Is heron's formula inaccurate?

Question

Here's one example of inaccuracy :-

Suppose a triangle $XYZ$ with sides $a=13$, $b=15$ and $c= 14$. We have to find a perpendicular to side $c$ passing from point $X$.

Image Link : https://i.stack.imgur.com/dmFSL.jpg

According to heron's formulae

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$

$A = \sqrt{21 \times 8 \times 7 \times 6}$

$A = 7 \times 3 \times 4$

$A = 84$

Now

$1/2 × \text{base} × \text{height} = 84$

$1/2 × 14 × h = 84$

$H = \dfrac{84}7$

$H = 12$

But by Pythagoras theorem :-

$H ^2 = P^2 + B^2$

Now let's suppose our expected answer is $d$ then (Refer image above ),

$(a+b)^2 = (2d)^2 + c^2$

$28 ^2 = (2d)^2 + 14^2$

$784 = 4d^2 + 196$

$4d^2 = 784 - 196$

$d^2 = 147$

$d = \sqrt{147}$

Which is not equal to $12$ Which proves inaccuracy

Answer 1

Here's what you mean:

You said:

$(a+b)^2=(2d)^2+c^2$

$\Leftrightarrow (XY+XZ)^2=(XH+XH)^2+(HY+HZ)^2$

$\Leftrightarrow XY^2+XZ^2+2XY.XZ=XH^2+XH^2+2XH^2+HY^2+HZ^2+2HY.HZ$

$\Leftrightarrow XY^2+XZ^2+2XY.XZ=(XH^2+HY^2)+(XH^2+HZ^2)+2(XH^2+HY.HZ)$

$\Leftrightarrow XY^2+XZ^2+2XY.XZ=XY^2+XZ^2+2(XH^2+HY.HZ)$

$\Leftrightarrow XY.XZ=XH^2+HY.HZ$

This equality obviously is not true for $XH=12;XY=13;XZ=15;HY=5;HZ=9$, because it is impossible to prove the last equality in general, which means it is false.

Answer 2

The inaccuracy is, unfortunately, yours. The step $(a+b)^2=(2d)^2+c^2$ is not a proper way of handling the Pythagorean relations.

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If the altitude (of length $d$) separates the base into parts $p$ and $q$, then Pythagoras lets us write $$a^2 = d^2 + p^2 \qquad\text{and}\qquad b^2 = d^2 + q^2 \tag{1}$$ We can use the relation $p+q=c$ to combine these into an equation that allows us to solve for $d$.

Your error, I believe, was in attempting to combine the squared elements in $(1)$, term-by-term: $$(a+b)^2 = (d+d)^2 + (p+q)^2 \qquad(\text{error!})$$ This would be a very convenient algebraic trick, if it worked; but it doesn't. For example, $$5^2=3^2+4^2 \qquad 13^2=5^2+12^2 \qquad\text{but}\qquad \underbrace{(5+13)^2}_{324}\neq\underbrace{(3+5)^2+(4+12)^2}_{320}$$

Instead, we have to work quite a bit harder. We can start by replacing $q$ in $(1)$ with $c-p$: $$\begin{align} b^2 &= d^2 + (c-p)^2 \tag{2}\\ &= d^2 + c^2 + p^2 - 2 c p \tag{3}\\ &= d^2 + c^2+(a^2-d^2) - 2 c p \tag{4}\\ &=a^2 + c^2 - 2 cp \tag{5}\\ 2 cp&= a^2 -b^2 + c^2 \tag{6}\\ 4c^2p^2 &= ( a^2 -b^2 + c^2)^2 \tag{7}\\ 4c^2(a^2-d^2) &= a^4 + b^2 + c^4 - 2 a^2 b^2+2a^2c^2-2b^2c^2 \tag{8} \end{align}$$

For the problem at hand, we can substitute $a=13$, $b=15$, $c=14$ to get $$4\cdot 196 \cdot( 169 - d^2 ) = 19600 \quad\to\quad 169 - d^2 = 25 \quad\to\quad d^2 = 144 \quad\to\quad d = \pm 12$$ where we discard the extraneous negative option. Thus, $d=12$, as expected. $\square$

I could stop here, but I won't. Bear with me as I continue manipulating $(8)$ ...

$$\begin{align} 4a^2c^2-4c^2d^2 &= a^4 + b^2 + c^4 - 2 a^2 b^2-2a^2c^2-2b^2c^2 \tag{9} \\[4pt] -4c^2d^2 &= a^4 + b^2 + c^4 - 2 a^2 b^2-2a^2c^2-2b^2c^2 \tag{10}\\[4pt] 4c^2d^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \qquad(\text{trust me}) \tag{11}\\[4pt] \frac14 c^2d^2 &= \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{12}\\[4pt] \left(\frac12 c d\right)^2 &= \frac{a+b+c}{2} \cdot \frac{-a+b+c}{2} \cdot \frac{a-b+c}{2}\cdot \frac{a+b-c}{2} \tag{13} \end{align}$$ Interestingly, if we define $s = (a+b+c)/2$, we have $$s-a = \frac12(a+b+c)-a = \frac12(a+b+c-2a)=\frac{-a+b+c}{2} \tag{14}$$ Also, $$s- b = \frac{a-b+c}{2} \qquad s-c = \frac{a+b-c}{2} \tag{15}$$ Thus, $(13)$ becomes $$\left(\frac12cd\right)^2 = s(s-a)(s-b)(s-c) \tag{16}$$

But $\frac12 cd$ is the area of the triangle! Therefore,

$$\text{area}= \sqrt{s(s-a)(s-b)(s-c)} \tag{$\star$}$$

We have re-proven Heron's formula! $\square$