## (CLASS 9)Polynomial -2 , Coordinate Geometry, Linear Equation In Two Variables, Euclid's Geometry, Lines And Angles

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Lines and Angles Class-IX In Fig. 6.13, lines AB and CD intersect at O. If LAOC + LBOE — - 700 and LBOD - — 400, find 1. 2. LBOE and reflex LCOE. c Answer Given, o Fig. 6.13 D LAOC + LBOE = 700 and LBOD = 400 LAOC + LBOE +LCOE = 1800 (Forms a straight line) 700 = 1800 LCOE = 1100 also, LCOE +LBOD + LBOE = 1800 (Forms a straight line) 1100 +400 + LBOE = 1800 + = 1800 LBOE = 300 In Fig. 6.14, lines XY and MN intersect at O. If LPOY = 900 and a : 2 : 3, find c. h x Fig. 6.14 Answer
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Given, LPOY = 900 and a • b = . 2:3 LPOY + + b = 1800 900 1800 a + b 900 Let a be 2x then will be 3x 2x + 3x = 900 5 X = 900 x 180 a = = 360 and b = = 540 also, b + c = 1800 (Linear Pair) 540 + C = 1800 C = 1260 3. In Fig. 6.15, LPQR = LPRQ, then prove that LPQS = Answer Given, LPQR = LPRQ To prove, LPQS = LPRT LPQS = 1800 - LPQR --- (i) also, LPRT = 1800 - LPRQ From (i) and (ii) LPQS = LPRT = 1800 - LPQR 4. LPRT. s LPQR +LPQS = 1800 (Linear Pair) LPRQ +LPRT = 1800 (Linear Pair) Q Fig. 6.15 LPRQ = 1800 - LPQR --- (ii) (LPQR = LPRQ) Therefore, LPQS = LPRT In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
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5. y o D Fig. 6.16 Answer Given, To Prove, AOB is a line or x + y = 1800 (linear pair.) x + y + w + z = 3600 (Angles around a point.) (x + y) + (w + z) = 3600 (x + y) + (x + y) = 3600 (Given x + y = w + z) 2(x + y) = 3600 (x + Y) = 1800 Hence, x + y makes a linear pair. Therefore, AOB is a staright line. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that LROS = 1/2(LQOS — LPOS). S Q Fig. 6.17 Answer Given, OR is perpendicular to line PQ To prove, LROS = 1/2(LQOS - LPOS) LPOR = LROQ = 900 (Perpendicular) LQOS = LROQ + LROS = 900 + LROQ --- (i) LPOS = LPOR - LROS = 900 - LROQ --- (ii) Subtracting (ii) from (i) LQOS - LPOS = 900 + LROQ - (900 - LROQ) LQOS - LPOS = 900 + LROQ - 900 + LROQ
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6. 7. LQOS - LPOS = 2LROQ LROS = 1/2(LQOS - LPOS) Hence, Proved. It is given that LXYZ = 640 and XY is produced to point P. Draw a figure from the given information. If ray Y Q bisects LZYP, find LXYQ and reflex LQYP. Answer Given, LXYZ = 640 YQ bisects LZYP z 640 x LXYZ +LZYP = 1800 (Linear Pair) 640 + L z Y P = u Y P = 1160 also, LZYP = LZYQ + LQYP LZYQ = LQYP (YQ bisects LZYP) LZYP = 2LZYQ 2LZYQ = 1160 LZYQ = 580 = LQYP Now, LXYQ = LXYZ + LZYQ LXYQ = 640 + 580 LXYQ = 1220 also, reflex LQYP = 1800 + LXYQ LQYP = 1800 + 1220 LQYP = 3020 In Fig. 6.28, find the values of x and y and then show that AB Il CD.
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500 c L) 1300 8. Fig. 6.28 Answer x + 500 = 1800 (Linear pair) X = 1300 also, y = 1300 (Vertically opposite) Now, x = y = 1300 (Alternate interior angles) Alternate interior angles are equal. Therefore, AB Il CD. In Fig. 6.29, if AB Il CD, CD Il EF and y : z = 3 : 7, find x. A c E Answer Given, AB Il CD and CD Il EF Now, B D Fig. 6.29 x + y = 1800 (Angles on the same side of transversal.) also, LO = z (Corresponding angles) and, y + LO = 1800 (Linear pair) Y = 3w and z = 7w 3w + 7w = 1800
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9. 1800 w 180 • Y = - 540 and, z = = 1260 Now, x + y = 1800 X + 540 = 1800 X = 1260 In Fig. 6.30, if AB Il CD, EF CD and LGED = 1260, find LAGE, LGEF and LFGE. AG c Answer Given, AB II CD EF CD LGED = 1260 LFED = 900 (EF CD) Now, D Fig. 6.30 LAGE = LGED (Since, AB II CD and GE is transversal. Alternate interior angles.) . LAGE = 1260 Also, LGEF = LGED - LFED LGEF = 1260 _ 900 LGEF = 360 Now, LFGE +LAGE - — 1800 (Linear pair) LFGE = 1800 - 1260 LFGE = 540 10. In Fig. 6.31, if PQ Il ST, LPQR = 1100 and LRST = 1300, find LQRS. [Hint : Draw a line parallel to ST through point R.]
