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- 1. Paper - I Q.1. Solve the following : 3 (i) Ray YM is the angle bisector of XYZ, where XY = YZ. Find the relation between XM and MZ. (ii) O is the centre of the circle. If m ABC = 80º, the find m (arc AC) and m (arc ABC). (iii) The circumcentre and incentre of ............... triangle are at the same point. Q.2. Solve the following : 6 (i) ABC ~ DEF, if AB = 2.4 cm, DE = 1.6 cm, find the ratio of the area of ABC and DEF. (ii) In the adjoining figure, if m (arc APC) = 60º and m BAC = 80º Find (a) ABC (b) m (arc BQC). (iii) Draw an arc with seg AB = 6.3 cm, inscribing ACB = 65º. S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min.GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS X M Y Z • • B O A C B A C80º • Q •P
- 2. Paper - I Q.3. Solve the following : 9 (i) Construct a right angled triangle PQR where PQ = 6 cm, QPR = 40º, PRQ = 90º. Draw circumcircle of PQR. (ii) In PQR, if QS is the angle bisector of Q then, show that A ( PQS) A ( QRS) = PQ QR (iii) Three congruent circles with centres A, B and C and with radius 5 cm each, touch each other in points D, E, G as shown in (a) What is the perimeter of ABC ? (b) What is the length of side DE of DEF ? Q.4. Solve the following : 12 (i) In the adjoining figure, DEFG is a square and BAC = 90º Prove that : (a) AGF ~ DBG (b) AGF ~ EFC (c) DBG ~ EFC (d) DE2 = BD . EC (ii) In a cyclic quadrilateral ABCD, the bisectors of opposite angles A and C meet the circle at P and Q respectively. Prove that PQ is a diameter of the circle. (iii) Construct SAB such that SB = 7.6 cm, SAB = 50º seg AD is median and AD = 5 cm. P Q S R A B C D EF A G F B D E C D P Q A C B • • × × Best Of Luck ... 2 ...
- 3. Paper - I A.1. Solve the following : (i) In XYZ, ray YM bisects XYZ [Given] XY YZ = XM MZ [Property of ½ angle bisector of a triangle] 1 = XM MZ [ XY = YZ] XM = MZ ½ (ii) m ABC = 1 2 m (arc AC) [By Inscribed angle theorem] 80º = 1 2 m (arc AC) m (arc AC) = 160º ½ m (arc ABC)= 360º – m (arc AC) = 360 – 160 m (arc ABC)= 200º ½ (iii) The circumcentre and incentre of an equilateral triangle are 1 at the same point. A.2. Solve the following : (i) ABC ~ DEF [Given] A ( ABC) A ( DEF) = AB DE 2 2 [Areas of similar triangles] ½ A ( ABC) A ( DEF) = (2.4) (1.6) 2 2 [Given] A ( ABC) A ( DEF) = 5.76 2.56 ½ MODEL ANSWER PAPER S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min. GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS X M Y Z • • B O A C
- 4. Paper - I... 2 ... A ( ABC) A ( DEF) = 576 256 ½ A ( ABC) A ( DEF) = 9 4 A (ABC) : A (DEF) = 9 : 4 ½ (ii) (a) m ABC = 1 2 m(arc APC) [Inscribed angle theorem] ½ m ABC = 1 2 × 60 m ABC = 30º ½ (b) m BAC = 1 2 m(arc BQC) [Inscribed angle theorem] ½ 80 = 1 2 m(arc BQC) m(arc BQC) = 80 × 2 m(arc BQC) = 160º ½ (iii) B A C80º • Q • P ½ mark for rough figure ½ mark for drawing base angles ½ mark for drawing an arc ½ mark for drawing ACB A 25º 6.3 cm B C 65º 25º 130º O A 6.3 cm B C 65º (Rough Figure)
