Hypotenuse and area given, find the radius of a circle inscribed in a right triangle?

Question

If the hypotenuse is given to be $13 $ cm, and the triangle's area is $30 $ cm$^2$ then how do I find the radius of a circle inscribed in the right triangle?

Answer 1

In order to find the lengths of the legs, we'll refer to this post:

Calculate sides of right triangle with hypotenuse and area or perimeter

In which they use the Pythagorean Theorem and the area formula. In your problem, you'd find:

$a = 5 \:\mathrm{cm}$

$b = 12 \:\mathrm{cm}$

To find the radius of the incircle (circle inscribed in the triangle), we can use the law of cotangents:

$1 / r = {\cot(A/2) \over s-a} = {\cot(B/2) \over s-b} = {\cot(C/2) \over s-c}$

Where $s$ is the semiperimeter:

$s = {a+b+c \over 2}$

Letting $c$ be the hypotenuse, we can rearrange this to be:

$r = (s - c)\tan(C/2)$

And we know that $C = 90$ degrees. Then:

$\tan(90/2) = \tan(45) = 1$

Finally:

$r = (s - c) = (a + b - c) / 2 = (5 + 12 - 13) / 2 = 2 \:\mathrm{cm}$

Answer 2

Let length and base be $x,y$ then $xy=60$ and $x^2+y^2=169$ solving these two gives us $(x,y)= (5,12)$.

Note that area of triangle equals $r(x+y+z)/2$ where $r$ is inradius and $x,y,z$ are sides of triangle. Thus $30= r(5+12+13)/2 \implies r=2$

Answer 3

Using the formula $$b+a=c+2r$$ we get by squaring $$b^2+a^2+2ab=4r^2+c^2+4rc$$ since $$2S=ab$$ and $$c^2=a^2+b^2$$ we get $$c^2+4S=(c+2r)^2$$ Can you finish?

Answer 4

If hypotenuse is 13, first thought comes is about Pythagorean triplet $ (5,12,13) $. Since area tallies as 30 square units it is determined there is no need to further go for Heron's formula.

We can use the in-radius formula with area and semi-perimeter:

$$ r = \frac{\Delta}{s} = \frac{ \dfrac12\cdot 5 \cdot 12}{\dfrac{5+12+13}{2}}= 2. $$

Answer 5

Let the side lengths be $p, q$ and the radius $r$.

With the axis along these sides, the equation of the hypotenuses is

$$qx+py=pq$$ and the center of the circle, $(r,r)$ is at distance

$$\frac{pq-qr-pr}{\sqrt{q^2+p^2}}=r$$ of the hypothenuses.

Hence

$$r=\frac{pq}{p+q+h}=\frac{pq}{\sqrt{p^2+q^2+2pq}+h}=\frac{2a}{\sqrt{h^2+4a}+h}.$$