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 #56 #. 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/2 #, vertex B has an angle of #( pi)/3  #, and the triangle's area is #56 #.  What is the area of the triangle's incircle?

Sophia Alfred · Accepted Answer

The area of the incircle is #=27.4u^2#

The area of the triangle is #A=56#

The angle #hatA=1/2pi#

The angle #hatB=1/3pi#

The angle #hatC=pi-(1/2pi+1/3pi)=1/6pi#

The sine rule is

#a/sinA=b/sinB=c/sinC=k#

So,

#a=ksinA#

#b=ksinB#

#c=ksinC#

Let the height of the triangle be #=h# from the vertex #A# to the opposite side #BC#

The area of the triangle is

#A=1/2a*h#

But,

#h=csinB#

So,

#A=1/2ksinA*csinB=1/2ksinA*ksinC*sinB#

#A=1/2k^2*sinA*sinB*sinC#

#k^2=(2A)/(sinA*sinB*sinC)#

#k=sqrt((2A)/(sinA*sinB*sinC))#

#=sqrt(112/(sin(pi/2)*sin(1/3pi)*sin(1/6pi)))#

#=10.58/0.66=16.03#

Therefore,

#a=16.03sin(1/2pi)=16.03#

#b=16.03sin(1/3pi)=13.88#

#c=16.03sin(1/6pi)=8.02#

The radius of the incircle is #=r#

#1/2*r*(a+b+c)=A#

#r=(2A)/(a+b+c)#

#=112/(37.93)=2.95#

The area of the incircle is

#area=pi*r^2=pi*2.95^2=27.4u^2#

Jayden Jackson · Answer

undefined To find the area of the triangle's incircle, we can use the formula:

$$ \text{Area of incircle} = \pi r^2 $$

where $ r $ is the radius of the incircle.

To find $ r $, we can use the formula:

$$ \text{Area of triangle} = rs $$

where $ s $ is the semi-perimeter of the triangle, and $ r $ is the inradius.

Given that the triangle has vertices A, B, and C with angles $ \frac{\pi}{2} $, $ \frac{\pi}{3} $, and area 56, respectively, and using the fact that the angles at the center of a circle are twice the angles at the circumference, we can determine the triangle's side lengths.

1. Angle at vertex A is $ \frac{\pi}{2} $, so side $ BC $ is the diameter of the incircle.
2. Angle at vertex B is $ \frac{\pi}{3} $, so side $ AC $ is the radius of the incircle.
3. Angle at vertex C is $ \frac{\pi}{6} $, so side $ AB $ is the radius of the incircle.

Since the area of the triangle is given as 56, and the semi-perimeter $ s = \frac{AB + BC + AC}{2} $, we can use the formula for the area of a triangle to find $ s $.

$$ 56 = \frac{AB \cdot BC}{2} $$

Now, using the fact that $ AB = AC = r $ and $ BC = 2r $ (as $ BC $ is the diameter of the incircle), we can solve for $ r $.

Once we find $ r $, we can plug it into the formula for the area of the incircle to find the answer.

Sophia Alfred · Answer

undefined To find the area of the triangle's incircle, you can use the formula:

$$Area = s \cdot r$$

where $s$ is the semi-perimeter of the triangle and $r$ is the radius of the incircle.

The semi-perimeter $s$ can be calculated as the sum of the lengths of the sides divided by 2.

$$s = \frac{AB + BC + CA}{2}$$

Using the given angles, you can find the lengths of the sides of the triangle using trigonometric ratios and the fact that the triangle is a right triangle (as angle $A$ is $π/2$ radians).

Once you have the lengths of the sides, you can find $s$ and then use the formula for the area of the incircle to find $r$. Then, you can calculate the area of the incircle using the formula:

$$Area = π \cdot r^2$$