- 5. Paper - I... 3 ... A.3. Solve the following : (i) (ii) In PQR ray QS bisects PQR [Given] PQ QR = PS SR ......(i) [By property of angle bisector of a triangle] 1 PQS and QRS have a common vertex Q and their bases PS and SR lie on the same line PR. Their heights are equal A ( PQS) A ( QRS) = PS SR [Triangles having equal heights] 1 A ( PQS) A ( QRS) = PQ QR [From (i)] 1 ½ mark for rough figure 1 mark for drawing triangle 1 mark for drawing perpendicular bisectors ½ mark for drawing circle R P6 cm 40º Q O R P6 cm 40º Q(Rough Figure) P Q S R
- 6. Paper - I (iii) (a) A - D - B B - E - C ½ A - F - C AD = DB = BE = EC = AF = FC = 5cm .......(i) [Radii of congruent circles and given] AB = AD + BD [ A - D - B] AB = 5 + 5 [From (i)] ½ AB = 10 cm ......(ii) Similarly, BC = 10 cm ......(iii) AC= 10 cm .......(iv) Perimeter of ABC = AB + BC + AC ½ = 10 + 10 + 10 [From (ii), (iii) and (iv)] Perimeter of ABC = 30 cm ½ (b) In ABC, D and E are mid-points of sides AB and BC respectively. [From (i)] DE = 1 2 AC [By mid-point theorem] ½ DE = 1 2 × 10 [From (iv)] DE = 5 cm. ½ A.4. Solve the following : (i) (a) DEFG is a square [Given] DE = EF = GF = DG .......(i) [Sides of a square] seg GF || seg DE [Opposite side of a square] ½ seg GF || seg BC [B - D - E - C) On transversal AB, AGF ABC .......(ii) [Converse of corresponding ½ angles test] On transversal AC, A B C D EF[If two circles are touching circles then the common point lies on the line joining their centres] A G F B D E C ... 4 ...
- 7. Paper - I AFG ACB ......(iii) In AGF and DBG, AGF GBD [From (ii) and A - G - B, B - D - C] GAF BDG [ each is 90º] AGF DBG ......(iv) [By AA test of similarity] 1 (b) In AGF and EFC, AFG FCE [From (iii) and A - F - C, C - E - A] GAF FEC [ Each is 90º] AGF ~ EFC ......(v) [By AA test of similarity] 1 (c) DBG ~ EFC [From (iv) and (v)] (d) BD EF = DG EC [c.s.s.t.] EF × DG = BD × EC DE × DE = BD × EC [From (i)] DE2 = BD × EC 1 (ii) Proof : DAP BAP [ ray AP bisects DAB] ½ Let m DAP = m BAP = xº ....(i) DCQ BCQ [ ray CQ bisects DCB] ½ Let, m DCQ = m BCQ = yº .....(ii) ABCD is cyclic [Given] m DAB + m DCB = 180º [Opposite angles of a cyclic quadrilateral are supplementary] m DAP + m BAP + DCQ + m BCQ = 180º [Angle addition property] x + x + y + y = 180º [From (i) and (ii)] 1 2x + 2y = 180º .....(iii) m DAP = 1 2 m (arc DP) [Inscribed angle theorem] x = 1 2 m (arc DP) [From (i)] m (arc DP) = 2xº .....(iv) ½ D P Q A C B • • × × ... 5 ...
- 8. Paper - I m DCQ = 1 2 m (arc DQ) [Inscribed angle theorem] y = 1 2 m (arc DQ) [From (ii)] m (arc DQ) = 2yº ......(v) ½ m (arc DP) + m (arc DQ) = 2x + 2y [Adding (iv) and (v)] m (arc PDQ) = 180º [Arc addition property and from (iii)] ½ Arc PDQ is a semicircle seg PQ is a diameter of the circle. ½ (iii) S B A 7.6 cm A 40º 40º 5 cm D O 100º S B 7.6 cm A 5 cm (Rough Figure) D 50º ½ mark for rough figure ½ mark for drawing centre O 1 mark for drawing the perpendicular bisector 1 mark for locating point A 1 mark for drawing SAB ... 6 ...
