The Triangle and its Properties

Welcome dear students! Today we are going to learn about The Triangle and its Properties from Class 7 Maths.

A triangle is a simple closed curve made of three line segments. It has three vertices, three sides and three angles. Here is ∆ABC. It has sides: AB, BC, CA. It has angles: ∠BAC, ∠ABC, ∠BCA. Its vertices are A, B, C. The side opposite to vertex A is BC. Can you name the angle opposite to side AB? It is ∠C, or ∠BCA. You know how to classify triangles based on sides and angles. Based on sides, we have scalene, isosceles and equilateral triangles. Based on angles, we have acute-angled, obtuse-angled and right-angled triangles. Make paper-cut models of these triangular shapes. Compare your models with those of your friends and discuss about them.

Let us try these questions. First, write the six elements of ∆ABC. They are the three sides AB, BC, CA and the three angles ∠BAC, ∠ABC, ∠BCA. Second, write the side opposite to vertex Q of ∆PQR. It is side PR. The angle opposite to side LM of ∆LMN is ∠N. The vertex opposite to side RT of ∆RST is vertex S. Third, look at the figures and classify each triangle by sides and angles. The first triangle has sides 6 cm, 10 cm, and 8 cm. Since all sides are different, it is scalene. Since 6² + 8² = 10², it is right-angled. The second triangle has two sides of 7 cm each. It is isosceles. The third triangle has a right angle, so it is right-angled.

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Now, let us explore something more about triangles. Section 6.2 covers 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. Consider any one of its sides, say BC. By paper-folding, locate the perpendicular bisector of BC. The folded crease meets BC at D, its mid-point. Join AD. The line segment AD, joining the mid-point of BC to its opposite vertex A is called a 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. Think, discuss and write. How many medians can a triangle have? A triangle has three medians, one from each vertex. Does a median lie wholly in the interior of the triangle? Yes, a median always lies completely inside the triangle.

Section 6.3 covers 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 to the base BC. From A to BC, you can think of many line segments. Which among them will represent its height? The height is given by the line segment that starts from A, comes straight down to BC, and is perpendicular to BC. This line segment 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. How many altitudes can a triangle have? A triangle has three altitudes. Draw rough sketches of altitudes from A to BC for acute-angled, right-angled, and obtuse-angled triangles. In an acute triangle, the altitude falls inside. In a right triangle, the altitude from the right angle vertex falls inside, while the other two are the legs themselves. In an obtuse triangle, the altitude from the obtuse vertex falls inside, but the altitudes from the acute vertices fall outside the triangle. Will an altitude always lie in the interior of a triangle? No, in an obtuse-angled triangle, two altitudes lie outside. Can you think of a triangle in which two altitudes of the triangle are two of its sides? Yes, in a right-angled triangle, the two legs are the altitudes corresponding to each other. Can the altitude and median be same for a triangle? Yes, in an isosceles or equilateral triangle, the altitude and median from the vertex angle to the base are the same line segment. Do this activity. Take several cut-outs of an equilateral triangle, an isosceles triangle, and a scalene triangle. Find their altitudes and medians. You will notice that in equilateral and isosceles triangles, the altitude and median from the apex coincide, while in a scalene triangle they generally do not.

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Let us solve Exercise 6.1. Question 1. In ∆PQR, D is the mid-point of QR. PM is an altitude. PD is a median. Is QM = MR? No, QM is not necessarily equal to MR unless PM is also a median, which is only true for specific triangles like isosceles triangles where PQ = PR. Question 2. Draw rough sketches. For part a, in ∆ABC, BE is a median. Draw ∆ABC. Mark E as the mid-point of AC. Join B to E. For part b, in ∆PQR, PQ and PR are altitudes. Draw a right-angled triangle at P. The sides PQ and PR are perpendicular to each other, so they act as altitudes. For part c, in ∆XYZ, YL is an altitude in the exterior of the triangle. Draw an obtuse triangle XYZ with obtuse angle at X. Extend base XZ. Drop a perpendicular from Y to the extended line XZ. The foot is L. YL is the exterior altitude. Question 3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same. Yes. Draw an isosceles triangle ABC with AB = AC. Draw the median from A to the mid-point D of BC. Because the triangle is isosceles, this median AD is also perpendicular to BC, making it the altitude as well.