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1300 1100 Fig. 6.31 Answer Given, PQ Il ST, LPQR = 1100 and LRST = 1300 Construction, A line XY parallel to PQ and ST is drawn. s 130' p x LPQR + LQRX = 1800 (Angles on the same side of transversal.) + = 1800 LQRX = 700 Also, LRST + LSRY = 1800 (Angles on the same side of transversal.) + = 1800 LSRY = 500 Now, LQRX +LSRY + LQRS = 1800 700 + 500 + = 1800 LQRS = 600 In Fig. 6.32, if AB Il CD, LAPQ = 500 and LPRD = 1270, 11. find x and y. 1270 c 500 Q Fig. 6.32 Answer Given, AB Il CD, LAPQ - - 500 and LPRD = 1270
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x = 500 (Alternate interior angles.) LPRD + LRPB = 1800 (Angles on the same side of transversal.) 1270 + = 1800 LRPB = 530 Now, y + 500 + LRPB = 1800 (AB is a straight line.) + 500 + 530 — 1800 1030 — 1800 Y 770 12 In Fig. 6.39, sides QP and RQ of APQR are produced to points S and T respectively. If LSPR = 1350 and LPQT = 1100, find LPRQ. S p 1350 I IOO Q Fig. 6.39 R Answer Given, LSPR = 1350 and LPQT = 1100 LSPR +LQPR = 1800 (SQ is a straight line.) LQPR = 450 also, LPQT +LPQR = 1800 (TR is a straight line.) = 1800 LPQR = 700 Now, LPQR +LQPR + LPRQ = 1800 (Sum of the interior angles of the triangle.) 700 + 450 + = 1800 + = 1800 LPRQ = 650 13 In Fig. 6.40, LX = 620, LXYZ = 540. If YO and ZO are the bisectors of LXYZ and LXZY respectively of A xyz, find LOZY and LYOZ.
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x 620 540 z Fig. 6.40 Answer Given, LX = 620, LXYZ = 540 YO and ZO are the bisectors of LXYZ and LXZY respectively. LX +LXYZ + LXZY = 1800 (Sum of the interior angles of the triangle.) 620 + 540 + LXZY = 1800 + = 1800 LXZY = 640 Now, LOZY = 1/2LXZY (ZO is the bisector.) LOZY = 320 also, LOYZ = 1/2LXYZ (YO is the bisector.) LOYZ = 270 Now, LOZY +LOYZ + LO = 1800 (Sum of the interior angles of the triangle.) 14. In Fig. 6.41, if AB Il DE, LBAC = 350 and LCDE = 530, find LDCE. 320 + 270 + LO = 1800 590 + = 1800 = 1210 Answer Given, 350 530 Fig. 6.41 AB Il DE, LBAC = 350 and L CDE = 530 LBAC = LCED (Alternate interior angles.) . LCED = 350 Now, LDCE +LCED + LCDE = 1800 (Sum of the interior angles of the triangle.)
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LDCE + 350 + 530 = 180 LDCE + 880 = 1800 LDCE = 920 In Fig. 6.42, if lines PQ and RS intersect at point T, such that LPRT — 15. - 400, LRPT - — 950 and LTSQ = 750, find LSQT. 950 400 Fig. 6.42 Answer Given, LPRT = 400, LRPT = 950 and LTSQ = 750 s 750 Q LPRT +LRPT + LPTR = 1800 (Sum of the interior angles of the triangle.) 400 + 950 + = 400 + 950 + = LPTR = 450 LPTR = LSTQ = 450 (Vertically opposite angles.) Now, LTSQ +LPTR + LSQT = 1800 (Sum of the interior angles of the triangle.) 750 + 450 + = 1800 1200 + LSQT = 1800 LSQT = 600 16. In Fig. 6.43, if PQ ps, PQ Il SR, LSQR = 280 and LQRT - — 650, then find the values of x and Q x Fig. 6.43 Answer Given,
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— 1800 (Linea pair) LP + LQ+ LR +LS = 3600 (Sum of the angles in a quadrilateral.) 900 + 650 + = 3600 + Y + = 3600 2700 + y + 370 = 3600 + = 3600 530 17. In Fig. 6.44, the side QR of APQR is produced to a point S. If the bisectors of LPQR and LPRS meet at point T, then prove that LQTR = 1/2LQPR. PQ ps, PQ Il SR, LSQR = 280 and LQRT = 650 Fig. 6 44 x +LSQR = X + 28 X 370 also, LQRT (Alternate angles as QR is transveersal.) o = 650 LQSR = x LQSR = 370 also, LQRS +LQRT - LQRS + 650 = 180 LQRS = 1150 Now, Q Answer Given, Bisectors of LPQR and LPRS meet at point T. To prove, LQTR = 1/2LQPR. Proof, R S LTRS = LTQR +LQTR (Exterior angle of a triangle equals to the sum of the two interior angles.) LQTR = LTRS - LTQR --- (i) also, LSRP = LQPR + LPQR 2LTRS = LQPR + 2LTQR LQPR = 2LTRS - 2LTQR
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1/2LQPR = LTRS - LTQR --- (ii) Equating (i) and (ii) LQTR - LTQR = 1/2LQPR Hence proved.

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