- 9. Paper - II Q.1. Solve the following : 3 (i) The sides of two similar triangles are 4 : 9. What is the ratio of their area ? (ii) What is the relation between ABC and ADC of cyclic ABCD ? (iii) If the circumcentre lies in the exterior of the triangle, then it is .......... angled triangle. Q.2. Solve the following : 6 (i) A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time a tower casts the shadow 40 m on the ground. Determine the height of the tower. (ii) If two circles with radii 8 and 3 respectively touch internally then show that the distance between their centers is equal to the difference of their radii, find that distance. (iii) Construct an arc PQM such that seg PM of length 6.2 cm subtends an angle of 40º on it. Q.3. Solve the following : 9 (i) Construct the incircle of RST in which RS = 6 cm, ST = 7 cm and RT = 6.5 cm. (ii) Triangle ABC has sides of length 5, 6 and 7 units while PQR has perimeter of 360 units. If ABC is similar to PQR then find the sides of PQR. S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min.GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS A B C D
- 10. Paper - II (iii) In the adjoining figure, in two chords AB and CD of the same circle are parallel to each other. P is the centre of the circle. Prove : m CPA = m DPB. Q.4. Solve the following : 12 (i) In ABC, seg DE || side BC. If 2A (ADE) = A (DBCE), find AB : AD and show BC = 3.DE (ii) ABC is inscribed in a circle with centre O, seg AX is a diameter of the circle with radius r. seg AD seg BC. Prove that (a) ABX ~ ADC, (b) A (ABC) = abc 4r . (a is side opposite to A, ...) (iii) Construct DEF such that DF = 6.2 cm, DEF = 60º, EM DF and EM = 4.4 cm. Best Of Luck A B D C O X ... 2 ... C D A B P
- 11. Paper - II A.1. Solve the following : (i) Let ABC ~ DEF, AB DE = 4 9 [Given] ABC ~ DEF A ( ABC) A ( DEF) = AB DE 2 2 [Areas of similar triangles] ½ A ( ABC) A ( DEF) = AB DE 2 A ( ABC) A ( DEF) = 4 9 2 A ( ABC) A ( DEF) = 16 81 A (ABC) : A (DEF) = 16 : 81 ½ (ii) ABCD is cyclic [Given] m ABC + m ADC = 180º [Opposite angles of quadrilateral are supplementary] ABC and ADC are supplementary. 1 (iii) If the circumcentre lies in the exterior of the triangle, then it 1 is an obtuse angled triangle. A.2. Solve the following : (i) In the adjoining figure, seg AB and seg PQ represents the vertical ½ stick and the tower respectively and seg BC and seg QR represents the shadow cast by them respectively. ½ ABC ~ PQR MODEL ANSWER PAPER S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min. GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS A B C D P Q R ? 40 m A B C8 m 12 m