Section 6.4 covers Exterior Angle of a Triangle and Its Property. Draw a triangle ABC and produce one of its sides, say BC as shown in the figure. Observe the angle ACD formed at the point C. This angle lies in the exterior of ∆ABC. We call it an exterior angle of ∆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 ∠A and ∠B and place them adjacent to each other. Do these two pieces together entirely cover ∠ACD? Can you say that m∠ACD = m∠A + m∠B? Yes. 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. You will observe that ∠ACD = ∠A + ∠B. 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 CE parallel to BA. Justification steps and reasons follow. Step a: ∠1 = ∠x. Reason: BA || CE and AC is a transversal. Therefore, alternate angles should be equal. Step b: ∠2 = ∠y. Reason: BA || CE and BD is a transversal. Therefore, corresponding angles should be equal. Step c: ∠1 + ∠2 = ∠x + ∠y. Step d: Now, ∠x + ∠y = m∠ACD from the figure. Hence, ∠1 + ∠2 = ∠ACD. The above relation between an exterior angle and its two interior opposite angles is referred to as the Exterior Angle Property of a triangle.

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Think, discuss and write. Exterior angles can be formed for a triangle in many ways. Three of them are shown in the figure. There are three more ways of getting exterior angles. Try to produce those rough sketches by extending the other sides at each vertex. Are the exterior angles formed at each vertex of a triangle equal? No, they are generally different unless the triangle is equilateral. What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle? They form a linear pair, so their sum is always 180°.

Example 1. Find angle x in the figure. The figure shows a triangle with interior angles 50° and x. The exterior angle adjacent to the third interior angle is 110°. Solution: Sum of interior opposite angles = Exterior angle. So 50° + x = 110°. Therefore, x = 60°.

Try these. Question 1. An exterior angle of a triangle is of measure 70° and one of its interior opposite angles is of measure 25°. Find the measure of the other interior opposite angle. Let the unknown angle be y. Using the exterior angle property, 25° + y = 70°. So y = 45°. Question 2. The two interior opposite angles of an exterior angle of a triangle are 60° and 80°. Find the measure of the exterior angle. Using the property, the exterior angle = 60° + 80° = 140°. Question 3. Is something wrong in this diagram? The diagram shows an exterior angle of 50°, and interior opposite angles of 60° and 70°. Yes, something is wrong. The sum of the interior opposite angles is 60° + 70° = 130°. This does not equal the exterior angle of 50°. So the diagram violates the exterior angle property.

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Exercise 6.2. Question 1. Find the value of the unknown exterior angle x in the following diagrams. Diagram 1 shows interior angles 50° and 70°. So x = 50° + 70° = 120°. Diagram 2 shows interior angles 65° and 45°. So x = 65° + 45° = 110°. Diagram 3 shows interior angles 30° and 40°. So x = 30° + 40° = 70°. Diagram 4 shows interior angles 50° and 60°. So x = 50° + 60° = 110°. Diagram 5 shows interior angles 50° and 50°. So x = 50° + 50° = 100°. Diagram 6 shows interior angles 60° and 30°. So x = 60° + 30° = 90°. Question 2. Find the value of the unknown interior angle x in the following figures. Figure 1 shows exterior angle 115° and one interior opposite angle 50°. So 50° + x = 115°. x = 65°. Figure 2 shows exterior angle 100° and one interior opposite angle 70°. So 70° + x = 100°. x = 30°. Figure 3 shows exterior angle 125° and one interior opposite angle 90°. So 90° + x = 125°. x = 35°. Figure 4 shows exterior angle 120° and one interior opposite angle 60°. So 60° + x = 120°. x = 60°. Figure 5 shows exterior angle 110° and one interior opposite angle 50°. So 50° + x = 110°. x = 60°. Figure 6 shows exterior angle 120° and one interior opposite angle 50°. So 50° + x = 120°. x = 70°.

Section 6.5 covers 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 four activities. Activity 1. Draw a triangle. Cut on the three angles. Rearrange them as shown. 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°. Activity 2. The same fact you can observe in a different way. Take three copies of any triangle, say ∆ABC. Arrange them as in the figure. What do you observe about ∠1 + ∠2 + ∠3? They form a straight angle, so their sum is 180°. Do you also see the exterior angle property? Yes, it is visible in the arrangement. Activity 3. Take a piece of paper and cut out a triangle, say ∆ABC. 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. 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°. Activity 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. For ∆ABC, measure angles A, B, and C, and add them. For ∆PQR, measure angles P, Q, and R, and add them. For ∆XYZ, measure angles X, Y, and Z, and add them. Allowing marginal errors in measurement, you will find that the last column always gives 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. ∠4 is the exterior angle when BC is extended to D. Justification: ∠1 + ∠2 = ∠4 by exterior angle property. ∠1 + ∠2 + ∠3 = ∠4 + ∠3 by adding ∠3 to both the sides. But ∠4 and ∠3 form a linear pair so it is 180°. Therefore, ∠1 + ∠2 + ∠3 = 180°.

