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### sagar

1. 1. NAME = SAGAR MANDAL CLASS = X - A SUBJECT = MATHS SCHOOL = K.V A.F.S T.K.D
2. 2. A Triangle is formed by three lines segments obtained by joining three pairs of points taken from a set of three non collinear points in the plane. In fig 1.1 , three non collinear points A, B, C have been joined and the figure ABC, enclosed by three line segments , AB,BC, and CA is called a triangle. The symbol ▲ (delta) is used to denote a triangle. The three given points are called the vertices of a triangle. The three line segments are called the sides of a triangle. The angle made by the line segment at the vertices are called the angles of a triangle. If the sides of a ▲ABC are extended in order, then the angle between the extended and the adjoining side is called the exterior angles of the triangle. In the fig. 1.2 1 ,2 and 3 are the exterior angles while 4 ,5,and 6 are the interior angles of the triangles. 1 4 5 2 6 3 A B B C A CB Fig. 1.1 Fig. 1.2
3. 3. • ON THE BASIS OF SIDES 1. SCALENE TRIANGLE – If all the sides of triangle are unequal then it is a scalene tirangle. 2. ISOSCELES TRIANGLE –If any two sides of a triangle are equal then it is a isosceles triangle. 3. EQUILATERAL TRIANGLE- If all the sides of a triangle are equal then it is an equilateral triangle. • ON THE BASIS OF ANGLES 1. ACUTE ANGLED TRIANGLE –If all the three angles of a triangle are less than 900 then it is an acute anglesd triangle. 2. RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal to 900 then it is a right angled triangle. 3. OBTUSE ANGLED TRIANGLE- If one angle of a triangle is greater then 900 then it is a obtuse angled triangle.
4. 4. Two triangles are congruent if three sides and three angles of one triangle are equal to the corresponding sides and angles of other triangle. The congruence of two triangles follows immediately from the congruence of three lines segments and three angles.
5. 5. 1. SIDE – ANGLE – SIDE RULE (SAS RULE) Two triangles are congruent if any two sides and the includes angle of one triangle is equal to the two sides and the included angle of other triangle. EXAMPLE :- (in fig 1.3) GIVEN: AB=DE, BC=EF , B= E SOLUTION: IF AB=DE, BC=EF , B= E then by SAS Rule ▲ABS = ▲DEF 4 cm4 cm 600 600 A B C D FE Fig. 1.3
6. 6. 2. ANGLE – SIDE – ANGLE RULE (ASA RULE ) Two triangles are congruent if any two angles and the included side of one triangle is equal to the two angles and the included side of the other triangle. EXAMPLE : (in fig. 1.4) GIVEN: ABC= DEF, ACB= DFE, BC = EF TO PROVE : ▲ABC = ▲DEF ABC = DEF, (GIVEN) ACB = DFE, (GIVEN) BS = EF (GIVEN) ▲ABC = ▲DEF (BY ASA RULE) A B C D E F Fig. 1.4
7. 7. 3. ANGLE – ANGLE – SIDE RULE (AAS RULE) Two triangles are congruent if two angles and a side of one triangle is equal to the two angles and one a side of the other. EXAMPLE: (in fig. 1.5) GIVEN: IN ▲ ABC & ▲DEF B = E A= D BC = EF TO PROVE :▲ABC = ▲DEF B = E A = D BC = EF D E F A B C Fig. 1.5
8. 8. 4. SIDE – SIDE – SIDE RULE (SSS RULE) Two triangles are congruent if all the three sides of one triangle are equal to the three sides of other triangle. Example: (in fig. 1.6) Given: IN ▲ ABC & ▲DEF AB = DE , BC = EF , AC = DF TO PROVE : ▲ABC = ▲DEF AB = DE (GIVEN ) BC = EF (GIVEN ) AC = DF (GIVEN ) ▲ABC = ▲DEF (BY SSS RULE) D E F A B C Fig. 1.6