- 12. Paper - II... 2 ... AB PQ = BC QR [c.s.s.t.] ½ 12 PQ = 8 40 [Given] PQ = 40 ×12 8 PQ = 60 Height of the tower is 60 m. ½ (ii) Let two circles with centres O and A touch each other internally at point T. O - A - T [If two circles are touching 1 circles then the common point lies a the line joining their centres] OT = OA + AT [O - A - T] OA = OT – AT ½ OT = 8 units, AT = 3 units [Given] OA = 8 – 3 OA = 5 units The distance between the centres is 5 units. ½ (iii) O A T ½ mark for rough figure ½ mark for drawing base angles ½ mark for drawing an arc ½ mark for drawing PQM MP 6.2 cm 40º Q (Rough Figure) MP 6.2 cm 50º 50º 40º Q O 80º
- 13. Paper - II A.3. Solve the following : (i) (ii) In ABC, AB = 5 units BC = 6 units [Given] AC = 7 units Perimeter of PQR = 360 units [Given] ½ PQ + QR + PR = 360 ......(i) ABC ~ PQR [Given] AB PQ = BC QR = AC PR [c.s.s.t.] 5 PQ = 6 QR = 7 PR 5 PQ = 6 QR = 7 PR = 5 6 7 PQ QR PR [By theorem on equal ratios] ½ 5 PQ = 6 QR = 7 PR = 18 360 [From (i)] 5 PQ = 6 QR = 7 PR = 1 20 .....(ii) ½ ½ mark for rough figure ½ mark for drawing RST 1 mark for drawing the angle bisectors 1 mark for drawing the incircle ... 3 ... S 7 cm 6 cm 6.5 cm R T × × • • O S 7 cm 6 cm 6.5 cm R T (Rough Figure)
- 14. Paper - II 5 PQ = 1 20 [From (ii)] PQ = 100 units ½ 6 QR = 1 20 [From (ii)] QR = 120 units ½ 7 PR = 1 20 [From (ii)] PR = 140 units ½ (iii) Construction : Draw seg BC. ½ Proof : m CPA = m (arc CA) .......(i) ½ m DPB = m (arc DB) ......(ii) m ABC = 1 2 m (arc CA) ......(iii) [Inscribed angle theorem] ½ m BCD = 1 2 m (arc DB) ......(iv) chord CD || chord AB [Given] On transversal BC, ABC BCD ......(v) [Converse of alternate angles test] ½ 1 2 m (arc CA) = 1 2 m(arc DB) [From (iii), (iv) and (v)] ½ m (arc CA) = m (arc DB) ......(vi) ½ m CPA = m DPB [From (i), (ii) and (vi)] A.4. Solve the following : (i) seg DE || side BC [Given] On transversal AB, ABC ADE .....(i) [Converse of ½ corresponding angles test] In ABC and ADE, ABC ADE [From (i)] BAC DAE [Common angle] ABC ~ ADE [By AA test of similarity] ½ [Definition of measure of minor arc] A D E B C ... 4 ... C D A B P
- 15. Paper - II AB AD = BC DE = AC AE .....(ii) [c.s.s.t.] ½ A ( ABC) A ( ADE) = AB AD 2 2 .....(iii) [Areas of similar triangles] ½ A (ABC) = A (ADE) + A (DBCE) [Area addition property] = A (ADE) + 2A (ADE) [ A (DBCE) = 2A (ADE)] A (ABC) = 3A (ADE) A ( ABC) A ( ADE) = 3 1 .....(iv) ½ 3 1 = AB AD 2 2 [From (iii) and (iv)] AB AD = 3 1 ......(v) [Taking square roots] ½ AB : AD = 3 : 1 BC DE = AB AD [From (ii)] ½ BC DE = 3 1 [From (v)] BC = 3 × DE ½ (ii) seg AX is a diameter [Given] arc ACX is a semi circle. m ABX = 90º [Angle subtended by ½ a semicircle] In ABX and ADC ABX ADC [Each is 90º] AXB ACD [Angles inscribed in the same arc and C - D - B] ABX ~ ADC [By AA test of similarity] 1 BC = a, AB = c, AC = b [Given] AB AD = AX AC [c.s.s.t.] ½ c AD = 2r b AD = bc 2r .......(i) ½ A B D C O X ... 5 ...