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Let us see how we can use this property. Example 2. In the given figure find m∠P. The figure shows ∆PQR with ∠Q = 47° and ∠R = 52°. Solution: By angle sum property of a triangle, m∠P + 47° + 52° = 180°. Therefore, m∠P = 180° – 47° – 52° = 180° – 99° = 81°.

Exercise 6.3. Question 1. Find the value of the unknown x in the following diagrams. Diagram 1: x + 50° + 60° = 180°. So x = 70°. Diagram 2: x + 90° + 30° = 180°. So x = 60°. Diagram 3: x + 30° + 110° = 180°. So x = 40°. Diagram 4: x + x + 50° = 180°. So 2x = 130°. x = 65°. Diagram 5: x + x + x = 180°. So 3x = 180°. x = 60°. Diagram 6: x + 2x + 90° = 180°. So 3x = 90°. x = 30°. Question 2. Find the values of the unknowns x and y in the following diagrams. Diagram 1: y + 120° = 180°, so y = 60°. Then x + 60° + 50° = 180°. So x = 70°. Diagram 2: x + y + 90° = 180°. Also x = y. So 2x = 90°. x = 45°, y = 45°. Diagram 3: y + 110° = 180°, so y = 70°. Then x + 70° + 50° = 180°, so x = 60°. Diagram 4: x + y + 120° = 180°. Also x = y. So 2x = 60°, x = 30°, y = 30°. Diagram 5: x + x + 120° = 180°. So 2x = 60°, x = 30°. Then y + 30° = 180°, so y = 150°. Diagram 6: x + y + 110° = 180°. Also x = y. So 2x = 70°, x = 35°, y = 35°.

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Try these. Question 1. Two angles of a triangle are 30° and 80°. Find the third angle. Let it be z. 30° + 80° + z = 180°. So z = 70°. Question 2. One of the angles of a triangle is 80° and the other two angles are equal. Find the measure of each of the equal angles. Let each be x. 80° + x + x = 180°. 2x = 100°. x = 50°. Question 3. The three angles of a triangle are in the ratio 1:2:1. Find all the angles. Let them be x, 2x, x. Sum is 4x = 180°. x = 45°. So angles are 45°, 90°, 45°. Classify the triangle in two different ways. It is a right-angled triangle because one angle is 90°. It is an isosceles triangle because two angles are equal. Think, discuss and write. Can you have a triangle with two right angles? No, because two right angles sum to 180°, leaving 0° for the third. Can you have a triangle with two obtuse angles? No, because two obtuse angles sum to more than 180°. Can you have a triangle with two acute angles? Yes, in fact, every triangle must have at least two acute angles. Can you have a triangle with all three angles greater than 60°? No, because the sum would exceed 180°. Can you have a triangle with all three angles equal to 60°? Yes, that is an equilateral triangle. Can you have a triangle with all three angles less than 60°? No, because the sum would be less than 180°.

Section 6.6 covers 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. Keep one fixed. Place the second on it. It fits exactly. Turn it round in any way and still they fit 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, and 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. 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: two sides have same length, and base angles opposite to the equal sides are equal. Try these. Question 1. Find angle x in each figure. Figure 1 shows an isosceles triangle with base angles x and x, and vertex angle 50°. So x + x + 50° = 180°. 2x = 130°. x = 65°. Figure 2 shows an isosceles triangle with vertex angle x and base angles 65° each. So x + 65° + 65° = 180°. x + 130° = 180°. x = 50°. Figure 3 shows an isosceles triangle with vertex angle 100° and base angles x and x. So x + x + 100° = 180°. 2x = 80°. x = 40°. Question 2. Find angles x and y in each figure. Figure 1 shows an isosceles triangle with base angles x and x, vertex angle y. Also, an exterior angle at the base is 110°. So x + 110° = 180°. x = 70°. Then 70° + 70° + y = 180°. y = 40°. Figure 2 shows an isosceles triangle with vertex angle x and base angles y and y. Exterior angle at vertex is 120°. So x + 120° = 180°. x = 60°. Then 60° + y + y = 180°. 2y = 120°. y = 60°.

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Section 6.7 covers 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 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, 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 BC and CA but will prefer BA. This is because BC + CA > AB. By a similar argument, you find that CA + AB > BC. 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 of different lengths, say 6 cm, 7 cm, 8 cm, 9 cm, up to 20 cm. Take any three and try to form a triangle. 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 ∆ABC, ∆PQR and ∆XYZ in your notebook. Use your ruler to find the lengths of their sides and then tabulate your results. You will verify that for each triangle, the sum of any two sides is greater than the third, and the difference of any two sides is smaller than the third. 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 3. 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, it is 10.3. Is 5.8 + 10.2 > 4.5? Yes, it is 16. Is 10.2 + 4.5 > 5.8? Yes, it is 14.7. Therefore, the triangle is possible. Example 4. 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.

