The Triangle and its Properties

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 ∆ABC (Fig 6.1). It has

Sides: \(\overline{AB},\overline{BC},\overline{CA}\)

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 (i) sides (ii) angles.

(i) Based on Sides: Scalene, Isosceles and Equilateral triangles.

(ii) Based on Angles: Acute-angled, Obtuse-angled and Right-angled triangles. Make paper-cut models of the above triangular shapes. Compare your models with those of your friends and discuss about them.

TRY THESE

The Triangle and its Properties Chapter 6 1. Write the six elements (i.e., the 3 sides and the 3 angles) of ∆ABC. 2. Write the:

(i) Side opposite to the vertex Q of ∆PQR

(ii) Angle opposite to the side LM of ∆LMN

(iii) Vertex opposite to the side RT of ∆RST 3. Look at Fig 6.2 and classify each of the triangles according to its (a) Sides (b) Angles

The Triangle and its Properties

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,\(\overline{BC}\). By paper-folding, locate the perpendicular bisector of \(\overline{BC}\) . The folded crease meets \(\overline{BC}\) at D, its mid-point. Join AD .

The line segment AD, joining the mid-point of \(\overline{BC}\) to its opposite vertex A is called a median of the triangle.

Consider the sides \(\overline{AB}\) and \(\overline{CA}\) and find two more medians of the triangle.

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

The Triangle and its Properties

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 \(\overline{BC}\).

From A to \(\overline{BC}\), 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 \(\overline{BC}\), and is perpendicular to  \(\overline{BC}\).

This line segment  \(\overline{AL}\) 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

THINK, DISCUSS AND WRITE

1. How many altitudes can a triangle have?

2. Draw rough sketches of altitudes from A to BC for the following triangles (Fig 6.6):

3. 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.

4. Can you think of a triangle in which two altitudes of the triangle are two of its sides?

5. 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)

The Triangle and its Properties

EXTERIOR ANGLE OF A TRIANGLE AND ITS PROPERTY

DO THIS

1.Draw a triangle ABC and produce one of its sides, say BC as shown in Fig 6.7. Observe the angle ACD formed at the point C. This angle lies in the exterior of ∆ABC. We call it an exterior angle of the ∆ABC formed at vertex C.

Clearly ∠BCA is an adjacent angle to ∠ACD. The remaining two angles of the triangle namely ∠A and ∠B are called the two interior opposite angles or the two remote interior angles of ∠ACD. Now cut out (or make trace copies of) ∠A and ∠B and place them adjacent to each other as shown in Fig 6.8. Do these two pieces together entirely cover ∠ACD? Can you say that m ∠ACD = m ∠A + m ∠B?

2. As done earlier, draw a triangle ABC and form an exterior angle ACD. Now take a protractor and measure ∠ACD, ∠A and ∠B. Find the sum ∠A + ∠B and compare it with the measure of ∠ACD. Do you observe that ∠ACD is equal (or nearly equal, if there is an error in measurement) to ∠A + ∠B?

You may repeat the two activities as mentioned 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 ∆ABC. ∠ACD is an exterior angle.

To Show: m∠ACD = m∠A + m∠B

 Through C draw \(\overline{CE}\), parallel to \(\overline{BA}\).

Justification

Steps                                                        Reasons

(a) ∠1 = ∠x                                          \(\overline{BA}\) || \(\overline{CE}\) and AC is a transversal

Therefore, alternate angles should be equal.

(b) ∠2 = ∠y                                         \(\overline{BA}\) || \(\overline{CE}\) and BD is a transversal.

Therefore, corresponding angles should be equal.

(c) ∠1 + ∠2 = ∠x + ∠y

(d) Now, ∠x + ∠y = m ∠ACD From Fig 6.9

Hence, ∠1 + ∠2 = ∠ACD The above relation between an exterior angle and its two interior opposite angles is referred to as the ExteriorAngle Property of a triangle.

EXAMPLE : Find angle x in Fig 6.11.