- 16. Paper - II... 6 ... A (ABC) = 1 2 × base × height ½ = 1 2 × BC × AD = 1 2 × a × bc 2r [From (i)] ½ A (ABC) = abc 4r ½ (iii) MD E E F 30º 6.2 cm 60º 4.4cm 30º 4.4cm O D F E 60º 4.4cm M 6.2 cm (Rough Figure) ½ mark for rough figure 1 mark for locating centre ‘O’ 1 mark for locating point E 1 mark for drawing the altitude ½ mark for drawing DEF
- 17. Paper - III Q.1. Solve the following : 3 (i) In the adjoining figure, line l || line m ||line n. Lines p and q are transversals. From given information find ST. (ii) If two circles touch externally then show that the distance between their centers is equal to the sum of their radii. (iii) What is the point of concurrence of the angle bisectors of a triangle called ? Q.2. Solve the following : 6 (i) ABC is a right angled at B. D is any point on AB. DE AC. If AD = 6 cm, AB = 12 cm, AC = 18 cm, find AE. (ii) In the adjoining figure, m (arc XAZ) = m (arx YBW). Prove that XY || ZW (iii) Construct an arc DCV such that seg DV of length 9.5 cm subtends an angle of 135º on it. S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min.GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS p q R TC B S A l m n 8 10 B C E D A X Y A B Z W
- 18. Paper - III Q.3. Solve the following : 9 (i) Construct the circumcircle of KLM in which KM = 7 cm, K = 60º, M = 55º. (ii) In the adjoining figure, AB || DC. Using the information given find the value of x. (iii) Prove that in a cyclic trapezium angles at the base are congruent. Q.4. Solve the following : 12 (i) Two poles of height ‘a’ meters and ‘b’ meters are ‘p’ meters apart. Prove that the height ‘h’ drawn from of the point of intersection N of the lines joining the top of each pole to the foot of the opposite pole is ab a + b meters. (ii) In the adjoining figure, points P and Q are the centers of the circles. Radius QN = 3, PQ = 9. M is the point of contact of the circles. Line ND is tangent to the larger circle. Point C lies on the smaller circle. Determine NC, ND and CD. (iii) SHR ~ SVU, In SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and SH SV = 3 5 ; construct SVU. D C A O B 3 x – 33x – 19 x – 5 S N a b h x yT BA R p C D Q P N M ... 2 ...
- 19. Paper - III A.1. Solve the following : (i) line l || line m || line n [Given] On transversals p and q, AB BC = RS ST [By Property of Intercepts made by threeparallel lines] ½ 8 10 = 12 ST [Given] ST = 12 ×10 8 ST = 15 units ½ (iii) Let two circles with centres O and A touch each other internally at point T. ½ O - A - T [If two circles are touching circles then the common point lies a the line joining their centres] OT = OA + AT [O - A - T] OA = OT – AT ½ (iii) The point of concurrence of the angle bisectors of a triangle 1 called is called incircle. MODEL ANSWER PAPER S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min. GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS p q R TC B S A l m n 8 10 O A T
- 20. Paper - III... 2 ... A.2. Solve the following : (i) In ABC and AED, BAC DAE [Common angle] ABC AED [ Each is 90º] 1 ABC ~ AED [By AA test of similarity] ½ AB AE = AC AD [c.s.c.t.] 12 AE = 18 6 [Given] AE = 12 × 6 18 AE = 4 units ½ (ii) Construction : Draw seg XW ½ Proof : m XWZ = 1 2 m (arc XAZ) .......(i) ½ m WXY = 1 2 m (arc YBW) .......(ii) But, m (arc XAZ) = m (arc YBW) .......(iii) [Given] ½ m XWZ = m WXY [From (i), (ii) and (iii)] XY || ZW [Alternate angles test] ½ (iii) B C E D A X Y A B Z W [Inscribed angle theorem] D C V 9.5 cm 45º45º 135º D C V 9.5 cm 135º (Rough Figure) ½ mark for rough figure ½ mark for drawing base angles ½ mark for drawing an arc ½ mark for drawing DCV
