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?

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Answer 1 · 2018-05-17T11:31:45

Radius of the incircle is $2.0$ $2.0$ unit

In Right Triangle the legs are $A B = P = 5, B C = B = 12$ $AB=P=5, BC=B=12$

Then hypotenuse is $H = \sqrt{P^{2} + B^{2}} = \sqrt{25 + 144} = 13$ $H =sqrt(P^2+B^2)=sqrt(25+144)=13$

In Right Triangle incircle radius is

$r = \frac{P + B - H}{2} = \frac{5 + 12 - 13}{2} = 2.0$ $r = (P+B-H)/2=(5+12-13)/2=2.0$

Radius of the incircle is $2.0$ $2.0$ unit [Ans]

Answer 2 · 2018-05-17T11:44:09

$2$ $2$ .

Observe that, the hypotenuse $A C = 13$ $AC=13$ .

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 Respected Binayaka C. has readily derived!