Exercises — Heron's formula (derivation using trig)

WHAT: just add and halve. $s=\frac{6+8+10}{2}=\frac{24}{2}=12.$ Then $s-a=12-6=6,s-b=12-8=4,s-c=12-10=2.$ WHY these must be positive: each is half of a "two-sides-minus-the-third" quantity, e.g. $s-c=\frac{1}{2}(a+b-c)$. The Triangle Inequality guarantees $a+b>c$, so this is positive — a green light that the triangle is real.

$\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{12\cdot 6\cdot 4\cdot 2}.$ Multiply inside: $12\cdot 6=72$, $4\cdot 2=8$, $72\cdot 8=576$. $\text{Area}=\sqrt{576}=24.$ Sanity check: $6,8,10$ is a scaled $3\text{-}4\text{-}5$ right triangle (legs $6,8$), so area $=\frac{1}{2}(6)(8)=24$. ✅ Agrees, so the plug-in was clean.

Check it's a real triangle first: $9+10=19>17$. ✅ $s=\frac{9+10+17}{2}=\frac{36}{2}=18.$ $s-a=9,s-b=8,s-c=1.$ $\text{Area}=\sqrt{18\cdot 9\cdot 8\cdot 1}=\sqrt{1296}=36.$ WHY $s-c=1$ is small: side $c=17$ is almost as long as $a+b=19$, so the triangle is very "flat/thin." A near-zero $s-c$ warns you the triangle is close to degenerate.

Heron: $s=\frac{5+5+6}{2}=8,s-a=3,s-b=3,s-c=2.$ $\text{Area}=\sqrt{8\cdot 3\cdot 3\cdot 2}=\sqrt{144}=12.$ Base–height: drop the height onto the base $c=6$. By symmetry it splits $c$ into two halves of $3$. The height $h$ satisfies (right triangle with hypotenuse $5$): $h=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4.$ $\text{Area}=\frac{1}{2}\cdot 6\cdot 4=12.$ ✅ Both give $12$ — Heron is just the height method with the height already "solved out."

WHAT we're missing: side $c$. Perimeter gives it: $a+b+c=32\Rightarrow c=32-10-13=9.$ Then: $s=\frac{32}{2}=16$ (semi-perimeter is half the perimeter, so we don't even re-add). $s-a=6,s-b=3,s-c=7.$ $\text{Area}=\sqrt{16\cdot 6\cdot 3\cdot 7}=\sqrt{2016}=\sqrt{144\cdot 14}=12\sqrt{14}.$ Numerically $12\sqrt{14}\approx 44.90$. WHY leave it in surd form: $2016$ is not a perfect square, so the exact area is irrational; $12\sqrt{14}$ is the honest exact answer.

Step 1 — get $\sin C$ from the area. $C$ is the angle between sides $a=13$ and $b=14$: $84=\frac{1}{2}(13)(14)\sin C=91\sin C\Rightarrow \sin C=\frac{84}{91}=\frac{12}{13}.$ Step 2 — get $\cos C$. Since ${\sin }^{2}C+{\cos }^{2}C=1$ and $C$ is a triangle angle ($0<C<180^{\circ }$), $\cos C=\pm \sqrt{1-{\left(\frac{12}{13}\right)}^2}=\pm \sqrt{1-\frac{144}{169}}=\pm \frac{5}{13}.$ Step 3 — pick the sign with the Law of Cosines (this settles the $\pm$): $\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{169+196-225}{2\cdot 13\cdot 14}=\frac{140}{364}=\frac{5}{13}.$ Positive → $C$ is acute → take $\cos C=+\frac{5}{13}$. ✅ Both methods agree, closing the loop between the trig-area start and the Law of Cosines used in the derivation.