9. 9. 5. RIGHT – HYPOTENUSE – SIDE RULE (RHS RULE ) Two triangles are congruent if the hypotenuse and the side of one triangle are equal to the hypotenuse and the side of other triangle. EXAMPLE : (in fig 1.7) GIVEN: IN ▲ ABC & ▲DEF B = E = 900 , AC = DF , AB = DE TO PROVE : ▲ABC = ▲DEF B = E = 900 (GIVEN) AC = DF (GIVEN) AB = DE (GIVEN) ▲ABC = ▲DEF (BY RHS RULE) D E F A B C 900 900 Fig. 1.7
10. 10. 1. The angles opposite to equal sides are always equal. Example: (in fig 1.8) Given: ▲ABC is an isosceles triangle in which AB = AC TO PROVE: B = C CONSTRUCTION : Draw AD bisector of BAC which meets BC at D PROOF: IN ▲ABC & ▲ACD AB = AD (GIVEN) BAD = CAD (GIVEN) AD = AD (COMMON) ▲ABD = ▲ ACD (BY SAS RULE) B = C (BY CPCT) A B D C Fig. 1.8
11. 11. 2. The sides opposite to equal angles of a triangle are always equal. Example : (in fig. 1.9) Given : ▲ ABC is an isosceles triangle in which B = C TO PROVE: AB = AC CONSTRUCTION : Draw AD the bisector of BAC which meets BC at D Proof : IN ▲ ABD & ▲ ACD B = C (GIVEN) AD = AD (GIVEN) BAD = CAD (GIVEN) ▲ ABD = ▲ ACD (BY ASA RULE) AB = AC (BY CPCT) A B D C Fig. 1.9
12. 12. When two quantities are unequal then on comparing these quantities we obtain a relation between their measures called “ inequality “ relation.
13. 13. Theorem 1 . If two sides of a triangle are unequal the larger side has the greater angle opposite to it. Example: (in fig. 2.1) Given : IN ▲ABC , AB>AC TO PROVE : C = B Draw a line segment CD from vertex such that AC = AD Proof : IN ▲ACD , AC = AD ACD = ADC --- (1) But ADC is an exterior angle of ▲BDC ADC > B --- (2) From (1) &(2) ACD > B --- (3) ACB > ACD ---4 From (3) & (4) ACB > ACD > B , ACB > B , C > B A B D C Fig. 2.1
14. 14. THEOREM 2. In a triangle the greater angle has a large side opposite to it Example: (in fig. 2.2) Given: IN ▲ ABC B > C TO PROVE : AC > AB PROOF : We have the three possibility for sides AB and AC of ▲ABC (i) AC = AB If AC = AB then opposite angles of the equal sides are equal than B = C AC ≠ AB (ii) If AC < AB We know that larger side has greater angles opposite to it. AC < AB , C > B AC is not greater then AB (iii) If AC > AB We have left only this possibility AC > AB A CB Fig. 2.2
15. 15. THEOREM 3. The sum of any two angles is greater than its third side Example (in fig. 2.3) TO PROVE : AB + BC > AC BC + AC > AB AC + AB > BC CONSTRUCTION: Produce BA to D such that AD + AC . Proof: AD = AC (GIVEN) ACD = ADC (Angles opposite to equal sides are equal ) ACD = ADC --- (1) BCD > ACD ----(2) From (1) & (2) BCD > ADC = BDC BD > AC (Greater angles have larger opposite sides ) BA + AD > BC ( BD = BA + AD) BA + AC > BC (By construction) AB + BC > AC BC + AC >AB A CB D Fig. 2.3
16. 16. THEOREM 4. Of all the line segments that can be drawn to a given line from an external point , the perpendicular line segment is the shortest. Example: (in fig 2.4) Given : A line AB and an external point. Join CD and draw CE AB TO PROVE CE < CD PROOF : IN ▲CED, CED = 900 THEN CDE < CED CD < CE ( Greater angles have larger side opposite to them. ) B A C ED Fig. 2.4
17. 17. 1. If the altitude from one vertex of a triangle bisects the opposite side, then the triangle is isosceles triangle. Example : (in fig.2.5) Given : A ▲ABC such that the altitude AD from A on the opposite side BC bisects BC i. e. BD = DC To prove : AB = AC SOLUTION : IN ▲ ADB & ▲ADC BD = DC ADB = ADC = 900 AD = AD (COMMON ) ▲ADB = ▲ ADC (BY SAS RULE ) AB = AC (BY CPCT) A CDB Fig. 2.5
