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Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Pioneer’s Revision Pack Lines and Angles 1. In the given figure, if x: y =1:4, then find the values of x and y, respectively. Solution. Given, x : y =1:4 Let x = a and y = 4a x + y = 180° [linear pair axiom] a+4a= 180° 5a =180° 1800 a 360 a 5 Hence, x – a = 36° and y = 4a = 4 × 36° = 144° 2. In the given figure, if AB x and y. CD, APQ =60°and PRD=137°, then find the values of Solution. AB CD [given] x = 60° [alternate interior angles] and x + y =137° (exterior angle of a triangle is equal to the sum of opposite interior angles) 60° + y =137° y =137° – 60° = 77° Hence, x = 60° and y = 77° www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 3 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes 4. In ABC, A + B = 122°and B + C =111°, find the measures of B and C. Solution. Given, ..(i) A + B = 122° and ..(ii) B + C=111° In ABC, ( A + B) + C =180° [sum of all angles of a triangle is 180°] 122° + C = 180° [from Eq. (i)] C=180°– 122° C = 58° On putting the value of C = 58° in Eq, (ii), we get B + 58° = 1110 B = 111° – 58° = 530 Hence, B = 53° and C = 58° In the given figure, AB CD EF. If CEF = 160° and BCE=10°, then find x. 5. Solution. In given Figure, AB CD EF and CE is a transversal. Therefore, sum of two interior angles is 180° FEC+ ECD=180° i.e., 160°+ ECD=180° [ FEC =160°, given] ECD=20° Since, AB CD and AB is a transversal ABC = BCD [alternate angles] x= BCE+ ECD . = 10° + 20° [ BCE =10°, given] = 30° In the given figure, XOY is a straight line and OQ XY at O. 3. Show that 2 QOP = YOP – XOP www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 4 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Solution. Given OQ XY To prove 2 QOP = YOP – XOP Proof OQ XOY [given] XOQ = QOY = 90° QOP + XOP = QOY QOP = QOY – XOP QOP + YOP = QOP – XOP QOP + QOP = YOP – XOP 2 QOP = YOP – XOP 6. Hence proved. In the given figure, AOC = ACO and BDO = BOD. Show that AC BD. Solution. Given AOC = ACO and BDO = BOD To prove AC BD Proof AOC = ACO [given] AOC = BOD [vertically opposite angles] ACO = BOD ..(i) Also, BOD = BDO ... (ii) [given] From Eqs. (i) and (ii), we get ACO = BDO i.e., angles are alternate interior angles. AC BD Hence proved. 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 In the given figure, EF DQ and AB CD. If FEB = 64° and PDC=27°, then find PDQ, AED and DEF. Solution. EF DQ and AB CD AED = PDC [corresponding angles] AED = 270 [given] AED + DEF + EES = 180° [linear pair axiom] 27° + DEF + 64° = 180° [ FEB = 64°, given] 91° + DEP =180° DEF =180° – 91° DEF = 89° PDQ = DEF [corresponding angles] PDQ = 89° Hence, PDQ = 89°, AED = 27° and DEF = 89°. 8. ABCDE is a regular pentagon and bisector of BAE meets CD at M. If bisector of BCD meets AM at P, then find CPM. Solution. Given, a regular pentagon bisector of A meets CD at M and bisector of BCD meets AM at P. Since, each angle of a regular pentagon www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 6 Pioneer Education {The Best Way To Success} = 0 540 1080 5 IIT – JEE /AIPMT/NTSE/Olympiads Classes [put n = 5] So, AM is the bisector of A. 1 1 BAM A 1080 540 2 2 Now, in quadrilateral ABCM, BAM + C + B + AMC = 360° 54° +108° +108° + AMC – 360° [ B = C =108°] 270° + AMC = 360° AMC = 360° –270° AMC = 90° PMC = AMC = 90° Since, CP is the bisector of BCD. 1 1 PCM C 1080 540 2 2 Now, in PCM, PCM + PMC + CPM = 180° [by angle sum property of a triangle is 180°] 54° + 90° + CPM = 180° CPM = 180° – 144 CPM = 36°. Hence, CPM is 360. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 7 Pioneer Education {The Best Way To Success} 9. In the given figure, AB IIT – JEE /AIPMT/NTSE/Olympiads Classes DE, find the value of BCD. Solution. Since, AB DE and DP is a transversal. EDF = ZDFB = 130° [alternate angles] ...(i) In a figure, DFC is a straight line. DPB + BFC = 180° [linear pair axiom] 130° + BFC=1800 [from Eq. (i)] BFC = 50° ...(ii) In BPC, ABC = BCF + BFC [exterior angle is equal to the sum of two interior angles] 80° = BCF + 50° [from Eq. (ii)] BCP = 30° 10. If the sides of an angle are respectively parallel to the sides of another angle, then prove that these angles are either equal or supplementary. Solution. Case I As QR BC 2 = 3 [corresponding angles] ...(i) and PQ AB, then www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 8 Pioneer Education {The Best Way To Success} 1 = 2 IIT – JEE /AIPMT/NTSE/Olympiads Classes [corresponding angles]...(ii) From Eqs. (i) and (ii), we get 1 = 2 = 3 Case II QR BC 1 3 [corresponding angles] But 1 + 2 = 180°, then 3 + 2 = 180°, supplementary. 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 Chapter Test {Lines and Angles} M: Marks: 40 1. 2. M: Time: 40 Min. In the given figure, lines XY and MN intersect at O. If POY = 900 and a : b = 2 : 3, then find the value of c. [4] Solution. Given, POY = 90° [linear pair axiom] POX = 90° 0 XOM + MOP = 90 ...(i) a + b = 90° Also, given a : b = 2 :3 Let a = 2k and b = 3k We have, a + b = 90° [from Eq.(i)] 2k + 3k = 90° 5k = 90° 900 k 5 k = 180 a = 2 × 180 = 360 and b = 3 × 180 = 540 Now, [Linear pair axiom] MOX + XON = 180° b + c = 180° [put b = 54°] 54° + C = 180° c = 180° – 54° c = 126° Two plane mirrors m and n are placed perpendicular to each other, as shown in figure. An incident ray AB to the first mirror is first reflected in the direction of BC and then reflected by the second mirror in the direction of CD. Prove that AB CD. [4] www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 10 Pioneer Education {The Best Way To Success} 3. IIT – JEE /AIPMT/NTSE/Olympiads Classes Solution. Let BM and CN be the normals to mirrors m and n, respectively and BM and CN intersect at O. Now, BM m, CN n and m n BM CN BOC = 90° ...(i) 2 + 3 = 90° By the laws of reflection, we have 1 = 2 and 3 = 4 [ angle of incidence = angle of reflection] 1 + 4 = 2 + 3 ( 2 + 3) 1 + 2 + 3 + 4 = 2 [ adding 2 and 3 both sides] ( 1 + 2) + ( 3 + 4) = 2(90°) = 180° [ from Eq. (i)] ABC + BCD = 180° But ABC and BCD are consecutive interior angles formed, when the transversal BC cuts AB and CD. AB CD Hence proved. In the given figure, m and n are two plane mirrors parallel to each other. Show that the incident ray CA is parallel to the reflected ray BD. [4] Solution. Given, two plane mirrors m and n, such that m n. An incident ray CA after reflection takes the path AB and then BD. AP and BQ are the normals to the plane mirrors m and n, respectively. Since, AP m, BQ n and m n AP n, BQ m = AP BQ Thus, AP and BQ are two parallel lines and a transversal AB cuts them at A and B, respectively. 2= 3 ...(i) www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 11 Pioneer Education {The Best Way To Success} 4. IIT – JEE /AIPMT/NTSE/Olympiads Classes [alternate interior angles] But ...(ii) 1 = 2 and ...