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Exercise 6.4. Question 1. Is it possible to have a triangle with the following sides? Part i: 2 cm, 3 cm, 5 cm. 2 + 3 = 5, which is not greater than 5. So no. Part ii: 3 cm, 6 cm, 7 cm. 3 + 6 = 9 > 7. 3 + 7 = 10 > 6. 6 + 7 = 13 > 3. So yes. Part iii: 6 cm, 3 cm, 2 cm. 3 + 2 = 5, which is not greater than 6. So no. Question 2. Take any point O in the interior of a triangle PQR. Is OP + OQ > PQ? Yes, because O, P, Q form a triangle. Is OQ + OR > QR? Yes, because O, Q, R form a triangle. Is OR + OP > RP? Yes, because O, R, P form a triangle. Question 3. AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM? Consider triangles ABM and AMC. In ∆ABM, AB + BM > AM. In ∆AMC, AC + MC > AM. Adding both, AB + AC + BM + MC > 2 AM. But BM + MC = BC. So AB + BC + CA > 2 AM. Yes. Question 4. ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD? In ∆ABC, AB + BC > AC. In ∆ADC, AD + DC > AC. In ∆ABD, AB + AD > BD. In ∆BCD, BC + CD > BD. Adding all four inequalities: 2(AB + BC + CD + DA) > 2(AC + BD). Dividing by two, AB + BC + CD + DA > AC + BD. Yes. Question 5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)? In ∆AOB, OA + OB > AB. In ∆BOC, OB + OC > BC. In ∆COD, OC + OD > CD. In ∆DOA, OD + OA > DA. Adding them: 2(OA + OB + OC + OD) > AB + BC + CD + DA. But OA + OC = AC, and OB + OD = BD. So 2(AC + BD) > AB + BC + CD + DA. Yes. Question 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? It must be greater than 15 – 12 = 3 cm, and less than 15 + 12 = 27 cm. So between 3 cm and 27 cm. Think, discuss and write. Is the sum of any two angles of a triangle always greater than the third angle? Yes. Let angles be A, B, C. A + B + C = 180°. So A + B = 180° – C. Since C < 180°, 180° – C > C. So A + B > C. Similarly for other pairs.

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Section 6.8 covers 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. The property is named after him. The Indian mathematician Baudhayan has also given an equivalent form. In a right-angled triangle, the side opposite to the right angle is called the hypotenuse; the other two sides are known as the legs. In ∆ABC, the right-angle is at B. So AC is the hypotenuse. AB and BC are the legs. Make eight identical copies of a right angled triangle. For example, hypotenuse a units, legs b units and c units. Draw two identical squares with sides of lengths b + c. Place four triangles in one square and the remaining four in the other square. The squares are identical; the eight triangles are identical. Hence the uncovered area of square A equals uncovered area of square B. That is, area of inner square of square A equals the total area of two uncovered squares in square B. So a² = b² + c². This is Pythagoras property. In a right-angled triangle, the square on the hypotenuse equals sum of the squares on the legs. 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. If the Pythagoras property holds for some triangle, will the triangle be right-angled? We will show that if sum of squares on two sides equals square of third side, it must be right-angled. Do this activity. Have cut-outs of squares with sides 4 cm, 5 cm, 6 cm. Arrange them. You find no right angle. Each angle is acute. Note that 4² + 5² ≠ 6². Repeat with squares of sides 4 cm, 5 cm and 7 cm. You get an obtuse-angled triangle! Note that 4² + 5² ≠ 7². 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 5. Determine whether the triangle whose lengths of sides are 3 cm, 4 cm, 5 cm is a right-angled triangle. Solution: 3² = 9. 4² = 16. 5² = 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 6. ∆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. By Pythagoras property, AB² = AC² + BC². AB² = 5² + 12². AB² = 25 + 144. AB² = 169. AB² = 13². So AB = 13. The length of AB is 13 cm. Note: To identify perfect squares, you may use prime factorisation technique. Try these. Find the unknown length x in the following figures. Figure i: right triangle with legs 12 and 5, hypotenuse x. x² = 12² + 5² = 144 + 25 = 169. x = 13. Figure ii: right triangle with legs 8 and 15, hypotenuse x. x² = 8² + 15² = 64 + 225 = 289. x = 17. Figure iii: right triangle with hypotenuse 37, one leg 12, other leg x. x² = 37² – 12² = 1369 – 144 = 1225. x = 35. Figure iv: right triangle with legs 24 and 7, hypotenuse x. x² = 24² + 7² = 576 + 49 = 625. x = 25. Figure v: right triangle with hypotenuse 25, one leg 24, other leg x. x² = 25² – 24² = 625 – 576 = 49. x = 7.

