## Introduction

- Length is the distance between two points. The SI unit of length is metres. Conversion of units of length.

1 kilometer (km) = 1000metres

1 hectometer (hm) = 100metres

1 decameter (Dm) =10 metres

1 decimeter (dm) =^{1}/_{10}metres

1 centimeter (cm) =^{1}/_{100}metres

1 millimeter (mm) =^{1}/_{1000}metres - The following prefixes are often used when referring to length:

Mega – 1000 000

Kilo – 1000

Hecto – 100

Deca – 1 0

Deci –^{1}/_{10}Centi –^{1}/_{100}

Milli –^{1}/_{1000}

Micro –^{1}/_{1000 000}

## Significant Figures

- The accuracy with which we state or write a measurement may depend on its relative size. It would be unrealistic to state the distance between towns A and B as 158.27 km. a more reasonable figure is 158 km. 158.27 km is the distance expressed to 5 significant figures and 158 km to 3 significant figures.

**Example**

Express each of the following numbers to 5, 4, 3, 2, and 1 significant figures:

- 906 315
- 0.08564
- 40.0089
- 15 600

**Solution**

number | 5 s.f | 4 s.f. | 3 s.f. | 2 s.f | 1 s.f. | |

(a) | 906 315 | 906 320 | 906 300 | 906 000 | 91 0 000 | 900 000 |

(b) | 0.085641 | 0.085641 | 0.08564 | 0.0856 | 0.085 | 0.09 |

(c) | 40.0089 | 40.009 | 40.01 | 40.0 | 40 | 40 |

(d) | 156 000 | 156 000 | 156 000 | 156 000 | 160 000 | 200 000 |

The above example show how we would round off a measurement to a given number of significant figures

Zero may not be a significant. For example:

- 0.085 - zero is not significant therefore, 0.085 is a two- significant figure.
- 2.30 – zero is significant. Therefore 2.30 is a three-significant figure.
- 5000 – zero may or may not be significant figure. Therefore, 5 000 to three significant figure is 500 (zero after 5 is significant). To one significant figure is 5 000. Zero after 5 is not significant.
- 31.805 Or 305 – zero is significant, therefore 31 .805 is five significant figure. 305 is three significant figure.

## Perimeter

- The perimeter of a plane is the total length of its boundaries. Perimeter is a length and is therefore expressed in the same units as length.

### Square Shapes

- Its perimeter is 5+5+5+5= 2(5+5)

=2(10)

=20cm

Hence 5 x 4 = 20

So perimeter of a square = Sides x 4

### Rectangular Shapes

Figure 12.2 is a rectangle of length 5cm and breadth 3cm.

Its perimeter is 5+3+5+3 =2(5+3)cm

= 2 x 8

= 16 cm

Hence perimeter of a rectangle p=2(L+ W)

### Triangular Shapes

- To find the perimeter of a triangle add all the three sides.

Perimeter = (a + b + c) units, where a, b and c are the lengths of the sides of the triangle.

### The Circle

The circumference of a circle = 2πr or πD

**Example**

**Find the circumference of a circle of a radius 7cm.**- The circumference of a bicycle wheel is 1 40 cm. find its radius.

**Solution**

- C= πd

=^{22}/_{7}x 7

=44 cm - C=πd

=2πr

=2x22/7xr

=140 ÷^{44}/_{7}

=22.27 cm

#### Length of an Arc

- An arc of a circle is part of its circumference. Figure 1 2.1 0 (a) shows two arcs AMB and ANB. Arc AMB, which is less than half the circumference of the circle, is called the minor arc, while arc ANB, which is greater half of the circumference is called the major arc. An arc which is half the circumference of the circle is called a semicircle.

**Example**

**An arc of a circle subtends an angle 60 at the centre of the circle. Find the length of the arc if the radius of the circle is 42 cm. (π=**^{22}/_{7}).

**Solution**

The length, l, of the arc is given by:

L =^{θ}/_{360} x 2πr.

θ=60, r=42 cm

Therefore, L =^{60}/_{360} x2 x ^{22}/_{7} x 42

= 44 cm

**Example**

**The length of an arc of a circle is 62.8 cm. find the radius of the circle if the arc subtends an angle 144 at the centre, (take π=3.142).**

**Solution**

L =^{θ}/_{360} x 2πr = 62.8 and θ= 144

Therefore, ^{144}/_{360} x 2 x 3.142 x r = 62.8

R= 62.8 x ^{360}/_{144} x 2 x 3.142

=24.98 cm

**Example**

Find the angle subtended at the centre of a circle by an arc of length 11 cm if the radius of the circle is 21 cm.

**Solution**

L= ^{θ}/_{360} x 2 xπr =11 cm and r =21 m

11 =^{θ}/_{360} x 2 x ^{22}/_{7} x 21

Thus,

θ = 11 x 360 x 7

2 x 22 x 21

= 300

## Past KCSE Questions on the Topic

- Two coils which are made by winding copper wire of different gauges and length have the same mass. The first coil is made by winding 270 metres of wire with cross sectional diameter 2.8mm while the second coil is made by winding a certain length of wire with cross-sectional diameter 2.1 mm. Find the length of wire in the second coil .
- The figure below represents a model of a hut with HG = GF = 10cm and FB = 6cm. The four slanting edges of the roof are each 12cm long.

Calculate- Length DF.
- Angle VHF

- The length of the projection of line VH on the plane EFGH.
- The height of the model hut.
- The length VH.

- The angle DF makes with the plane ABCD.

- A square floor is fitted with rectangular tiles of perimeters 220 cm. each row (tile length wise) carries 20 less tiles than each column (tiles breadth wise). If the length of the floor is 9.6m. Calculate:
- The dimensions of the tiles
- The number of tiles needed
- The cost of fitting the tiles, if tiles are sold in dozens at sh. 1 500 per dozen and the labour cost is sh. 3000

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