(iii) 3 = 4 [by the laws of reflection] 1 + 2 = 2 + 2 [adding 2 both sides of Eq. (ii)] and 3 + 4 = 3 + 3 [adding 3 both sides of Eq. (iii)] 1 + 2 = 2( 2) and 3 + 4 = 2( 3) 1+ 2 = 3 + 4 [ 2 = 3 2( 2) = 2( 3)] CAB = ABD Thus, lines CA and BD are intersected by transversal AB such that CAB = ABD. i.e., alternate interior angles are equal. Hence, AC BD In the figure given below, prove that: x = a + b + c. [4] Solution. In the given figure, join B and D and produce to E. 5. Here, 1 + 2 = b and 3 + 4 = x ...(i) Also, 3 = 1 + a ...(ii) [exterior angle is equal to the sum of opposite interior angles] 4 = 2 + c and ...(iii) [exterior angle is equal to the sum of opposite interior angles] On adding Eqs. (ii) and (iii), we get 3 + 4 = 1 + 2 + a + c x=b+a+c [from Eq. (i)] In the given figure, side BC of ABC is produced to form ray BD and CE Show that ACD = A+ B. Deduce that: A + B + C = 180°. BA. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 [4] 12 Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes Solution. Given, CE BA and AC is the transversal. ...(i) 4 =1 [alternate interior angles] Again, CE BA and BD is the transversal. ...(ii) [corresponding angles] 5 = 2 On adding Eqs. (i) and (ii), we get 4 + 5 = 1 + 2 Hence, ACD = A + B Also, 1 + 2 = 4 + 5 [exterior angle is equal to the sum of the opposite interior angles] 1 + 2+ 3 = 3 + 4 + 5 [adding 3 both sides] A + B + C = 180° [ 3 + 4 + 5 = BCD = a straight angle] 6. In the given figure, AB CD. Find the values of x, y and z. [4] Solution. Since AB CD (Alternate angles) y = 750 0 z + 75 = 1250 (by exterior angle property) z = 1250 – 750 z = 500 Also, x + y = 1800 (Linear pair) 0 0 x + 75 = 180 x = 1800 – 750 x = 1050 Hence, x = 1050, y = 750 and z = 500 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 13 Pioneer Education {The Best Way To Success} 7. 8. IIT – JEE /AIPMT/NTSE/Olympiads Classes In the given figure, PQ RS and EF QS. If PQS = 60°, then find the measure of RFE. [4] Solution. Since PQ RS PQS + RSQ = 1800 (co-interior angles) 600 + RSQ = 1800 RSQ = 1200 Also EF QS (corresponding angles) RFE = RSQ 0 RFE = 120 In the given figure, AB CD EF. Also, EA AB. = 70°, then find the values of p, q and r. [4] Solution. Since AB CD p = q (corresponding angles) Also, EF CD q + 700 = 1800 (co-interior angles) q = 1800 – 700 q = 1100 p = 1100 Since AB CD, CD EF AB EF (co-interior angles) EAB + FEA = 1800 900 + FEA = 1800 FEA = 900 r + 700 = 900 r = 200 Hence, p = 1100 , q = 1100 and r = 200. www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 14 Pioneer Education {The Best Way To Success} 9. IIT – JEE /AIPMT/NTSE/Olympiads Classes In the given figure, l p and m q and lines l and m are intersecting lines. Prove that lines p and q also intersects. [4] Solution. Given : l p, m q To prove : p and q intersects Proof : Since l and m are intersecting lines and l p p and m intersect. Now, p and m intersect and m q Hence, p and q intersect. 10. In the given figure, ABCD in a quadrilateral in which ABC = 73°, C = 97° and D = 110°. If AE DC and BE AD and AE intersects BC at F, then find the value of EBF. [4] Solution. Given : AE DC, BE AD, ABC = 730, C = 970 To find : EBF Since AE DC 1100 DAE 1800 (Co-interior 's ) DAF = 700 Also, BEA = DCF = 970 (Corresponding B 's ) 0 BEA = 97 Now, in AFB , FAB + ABF + AFB = 1800 (Angle sum property) FAB + 730 + 970 = 1800 FAB = 1800 – 1700 = 100 Now, AD BE (Given) DAB + ABE = 1800 (Co-interior angles) DAF + FAB + ABF + EBF = 1800 700 + 100 + 730 + EBE = 1800 FBE = 1800 – 1530 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 15 Pioneer Education {The Best Way To Success} FBE = IIT – JEE /AIPMT/NTSE/Olympiads Classes 270 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 16