NEED HELP trinagle

Ana.stasia

New member
Joined
Sep 28, 2020
Messages
38
The question is:
Calculate all sides of a triangle which has a 90 degree angle and it's area if you know that the circle drawn inside the triangle touches the side opposite to the 90 degree angle so that it divides the side into two part whose ratio is 2:3 and if you know that thr sum of all sides is 36.

I tried to get a formula that would get me a ratio of sorts for the sides so I can just divide 36 by the ration. However I was unable to do that.
Here is everything I could come up with.

What am I not seeing?

IMG_20201014_182726.jpg
IMG_20201014_182731.jpg
 

skeeter

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Joined
Dec 15, 2005
Messages
2,538
the circle center, the incenter, occurs at the intersection of the angle bisectors of the triangle’s three angles
 

Ana.stasia

New member
Joined
Sep 28, 2020
Messages
38
the circle center, the incenter, occurs at the intersection of the angle bisectors of the triangle’s three angles
Okay, that would tell me that x=b-r and y=c-r which brings me to the conclusion that r+ 5k =18. What next?
 

skeeter

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\(\displaystyle 10x + 2r = 36\)

\(\displaystyle (r+2x)^2 + (r+3x)^2 = (2x+3x)^2\)

try again ...


right_incenter.jpg
 

Ana.stasia

New member
Joined
Sep 28, 2020
Messages
38
\(\displaystyle 10x + 2r = 36\)

\(\displaystyle (r+2x)^2 + (r+3x)^2 = (2x+3x)^2\)

try again ...


View attachment 22311
I got r+5k=18 which is the same thing as 2r+10x=36. It's only multiplied by two. I don't understand how (r+2x)2+(r+3x)2=(2x+3x)2
What is (2x+3x)2 ?
 

skeeter

Elite Member
Joined
Dec 15, 2005
Messages
2,538
Pythagoras

base leg = r + 3x

vertical leg = r + 2x

hypotenuse = 2x + 3x = 5x
 
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