- 21. Paper - III... 3 ... A.3. Solve the following : (i) (ii) seg AB || seg DC [Given] On transversal BD, CDB ABD [Converse of ½ alternate angles test] CDO ABO .......(i) [D - O - B] In DOC and BOA, CDO ABO [From (i)] DOC BOA [Vertically opposite angles] 1 DOC ~ BOA [By AA test of similarity] DO BO = OC OA [c.s.s.t.] ½ 3 x – 3 = x – 5 3x –19 K 60º 55º 7 cm M L O K 60º 55º 7 cm M L (Rough figure) D C A O B 3 x – 33x – 19 x – 5 ½ mark for rough figure 1 mark for drawing LKM 1 mark for drawing perpendicular bisectors ½ mark for drawing circle
- 22. Paper - III 3(3x – 19) = (x – 5) (x – 3) 9x – 57 = x2 – 3x –5x + 15 9x – 57 = x2 – 8x + 15 ½ x2 – 8x – 9x + 15 +57 = 0 x2 – 17x + 72 = 0 x2 – 9x – 8x + 72 = 0 x (x – 9) – 8 (x – 9) = 0 (x – 9) (x – 8) = 0 x – 9 = 0 or x – 8 = 0 x = 9 or x = 8 ½ (iii) Given : ABCD is a cyclic trapezium. seg AB || seg BC ½ To prove : ABC DCB Proof : ABCD is cyclic [Given] m ABC + m ADC = 180º .....(i) [Opposite angles of a cyclic ½ quadrilateral are supplementary] seg AD || seg BC [Given] On transversal DC, m DCB + m ADC = 180º ......(ii) [Converse of interior angle ½ test] m ABC + m ADC = m DCB + m ADC [From (i) and (ii)] ½ m ABC = m DCB ABC DCB ½ A.4. Solve the following : (i) AB = AT + TB [ A - T - B] AB = (x + y) = p ......(i) ½ In ATN and ABR, TAN BAR [Common angle] ATN ABR [ each is 90º] ATN ABR [By AA test of similarity] ½ S N a b h x yT BA R p A D B C (½ mark for figure) ... 4 ...
- 23. Paper - III AT AB = TN BR [c.s.s.t] x p = h a ......(ii) [Given and from (i)] ½ In BTN and BAS, TBN ABS [Common angle] BTN BAS [ each is 90º] BTN BAS [By AA test of similarity] ½ BT AB = TN AS [c.s.s.t] y p = h b ......(iii) [Given and from (i)] ½ Adding (ii) and (iii), x y + p p = h h + a b ½ x + y p = 1 1 h + a b p p = a + b h ab [From (i)] ½ 1 = a + b h ab ab a + b = h h = ab a +b metres ½ (ii) Construction : Draw seg CM and seg DP ½ Sol. Q - M - P [If two circles are ½ touching circles the the common point lies on the line joining their centre] QN = 3 units [Given] PQ = 9 units PN = PQ + QN [ P - Q - N] PN = 9 + 3 PN = 12 units ½ QN = QM = 3 units [Radii of the same circle] MN = 6 units [ Diameter is twice the radius] C D Q P N M ... 5 ...
- 24. Paper - III PM + QM = PQ [ P - M - Q] PM + 3 = 9 [Given] PM = 9 – 3 PM = 6 units PM = PD = 6 units [Radii of the same circle] ½ In NDP, m NDP = 90º ......(i) [Radius is perpendiculer to the tangent] NP² = ND² + PD² [By Pythagoras theorem] 12² = ND² + 6² 144 = ND² + 36 ND² = 144 – 36 ND² = 108 ND = 108 ND = 36 3 ND = 6 3 units ½ Now, m NCM = 90º.....(ii) [Angle inscribed in a semicircle is a right angle] NCM NDP [From (i) and (ii)] seg CM || seg DP ....(iii) [By corresponding angles test] ½ In NDP, seg CM || seg DP, [From (iii)] NC CD = NM MP [By B.P.T.] NC CD = 6 6 NC CD = 1 NC = CD ½ C is the midpoint of seg ND NC = CD = 1 2 ND NC = CD = 1 6 3 2 NC = CD = 3 3 units. ½ ... 6 ...
- 25. Paper - III ½ mark for SHR 1 mark for constructing 5 congruent parts 1 mark for constructing VS5 S HS3 S 1 mark for constructing UVS RHS ½ mark for required SVU U R S H V 5.2cm 4.5 cm S1 S2 S3 S4 S5 ×× 5.8cm • • U S H V R (Rough Figure)(iii) ... 7 ...
