Fun with the right trianlge.

**ThePerfectHacker** · January 23rd 2006, 11:48 AM

The other day I saw something interesting,
Take any right triangle. Then the products of the segments of the hypotenuse divided by the inscribed circle is its area.

**earboth** · January 24th 2006, 10:42 AM

Originally Posted by ThePerfectHacker

The other day I saw something interesting,
Take any right triangle. Then the products of the segments of the hypotenuse divided by the inscribed circle is its area.

Hello,

I'm a little bit puzzled about your question: What are the segments of the hypotenuse and what exactly do you take as divisor? The radius, the perimeter or the area of the inscribed circle?

I've attached a drawing of a right triangle so you can check which names I use for the variables.

Heron of Alexandra had found that in a right triangle is $\rho=\frac{a \cdot b}{a+b+c}$

The three bisectors of a triangle divide it into three smaller triangles. The height in every of these smaller triangles is always ρ. That's true with every triangle.

So you get:
$A=\frac{1}{2} \cdot \rho \cdot a+\frac{1}{2} \cdot \rho \cdot b+\frac{1}{2} \cdot \rho \cdot c=\frac{1}{2} \cdot \rho \cdot (a+b+c)$

Heron set $\frac{1}{2} \cdot (a+b+c)=s$ and you get: $A=\rho \cdot s$

With a right triangle you've additional conditions:
$A=\frac{1}{2} \cdot c \cdot h=\frac{1}{2} \cdot a \cdot b=\frac{1}{2} \cdot \rho \cdot (a+b+c)$

and therefore $\rho=\frac{a\cdot b}{a+b+c}$

With $a^2=p \cdot c$ and $b^2=q \cdot c$ you can change $p \cdot q=\frac{a^2}{c} \cdot \frac{b^2}{c}=\frac{a^2 \cdot b^2}{c^2}$

Now you want to divide this term by a part of the inscribed circle. But I cann't help you any further because I don't know which part you want to take. But maybe you can use my remarks to get a little bit ahead.

Bye

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