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# 6. The Triangles and its Properties

## 6.1 INTRODUCTION

A triangle, you have seen, is a simple closed curve made of three line segments. It has three vertices, three sides and three angles. Here is ΔABC (Fig 6.1). It has

**Sides**: , ,

**Angles**: ∠BAC, ∠ABC,∠BCA

**Vertices**: A, B, C

The side opposite to the vertex A is BC. Can you name the angle opposite to the side AB? You know how to classify triangles based on the sides and angles.

- Based on Sides: Scalene, Isosceles and Equilateral triangles.
- Based on Angles: Acute-angled, Obtuse-angled and Right-angled triangles

## 6.2 MEDIANS OF A TRIANGLE

Given a line segment, you know how to find its perpendicular bisector by paper folding. Cut out a triangle ABC from a piece of paper (Fig 6.3). Consider any one of its sides, say, . By paper-folding, locate the perpendicular bisector of. The folded crease meets at D, its mid-point. Join .

**median**of the triangle. Consider the sides AB and CA and find two more medians of the triangle.

**A median connects a vertex of a triangle to the mid-point of the opposite side.**

- How many medians can a triangle have?
- Does a median lie wholly in the interior of the triangle? (If you think that this is not true, draw a figure to show such a case).

## 6.3 ALTITUDES OF A TRIANGLE

Make a triangular shaped cardboard ABC. Place it upright, on a table. How “tall” is the triangle? The **height **is the distance from vertex A (in the Fig 6.4) to the base .

From A to you can think of many line segments (see the next Fig 6.5). Which among them will represent its height?

The **height** is given by the line segment that starts from A, comes straight down to , and is perpendicular to . This line segment is an **altitude** of the triangle.**An altitude has one end point at a vertex of the triangle and the other on the line containing the opposite side.** Through each vertex, an altitude can be drawn.

- How many altitudes can a triangle have?
- Draw rough sketches of altitudes from A to for the following triangles (Fig 6.6):
- Will an altitude always lie in the interior of a triangle? If you think that this need not be true, draw a rough sketch to show such a case.
- Can you think of a triangle in which two altitudes of the triangle are two of its sides?
- Can the altitude and median be same for a triangle?

(Hint: For Q.No. 4 and 5, investigate by drawing the altitudes for every type of triangle).

## 6.4 EXTERIOR ANGLE OF A TRIANGLE AND ITS PROPERTY

You may repeat the above two activities by drawing some more triangles along with their exterior angles. Every time, you will find that the exterior angle of a triangle is equal to the sum of its two interior opposite angles.

A logical step-by-step argument can further confirm this fact.

**An exterior angle of a triangle is equal to the sum of its interior opposite angles. **

**Given**Consider ΔABC.

∠ACD is an exterior angle.

**To Show**: m∠ACD = m∠A + m∠B

Through C draw , parallel to .

### Justification

Steps |
Reasons | |

(a) | ∠1 = ∠x | || and is a transversal. Therefore, alternate angles should be equal. |

(b) | ∠2 = ∠y | || and is a transversal. Therefore, corresponding angles should be equal. |

(c) | ∠1 + ∠2 = ∠x + ∠y | |

(d) | Now, ∠x + ∠y = m ∠ACD Hence, ∠1 + ∠2 = ∠ACD | From Fig 6.9 |

The above relation between an exterior angle and its two interior opposite angles is referred to as the **Exterior Angle Property of a triangle.**

- Exterior angles can be formed for a triangle in many ways. Three of them are shown here (Fig 6.10)
- Are the exterior angles formed at each vertex of a triangle equal?
- What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle?

Find angle x in Fig 6.11.

Sum of interior opposite angles = Exterior angle

Or 50° + x = 110°

Or x = 60°

- What can you say about each of the interior opposite angles, when the exterior angle is
- a right angle?
- an obtuse angle?
- an acute angle?

- Can the exterior angle of a triangle be a straight angle?

## 6.5 ANGLE SUM PROPERTY OF A TRIANGLE

There is a remarkable property connecting the three angles of a triangle. You are going to see this through the following four activities.

- Draw a triangle. Cut on the three angles. Rearrange them as shown in Fig 6.13 (i), (ii). The three angles now constitute one angle. This angle is a straight angle and so has measure 180°.
- The same fact you can observe in a different way also. Take three copies of any triangle, say ΔABC (Fig 6.14).
- . Take a piece of paper and cut out a triangle, say, ΔABC (Fig 6.16).

Make the altitude AM by folding ΔABC such that it passes through A.

