# TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions

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1 CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii) SSS (iv) RHS AAS criterion for congruence of triangles as a particular case of ASA criterion. Angles opposite to equal sides of a triangle are equal, Sides opposite to equal angles of a triangle are equal, A point equidistant from two given points lies on the perpendicular bisector of the line-segment joining the two points and its converse, A point equidistant from two intersecting lines lies on the bisectors of the angles formed by the two lines, In a triangle (i) side opposite to the greater angle is longer (ii) (iii) angle opposite the longer side is greater the sum of any two sides is greater than the third side. (B) Multiple Choice Questions Write the correct answer : Sample Question 1 : If Δ ABC Δ PQR and Δ ABC is not congruent to Δ RPQ, then which of the following is not true: (A) BC = PQ (B) AC = PR (C) QR = BC (D) AB = PQ Solution : Answer (A)

2 64 EXEMPLAR PROBLEMS EXERCISE 7.1 In each of the following, write the correct answer: 1. Which of the following is not a criterion for congruence of triangles? (A) SAS (B) ASA (C) SSA (D) SSS 2. If AB = QR, BC = PR and CA = PQ, then (A) Δ ABC Δ PQR (B) Δ CBA Δ PRQ (C) Δ BAC Δ RPQ (D) Δ PQR Δ BCA 3. In Δ ABC, AB = AC and B = 50. Then C is equal to (A) 40 (B) 50 (C) 80 (D) In Δ ABC, BC = AB and B = 80. Then A is equal to (A) 80 (B) 40 (C) 50 (D) In Δ PQR, R = P and QR = 4 cm and PR = 5 cm. Then the length of PQ is (A) 4 cm (B) 5 cm (C) 2 cm (D) 2.5 cm 6. D is a point on the side BC of a Δ ABC such that AD bisects BAC. Then (A) BD = CD (B) BA > BD (C) BD > BA (D) CD > CA 7. It is given that Δ ABC Δ FDE and AB = 5 cm, B = 40 and A = 80. Then which of the following is true? (A) DF = 5 cm, F = 60 (B) DF = 5 cm, E = 60 (C) DE = 5 cm, E = 60 (D) DE = 5 cm, D = Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be (A) 3.6 cm (B) 4.1 cm (C) 3.8 cm (D) 3.4 cm 9. In Δ PQR, if R > Q, then (A) QR > PR (B) PQ > PR (C) PQ < PR (D) QR < PR 10. In triangles ABC and PQR, AB = AC, C = P and B = Q. The two triangles are (A) isosceles but not congruent (B) isosceles and congruent (C) congruent but not isosceles (D) neither congruent nor isosceles 11. In triangles ABC and DEF, AB = FD and A = D. The two triangles will be congruent by SAS axiom if (A) BC = EF (B) AC = DE (C) AC = EF (D) BC = DE

4 66 EXEMPLAR PROBLEMS 11. Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer. 12. Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer. (D) Short Answer Questions Sample Question 1 : In Fig 7.1, PQ = PR and Q = R. Prove that Δ PQS Δ PRT. Solution : In Δ PQS and Δ PRT, PQ = PR (Given) Q = R (Given) and QPS = RPT (Same angle) Therefore, Δ PQS Δ PRT (ASA) Sample Question 2 : In Fig.7.2, two lines AB Fig. 7.1 and CD intersect each other at the point O such that BC DA and BC = DA. Show that O is the midpoint of both the line-segments AB and CD. Solution : BC AD (Given) Therefore, CBO = DAO (Alternate interior angles) and BCO = ADO (Alternate interior angles) Also, BC = DA (Given) So, Δ BOC Δ AOD (ASA) Therefore, OB = OA and OC = OD, i.e., O is the mid-point of both AB and CD. Fig. 7.2 Sample Question 3 : In Fig.7.3, PQ > PR and QS and RS are the bisectors of Q and R, respectively. Show that SQ > SR. Solution : PQ > PR (Given) Therefore, R > Q(Angles opposite the longer side is greater) So, SRQ > SQR(Half of each angle) Therefore, SQ > SR(Side opposite the greater angle will be longer) Fig. 7.3

5 TRIANGLES 67 EXERCISE ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE. 2. In Fig.7.4, D and E are points on side BC of a Δ ABC such that BD = CE and AD = AE. Show that Δ ABD Δ ACE. 3. CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that Δ ADE Δ BCE. Fig. 7.4 E Fig In Fig.7.6, BA AC, DE DF such that BA = DE and BF = EC. Show that Δ ABC Δ DEF. 5. Q is a point on the side SR of a Δ PSR such that PQ = PR. Prove that PS > PQ. 6. S is any point on side QR of a Δ PQR. Show that: PQ + QR + RP > 2 PS. 7. D is any point on side AC of a Δ ABC with AB = AC. Show that CD < BD. 8. In Fig. 7.7, l m and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its end points on l and m, respectively. 9. Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M. Prove that MOC = ABC. Fig. 7.6 Fig. 7.7

7 TRIANGLES 69 BD = CD (Given) AD = ED (By construction) and ADB = EDC (Vertically opposite angles) Therefore, Δ ABD Δ ECD (SAS) So, AB = EC (1) (CPCT) and BAD = CED (2) Also, BAD = CAD (Given) Therefore, CAD = CED [From (2)] So, AC = EC [Sides opposite the equal angles] (3) Therefore, AB = AC [From (1) and (3)] Sample Question 4 : S is any point in the interior of Δ PQR. Show that SQ + SR < PQ + PR. Solution : Produce QS to intersect PR at T (See Fig. 7.11). From Δ PQT, we have PQ + PT > QT(Sum of any two sides is greater than the third side) i.e., PQ + PT > SQ + ST (1) From Δ TSR, we have ST + TR > SR (2) Adding (1) and (2), we get Fig PQ + PT + ST + TR > SQ + ST + SR i.e., PQ + PT + TR > SQ + SR i.e., PQ + PR > SQ + SR or SQ + SR < PQ + PR EXERCISE Find all the angles of an equilateral triangle. 2. The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in Fig Prove that the image is as far behind the mirror as the object is in front of the mirror. Fig. 7.12

9 TRIANGLES Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m. 16. Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC. 17. ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD. 18. ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC. 19. AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of B and D decide which is greater. 20. Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2 3 of a right angle. 21. ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.

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