See following link:

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

April 12, 2018 - 6:00am

#1
BobDH

Area of Right Triangle Using Radius of Incircle

See following link:

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

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- Finding the sum of sequence
- Fractional Parts of Measured Dimensions on Hypotenuses of Right Right Triangles
- Product of Areas of Three Dissimilar Right Triangles
- Area of Right Triangle Using Radius of Incircle
- The Formula for the Curve of a Trumpet Bell
- Perimeter of Right Triangle by Tangents
- Basic Cal.
- How can i compute the height of the ice cube to be put in a conical tank when it turns into a liquid
- Hansen's Theorem
- ADVANCED MATH

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 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 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 the new derivation. Nice presentation.

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