Fold now the three corners such that all the three vertices A, B and C touch at M. - Draw any three triangles, say ΔABC, ΔPQR and ΔXYZ in your notebook. Use your protractor and measure each of the angles of these triangles. Tabulate your results

What do you observe about ∠1 + ∠2 + ∠3?

(Do you also see the ‘exterior angle property’?)

Name of Δ |
Measures of Angles |
Sum of the Measures of the three Angles |
---|---|---|

ΔABC | ∠A = m∠B = m∠C = | m∠A + m∠B + m∠C = |

ΔPQR | m∠P = m∠Q = m∠R = | m∠P + m∠Q + m∠R = |

ΔXYZ | m∠X = m∠Y = m∠Z = | m∠X + m∠Y + m∠Z = |

Allowing marginal errors in measurement, you will find that the last column always gives 180° (or nearly 180°).

When perfect precision is possible, this will also show that the sum of the measures of the three angles of a triangle is 180°.

You are now ready to give a formal justification of your assertion through logical argument.

#### Statement

** The total measure of the three angles of a triangle is 180°.**
To justify this let us use the exterior angle property of a triangle.

**Given** ∠1, ∠2, ∠3 are angles of ΔABC (Fig 6.17).

∠4 is the exterior angle when BC is extended to D.

**Justification ** ∠1 + ∠2 = ∠4 (by exterior angle property)

∠1 + ∠2 + ∠3 = ∠4 + ∠3 (adding ∠3 to both the sides)

But ∠4 and ∠3 form a linear pair so it is 180°. Therefore, ∠1 + ∠2 + ∠3 = 180°.

Let us see how we can use this property in a number of ways.

In the given figure (Fig 6.18) find m∠P.

m∠P + 47^{o}+ 52^{o} = 180^{o}

Therefore m∠P = 180^{o} – 47^{o} – 52^{o}

= 180^{o} – 99^{o} = 81^{o}

- Can you have a triangle with two right angles?
- Can you have a triangle with two obtuse angles?
- Can you have a triangle with two acute angles?
- Can you have a triangle with all the three angles greater than 60º?
- Can you have a triangle with all the three angles equal to 60º?
- Can you have a triangle with all the three angles less than 60º?

## 6.6 TWO SPECIAL TRIANGLES : EQUILATERAL AND ISOSCELES

**A triangle in which all the three sides are of equal lengths is called an equilateral triangle.**
Take two copies of an equilateral triangle ABC (Fig 6.19). Keep one of them fixed. Place the second triangle on it. It fits exactly into the first. Turn it round in any way and still they fit with one another exactly. Are you able to see that when the three sides of a triangle have equal lengths then the three angles are also of the same size?

We conclude that in an equilateral triangle:

- all sides have same length.
- each angle has measure 60°.

**A triangle in which two sides are of equal lengths is called an isosceles triangle.**

Thus, in an isosceles triangle:

- two sides have same length.
- base angles opposite to the equal sides are equal.

## 6.7 SUM OF THE LENGTHS OF TWO SIDES OF A TRIANGLE

- Mark three non-collinear spots A, B and C in your playground. Using lime powder mark the paths AB, BC and AC. Ask your friend to start from A and reach C, walking along one or more of these paths. She can, for example, walk first along and then along to reach C; or she can walk straight along . She will naturally prefer the direct path AC. If she takes the other path ( and then ), she will have to walk more. In other words,
- AB + BC > AC Similarly, if one were to start from B and go to A, he or she will not take the route and but will prefer BA This is because
- BC + CA > AB By a similar argument, you find that
- CA + AB > BC

**the sum of the lengths of any two sides of a triangle is greater than the third side.** - Collect fifteen small sticks (or strips) of different lengths, say, 6 cm, 7 cm, 8 cm, 9 cm, ..., 20 cm. Take any three of these sticks and try to form a triangle. Repeat this by choosing different combinations of three sticks. Suppose you first choose two sticks of length 6 cm and 12 cm. Your third stick has to be of length more than 12 – 6 = 6 cm and less than 12 + 6 = 18 cm. Try this and find
out why it is so.

To form a triangle you will need any three sticks such that the sum of the lengths of any two of them will always be greater than the length of the third stick.

This also suggests that the sum of the lengths of any two sides of a triangle is greater than the third side. - Draw any three triangles, say ΔABC, ΔPQR and ΔXYZ in your notebook (Fig 6.22).