- 26. Paper - IV Q.1. Solve the following : 3 (i) ABC ~ PQR, A = 47º, Q = 83º. What is the measure of C ? (ii) A chord divides a circle into 2 arcs measuring 2x and 7x. What is the measure of the minor arc ? (iii) What is the point of concurrence of the altitudes of a triangle called ? Q.2. Solve the following : 6 (i) ABC ~ DEF, if A (ABC) = 9 cm2 , A (DEF) = 64 cm2 , DE = 5.6 cm then find AB. (ii) If two circles with radii 8 and 3 respectively touch internally then show that the distance between their centers is equal to the difference of their radii, find that distance. (iii) Draw an arc with seg MN = 8.9 cm, inscribing MPN = 125º. Q.3. Solve the following : 9 (i) Construct the incircle of STU in which, ST = 7 cm, T = 120º, TU = 5 cm. (ii) In the adjoining figure, ADB and CDB have the same base DB. If AC and BD intersect at O then prove that A ( ADB) AO = A ( CDB) CO . S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min.GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS A D N O M B C
- 27. Paper - IV (iii) Find the radius of the circle passing through the vertices of a right angled triangle when lengths of perpendicular sides are 6 and 8. Q.4. Solve the following : 12 (i) In ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC. If PQ divides ABC into two equal parts means equal in area, find BP AB . (ii) In a cyclic quadrilateral ABCD, the bisectors of opposite angles A and C meet the circle at P and Q respectively. Prove that PQ is a diameter of the circle. (iii) LMN ~ XYZ, In LMN, LM = 6 cm, MN = 6.8 cm, LN = 7.6 cm and LM XY = 4 3 ; construct XYZ. D P Q A C B • • × × ... 2 ... Best Of Luck
- 28. Paper - IV A.1. Solve the following : (i) ABC ~ PQR [Given] m A = m P = 47º [c.a.s.t. and given] ½ m B = m Q = 83º In ABC, m A + m B + m C = 180º [Angle sum property of a triangle] 47º + 83º + m C = 180º 130º + m C = 180º – 130º m C = 50º ½ (ii) The measure of minor arc is 2x and measure of major arc is 7x 2x + 7x = 360º [Measure of a circle is 360º] ½ 9x = 360 x = 360 9 x = 40 Measure of minor arc = 2x = 2x × 40 = 80º. ½ (iii) The point of concurrence of the altitudes of a triangle is called 1 orthocentre. A.2. Solve the following : (i) ABC ~ DEF [Given] A ( ABC) A ( DEF) = AB DE 2 2 [Areas of similar triangles] ½ 9 64 = AB (5.6) 2 2 [Given] 3 8 = AB 5.6 [Taking square roots] ½ AB = 3 5.6 8 AB = 3 × 0.7 ½ AB = 2.1 cm ½ MODEL ANSWER PAPER S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min. GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS
- 29. Paper - IV... 2 ... (ii) Let two circles with centres O and A touch each other internally at point T. O - A - T [If two circles are touching 1 circles then the common point lies a the line joining their centres] OT = OA + AT [O - A - T] OA = OT – AT ½ OT = 8 units, AT = 3 units [Given] OA = 8 – 3 OA = 5 units The distance between the centres is 5 units. ½ (iii) A.3. Solve the following : (i) O A T M N35º 125º P 35º 8.9 cm 110º O M N 125º P 8.9 cm (Rough Figure) ½ mark for rough figure ½ mark for drawing base angles ½ mark for drawing an arc ½ mark for drawing MPN 7 cm S T 5 cm U 120º (Rough Figure)
- 30. Paper - IV (ii) ADB and CDB have a common base BD, A( ADB) A( CDB) = AN CM ....(i) [Triangles with common base] 1 In ANO and CMO, ANO CMO [ Each is 90º] AON COM [Vertically opposite angles] 1 ANO ~ CMO [By AA test of similarity] AN CM = AO CO ......(ii) [c.s.s.t.] ½ A ( ADB) A ( CDB) = AO CO [From (i) and (ii)] ½ 7 cm ST 5 cm U 120º • × • × O ½ mark for rough figure ½ mark for drawing STU 1 mark for drawing the angle bisectors 1 mark for drawing the incircle A D N O M B C ... 3 ...
