Slide 1: INTRODUCTION NAME = AASTHA SHARMA CLASS = VIII - C SUBJECT = MATHS SCHOOL = D.P.S
Slide 2: USES OF TRIANGLES IN DAY TO DAY LIFE
Slide 3: 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. TRIANGLES A C B Fig. 1.1
Slide 4: The three sides AB, BC, AC and three angles angle A, angle B, angle C are called six elements of a triangle. (Fig 2.1) The three line segments are called the sides of a triangle. The three given points are called the vertices 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 Fig. 2.1 1,2,3 are called the exterior angles and 4,5,6 are called the interior angles 1 4 5 2 6 3 A B C Fig. 2.2 ELEMENTS OF A TRIANGLE A C B Fig. 2.1
Slide 5: 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. SCALENE TRIANGLE ISOSCELES TRIANGLE EQUILATERAL TRIANGLE
Slide 6: CLASSIFICATION ON THE BASIS OF ANGLES ACUTE ANGLED TRIANGLE – If all the three angles of a triangle are less than 90 0 then it is an acute angled triangle. RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal to 90 0 then it is a right angled triangle. OBTUSE ANGLED TRIANGLE- If one angle of a triangle is greater then 90 0 then it is a obtuse angled triangle. ACUTE ANGLED TRIANGLE RIGHT ANGLED TRIANGLE OBTUSE ANGLED TRIANGLE
Slide 7: 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 8: SIDE – ANGLE – SIDE (SAS) CONDITION 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 60 0 60 0 A B C D F E Fig. 1.3 CONGURENCE CONDITIONS
Slide 9: ANGLE – SIDE – ANGLE (ASA) CONDITION 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 10: 3. ANGLE – ANGLE – SIDE (AAS) CONDITION 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 11: 4. SIDE – SIDE – SIDE (SSS) CONDITION 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 12: 5. RIGHT – HYPOTENUSE – SIDE (RHS) CONDITION 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 = 90 0 , AC = DF , AB = DE TO PROVE : ▲ABC = ▲DEF B = E = 90 0 (GIVEN) AC = DF (GIVEN) AB = DE (GIVEN) ▲ABC = ▲DEF (BY RHS RULE) D E F A B C 90 0 90 0 Fig. 1.7
Slide 13: 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 14: 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 15: 4. 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 that AC > AB A C B Fig. 2.2
Slide 16: 5. 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 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