When you renovate the floors in a room, you need to know its surface very precisely, in order to estimate the supplies of mortars, leveling compounds, volume of screed, or **flooring**. This is a simple operation when the parts are **square or rectangular**, but which can become problematic in case of complex surfaces. Here is the manual to get you out of trouble.

## Reminder on the procurement rules for floor coverings

The laying of a floor covering inevitably involves a loss due to cuts and other parameters, such as breakage, the configuration of the part, the shape and surface of the materials or the packaging of the product. Usage empirically backs up the percentages of areas to be added to the area of the room, to cover the needs of the site:

**+ 8%**, for wood or laminate floors laid with broken sticks and**+12 to 15%**for other layouts,**+ 10%**for tiles and slabs of small and medium surfaces,**+ 15%**for larger formats,

For coatings packaged in boxes or bundles, you must reason in **number of packages** minimum for the area considered. Thus, if you need 15.85 m² of tiles sold in boxes of 1.25 m², you will need to supply: 15.85 m² ÷ 1.25 m² = 12.68 boxes, added to the upper unit, i.e. 13 boxes.

For a tiling of very large surface (> 1 m²), one counts per unit. It will be necessary to materialize the layout on a **scale plan** to define the exact number of tiles needed.

In certain situations, such as coatings in **rollers**, we no longer resonate on the surface, but in linear meters (**ml**). For example, to cover a room 3.15 m long by 2.15 wide with a lino vinyl in rolls 140 cm wide, two options are available to you:

- You have the
**bands in the width direction**The quantity of vinyl required is 3.15 ml ÷ 1.40 ml = 2.25 strips, i.e., compared to the upper unit, 3 strips of 2.15 ml =**6.45 ml of vinyl purchased**(the**fall**of**2.25 m²**, measures 2.15 ml x 1.05 ml). - You have the
**strips lengthwise**. The amount of material is 2.15 m ÷ 1.40 ml = 1.53 strips, i.e. 2 strips of 3.15 =**6.30 ml of vinyl purchased**(the**fall**of**2.05 m²**, measure 3.15 ml x 0.65 ml).

This simple calculation shows the interest of finding a Lino 1.60 m wide, in order to avoid disproportionate falls.

## Floor coverings: calculation of simple surfaces

** The rectangle** is a quadrilateral with 4 right angles and opposite sides of the same length. The area of a rectangle is obtained by multiplying the Length by the width.

**SURFACE of a RECTANGLE-shaped piece = L x W**

*Most of the rooms in a house are rectangular.*

**The square** is a quadrilateral comprising 4 right angles and 4 sides of the same length. The area of a square is obtained by multiplying the side by itself.

**SURFACE of a piece in the form of SQUARE = C x C or C²**

*Rarer than rectangular rooms, square spaces remain common*.

**The triangle **is a polygon with 3 sides and the sum of the angles of which is equal to 180 °. There are 4 triangular shapes: the **any triangle**, without any particular property, the **rectangle triangle** having a right angle, the **isosceles triangle** with 2 equal sides and 2 equal angles, the **equilateral triangle**, finally, of which the 3 sides and the 3 angles are equal. Note that the isosceles triangle can also be a right triangle. The height (**h**) of the triangle, is the line measured between an angle and its opposite side (**b**) or its extension and with which it forms a right angle. The area of the triangle is obtained by dividing by 2, the product of the base by the height.

**SURFACE of a TRIANGLE-shaped piece = (bxh) x 0.5**

*Triangular pieces are rare, but this shape is frequently found in complex surfaces.*

**The diamond** is a quadrilateral with opposite angles of the same value and 4 sides of the same length. The height (**h**) of the rhombus is the distance measured between two opposite sides (**b1 and b2**), along a line forming a right angle with each side. The area of the rhombus is obtained by multiplying the side by the height.

**SURFACE of a piece in the shape of a LOSANGE = (bxh) x 0.5**

*In buildings, the extremely rare diamond-shaped pieces can be found on terraces. *

**The trapeze** is a quadrilateral of which 2 sides are parallel, but of unequal lengths. The trapezoid has interesting properties: it can be inscribed in a circle or contain 3 equal isosceles triangles, 4 triangles including 2 of the same area, 4 right trapezoids or 4 isosceles trapezoids. The base (**B) **of the trapezoid is the longest parallel side. The base (**b)** is the parallel, short, opposite side. The height (**h**) of the trapezoid is the line measured between the parallel sides and forming a right angle with them. If one of the sides forms a right angle with the parallel sides, it is a right-angled trapezoid. The area of the trapezoid is equal to the sum of the parallel sides multiplied by the height, then divided by 2.

**SURFACE of a TRAPEZOUS piece = (B + b) xh X 0.5**

*Rooms with a more or less pronounced trapezoidal shape are common in older houses.*

**Circles or ½ circles** add a particular difficulty to the calculation of surfaces of floor coverings, the falls are more important in the round parts of small sizes than in those of large size. Count 15% of offcuts when supplying the flooring. The area of the circle is obtained by multiplying the radius (R) by the radius and by **π.**

**SURFACE of a CIRCLE-shaped piece = R² x 3.1416**

*This architectural form is little used nowadays.*

## Floor coverings: calculation of complex surfaces

We consider “complex”, the area of a part which does not fall into the 6 preceding categories (L-shaped, multi-polygonal parts, etc.). To calculate this type of surface, it is necessary to **cut out drawing**, in a succession of **simple surfaces**. The area of the part is equal to the sum of the simple areas.

As an example, let us take the case of a pentagon roughly representing the facade of a house with a roof with 2 slopes of unequal lengths. This drawing can be cut into 3 simple surfaces:

- starting from the lowest angle (outside the angles of the base), draw a first line parallel to the base. You have just created a first trapezoid, which we call A.
- Draw a second line from the second angle. You have created 1 new trapezoid in the lower part (B) and 1 triangle in the upper part (C).

These 3 new simple surfaces are easy to calculate.

**TOTAL AREA of the PENTAGON = A + B + C**

We cut, just as simply, a part **in L**, in **2 rectangles**.