From the parent note: $s=21$ and $\text{Area}=84$. $r=\frac{\text{Area}}{s}=\frac{84}{21}=4.$ WHY this formula holds: connect each vertex to the incentre, splitting the triangle into three smaller triangles, each with base a side and height $r$. Their areas sum to $\frac{1}{2}r(a+b+c)=rs$. Setting that equal to the whole area gives $r=\text{Area}/s$. See Semi-perimeter and Incircle (r = Area/s) — the same $s$ from Heron reappears.

Step 1 — scale factor. The ratio parts sum to $5+12+13=30$. Perimeter $60$ means each "part" is worth $60/30=2$. So $a=10,b=24,c=26.$ Step 2 — right angle? Check $10^2+24^2=100+576=676=26^2$. ✅ Right-angled at the vertex opposite $c=26$. Step 3 — area (Heron, to practise it): $s=30,s-a=20,s-b=6,s-c=4.$ $\text{Area}=\sqrt{30\cdot 20\cdot 6\cdot 4}=\sqrt{14400}=120.$ Cross-check with legs: $\frac{1}{2}(10)(24)=120.$ ✅ Consistent.

General argument: if $c=a+b$ then $a+b-c=0$, and recall $s-c=\frac{1}{2}(a+b-c)=0$. One factor under the root vanishes, so the whole product is $0$: $\text{Area}=\sqrt{s\cdot (s-a)(s-b)\cdot 0}=0.$ Concrete case $a=3,b=4,c=7$: $s=\frac{14}{2}=7$, so $s-c=7-7=0$. $\text{Area}=\sqrt{7\cdot 4\cdot 3\cdot 0}=0.$ WHAT IT LOOKS LIKE (see figure): the "triangle" has collapsed onto a straight segment — the point that would be the apex has flattened into the base. Zero height ⇒ zero area, and Heron reports it faithfully at the boundary between "real triangle" ($c<a+b$) and "impossible" ($c>a+b$).

Setup. Fix $s=\frac{P}{2}$ (constant). Heron gives ${\text{Area}}^2=s(s-a)(s-b)(s-c).$ Since $s$ is fixed, maximising the area means maximising the product $X=(s-a)(s-b)(s-c)$. Constraint on the three factors. Their sum is fixed: $(s-a)+(s-b)+(s-c)=3s-(a+b+c)=3s-2s=s.$ So we're maximising a product of three positive numbers whose sum is fixed at $s$. AM–GM. For positive numbers, the product is largest when they are all equal: $\sqrt[3]{XYZ}\le \frac{X+Y+Z}{3},\text{equality iff }X=Y=Z.$ Here equal means $s-a=s-b=s-c$, i.e. $a=b=c$ — the equilateral triangle. Value at the max. Each factor $=\frac{s}{3}$, so ${\text{Area}}^2=s{\left(\frac{s}{3}\right)}^3=\frac{s^4}{27},\text{Area}=\frac{s^2}{3\sqrt{3}}.$ Sanity check with side length. For an equilateral triangle side $=\frac{2s}{3}$, and the parent note's result $\frac{\sqrt{3}}{4}{a}^2$ gives $\frac{\sqrt{3}}{4}{\left(\frac{2s}{3}\right)}^2=\frac{\sqrt{3}}{4}\cdot \frac{4s^2}{9}=\frac{\sqrt{3}{s}^2}{9}=\frac{s^2}{3\sqrt{3}}.$ ✅ Same value. Equilateral wins.

WHY this form exists: it is Heron before substituting $s$ — the exact line from Step 5 of the derivation, multiplied through by $16$. Plug in $a=13,b=14,c=15$:

• $a+b+c=42$
• $-a+b+c=-13+14+15=16$
• $a-b+c=13-14+15=14$
• $a+b-c=13+14-15=12$ $42\cdot 16\cdot 14\cdot 12=112896.$ Divide by $16$: ${\text{Area}}^2=\frac{112896}{16}=7056$, so $\text{Area}=\sqrt{7056}=84.$ ✅ This is the same $84$ — the "16 form" and the "$s$ form" are one formula wearing two outfits.