Quadrilaterals and polygons

A tri­an­gle has three sides. Add a fourth and you get a quadri­lat­eral: the shape of a page, a door, a kite or a slice of a dia­mond. Add more sides and you get pen­tagons, hexa­gons and other poly­gons. In this les­son you will learn the names of these shapes, how to find the sum of their angles, and the spe­cial prop­er­ties of the quadri­lat­er­als you meet most often.

Poly­gons#

A poly­gon is a closed fig­ure made of three or more line seg­ments, where each seg­ment meets exactly two oth­ers at its ends. The seg­ments are its sides and the cor­ners are its ver­tices.

A poly­gon with all its sides equal and all its angles equal is called a reg­u­lar poly­gon. An equi­lat­eral tri­an­gle and a square are reg­u­lar poly­gons.

Diag­o­nals#

A diag­o­nal is a line seg­ment that joins two ver­tices that are not next to each other.

A quadri­lat­eral $ABCD$ has 2 diag­o­nals, $\overline{AC}$ and $\overline{BD}$. A tri­an­gle has none, because every ver­tex is next to both the oth­ers.

From each ver­tex of a poly­gon with $n$ sides, you can draw diag­o­nals to all the other ver­tices except itself and its two neigh­bours, that is, to $n - 3$ ver­tices. Each diag­o­nal gets counted twice this way, once from each end. So

$$\displaystyle \text{number of diagonals} = \frac{n(n - 3)}{2}$$

A pen­ta­gon has $\displaystyle \frac{5 \times 2}{2} = 5$ diag­o­nals, and a hexa­gon has $\displaystyle \frac{6 \times 3}{2} = 9$.

The angle sum of a quadri­lat­eral#

Draw the diag­o­nal $\overline{AC}$ of a quadri­lat­eral $ABCD$. It cuts the quadri­lat­eral into two tri­an­gles, $ABC$ and $ACD$. The angles of the two tri­an­gles together make up the four angles of the quadri­lat­eral. Each tri­an­gle gives $180^\circ$, so

The four angles of a quadri­lat­eral add up to $360^\circ$.

Exam­ple 1#

Three angles of a quadri­lat­eral are $80^\circ$, $95^\circ$ and $105^\circ$. Find the fourth.

$$360^\circ - 80^\circ - 95^\circ - 105^\circ = 80^\circ$$

Exam­ple 2#

The angles of a quadri­lat­eral are in the ratio $1 : 2 : 3 : 4$. Find them.

$$k + 2k + 3k + 4k = 360 \quad \text{so} \quad 10k = 360 \quad \text{and} \quad k = 36$$

The angles are $36^\circ$, $72^\circ$, $108^\circ$ and $144^\circ$.

The angle sum of any poly­gon#

The same trick works for any poly­gon. From one ver­tex, draw all the diag­o­nals. A poly­gon with $n$ sides is cut into $n - 2$ tri­an­gles. So

$$\text{sum of the angles of a polygon with } n \text{ sides} = (n - 2) \times 180^\circ$$

Check: for a tri­an­gle, $n = 3$ gives $1 \times 180^\circ$. For a quadri­lat­eral, $n = 4$ gives $2 \times 180^\circ = 360^\circ$.

In a reg­u­lar poly­gon all the angles are equal, so each angle is the sum divided by $n$.

Exte­rior angles. If you walk all the way round a poly­gon, turn­ing at each cor­ner, you turn through one com­plete turn in all. So the exte­rior angles of any poly­gon add up to $360^\circ$. For a reg­u­lar poly­gon with $n$ sides, each exte­rior angle is $\displaystyle \frac{360^\circ}{n}$.

Exam­ple 3#

Find the sum of the angles of an octa­gon, and each angle of a reg­u­lar octa­gon.

$$\text{sum} = (8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ$$

$$\text{each angle of a regular octagon} = 1080^\circ \div 8 = 135^\circ$$

Check with the exte­rior angle: $360^\circ \div 8 = 45^\circ$, and $180^\circ - 45^\circ = 135^\circ$.

Exam­ple 4#

Each exte­rior angle of a reg­u­lar poly­gon is $40^\circ$. How many sides does it have?

$$\displaystyle n = \frac{360^\circ}{40^\circ} = 9$$

It has 9 sides. It is a reg­u­lar nonagon.

Spe­cial quadri­lat­er­als#

Prop­er­ties of a par­al­lel­o­gram#

In a par­al­lel­o­gram $ABCD$:

1. Oppo­site sides are equal: $AB = CD$ and $BC = DA$.
2. Oppo­site angles are equal: $\angle A = \angle C$ and $\angle B = \angle D$.
3. Any two angles next to each other add up to $180^\circ$. (They are co-inte­rior angles between par­al­lel sides.)
4. The diag­o­nals cut each other in half. We say they bisect each other.

Rec­tan­gles, rhom­buses and squares are all par­al­lel­o­grams, so they have all these prop­er­ties too. They have some extra ones:

• Rec­tan­gle: the diag­o­nals are equal in length.
• Rhom­bus: the diag­o­nals cross at right angles.
• Square: the diag­o­nals are equal and cross at right angles.

Exam­ple 5#

One angle of a par­al­lel­o­gram is $70^\circ$. Find the other three.

The angle oppo­site it is also $70^\circ$. Each angle next to it is $180^\circ - 70^\circ = 110^\circ$. Going round, the angles are $70^\circ$, $110^\circ$, $70^\circ$ and $110^\circ$. Check: they add up to $360^\circ$.

Exam­ple 6#

Two angles next to each other in a par­al­lel­o­gram are $(3x + 20)^\circ$ and $(2x + 10)^\circ$. Find them.

$$(3x + 20) + (2x + 10) = 180 \quad \text{so} \quad 5x + 30 = 180 \quad \text{and} \quad x = 30$$

The angles are $110^\circ$ and $70^\circ$.

Prac­tice#

1. Three angles of a quadri­lat­eral are $90^\circ$, $75^\circ$ and $110^\circ$. Find the fourth.
2. The angles of a quadri­lat­eral are in the ratio $2 : 3 : 3 : 4$. Find them.
3. How many diag­o­nals does an octa­gon have?
4. Find the sum of the angles of a hexa­gon, and each angle of a reg­u­lar hexa­gon.
5. Find the sum of the angles of a decagon.
6. Each exte­rior angle of a reg­u­lar poly­gon is $30^\circ$. How many sides does it have?
7. One angle of a par­al­lel­o­gram is $115^\circ$. Find the other three.
8. In a par­al­lel­o­gram, one side is 7 cm and the side next to it is 4 cm. Find the perime­ter, that is, the total length of all four sides.
9. Which quadri­lat­er­als have diag­o­nals that cross at right angles: rec­tan­gle, rhom­bus, square?

Answers#