Name of Δ |
Lengths of Sides |
Is this True? |
Option |
---|---|---|---|

ΔABC | AB___ BC___ CA___ |
AB – BC < CA ___ + ___ > ___ BC – CA < AB ___ + ___ > ___ CA – AB < BC ___ + ___ > ___ |
(Yes/No) (Yes/No) (Yes/No) |

ΔPQR | PQ___ QR ___ RP ___ |
PQ – QR < RP ___ + ___ > ___ QR – RP < PQ ___ + ___ > ___ RP – PQ < QR ___ + ___ > ___ |
(Yes/No) (Yes/No) (Yes/No) |

ΔXYZ | XY___ YZ___ ZX___ |
XY – YZ < ZX ___ + ___ > ___ YZ – ZX < XY ___ + ___ > ___ ZX – XY < YZ ___ + ___ > ___ |
(Yes/No) (Yes/No) (Yes/No) |

**sum of the lengths of any two sides of a triangle is greater than the length of the third side.**

We also find that the difference between the length of any two sides of a triangle is smaller than the length of the third side.

Is there a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm?

Suppose such a triangle is possible. Then the sum of the lengths of any two sides would be greater than the length of the third side. Let us check this.

Is 4.5 + 5.8 > 10.2? Yes

Is 5.8 + 10.2 > 4.5? Yes

Is 10.2 + 4.5 > 5.8? Yes

Therefore, the triangle is possible.

The lengths of two sides of a triangle are 6 cm and 8 cm. Between which two numbers can length of the third side fall?

We know that the sum of two sides of a triangle is always greater than the third.

Therefore, one-third side has to be less than the sum of the two sides. The third side is thus less than 8 + 6 = 14 cm.

The side cannot be less than the difference of the two sides. Thus the third side has to
be more than 8 – 6 = 2 cm.

The length of the third side could be any length greater than 2 and less than 14 cm.

- Is the sum of any two angles of a triangle always greater than the third angle?

## 6.8 RIGHT-ANGLED TRIANGLES AND PYTHAGORAS PROPERTY

Pythagoras, a Greek philosopher of sixth century B.C. is said to have found a very important and useful property of right-angled triangles given in this section. The property is hence named after him. In fact, this property was known to people of many other countries too. The Indian mathematician Baudhayan has also given an equivalent form of this property. We now try to explain the Pythagoras property.

In a right angled triangle, the sides have some special names. The side opposite to the right angle is called the **hypotenuse**; the other two sides are known as the **legs** of the right-angled triangle.

In ΔABC (Fig 6.23), the right-angle is at B. So, AC is the hypotenuse. and are the legs of ΔABC.

You are to place four triangles in one square and the remaining four triangles in the other square, as shown in the following diagram (Fig 6.25).

Hence the uncovered area of square A = Uncovered area of square B.

i.e., Area of inner square of square A = The total area of two uncovered squares in square B.

a

^{2}= b

^{2}+ c

^{2}This is Pythagoras property. It may be stated as follows:

Pythagoras property is a very useful tool in mathematics. It is formally proved as a theorem in later classes. You should be clear about its meaning.

It says that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the legs.

Draw a right triangle, preferably on a square sheet, construct squares on its sides, compute the area of these squares and verify the theorem practically (Fig 6.26).

If you have a right-angled triangle, the Pythagoras property holds. If the Pythagoras property holds for some triangle, will the triangle be rightangled? (Such problems are known as converse problems). We will try to answer this. Now, we will show that, if there is a triangle such that sum of the squares on two of its sides is equal to the square of the third side, it must be a right-angled triangle.

This shows that Pythagoras property holds if and only if the triangle is right-angled. Hence we get this fact:

Determine whether the triangle whose lengths of sides are 3 cm, 4 cm, 5 cm is a right-angled triangle.

3^{2} = 3 × 3 = 9; 4^{2} = 4 × 4 = 16; 5^{2} = 5 × 5 = 25

We find 3^{2} + 4^{2} = 5^{2}.

Therefore, the triangle is right-angled.

**Note**: In any right-angled triangle, the hypotenuse happens to be the longest side. In this example, the side with length 5 cm is the hypotenuse.

ΔABC is right-angled at C. If AC = 5 cm and BC = 12 cm find the length of AB.

A rough figure will help us (Fig 6.28).

By Pythagoras property,

AB^{2} = AC^{2} + BC^{2}

= 5^{2} + 12^{2} = 25 + 144 = 169 = 132

or AB^{2}= 132. So, AB = 13 or the length of AB is 13 cm.

**Note**: To identify perfect squares, you may use prime factorisation technique.

- Which is the longest side in the triangle PQR right angled at P?
- Which is the longest side in the triangle ABC right angled at B?
- Which is the longest side of a right triangle?
- “The diagonal of a rectangle produce by itself the same area as produced by its length and breadth”– This is Baudhayan Theorem. Compare it with the Pythagoras property