Basic Mensuration

We start by considering the area of a rectangle. If a rectangle has base $b$ (units) long and height $h$ , its area is $b \times h \equiv bh$ .

In fact, we used the rectangle intuition to list out the properties of real numbers, since we regarded $b, h$ as real numbers.

Theorem 1. Any triangle with base $b$ and height $h$ has area $\frac 12 bh$ .

Proof. We first suppose the triangle is right-angled:

By duplicating the triangle and rotating it, we obtain a rectangle with base $b$ and height $h$ :

Denoting the area of the triangle by $A$ ,

$2A = A + A = bh \quad \Rightarrow \quad A = \frac 12 bh.$

Now suppose the altitude lies inside the triangle:

Adding the areas, we obtain a total area of

$\frac 12 b_1 h + \frac 12 b_2 h = \frac 12 (b_1 + b_2)h = \frac 12 bh.$

Now we consider the case when the altitude lies outside of the triangle:

Denote the area of the original triangle by $A_1$ . Consider the new triangle with area $A_2$ as follows:

Since areas are additive, use the first case to obtain

$A_1 + \frac 12 ch = A_1 + A_2 = \frac 12 (b+c) h.$

Performing algebruh,

$A_1 = \frac 12 (b+c) h - \frac 12 c h = \frac 12 ((b+c)-c)h = \frac 12 bh.$

Theorem 2. Any trapezium with bases $b_1, b_2$ and height $h$ has area $\frac 12 (b_1 + b_2)h$ .

Proof. Consider the following trapezium without loss of generality:

Split the trapezium into two identical triangles:

Using Theorem 1, the area of the trapezium is given by

$\frac 12 b_1 h+\frac 12 b_2 h = \frac 12 (b_1 + b_2) h.$

Theorem 3. Any parallelogram with base $b$ and height $h$ has area $bh$ .

Proof. Consider the parallelogram below without loss of generality:

Since this parallelogram is a trapezium with bases $b_1 = b_2 = b$ , by Theorem 2, its area is given by

$\frac 12 (b_1 + b_2) h = \frac 12 (b+b)h = \frac 12 \cdot 2b \cdot h = bh.$

Most high school geometry problems boils down to applying these formulas one after another. However, they also include more “curvy” areas like circles. How do we compute such areas? We could memorise their formulas like $\pi r^2$ , but how do we get that formula in the first place?

Definition 1. Given a circle with circumference $C$ and radius $r$ , we define $\pi$ to be the constant

$\displaystyle \pi := \frac C{2r}.$

Equivalently, we recover the circumference formula $C = 2 \pi r$ . Numerically, $\pi \approx 3.142$ .

Theorem 4. The area of a circle with radius $r$ is $\pi r^2$ .

Proof Sketch. We will subdivide the circle into a collection of “curved triangles” or more technically, sectors:

Here, we use $12$ sub-divisions for convenience. We notice that the areas total to the circle area. By rearranging these sectors, we recover an approximate rectangle with approximate base $\pi r$ and approximate height $r$ :

Therefore, using the area of a rectangle, the circle has approximate area $\pi r \cdot r = \pi r^2$ . Increasing the number of subdivisions improves the approximation, and in the limit, we get an exact area of $\pi r^2$ .

Remark 1. A more complete construction can be formalised using integral calculus.

Finally, lets re-visit angles for completeness. Previously, we have usually used degrees— $360^\circ$ (i.e. $360$ degrees) to denote the angle of a “complete turn”. The reason for $360$ is convenience—the number $360$ has the factors $1,2,3,4,5,6$ that find many uses in astronomy and geology. In particular, $90^\circ$ denotes a $\frac 14$ -turn, $45^\circ$ denotes a $\frac 18$ -turn, so on and so forth.

Rather than use the convenience of $360$ , we might find it helpful to use the circumference of a circle. If a circle has radius $1$ , then its circumference is $2\pi$ :

Here’s our intuition: an angle of $2 \pi$ denotes a complete turn. In particular, $2 \pi = 360^\circ$ . In this case, we say that the angle is $2\pi$ radians, and automatically refer to “radians” when we do not write any unit.

Definition 2. An angle is said to be of $\theta$ radians if the corresponding curved segment (i.e. arc) in the unit circle has a length of $\theta$ .

Therefore, an angle of $2\pi$ radians corresponds to the whole circumference.

Lemma 1. Given any circle with radius $r$ , the angle corresponding to an arc with length $s$ is given by

$\displaystyle \theta = \frac sr.$

Equivalently, $s = r \theta$ . In particular, $C = r \cdot 2\pi = 2\pi r$ , agreeing with our prior intuitions on the circumference.

Proof Sketch. Consider the diagram below.

By regarding arcs as limits of straight lines, we can pass the intercept theorem for similar triangles to the limit and obtain

$\displaystyle \frac 1r = \frac{\theta}{s} \quad \Rightarrow \quad s =r \theta.$

Therefore, radians work for angles regardless of radius size.

Lemma 2. $2 \pi = 360^\circ$ . By algebra,

$\displaystyle 1 = \frac{1}{2\pi} \cdot 360^\circ, \quad 1^\circ = \frac{2\pi}{360}.$

Here, of course, $N^\circ := N \cdot 1^\circ$ .

Theorem 5. A sector with radius $r$ and interior angle $\theta$ (radians) has area $\frac 12 \theta r^2$ . In particular, with a full circle, we obtain the expected area $\pi r^2$ .

Proof. Since the sector sweeps $\theta/ 2\pi$ of the original circle, its area would correspond to

$\displaystyle \frac{ \theta }{ 2 \pi } \cdot \pi r^2 = \frac { \theta r^2 }2.$

For a full circle, set $\theta = 2\pi$ to obtain an area of $2\pi r^2/2 = \pi r^2$ .

Having explored areas, we need to think deeper. Literally. Next time, we will look at volumes and surface areas, the three-dimensional version of areas and perimeters.

Oh, let’s just conclude with perimeters:

Definition 3. The perimeter of a shape is the total length of its boundary.

Example 1. A rectangle with base $b, h$ has perimeter $2(b+h)$ :