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Slide 1:

TrIaNgLeS Maths Project Based on CH-7

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Contents :- Introduction Corresponding Parts (Matching Parts) Congruent Figures Congruence Of The Triangles Let Us Do Criteria Of Congruence Of The Triangles Conditions For Congruence Of Triangles SAS Congruence Condition ASA Congruence Condition SSS Congruence Condition RHS Congruence Condition Properties Of Triangles Thanx For Being Patient

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Introduction We have studied about triangles and their various properties in our earlier classes. We know that a closed figure formed by three intersecting lines is called a triangle. (‘Tri’ means ‘three’). A triangle has three sides, three angles and three vertices. For example, in triangle ABC, denoted as Δ ABC ; AB, BC, CA are the three sides, ∠ A, ∠ B, ∠ C are the three angles and A, B, C are three vertices. We have also studied some properties of triangles. In this chapter, we will study in details about the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle.

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The two triangles coincide when we make the matching ABC<->DEF. So we say that under the matching ABC<->DEF, ∆ABC is congruent to ∆DEF. The other matching do not give a congruence. ABC<->DEF ABC<->DFE ABC<->EDF ABC<->EFD ABC<->FDE ABC<->FED The two triangles cover each other exactly. It follows immediately that: AB=DE BC=EF AC=DF ∟A= ∟D ∟B=∟E ∟C=∟F CORRESPONDING PARTS (MATCHING PARTS)

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Figures, which have same shape and size are called congruent figures. A plane figure F 1 is said to be congruent to F 2 , if F 1 ,when superposed over F 2 , covers it exactly F 1 F 2 F 1 F 2 F 1 F 2 CONGRUENT FIGURES

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Congruence Of The Triangle You must have observed that two copies of your photographs of the same size are identical. Similarly, two bangles of the same size, two ATM cards issued by the same bank are identical. You may recall that on placing a one rupee coin on another minted in the same year, they cover each other completely. Do you remember what such figures are called? Indeed they are called congruent figures (‘congruent’ means equal in all respects or figures whose shapes and sizes are both the same). For Example : We all must have seen the ice tray in our refrigerator. We to observe that the moulds for making ice are all congruent. The cast used for moulding in the tray also has congruent depressions So, whenever identical objects have to be produced, the concept of congruence is used in making the cast.

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A B C D E F A(D) B(E) C(F) Congruence of triangles

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Let Us Do :- Now, let us see which of the triangles given below are congruent to triangle ABC in Fig. 7.4 (i)? . Observe that triangles in Fig. 7.4 (ii), (iii) and (iv) are congruent to Δ ABC while Δ TSU of Fig 7.4 (v) is not congruent to Δ ABC. If Δ PQR is congruent to Δ ABC, we write Δ PQR ≅ Δ ABC. Notice that when Δ PQR ≅ Δ ABC, then sides of Δ PQR fall on corresponding equal sides of Δ ABC and so is the case for the angles. That is, PQ covers AB, QR covers BC and RP covers CA; ∠ P covers ∠ A, ∠ Q covers ∠ B and ∠ R covers ∠ C. Also, there is a one-one correspondence between the vertices. That is, P corresponds to A, Q to B, R to C and so on which is written as P ↔ A, Q ↔ B, R ↔ C. Note that under this correspondence, Δ PQR ≅ Δ ABC; but it will not be correct to write ΔQRP ≅ Δ ABC. Similarly, for Fig. 7.4 (iii),

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FD ↔ AB, DE ↔ BC and EF ↔ CA and F ↔ A, D ↔ B and E ↔ C So, Δ FDE ≅ Δ ABC but writing Δ DEF ≅ Δ ABC is not correct. Give the correspondence between the triangle in Fig. 7.4 (iv) and Δ ABC. So, it is necessary to write the correspondence of vertices correctly for writing of congruence of triangles in symbolic form. Note that in congruent triangles corresponding parts are equal and we write in short ‘CPCT’ for corresponding parts of congruent triangles.

