A triangle has corners at #(1 , 5 )#, #(9 ,4 )#, and #(1 ,8 )#. What is the radius of the triangle's inscribed circle?

Question

A triangle has corners at #(1  , 5    )#, #(9   ,4  )#, and #(1 ,8   )#.  What is the radius of the triangle's inscribed circle?

Avery Allen · Accepted Answer

#1.200#

Refer to figure below. (Obs.: D is the circle's center)

With #A(1,5), B(9,4) and C(1,8)#
#AB=sqrt((9-1)^2+(4-5)^2)=sqrt(64+1)=sqrt(65)~=8.1#
#BC=sqrt((1-9)^2+(8-4)^2)=sqrt(64+16)=sqrt(80)=4sqrt(5)~=8.9#
#CA=sqrt((1-1)^2+(5-8)^2)=3#

Using the variables #m, n and o#
#m+n=4sqrt(5)#
#m+o=sqrt(65)#
#n+o=3# => #o=3-n#
=> #m+3-n=sqrt(65)# => #m-n=sqrt(65)-3#
Adding the last with the first equation, we get
#2m=4sqrt(5)+sqrt(65)-3# => #m=2sqrt(5)+sqrt(65)/2-1.5#

In the figure we can see that
#tan alpha=r/m# => #r=m*tan alpha#

We only need to know #alpha#

Using vectors
#BvecA=(1-9)hat i+(5-4)hat j=-8hat i+hat j#
#BvecC=(1-9)hat i+(8-4)hat j=-8hat i+4hat j#

#cos (2alpha)=(BvecA*BvecC)/(|BA|*|BC|)=(8*8+1*4)/(sqrt(65)*4sqrt(5))=68/(20sqrt(13))=17/(5sqrt(13))#
=> #2alpha=19.440^@# => #alpha=9.720^@#

Using the Law of Sines
#3/sin(180^@-2alpha)=sqrt(65)/sin(180^@-2beta)=(4sqrt(5))/sin(180^@-2gamma)#
Since sines of supplementary angles are equal we can rewrite the previous equation as
#3/(sin 2alpha)=sqrt(65)/(sin 2beta)=(4sqrt(5))/(sin 2gamma)#
#3/(sin 2alpha)=sqrt(65)/(sin 2beta)# => #sin 2beta=sqrt(65)/3*sin 2alpha#

We know also that #2alpha+2beta+2gamma=180^@# => #2gamma=180^@-2alpha-2beta#
So
#3/sin(2alpha)=(4sqrt(5))/sin(180^@-2alpha-2beta)=(4sqrt(5))/sin (2alpha+2beta)=(4sqrt(5))/(sin 2alpha *cos 2beta+sin 2beta *cos 2alpha)#

#-> 3/cancel (sin2alpha)=##(4sqrt(5))/(cancel(sin 2alpha)*sqrt(1-65/9*sin^2 2alpha)+sqrt(65)/3*cancel(sin 2alpha) *sqrt(1-sin^2 2alpha)#

#-> cancel(3)*sqrt(9-65sin^2 2alpha)/cancel(3)+cancel(3)*sqrt(65)/cancel(3)*sqrt(1-sin^2 2alpha)=4sqrt(5)#
#-> (sqrt(9-65sin^2 2alpha))^2=(4sqrt(5)-sqrt(65)*sqrt(1-sin^2 2alpha))^2#
#-> 9-cancel(65sin^2 2alpha)=80-8*5*sqrt(13)*sqrt(1-sin^2 2alpha)+65-cancel(65sin^2 2alpha)#

#40sqrt(13)*sqrt(1-sin^2 2alpha)=136#
#1300-1300*sin^2 2alpha=1156# => #sin^2 2alpha=144/1300# => #sin 2alpha=6/(5sqrt(13))# => #2alpha=19.440^@# => #alpha=9.720^@#

Finally,
#r=m*tan alpha=(2sqrt(5)+sqrt(65)/2-1.5)*tan 9.720^@# => #r=1.220#

Aaliyah Anderson · Answer

undefined To find the radius of the inscribed circle of a triangle, you can use the formula:

$$r = \frac{2 \times \text{Area of Triangle}}{\text{Perimeter of Triangle}}$$

Given the coordinates of the triangle's vertices, you can calculate the lengths of its sides using the distance formula. Then, use Heron's formula to find the area of the triangle. Finally, compute the perimeter of the triangle. Substitute these values into the formula above to find the radius of the inscribed circle.

Aurora Ayala · Answer

undefined To find the radius of the triangle's inscribed circle, you can use the formula for the radius of the inscribed circle of a triangle, which is given by:

$$ r = \frac{2 \cdot \text{Area of the Triangle}}{\text{Perimeter of the Triangle}} $$

First, calculate the area of the triangle using the coordinates of its vertices and the formula for the area of a triangle given its coordinates.

Then, find the perimeter of the triangle by summing the lengths of its sides.

Finally, plug these values into the formula for the radius of the inscribed circle to find the radius $ r $.