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Exercise 6.5. Question 1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR. By Pythagoras property, QR² = PQ² + PR². QR² = 10² + 24² = 100 + 576 = 676. QR = 26 cm. Question 2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC. By Pythagoras property, AB² = AC² + BC². 25² = 7² + BC². 625 = 49 + BC². BC² = 576. BC = 24 cm. Question 3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall. The ladder, wall, and ground form a right triangle. Hypotenuse is 15 m, height is 12 m, base is a. So 15² = 12² + a². 225 = 144 + a². a² = 81. a = 9 m. Question 4. Which of the following can be the sides of a right triangle? Part i: 2.5 cm, 6.5 cm, 6 cm. Check: 2.5² + 6² = 6.25 + 36 = 42.25. 6.5² = 42.25. Yes, it is a right triangle. The right angle is opposite the 6.5 cm side. Part ii: 2 cm, 2 cm, 5 cm. 2 + 2 = 4 < 5. Cannot form a triangle at all. Part iii: 1.5 cm, 2 cm, 2.5 cm. 1.5² + 2² = 2.25 + 4 = 6.25. 2.5² = 6.25. Yes, it is a right triangle. The right angle is opposite the 2.5 cm side. Question 5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree. The broken part, standing part, and ground form a right triangle. Standing part is 5 m, distance is 12 m, broken part is hypotenuse. Hypotenuse² = 5² + 12² = 25 + 144 = 169. Hypotenuse = 13 m. Original height = 5 + 13 = 18 m. Question 6. Angles Q and R of ∆PQR are 25° and 65°. Find ∠P. ∠P + 25° + 65° = 180°. ∠P + 90° = 180°. ∠P = 90°. So it is right-angled at P. Therefore, PQ² + RP² = QR². Option ii is true. Question 7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. Let breadth be b. Diagonal² = length² + breadth². 41² = 40² + b². 1681 = 1600 + b². b² = 81. b = 9 cm. Perimeter = 2(40 + 9) = 2(49) = 98 cm. Question 8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter. Diagonals of a rhombus bisect each other at right angles. So they form four right triangles with legs 8 cm and 15 cm. Side² = 8² + 15² = 64 + 225 = 289. Side = 17 cm. Perimeter = 4 × 17 = 68 cm.

Think, discuss and write. Which is the longest side in ∆PQR right-angled at P? The side opposite P, which is QR. Which is the longest side in ∆ABC right-angled at B? The side opposite B, which is AC. Which is the longest side of a right triangle? The hypotenuse. Baudhayan Theorem says the diagonal of a rectangle produces by itself the same area as produced by its length and breadth. This is exactly the Pythagoras property applied to a rectangle, where the diagonal² = length² + breadth². Enrichment activity: There are many proofs for Pythagoras theorem using dissection and rearrangement. Try to collect a few and draw charts explaining them.

What have we discussed? 1. The six elements of a triangle are its three angles and the three sides. 2. The line segment joining a vertex of a triangle to the mid point of its opposite side is called a median of the triangle. A triangle has 3 medians. 3. The perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of the triangle. A triangle has 3 altitudes. 4. An exterior angle of a triangle is formed when a side of a triangle is produced. At each vertex, you have two ways of forming an exterior angle. 5. A property of exterior angles: The measure of any exterior angle of a triangle is equal to the sum of the measures of its interior opposite angles. 6. The angle sum property of a triangle: The total measure of the three angles of a triangle is 180°. 7. A triangle is said to be equilateral, if each one of its sides has the same length. In an equilateral triangle, each angle has measure 60°. 8. A triangle is said to be isosceles, if atleast any two of its sides are of same length. The non-equal side of an isosceles triangle is called its base; the base angles of an isosceles triangle have equal measure. 9. Property of the lengths of sides of a triangle: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. The difference between the lengths of any two sides is smaller than the length of the third side. This property is useful to know if it is possible to draw a triangle when the lengths of the three sides are known. 10. In a right angled triangle, the side opposite to the right angle is called the hypotenuse and the other two sides are called its legs. 11. Pythagoras property: In a right-angled triangle, the square on the hypotenuse equals the sum of the squares on its legs. If a triangle is not right-angled, this property does not hold good. This property is useful to decide whether a given triangle is right-angled or not.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]