# Area of Right Triangle Using Radius of Incircle

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BobDH
Area of Right Triangle Using Radius of Incircle
Jhun Vert

Good day sir. I made the attempt to trace the formula in your link, $A = R(a + b - c)$, but with no success. I notice however that at the bottom there is this line, $R = (a + b - c)/2$.

The radius of inscribed circle however is given by $R = (a + b + c)/2$ and this is true for any triangle, may it right or not. I have this derivation of radius of incircle here: https://www.mathalino.com/node/581.

I think that is the reason why that formula for area don't add up.

BobDH

Thank you for reviewing my post. I think, if you'll look again, you'll find my formula for the area of a right triangle is A = R (a + b - R), not A = R (a+ b - c).

Also, by your formula, R = (a + b + c) / 2 would mean that R for a 3, 4, 5 triangle would be 6.00, whereas, mine R = (a + b - c) /2 gives a R of 1.00.

See link below for another example:

http://mathforum.org/library/drmath/view/54670.html

Jhun Vert

My bad sir, I was not so keen in reading your post, even my own formula for R is actually wrong here. It should be $R = A_t / s$, not $R = (a + b + c)/2$ because $(a + b + c)/2 = s$ in the link I provided.

Anyway, thank again for the link to Dr. Math page. I never look at the triangle like that, the reason I was not able to arrive to your formula. Though simpler, it is more clever. I will add to this post the derivation of your formula based on the figure of Dr. Math. Thanks.

BobDH

No problem. Thanks for adding the new derivation. Nice presentation.

Jhun Vert

For the convenience of future learners, here are the formulas from the given link:
$A = r(a + b - r)$

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

Derivation:
From the figure below, AD is congruent to AE and BF is congruent to BE. Hence:
Area ADO = Area AEO = A2
Area BFO = Area BEO = A3

Area of triangle ABC
$A = A_1 + 2A_2 + 2A_3$

$A = r^2 + 2\left[ \dfrac{r(b - r)}{2} \right] + 2\left[ \dfrac{r(a - r)}{2} \right]$

$A = r^2 + (br - r^2) + (ar - r^2)$

$A = br + ar - r^2$

$A = r(a + b - r)$     ←   the formula

$AE + EB = AB$
$(b - r) + (a - r) = c$
$a + b - c = 2r$
$r = \dfrac{a + b - c}{2}$     ←   the formula