triangles

Information about triangles

Published on December 30, 2009

Author: deepakks1995

Source: authorstream.com

Content

Slide 1: INTRODUCTION NAME = DEEPAK SHARMA CLASS = IX - B SUBJECT = MATHS SCHOOL = KAPIL GYANPEETH Slide 2: TRIANGLES 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 C B Fig. 1.1 Fig. 1.2 Slide 3: CLASSIFICATION ON THE BASIS OF SIDES SCALENE TRIANGLE – If all the sides of triangle are unequal then it is a scalene tirangle. ISOSCELES TRIANGLE –If any two sides of a triangle are equal then it is a isosceles triangle. EQUILATERAL TRIANGLE- If all the sides of a triangle are equal then it is an equilateral triangle. ON THE BASIS OF ANGLES ACUTE ANGLED TRIANGLE –If all the three angles of a triangle are less than 900 then it is an acute anglesd triangle. RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal to 900 then it is a right angled triangle. OBTUSE ANGLED TRIANGLE- If one angle of a triangle is greater then 900 then it is a obtuse angled triangle. Slide 4: CONGRUENCE OF TRIANGLES 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. Slide 5: ANXIOMS 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 cm 4 cm 600 600 A B C D F E Fig. 1.3 Slide 6: 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 Slide 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 ▲ABC = ▲DEF (BY AAS RULE) D E F A B C Fig. 1.5 Slide 8: SSS RULE 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 Slide 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 Slide 10: SOME PROPERTIES OF TRIANGLE 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 Slide 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 Slide 12: INEQUALITIES When two quantities are unequal then on comparing these quantities we obtain a relation between their measures called “ inequality “ relation. Slide 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 Slide 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 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 If AC > AB We have left only this possibility AC > AB A C B Fig. 2.2 Slide 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 C B D Fig. 2.3 Slide 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 E D Fig. 2.4 Slide 17: SOME OTHER APPLICATIONS OF ASA AND AAS CONGRUENCE CRITERIA 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 C D B Fig. 2.5 Slide 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 C D B Fig. 2.6 Slide 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 C B A Fig. 2.7 Slide 20: EXAMPLES 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 Slide 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 Slide 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 Slide 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 Slide 24: 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 Slide 25: MADE BY :- DEEPAK SHARMA CLASS :- IX - B SCHOOL = KAPIL GYANPEETH SUBMITTED TO : SURESH SIR

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