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Criteria for Congruence of Triangles Draw two triangles with one side 3 cm. Are these triangles congruent? Observe that they are not congruent (see Fig. 7.5). Now, draw two triangles with one side 4 cm and one angle 50° (see Fig. 7.6). Are they congruent? See that these two triangles are not congruent

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So, equality of one pair of sides or one pair of sides and one pair of angles is not sufficient to give us congruent triangles. What would happen if the other pair of arms (sides) of the equal angles are also equal? In Fig 7.7, BC = QR, ∠ B = ∠ Q and also, AB = PQ. Now, what can you say about congruence of Δ ABC and Δ PQR? Do you observe that the equality of two sides and the included angle is enough for the congruence of triangles? Yes, it is enough. This is the first criterion for congruence of triangle.

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SAS (SIDE-ANGLE-SIDE) ASA (ANGLE-SIDE-ANGLE) SSS (SIDE-SIDE-SIDE) RHS (RIGHT ANGLE-HYPOTENUSE-SIDE) CONDITIONS FOR CONGRUENCE OF TRIANGLES

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Two triangles are congruent if two sides and in the included angle of one triangle are respectively equal to the two sides and the included angle of the other. A B C D E F SAS Congruence Condition

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Two triangles are congruent, if two angles and the included side of one triangle are respectively equal to the two angles and the included side of the other triangle A B C D E F 6 CM. 6 CM. 35 º 50 º 50º 35º ASA Congruence Condition

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Two triangles are congruent, if the three sides of one triangle are respectively equal to the three corresponding sides of the other triangle. 5 CM. 5 CM. 4 CM. 4 CM. 3 CM. 3 CM. A B C D E F SSS Congruence Condition

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The two triangles are congruent, if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the one side of the other triangle. A B C D E F 5 CM. 13 CM. 5 CM. 13 CM. 90 º 90 º RHS Congruence Condition

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Theorams :- Properties Of Triangles :- Angles opposite to equal sides of an isosceles triangle are equal. This result can be proved in many ways. One of the proofs is given here. Proof : We are given an isosceles triangle ABC in which AB = AC. We need to prove that ∠ B = ∠ C. Let us draw the bisector of ∠ A and let D be the point of intersection of this bisector of ∠ A and BC (see Fig. 7.25). In Δ BAD and Δ CAD, AB = AC (Given) ∠ BAD = ∠ CAD (By construction) AD = AD (Common) So, Δ BAD ≅ Δ CAD (By SAS rule) So, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles. So, ∠ B = ∠

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The sides opposite to equal angles of a triangle are equal. Example : In Δ ABC, the bisector AD of ∠ A is perpendicular to side BC (see Fig. 7.27). Show that AB = AC and Δ ABC is isosceles. Solution : In ΔABD and ΔACD, ∠ BAD = ∠ CAD (Given) AD = AD (Common) ∠ ADB = ∠ ADC = 90° (Given) So, Δ ABD ≅ Δ ACD (ASA rule) So, AB = AC (CPCT) or, Δ ABC is an isosceles triangle

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If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater). Activity : Draw a line-segment AB. With A as centre and some radius, draw an arc and mark different points say P, Q, R, S, T on it. Join each of these points with A as well as with B (see Fig. 7.44). Observe that as we move from P to T, ∠ A is becoming larger and larger. What is happening to the length of the side opposite to it? Observe that the length of the side is also increasing; that is ∠ TAB > ∠ SAB > ∠ RAB > ∠ QAB > ∠ PAB and TB > SB > RB > QB > PB. Now, draw any triangle with all angles unequal to each other. Measure the lengths of the sides (see Fig. 7.45). Observe that the side opposite to the largest angle is the longest. In Fig. 7.45, ∠ B is the largest angle and AC is the longest side. Repeat this activity for some more triangles and we see that the converse of Theorem 7.6 is also true. In this way, we arrive at the following theorem:

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The sum of any two sides of a triangle is greater than the third side. In Fig. 7.46, observe that the side BA of Δ ABC has been produced to a point D such that AD = AC. Can you show that ∠ BCD > ∠ BDC and BA + AC > BC? Have you arrived at the proof of the above theorem Example : D is a point on side BC of Δ ABC such that AD = AC (see Fig. 7.47). Show that AB > AD. Solution : In Δ DAC, AD = AC (Given) So, ∠ ADC = ∠ ACD (Angles opposite to equal sides) Now, ∠ ADC is an exterior angle for ΔABD. So, ∠ ADC > ∠ ABD or, ∠ ACD > ∠ ABD or, ∠ ACB > ∠ ABC So, AB > AC (Side opposite to larger angle in Δ ABC) or, AB > AD (AD = AC)

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