18. 18. THEOREM 2. In a isosceles triangle altitude from the vertex bisects the base . EXAMPLE: (in fig. 2.6) GIVEN: An isosceles triangle AB = AC To prove : D bisects BC i.e. BD = DC Proof: IN ▲ ADB & ▲ADC ADB = ADC AD = AD B = C ( AB = AC ; B = C) ▲ADB = ▲ ADC BD = DC (BY CPCT) A CDB Fig. 2.6
19. 19. THEOREM 3. If the bisector of the vertical angle of a triangle bisects the base of the triangle, then the triangle is isosceles. EXAMPLE: (in fig. 2.7) GIVEN: A ▲ABC in which AD bisects A meeting BC in D such that BD = DC, AD = DE To prove : ▲ABC is isosceles triangle . Proof: In ▲ ADB & ▲ EDC BD = DC AD = DE ADB = EDC ▲ADB = ▲EDC AB = EC BAD = CED (BY CPCT) BAD = CAD (GIVEN) CAD = CED AC = EC (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL) AC = AB , HENCE ▲ABC IS AN ISOSCELES TRIANGLE. E D CB A Fig. 2.7
20. 20. QUES. IN FIG. AB = AC, D is the point in interior of ▲ABC such that DBC = DCB. Prove that AD bisects BAC of ▲ ABC SOLUTION: IN ▲BDC, DBC = DCB (GIVEN) DC = DB (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL.) IN ▲ ABD & ▲ ACD AB = AC (GIVEN) BD = CD (GIVEN) AD = AD (COMMON) ▲ABD = ▲ACD (BY SSS RULE) BAD = CAD (BY CPCT) Hence, AD is the bisector of ▲BAC A D B C
21. 21. QUES. In fig. PA AB , QB AB & PA = QB.If PQ intersects AB at O, show that o is the mid point of AB as well as that of PQ SOLUTION: In ▲OAP & ▲OBQ PA = QB (GIVEN) PAO = OBQ = 900 & AOP = BOQ ▲ OAP = ▲OBQ (BY AAS RULE) OA = OB ( BY CPCT) OP = OQ ( BY CPCT) O B A P Q
22. 22. QUES. In fig. PS = PR, TPS = QPR.Prove that PT = PQ SOLUTION: IN ▲PRS PS = PR PRS = PSR (ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL) 1800 - PRS = 1800 - PSR PRQ = PST Thus , in ▲ PST & ▲PRQ TPS = QTR PS = PR PST = PRQ ▲PRT = ▲PSQ ( BY ASA RULE) PT = PQ (BY CPCT) T S R Q P
23. 23. QUES. In fig. BM and DN are both perpendicular to the line segment AC and BM = DN. Prove that AC bisects BD. SOLUTION IN ▲BMR and ▲DNR BMR = DNR (GIVEN) BRM = DRN (VERTICALLY OPPOSITE ANGLES) BM = DN (GIVEN) ▲ BMR = ▲DNR (BY AAS RULE) BR = DR R is the mid point of BD Hence AC bisects BD B A C D M N R
24. 24. Deduction 1: All angles at the circumference on the same arc are equal in measure. To prove: ∠bac = ∠bdc Proof: ∠3 = 2 ∠1 Angle at the centre is twice the angle on the circumference (both on the arc bc) ∠3 = 2 ∠2 Angle at the centre is twice the angle on the circumference (both on arc bc) ∴ 2 ∠1 = 2 ∠2 ∴ ∠1 = ∠2 i.e. ∠bac = Q.E.D. a b c d 3 1 2 .o
25. 25. QUES. In fig . It is given that BC = CE and 1 = 2. Prove that ▲ GCB = ▲ DCE SOLUTION: IN ▲ GCB & ▲ DCE 1 + GBC = 1800 (LINEAR PAIR ) 2 + DEC = 1800 ( LINEAR PAIR) 1 + GBC = 2 + DEC -- (1) GBC = DEC (From (1) ) GBC = DEC (GIVEN) BC = CE (GIVEN) GCB = DCE (VERTICALLY OPPOSITE ANGLES ) ▲GCB = ▲ DCE (BY ASA RULE) *----- *---- *------ * -----* ----- * ----* ---- * A B C F G E D 1 2
26. 26. a b c Ld ∟ ∟. Theorem: A line through the centre of a circle perpendicular to a chord bisects the chord. Given : Circle, centre c, a line L containing c, chord [ab], such that L ⊥ab and L∩ ab = d. To prove: ad = bd Constructio n: Label right angles 1 and 2. Proof : ∠1 = ∠2 = 90° Given ca = cb Both radii cd = cd commo nR H S Corresponding sides Consider cda and cdb: cda cdb∴ ad = bd ∴ 1 2 Q.E.D.
