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Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Revision Question Bank Triangles 1. 2. In the given figure, find the values of x and y. Solution. Since, AB = AC C =B [angles opposite to the equal sides are equal] x = 50° Also, the sum of all angles of a triangle is 180°. 50° +x + y=180° [put x = 50°] 50° + 50° + y = 180° y= 180° – 100° = 80° Hence, x = 50° and y = 80° In the given figure, PQ > PR, QS and RS are the bisectors of Q and R, respectively. Then, prove that SQ > SR Solution. Given, PQ > PR PRQ > PQR [angle opposite to the longer angle is longer] 1 1 PRO > PQR 2 2 [dividing both sides by 2] SRQ > SQR SQ > SR [side opposite to the longer angle is longer] www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 3 Pioneer Education {The Best Way To Success} 3. 4. IIT – JEE /AIPMT/NTSE/Olympiads Classes In the given figure, BA AC,DE DF such that BA=DE and BF = EC. Show that ABC DEF Solution. In ABC and DEF, BA =DE [given] BF = EC [given] BF + FC = EC + CF [adding both sides by FC] BC = EF Also, CAB = FDF [each 90°] [by RHS congruence rule] ABC DEF In figure, AB = DE, BC=EF and median AP = median DQ. Prove that B = E. Solution. Given ABC and DEF in which AB = DE, BC = EF and median AP = median DQ. To prove B = E Proof BC = EF [given] 1 1 BC EF [dividing both sides by 2] 2 2 BP = EQ [ AP and DQ are medians] In ABP and DEQ, AB = DE [given] AP=DQ [given] and BP = EQ [proved above] www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 4 Pioneer Education {The Best Way To Success} ABP DEQ 5. 6. IIT – JEE /AIPMT/NTSE/Olympiads Classes [by SSS congruence rule] [by CPCT] B =E Hence proved. Show that a median of a triangle divides it into two triangles of equal areas. Solution. Given ABC in which AD is the median i.e., BD = CD To prove ar( ABD) = ar( ACD) Construction Draw AE BC 1 Proof ar( ABD) × BD × AE ...(i) 2 1 and ar ( ACD) = CD × AE 2 1 = BD × AE ...(ii) 2 [ CD = BD, given] From Eqs. (i) and (ii), we get ar( ABD) = ar( ACD) Hence proved. In the given figure, XYZ and PYZ are two isosceles triangles on the base YZ with XY = XZ and PY= PZ. If P=120° and XYP=40°, then find YXZ. Solution. Given, in PXZ, PY = PZ PZY = PYZ [opposite angles of equal sides are equal] www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 5 Pioneer Education {The Best Way To Success} 7. IIT – JEE /AIPMT/NTSE/Olympiads Classes Let PYZ = PZY =x P + PYZ + PZY = 180° [ the sum of all angles of a triangle is 180°] 120°+x + x = 180° 2x = 180° – 120° 2x = 60° x = 30° PYZ = PZY = 300 Now, XYZ = XYP + PYZ = 40° + 30° = 70° In XYZ, YXZ + XYZ + XZY = 180° [ sum of all angles of a triangle is 180°] YXZ + 70° + 70° = 180° YXZ = 180° – 140° = 40° In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side. Solution. We have, AABC in which B is right angle and C is double A. Produce CB upto D such that BC = BD and join AD. In ABD and ABC, BD = BC [by construction] [each 90°] ABD = ABC and AB = AB [common side] ABD ABC [ by SAS Congruence rule] AD = AC and DAB = CAB = x [CPCT] [by CPCT] www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 6 Pioneer Education {The Best Way To Success} 8. IIT – JEE /AIPMT/NTSE/Olympiads Classes Now, DAC = DAB + CAB DAC = x + x = 2x DAC = ACB [ ACB = 2x] DC = AD [ sides opposite to equal angles are equal] [ BC = DB] 2BC = AD = AC Hence proved. In the given figure, RS = QT and QS=RT. Prove that PQ =PR. Solution. Given In PQR, RS = QT and QS = RT To prove PQ = PR Proof In RESQ and QTR, RS = QT .. (i) [given] SQ = RT ..(ii) [given] and RQ = RQ ...(iii) [common side] RSQ QTR [by SSS congruence rule] ...(i) QRS = RQT [by CPCT] and [by CPCT] ...(ii) QRT = RQS On subtracting Eq. (ii) from Eq. (i), we get QRS – QRT = RQT – RQS TRS = SQT Also, RSQ = QTR On multiplying both sides by -1 and then adding 180°, we get 180° – RSQ =180°– QTR QSP = RTP RPT QPS [by ASA congruence rule] 9. PQ = PR Hence proved. ABC is a triangle in which B = 2 C. D is a point on side BC such that AD bisects BAC and AB = CD. Prove that BAC=72°. Solution. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 7 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Given B = 2 C AD is the bisector of A and AB = DC To prove BAC =72° Construction Draw BE as angle bisector of B. Join DE. If C = x, then ABE = EBC = x Proof Since, AD is the bisector of A. [let] BAD = CAD = y Now, EBC is. an isosceles triangle. So, EBC = ECB= x° EB = EC Consider ABE DCE [by SAS congruence rule] [by CPCT] ...(i) EDC = 2y Consider isosceles AED as EA = ED EA = ED [by CPCT] ... (ii) LDA = EAD = y From Eqs. (i) and (ii), we get ... (iii) ADC = y + 2y =3y or ... (iv) ADC = 2x + y [exterior angles of AABD ] From Eqs. (iii) and (iv), we get 3y = 2x + y ...(v) x = y Now, in ABC, A + B + C=180° [ sum of all angles of a triangle is 180°) 2y + 2x + x =180° [from Eq. (v)] 2y + 2y + y = 180° 5y =180° 1800 360 y 5 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 8 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes A = 2y = 2 × 36° = 72° Hence, BAC = 72° Hence proved. 10. In the given figure, if E > A, C > D, then AD > EC. Is it true? Solution. True, In AEB, E > A ...(i) AB > BE In BDC, C > D ...(ii) BD > BC On adding Eqs. (i) and (ii), we get AB + BD > BE + BC AD > EC Chapter Test {Triangles} M: Marks: 40 1. M: Time: 40 BE and CF are two equal altitudes of a ABC . Using RHS congruence rule, prove that [4] ABC is an isosceles triangle. Solution. Given In a ABC, BE and CF are two equal altitudes of a ABC. To prove ABC is an isosceles triangle. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 9 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Proof In ABC and CFB, BE=CF BEC = CFB and BC = BC BEC CFB [given, equal altitudes] [each 90°] [common side] [by RHS congruence rule] [by CPCT] ECB = FBC or ACB = ABC Hence proved. 2. In figure, S is any point on the side QR of PQR. Prove that PQ + QR+ RP>2PS. [4] Solution. We know that, sum of the two sides of a triangle is greater than the third side. In PQS, PQ + QS>PS ...(i) and in PRS, PR + SR>PS ...(ii) On adding Eqs. (i) and (ii), we get (PQ + QS) + (PR + SR)> 2PS PQ + (QS + SR) +PR > 2PS PQ + QR + RP>2PS Hence proved. 3. In figure, AB = AC, D is a point in the interior of ABC such that DBC = DCB. Prove that AD bisects BAC of ABC. [4] Solution. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 10 Pioneer Education {The Best Way To Success} In BDC, we have DBC = DCB DC = DB ..(i) IIT – JEE /AIPMT/NTSE/Olympiads Classes [Given] Sides opposite to equal angles of DBC are equal Now, in ABD and ACD, we have AB = AC [Given] BD = CD [From (i)] and, AD = AD [Common side] So, by SSS congruency criterion, we have ABD ACD BAD = CAD Hence, AD is the bisector of BAC. 4. [c.p.c.t.] In a ABC, AB = AC and AD is bisector of BAC. Anju has to prove that: ABD ACD, which as follows: Here, AB = AC [given] [given] BAD = CAD and B = C [corresponding angles of a equal side are equal) ABD ACD [by ASA congruence rule] Further Anju shows this proof to his classmate Amita and she find that there is some errors in the proof. (i) Write the correct proof. (ii) What is the mistake in Anju's proof? (iii) Which value is depicted from this action? [4] Solution. (i) Given In ABC, AB = AC and AD is bisector of BAC. To prove ABD ACD Proof In ABD and ACD AB = AC [given] BAD = CAD [given] and AD = AD [common side] ABD ACD [by SAS congruence rule] (ii) Anju has used the result B = C for proving ABD = ACD, which is wrong. Because for proving B = C, firstly we prove that ABD ACD. (iii) The value depicted from this action, it is cooperative and learning among www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 11 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes students without any religion bias. 5. In the adjoining figure, ABC is a triangle and D is any point in its interior. Show that BD + DC < AB + AC. [4] Produced BD to meet AC in E. In ABE, AB + AE>BE [ sum of two sides is greater than the third side] ...