SOLUTION Sum of interior opposite angles = Exterior angle

or 50° + x = 110°

 or x = 60°

The Triangle and its Properties

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.

1. 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°

Thus, the sum of the measures of the three angles of a triangle is 180°.

2. The same fact you can observe in a different way also. Take three copies of any triangle, say ∆ABC (Fig 6.14).

Arrange them as in Fig 6.15. What do you observe about ∠1 + ∠2 + ∠3? (Do you also see the ‘exterior angle property’?)

3. Take a piece of paper and cut out a triangle, say, ∆ABC (Fig 6.16).

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

Fold now the three corners such that all the three vertices A, B and C touch at M. (i) (ii) (iii) Fig 6.16

You find that all the three angles form together a straight angle. This again shows that the sum of the measures of the three angles of a triangle is 180°.

4. Draw any three triangles, say ∆ABC, ∆PQR and ∆XYZ in your notebook. Use your protractor and measure each of the angles of these triangles.

Tabulate your results

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 ∆ABC (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.

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

SOLUTION :By angle sum property of a triangle, m∠P + 47° + 52° = 180°

Therefore m∠P = 180° – 47° – 52° = 180° – 99° = 81°

The Triangle and its Properties

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:

(i) all sides have same length.

(ii) each angle has measure 60°

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

From a piece of paper cut out an isosceles triangle XYZ, with XY=XZ (Fig 6.20). Fold it such that Z lies on Y. The line XM through X is now the axis of symmetry (which you will read in Chapter 14). You find that ∠Y and ∠Z fit on each other exactly. XY and XZ are called equal sides; YZ is called the base; ∠Y and ∠Z are called base angles and these are also equal. Thus, in an isosceles triangle:

 (i) two sides have same length.

 (ii) base angles opposite to the equal sides are equal.

The Triangle and its Properties

SUM OF THE LENGTHS OF TWO SIDES OF A TRIANGLE

1. 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 \(\overline{AB}\) and then along \(\overline{BC}\) to reach C; or she can walk straight along AC. She will naturally prefer the direct path \(\overline{AC}\). If she takes the other path (\(\overline{AB}\) and then \(\overline{BC}\) ), she will have to walk more. In other words,

AB + BC > AC………(i)

Similarly, if one were to start from B and go to A, he or she will not take the route \(\overline{BC}\) and \(\overline{CA}\) but will prefer \(\overline{BA}\) This is because

BC + CA > AB ……..(ii)

By a similar argument, you find that

CA + AB > BC ………(iii)

These observations suggest that the sum of the lengths of any two sides of a triangle is greater than the third side.

2. 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.

3. Draw any three triangles, say ∆ABC, ∆PQR and ∆XYZ in your notebook (Fig 6.22)

Use your ruler to find the lengths of their sides and then tabulate your results as follows:

This also strengthens our earlier guess. Therefore, we conclude that 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.

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

 SOLUTION 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.

EXAMPLE  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?

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

Therefore, 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.

The Triangle and its Properties

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 ∆ABC (Fig 6.23), the right-angle is at B. So, AC is the hypotenuse. \(\overline{AB}\) and \(\overline{BC}\) are the legs of ∆ABC.

Make eight identical copies of right angled triangle of any size you prefer. For example, you make a right-angled triangle whose hypotenuse is a units long and the legs are of lengths b units and c units (Fig 6.24)

Draw two identical squares on a sheet with sides of lengths b + c.

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).

The squares are identical; the eight triangles inserted are also identical.

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:

In a right-angled triangle, the square on the hypotenuse = sum of the squares on the legs.

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:

If the Pythagoras property holds, the triangle must be right-angled.

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

SOLUTION :3 ² = 3 × 3 = 9; 4² = 4 × 4 = 16; 5² = 5 × 5 = 25

We find 3² + 4² = 5² .

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

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

SOLUTION: A rough figure will help us (Fig 6.28). By Pythagoras property,

AB² = AC² + BC² = 5² + 12² = 25 + 144 = 169 = 132

or AB² = 13² . So, AB = 13 or the length of AB is 13 cm.

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