# Tilings · Grade 10 · 3 units

> Tilings, Grade 10 · 3 units: The angle sum of a polygon and more. Explanation, worked examples and free practice following the Israeli curriculum.

In Tilings (Grade 10 · 3 units) you practise 4 skills: The angle sum of a polygon · Angles of a regular polygon · Which polygons tile the plane · The cost of tiling.

A tiling covers the plane with shapes, with no gaps and no overlaps. Round every vertex of a tiling the angles fill a full turn, 360°.

The diagonals from one vertex split a polygon into triangles, two fewer than it has sides. So the angles of a polygon add up to the number of sides minus 2, times 180°.

In a regular polygon all the sides are equal and all the angles are equal. A regular polygon tiles the plane alone only when its angle goes into 360° a whole number of times: the equilateral triangle, the square and the regular hexagon.

Every triangle and every quadrilateral tiles the plane. To find how many tiles are needed, put the lengths in the same unit and work out how many tiles fit along each side.

**Tip:** Before you work anything out, check every length is in the same unit: one metre is 100 cm.

## What you learn

### The angle sum of a polygon

- Example: (A polygon with the diagonals from one vertex dashed) What is the sum of the interior angles of a polygon with 4 sides?
- Answer: 360 °
- Solution: The diagonals from one vertex make 2 triangles, so the sum is (4 − 2) × 180° = 360°

### Angles of a regular polygon

- Example: (A square) What is the size of each interior angle of a square?
- Answer: 90 °
- Solution: The angle sum: (4 − 2) × 180° = 360° . Each angle: 360° ÷ 4 = 90°

### Which polygons tile the plane

- Example: Each angle of a square is 90°. Can the plane be tiled with copies of it alone, with no gaps and no overlaps?
- Options: No · Yes
- Answer: Yes
- Solution: The quotient 360 ÷ 90 = 4 is a whole number, so it tiles.

### The cost of tiling

- Example: The floor of a rectangular room. The lengths are in metres. (A rectangular floor with the lengths of two neighbouring sides) It is tiled with square tiles of 20 cm by 20 cm. How many tiles are needed?
- Answer: 600
- Solution: Along one side: 800 ÷ 20 = 40 . Along the other side: 300 ÷ 20 = 15 . All the tiles: 40 × 15 = 600

## Not on the site yet

- Designing composite tilings
- Tiling by altering a polygon
- Explaining why a polygon tiles or does not

The exercises are made by rules and the answers are worked out in code. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 3 units (grades 10 and 11).

[Practise on the site](https://kixplo.com/il/en/) · [All Grade 10 · 3 units topics](https://kixplo.com/il/en/math/grade-10-3u/)