(i) AB+AE > BD + DE [ BE =BD + DE] In CDE, DE + EC > DC ...(ii) [ sum of two sides is greater than the third side] On adding Eqs. (i) and (ii), we get AB + AE +DE +EC > BD + DE + DC AB + (AE +EC)>BD + DC AB + AC>BD + DC Hence, BD + DC < AB + AC 6. In ABC, if AD is the bisector of A, then show that AB > BD and AC > DC. [4] Solution : In ABC, AD is the bisector of A ..(i) 1 = 2 Since exterior angle of a triangle is greater than each of interior opposite angle. Therefore, in ADC, we have Ext. ADC > 2 3 > 2 3 > 1 [Using (i)] Thus, in ABD, we have 3 > 1 [Side opp, to greater angle is larger] AB >BD www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 12 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Hence, AB > BD. Similarly, we can prove that AC < CD. 7. In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that (i) PT = PS (ii) PSR = 15° Solution Given : QRST is a square and PQR is an equilateral triangle. To prove : (i) PS = PS (ii) PSR = 150 Proof : Since QRST is a square and PQR is an equilateral triangle. TQR = 900 and PQR = 600 TQR + PQR = 900 + 600 TQP = 1500 Thus, in TQP and SRP, we have TQ = RS TQP = SRP = 1500 and QP = RP (by SAS congruence criteria) TQP SRP (by cpct) PT = PS (ii) In PSR SR = RP ( QR = SR) RPS = PSR = x (say) Now, RPS + PSR + SRP = 1800 x + x + 1500 = 1800 2x + 1500 = 1800 2x = 1800 – 1500 [4] (by angle sum property) www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 13 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes 2x = x = 150 PSR = 150 300 8. A plot is in the form of AEF. Owner of this plot wants to build old age home, health centre and dispensary for elderly people as shown in figure. In which ABCD is a parallelogram and AB = BE and AD = DF. (i) Prove that BEC DCF . (ii) What values are depicted here? [4] Solution : Given : ABCD is a parallelogram AB = BE and AD = DF To prove : – BEC DCF Proof : Since ABCD is a parallelogram AB = DC But AB = BE (given) DC = BE Also, AD = BC but AD = DF (given) BC = DF As ABCD is a parallelogram , ..(i) ..(ii) www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 14 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes AD BC and AE as a transversal, ..(iii) 1 = 2 (corresponding angles) Also, AB DC and AF as a transversal, ..(iv) 1 = 3 (corresponding angles) from (iii) and (iv), we get ..(v) 2 = 3 Therefore, in BEC and DCF, we have BE = DC (from (i) BC = DF (from (ii) (from (v) 2 = 3 (by SAS congruence rule) BEC DCF (ii) Kindness and respectful for elder people. 9. In the figure, ABCD is a quadrilateral in which AD = BC and DAB = CBA. Prove that (i) ABD BAC (ii) BD = AC [4] Solution : Given : AD = BC and DAB = CBA To prove : (i) ABD ABC (ii)BD = AC Proof : In ABD and BAC, we have : AD = BC (Given) DAB = CBA (Given) AB = AB (common) Therefore, ABD BAC (by SAS congruence rule) (ii) BD = AC (by cpct) 10. ABCD is a square. P and Q are points on DC and BC respectively, such that AP = DQ, prove that (i) ADP ≅ DCQ (ii) DMP = 90° [4] www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 15 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Solution : Given : ABCD is a square, AP = AQ To prove : (i) ADP DCQ (ii) DMP = 900 Proof : (i) In ADP and DCQ ADP = DCQ = 900 AP = DQ (Given) AD = CD ( ABCD is a square) ADP DCQ A (by RHS congruence rule) (ii) Since ADP DCQ (by (i)) DAP + DAP + DPA = 1800 (by angle sum property) 900 + DAP + DPA = 1800 DAP + DPA = 900 QDC + DPA = 900 MDP + DPM = 900 In MDP, MDP + DPM + PMD = 180 900 + PMD = 1800 PMD = 1800 – 900 PMD = 900 (by (i)) .. (ii) (by angle sum property by ii) www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 16 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 17