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Draw a triangle ABC and produce one of its sides, say BC as shown in . 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.
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
Consider ∆ABC. ∠ACD is an exterior angle.
To Show: m∠ACD = m∠A + m∠B
Through C draw CE, parallel to BA
|a)||∠1 = ∠x||BA || CE and AC is a transversal. Therefore, alternate angles should be equal.|
|b)||∠2 = ∠y||BA || CE and BD is a transversal.|
|c)||∠1 + ∠2 = ∠x + ∠y|
|d)||Now, ∠x + ∠y = m ∠ACD.
Hence, ∠1 + ∠2 = ∠ACD
|From the figure above.|
The above relation between an exterior angle and its two interior opposite angles is referred to as the Exterior Angle 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.
Thus, the sum of the measures of the three angles of a triangle is 180°.
Arrange them. What do you observe about ∠1 + ∠2 + ∠3? You will also see the ‘exterior angle property’.
Take a piece of paper and cut out a triangle, say, ΔABC . 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.
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°.
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
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. ∠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.
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. 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.
We conclude that in an equilateral triangle:
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. Fold it such that Z lies on Y. The line XM through X is now the axis of symmetry. 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:
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 AB and then along BC to reach C; or she can walk straight along AC . She will naturally prefer the direct path AC. If she takes the other path (AB and then 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 BC and CA but will prefer 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.
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 choosingdifferent 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.
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.
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.
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, the right-angle is at B. So, AC is the hypotenuse. AB and 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.
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.
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.
a2= b2 + c2
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.
If you have a right-angled triangle,the Pythagoras property holds. If thePythagoras property holds for sometriangle, will the triangle be rightangled?(Such problems are known asconverse problems). We will try toanswer this. Now, we will show that,if there is a trianglesuch that sum ofthe squares on two of its sides is equalto 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.
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