 180 - (22.5 + 30) = {127.5}^{0} : A_{t} = 22$ $/_C= 180-(22.5+30)=127.5^0: A_t=22$

We know Area , $A_{t} = \frac{1}{2} \cdot b \cdot c \cdot sin A$ $A_t= 1/2*b*c*sinA$ or

$b \cdot c = \frac{2 \cdot 22}{sin 22.5} \approx 114.98$ $b*c=(2*22)/sin22.5 ~~114.98$ , similarly , $a \cdot c = \frac{2 \cdot 22}{sin 30}$ $a*c=(2*22)/sin30$

$:.a*c = 88.0; a*b=(2*22)/sin127.5 ~~ 55.46$

$(a*b)*(b*c)*(c.a)=(abc)^2= (55.46*114.98*88.0)$

$=561165.55 :. abc =sqrt(561165.55)=749.11$

$a= (abc)/(bc)=749.11/114.98~~6.52$

$b= (abc)/(ac)=749.11/88.00~~8.51$

$c= (abc)/(ab)=749.11/55.46~~13.51$

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]

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

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1 Answer

Explanation:

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