27. 27. b a c d e f 2 1 2 3 1 3 Theorem: If two triangles are equiangular, the lengths of the corresponding sides are in proportion. Given : Two triangles with equal angles. To prove: |df| |ac| = |de| |ab| |ef| |bc| = Constructio n: On ab mark off ax equal in length to de. On ac mark off ay equal to df and label the angles 4 and 5. Proof: ∠1 = ∠4 [xy] is parallel to [bc] |ay| |ac| = |ax| |ab| As xy is parallel to bc. |df| |ac| = |de| |ab| Similarly |ef| |bc| = x y4 5 Q.E.D.
28. 28. Theorem: In a right-angled triangle, the square of the length of the side opposite to the right angle is equal to the sum of the squares of the other two sides. Q.E.D. c b a c b a b a c b a c 1 2 3 45 To prove that angle 1 is 90º Proof: ∠3+ ∠4+ ∠5 = 180º ……Angles in a triangle But ∠5 = 90º => ∠3+ ∠4 = 90º => ∠3+ ∠2 = 90º ……Since ∠2 = ∠4 Now ∠1+ ∠2+ ∠3 = 180º ……Straight line => ∠1 = 180º - ( ∠3+ ∠2 ) => ∠1 = 180º - ( 90º ) ……Since ∠3+ ∠2 already proved to be 90º => ∠1 = 90º
29. 29. The HypotenuseThe Hypotenuse  The hypotenuse of aThe hypotenuse of a right triangle is theright triangle is the triangle's longest side,triangle's longest side, i.e., the side oppositei.e., the side opposite the right angle.the right angle.
30. 30. To day we are TeachingTo day we are Teaching abouTabouT
31. 31. Pythagorean Theorem Pythagoras (~580-500 B.C.) He was a Greek philosopher responsible for important developments in mathematics, astronomy and the theory of music.
32. 32. 1. cut a triangle with base 4 cm and height 3 cm 0 1 2 3 4 5 4 cm 012345 3cm2. measure the length of the hypotenuse 0 1 2 3 4 5 Now take out a square paper and a ruler. 5 cm
33. 33. Consider a square PQRS with sides a + b a a a a b b b b c c c c Now, the square is cut into - 4 congruent right-angled triangles and - 1 smaller square with sides c Proof of Pythagoras’ Theorem P Q R S
34. 34. a + b a + b A B CD Area of square ABCD = (a + b) 2 b b a b b a a a c c c c P Q RS Area of square PQRS = 4 + c 2 ab 2       a 2 + 2ab + b 2 = 2ab + c 2 a 2 + b 2 = c 2
35. 35. Theorem states that: "The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides." The Pythagorean Theorem asserts that for a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: a2 + b2 = c2 The figure above at the right is a visual display of the theorem's conclusion. The figure at the left contains a proof of the theorem, because the area of the big, outer, green square is equal to the sum of the areas of the four red triangles and the little, inner white square: c2 = 4(ab/2) + (a - b)2 = 2ab + (a2 - 2ab + b2 ) = a2 + b2
36. 36. Animated Proof of the Pythagorean Theorem Below is an animated proof of the Pythagorean Theorem. Starting with a right triangle and squares on each side, the middle size square is cut into congruent quadrilaterals (the cuts through the center and parallel to the sides of the biggest square). Then the quadrilaterals are hinged and rotated and shifted to the big square. Finally the smallest square is translated to cover the remaining middle part of the biggest square. A perfect fit! Thus the sum of the squares on the smaller two sides equals the square on the biggest side. Afterward, the small square is translated back and the four quadrilaterals are directly translated back to their original position. The process is repeated forever.
