Solved problems strictly according to the syllabus.
1
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
2
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.
3
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
4
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.
5
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
6
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.]
7
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
8
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.
9
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.)
10
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,
11
— 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
12
1/2LQPR = LTRS - LTQR --- (ii) Equating (i) and (ii) LQTR - LTQR = 1/2LQPR Hence proved.
Discussion
Copyright Infringement: All the contents displayed here are being uploaded by our members. If an user uploaded your copyrighted material to LearnPick without your permission, please submit a Takedown Request for removal.
Need a Tutor or Coaching Class?
Post an enquiry and get instant responses from qualified and experienced tutors.
If you have your own Study Notes which you think can benefit others, please upload on LearnPick. For each approved study note you will get 25 Credit Points and 25 Activity Score which will increase your profile visibility.