Area of Right Triangle Using Radius of Incircle
See following link:
https://righttrianglecuriosities.quora.com/Area-of-a-Right-Triangle-Usi…
Active forum topics
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
- Sight Distance of Vertical Parabolic Curve
New forum topics
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
- Eliminate the Arbitrary Constants
Recent comments
- Yes.1 week 2 days ago
- Sir what if we want to find…1 week 2 days ago
- Hello po! Question lang po…3 weeks 6 days ago
- 400000=120[14π(D2−10000)]
(…2 months ago - Use integration by parts for…2 months 4 weeks ago
- need answer2 months 4 weeks ago
- Yes you are absolutely right…3 months ago
- I think what is ask is the…3 months ago
- $\cos \theta = \dfrac{2}{…3 months ago
- Why did you use (1/SQ root 5…3 months ago
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
In reply to Good day sir. I made the by Jhun Vert
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 reply to Thank you for reviewing my by BobDH
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.
No problem. Thanks for adding
In reply to My bad sir, I was not so keen by Jhun Vert
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, 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
Radius of inscribed circle:
$AE + EB = AB$
$(b - r) + (a - r) = c$
$a + b - c = 2r$
$r = \dfrac{a + b - c}{2}$ ← the formula