I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region.

Take such a region $D$ in $\mathbb{R}^2$, $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$

$$ \eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D) $$

exists. If this is the case, then we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.

The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region

$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace, $$

then $D(U,L)$ is measurable

$$\eA(D(U,L))=\int_a^b \bigl(\; U(x)-L(x)\;\bigr) dx. $$

The next thing to prove is a weak form of the inclusion-exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the grapf of a $C^1$-function, then $D_1\cup D_2$ is measurable and

$$\eA(D_1\cup D_2)=\eA(D_1)+\eA(D_2). $$

I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region.

Take such a region $D$ in $\mathbb{R}^2$, $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$

$$ \eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D) $$

exists. If this is the case, then we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.

The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region

$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace, $$

then $D(U,L)$ is measurable

$$\eA(D(U,L))=\int_a^b \bigl(\; U(x)-L(x)\;\bigr) dx. $$

The next thing to prove is a weak form of the inclusion-exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the graph of a $C^1$-function, then $D_1\cup D_2$ is measurable and

$$\eA(D_1\cup D_2)=\eA(D_1)+\eA(D_2). $$

I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region.

Take such a region $D$ in $\mathbb{R}^2$, $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$

$$ \eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D). $$

exists. If this is the case we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.

The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region

$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace, $$

then $D(U,L)$ is measurable

$$\eA(D(U,L))=\int_a^b \bigl(\, U(x)-L(x)\;\bigr) dx. $$

The next thing to prove is a weak form of the inclusion exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the grapf of a $C^1$-function, then $D_1\cup D_2$ is measurable and

$$\eA(D_1\cup D_2)=\eA_(D_1)+\eA(D_2). $$

I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region.

Take such a region $D$ in $\mathbb{R}^2$, $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$

$$ \eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D) $$

exists. If this is the case, then we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.

The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region

$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace, $$

then $D(U,L)$ is measurable

$$\eA(D(U,L))=\int_a^b \bigl(\; U(x)-L(x)\;\bigr) dx. $$

The next thing to prove is a weak form of the inclusion-exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the grapf of a $C^1$-function, then $D_1\cup D_2$ is measurable and

$$\eA(D_1\cup D_2)=\eA(D_1)+\eA(D_2). $$