# Area of Right Triangle Using Radius of Incircle

Submitted by BobDH on April 12, 2018 - 6:00am

See following link:

https://righttrianglecuriosities.quora.com/Area-of-a-Right-Triangle-Usin...

- Add new comment
- 780 reads

Submitted by BobDH on April 12, 2018 - 6:00am

See following link:

https://righttrianglecuriosities.quora.com/Area-of-a-Right-Triangle-Usin...

- Add new comment
- 780 reads

## Good day sir. I made the

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.

## Thank you for reviewing my

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

## My bad sir, I was not so keen

My bad sir, I was not so keen in reading your post, even my own formula for

Ris 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.

## No problem. Thanks for adding

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

## For the convenience of future

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,

ADis congruent toAEandBFis congruent toBE. Hence:Area

ADO= AreaAEO=A_{2}Area

BFO= AreaBEO=A_{3}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

Radius of inscribed circle:

$AE + EB = AB$

$(b - r) + (a - r) = c$

$a + b - c = 2r$

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

## Add new comment