37. 37. Pythagorean Theorem Over 2,500 years ago, a Greek mathematician named Pythagoras developed a proof that the relationship between the hypotenuse and the legs is true for all right triangles. In any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs." This relationship can be stated as: and is known as the  Pythagorean Theorem   a, b are legs. c is the hypotenuse (across from the right angle). There are certain sets of numbers that have a very special property.  Not only do these numbers satisfy the Pythagorean Theorem, but any multiples of these numbers also satisfy the Pythagorean Theorem For example:  the numbers 3, 4, and 5 satisfy the Pythagorean Theorem.  If you multiply all three numbers by 2  (6, 8, and 10), these new numbers ALSO satisfy the Pythagorean theorem.
38. 38. If we think about a right triangle we know of course that one of the angles is a right angle. We also know that the other two angles are acute angles (why?). In fact we know that the other two angles are complementary angles. Therefore there is a relationship between the sizes of the angles that the two acute angles have measures that add up to ninety degrees. What about sides? Is there a relationship between the sides of a right triangle? We know from previous lessons that if we have the lengths of just two of the sides we can construct the triangle so it is enough to know the lengths of two sides to determine the length of the third side. We shall now try to figure out the relationship. We shall, to make it easy to communicate assume that the length of the hypotenuse is c units and that the two legs are of length a and b units.So according to the Pythagorean Theorem, the area of square A, plus the area of square B should equal the area of square C. The special sets of numbers that possess this property are called  Pythagorean Triples. The most common Pythagorean Triples are: 3, 4, 5 5, 12, 13 8, 15, 17 The Pythagorean Theorem                                                                                                           The Pythagorean Theorem is one of Euclidean Geometry's most beautiful theorems. It is simple, yet obscure, and is used continuously in mathematics and physics. In short, it is really cool. This first method is one of the ways the Pythagoreans would have proved the theorem. Unfortunately, it lacks glamour. In the following picture let ABC be a right triangle and BD be a segment drawn perpendicular to AC. Since the triangles are similar, the sides must be of proportional lengths. •AB/AD=AC/AB, or AB x AB = AD x AC •BC/CD=AC/BC, or BC x BC= AC x CD •Then, adding the two together, BC^2 + AB^2 = (AD + DC) x AC= AC^2 •
39. 39. Pythagorean Theorem in text bookPythagorean Theorem in text book of 10of 10thth classclass Given:- ABC is a right angle TriangleGiven:- ABC is a right angle Triangle .. angle B =90angle B =9000 R.T.P:- ACR.T.P:- AC22 = AB= AB22 +BC+BC22 Construction:- To draw BDConstruction:- To draw BD ⊥⊥ AC .AC . A B C D Proof:- In Triangles ADB and ABC AngleA=Angle A (common) Angle ADB=Angle ABC (each 900 ) ADB ~ ABC ( A.A corollary ) So that AD/AB=AB/AC (In similar triangles corresponding sides are proportional ) AB2 = AD X AC _________(1) Similarly BC2 = DCXAC _________(2) Adding (1)&(2) we get AB2 +BC2 = AD X AC + DCXAC = AC (AD +DC) = AC . AC =AC2 There fore AB2 +BC2 =AC2
40. 40. Typical Examples
41. 41. Example 1. Find the length of AC. Hypotenus e AC2 = 122 + 162 (Pythagoras’ Theorem) AC2 = 144 + 256 AC2 = 400 AC = 20 A CB 16 12 Solution :
42. 42. Example 2. Find the length of diagonal d . 10 24 d Solution: d2 = 102 + 242 (Pythagoras’ Theorem) d = + = = 10 24 26 2 2 676
43. 43. 16km 12km 1.A car travels 16 km from east to west. Then it turns left and travels a further 12 km. Find the displacement between the starting point and the destination point of the car. N ? Application of Pythagoras’ Theorem
44. 44. 16 km 12 km A B C Solution : In the figure, AB = 16 BC = 12 AC2 = AB2 + BC2 (Pythagoras’ Theorem) AC2 = 162 + 122 AC2 = 400 AC = 20 The displacement between the starting point and the destination point of the car is 20 km
45. 45. 2. The height of a tree is 5 m. The distance between the top of it and the tip of its shadow is 13 m. Solution: 132 = 52 + L2 (Pythagoras’ Theorem) L2 = 132 - 52 L2 = 144 L = 12 Find the length of the shadow L. 5 m 13 m L