• MATHEMATICS EXERCISE 6.1 1. In Δ PQR, D is the mid-point of QR . PM is _________________. PD is _________________. Is QM = MR? Q MD2. Draw rough sketches for the following: (a) In ΔABC, BE is a median. (b) In ΔPQR, PQ and PR are altitudes of the triangle. (c) In ΔXYZ,YL is an altitude in the exterior of the triangle. 3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same. 6.4 EXTERIOR ANGLE OF A TRIANGLE AND ITS PROPERTY 1. Draw a triangleABC 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 ΔABC. We call it an exterior angle of the ΔABC formed at vertex C. Clearly ∠BCA is an adjacent angle to ∠ACD. The Fig 6.7 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 triangleABC and form an exterior angleACD. 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? Fig 6.8 THE TRIANGLE AND ITS PROPERTIES 2. Find the value of the unknown interior angle x in the following figures: 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. 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°. (i) (ii)Fig 6.13 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 ΔABC (Fig 6.14). Fig 6.14 MATHEMATICS 1. Can you have a triangle with two right angles? 2. Can you have a triangle with two obtuse angles? 3. Can you have a triangle with two acute angles? 4. Can you have a triangle with all the three angles greater than 60º? 5. Can you have a triangle with all the three angles equal to 60º? 6. 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: (i) all sides have same length. B (i) (ii)(ii) each angle has measure 60°. Fig 6.19 MATHEMATICS 6.7 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 fromAand 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,Fig 6.21 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. 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. THE TRIANGLE AND ITS PROPERTIES 3. Draw any three triangles, say ΔABC, ΔPQR and ΔXYZ in your notebook (Fig 6.22). (i) (ii) (iii)Fig 6.22 Use your ruler to find the lengths of their sides and then tabulate your results as follows: Name of Δ Lengths of Sides Is this True? Δ ABC AB ___ BC ___ CA ___ AB – BC < CA ___ + ___ > ___ BC – CA < AB ___ + ___ > ___ CA – AB < BC ___ + ___ > ___ (Yes/No) (Yes/No) (Yes/No) Δ PQR PQ ___ QR ___ RP ___ PQ – QR < RP ___ + ___ > ___ QR – RP < PQ ___ + ___ > ___ RP – PQ < QR ___ + ___ > ___ (Yes/No) (Yes/No) (Yes/No) Δ XYZ XY ___ YZ ___ ZX ___ XY – YZ < ZX ___ + ___ > ___ YZ – ZX < XY ___ + ___ > ___ ZX – XY < YZ ___ + ___ > ___ (Yes/No) (Yes/No) (Yes/No) 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. THE TRIANGLE AND ITS PROPERTIES 6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall? 1. 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. AB and BC are the legs of ΔABC. 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. Fig 6.23 Fig 6.24 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). Square A Square BFig 6.25 MATHEMATICS 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 squareA = The total area of two uncovered squares in square B. a2 = b2 + c2 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 right-angled? (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.

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