ABC is a Right Triangle ,right angled at B. C(o,r) is the incircle and o is the incenter of the triangle.AB=5,BC=12,what is the radius of incircle?

Question

Sadie Allen · Accepted Answer

Radius of the incircle is #2.0# unit

In Right Triangle the legs are #AB=P=5, BC=B=12# Then hypotenuse is # H =sqrt(P^2+B^2)=sqrt(25+144)=13 # The incircle radius in a right triangle is #r = (P+B-H)/2=(5+12-13)/2=2.0# Radius of the incircle is #2.0# unit [Ans]

Savannah Aguilar · Answer

# 2#.

Observe that, the hypotenuse #AC=13#. From Trigonometry, we know that, #Delta=rs#. Here, #Delta="the Area of the triangle ABC"#, #=1/2*AB*BC#, #=1/2*5*12#, #=30#, #r="the inradius, and, "# #s="semi-perimeter"#, #=1/2(AB+BC+AC)#, #=1/2(5+12+13)#, #15#. #:. r=Delta/s#, #=30/15#, #=2#, as the Honorable Binayaka C. has easily deduced!

Sadie Allen · Answer

undefined To find the radius of the incircle of a right triangle $ABC$, we can use the formula for the radius of the incircle in terms of the triangle's sides:

$$ r = \frac{a + b - c}{2} $$

where $ a $, $ b $, and $ c $ are the lengths of the sides of the triangle, and $ r $ is the radius of the incircle.

Given that $ AB = 5 $ and $ BC = 12 $, and since $ ABC $ is a right triangle, we can use the Pythagorean theorem to find the length of the third side, $ AC $.

$$ AC^2 = AB^2 + BC^2 $$
$$ AC^2 = 5^2 + 12^2 $$
$$ AC^2 = 25 + 144 $$
$$ AC^2 = 169 $$

So, $ AC = 13 $.

Now, we can substitute the values of $ AB $, $ BC $, and $ AC $ into the formula for the radius of the incircle:

$$ r = \frac{AB + BC - AC}{2} $$
$$ r = \frac{5 + 12 - 13}{2} $$
$$ r = \frac{4}{2} $$
$$ r = 2 $$

Therefore, the radius of the incircle of triangle $ ABC $ is $ 2 $ units.