Areas of Elementary Plane Figures: Examples and Counterexamples

Here $B-A=(4,0)$ and $C-A=(2,3)$, so $\det [B-AC-A]=12$ and [L1] gives content $12\mathrm{/}2=6$.

The chosen horizontal base has length $4$, and $C$ is at perpendicular distance $3$ from its line, so the base--height form in [L1] gives $\frac{1}{2}\cdot 4\cdot 3=6$ again.

The determinant is $3\cdot 3-1\cdot 1=8$, so [L1] gives content $8$.

The base length is $\sqrt{10}$. Since $⟨v,w⟩=6$, the residual after projecting $w$ onto $\mathbb{R}v$ is $(-4\mathrm{/}5,12\mathrm{/}5)$ and has norm $8\mathrm{/}\sqrt{10}$; the base--height product is therefore $\sqrt{10}\cdot 8\mathrm{/}\sqrt{10}=8$.

One has $C-A=3(B-A)$, so the three vertices are collinear.

By [L1] the content is zero; directly, the determinant of $(2,1)$ and $(6,3)$ is $6-6=0$, and the perpendicular height is zero as well.

Substitute $r=1$ in [L1] to obtain $\mathrm{cont}(D_1)=\pi$.

The same disc lies between $y=-\sqrt{1-x^2}$ and $y=\sqrt{1-x^2}$ on $[-1,1]$, so [L2] identifies that value with the established graph-area convention, including the endpoint sections of height zero.

Write the vertices as $v_0,\dots ,v_4$ in the displayed order. The five supporting-line calculations $\det (v_{i+1}-v_i,v_j-v_i)>0(j\mathrm{\notin }\{i,i+1\})$ give respectively the positive lists $(8,16,12)$, $(8,8,11)$, $(16,8,7)$, $(6,10,7)$, and $(12,15,6)$, with indices modulo five. Thus every other vertex lies strictly to the left of every directed boundary edge. The intersection of these five closed left half-planes is convex, has exactly the displayed boundary chain, and contains both diagonals from $v_0$; hence the chain is simple and counterclockwise and those diagonals triangulate it. The three triangle determinants are $8$, $16$, and $6$, so [L1] gives content $(8+16+6)\mathrm{/}2=15$.

The cyclic shoelace sum is $(0+8+20+6+0)-(0+0+4+0+0)=30$, so [L2] gives content $30\mathrm{/}2=15$.

The triangulation and shoelace computations agree at $15$.

The matrix of $T$ is $\mathrm{diag}(2,1)$, with inverse $\mathrm{diag}(1\mathrm{/}2,1)$ and determinant $2$, and $T(Q)=[0,2]\times [0,1]$.

Each rectangle itself is both a one-rectangle inner family and outer cover, so [L2] and [L3] give $\mathrm{cont}(Q)=1$ and $\mathrm{cont}(T(Q))=2$; equivalently [L1] gives the same scaling. Thus the bijection $T$ does not preserve Jordan content.