ABC is A right angled triangle right angled at B with AB = 12,BC = 5.Find the radius of incircle to the triang

Question

ABC is A right angled triangle right angled at B with AB = 12,BC = 5.Find the radius of incircle to the triangle.

Answer 1

radius of a incircle of a triangle is 2A/P
where 'A' is the area of the triangle
'P' is the perimeter of the triangle

in your case
area of the triangle=(5x12)/2
A =30
P = 5+12+13

Apply these value in the formula
r=2A/P
=(2x30)/30
r =2

Answer 2

...A
....|\
... |..\
12|....\ 13
... |......\
... |____\
...B..5....C

Just for fun heres a little picture of it. Now if your triangle is 12,5,13. you want to find a box that can fit in it that will be the diameter of the circle.

So a box, or square with a lenght of 2 will fit inside the triangle. Then you circumscribe a circle (draw inside) in the square. The length is the diameter. And the radius of the circumscribed circle of the triangle is 1.

[edit]
for those who keep saying 2 is the radius of the incircle, recheck it. If the triangle where graphed, the point (4,4) does not exist on the line AC thus proving that the circle's radius isn't 2.

Answer 3

construct a triangle A'B'C' having angles A' = 40 4, B'= 80 one and C' = fifty 5 stages. Draw the bisectors of each attitude which will be concurrent at factor I, the incenter of triangle A'B'C'. Now draw a circle with radius 2.6 inches and center I. Draw perpendiculars from I to AC at D, to BC at E, and to AB at F. Draw a line || to A'B' and tangent to circle I. Draw similar strains || to B'C' and A'B'. Label their intersections as A,B,C. The triangle ABC is resembling triangle A'B'C' and is the mandatory triangle. the perimeters can now be computed through determining that the size of each tangent from factor A to circle I is two.6cot 22. further the different tangent lengths are 2.6cot 27.5 and a pair of.6cot 40.5. the area of the triangle is two.6s the position s is the semiperimeter of the triangle.

Answer 4

2. There is a formula for this, but I forget what it is, so I guessed 2 and found that it worked. Different values don't work, so 2 should be correct. One starts by finding the hypotenuse, which is 13.
There is a flaw in the subsequent responder's analysis: an incribed square will not be the same size as an inscribed circle -- the tangent point to the hypotenuse does not fall on the side of the square. (It is slightly inside.)

Answer 5

6.5