- 31. Paper - IV (iii) Given : (a) In PQR, m PQR = 90º (b) PQ = 6 units, QR = 8 units. (c) Points P, Q and R lie on the circle. ½ To Find : radius of the circle. Sol. In PQR, m PQR = 90º [Given] PR² = PQ² + QR² [By Pythagoras theorem] PR² = 6² + 8² PR² = 36 + 64 PR² = 100 PR = 10 units [Taking square roots] ½ Let Y be a point on the circle as shown in the figure m PQR = 90º [Given] m PQR = 1 2 m(arc PYR) [Inscribed angle theorem] ½ 90º = 1 2 m(arc PYR) ½ m (arc PYR) = 180º arc PYR is a semicircle seg PR is the diameter. Diameter = 10 units. Radius = 5 units [ Radius is half of the diameter] ½ Radius of the circle is 5 units. A.4. Solve the following : (i) seg PQ divides ABC into two parts of equal areas [Given] A (APQ) = 1 2 A (ABC) A ( APQ) A ( ABC) = 1 2 ......(i) ½ seg PQ || side BC [Given] On transversal AC, AQP ACB ......(ii) [Converse of corresponding angles ½ test] In APQ and ABC, AQP ACB [From (ii)] PAQ BAC [Common angle] APQ ~ ABC [By AA test of similarity] ½ A ( APQ) A ( ABC) = AP AB 2 2 [Areas of similar triangles] ½ 1 2 = AP AB 2 2 [From (i)] 8 Y R Q P 6 • (½ mark for figure) B P A Q C ... 4 ...
- 32. Paper - IV 1 2 = AP AB [Taking square roots] ½ AP AB = 1 2 AB – BP AB = 1 2 [ A - P - B] ½ AB BP – AB AB = 1 2 BP 1 – AB = 1 2 ½ 1 1 – 2 = BP AB BP AB = 2 – 1 2 ½ (ii) Proof : DAP BAP [ ray AP bisects DAB] ½ Let m DAP = m BAP = xº ....(i) DCQ BCQ [ ray CQ bisects DCB] ½ Let, m DCQ = m BCQ = yº .....(ii) ABCD is cyclic [Given] m DAB + m DCB = 180º [Opposite angles of a cyclic quadrilateral are supplementary] m DAP + m BAP + DCQ + m BCQ = 180º [Angle addition property] x + x + y + y = 180º [From (i) and (ii)] 1 2x + 2y = 180º .....(iii) m DAP = 1 2 m (arc DP) [Inscribed angle theorem] x = 1 2 m (arc DP) [From (i)] m (arc DP) = 2xº .....(iv) ½ m DCQ = 1 2 m (arc DQ) [Inscribed angle theorem] y = 1 2 m (arc DQ) [From (ii)] D P Q A C B • • × × ... 5 ...
- 33. Paper - IV... 6 ... m (arc DQ) = 2yº ......(v) ½ m (arc DP) + m (arc DQ) = 2x + 2y [Adding (iv) and (v)] m (arc PDQ) = 180º [Arc addition property and from (iii)] ½ Arc PDQ is a semicircle seg PQ is a diameter of the circle. ½ (iii) Analysis : LMN ~ XYZ [Given] LM XY = MN YZ = LN XZ = 4 3 ...... (i) [c.s.s.t.] LM XY = 4 3 [From (i)] MN YZ = 4 3 [From (i)] LN XZ = 4 3 [From (i)] 6 XY = 4 3 6.8 YZ = 4 3 7.6 XZ = 4 3 18 4 = XY 20.4 4 = YZ 22.8 4 = XZ XY = 4.5 cm YZ = 5.1 cm XZ = 5.7 cm Information for constructing XYZ is complete. X Y Z 4.5cm 5.7 cm 5.1 cm (Required triangle) 7.6 cm 6cm 6.8 cm L M N (Given triangle) 2 marks for analysis 1 mark for LMN 1 mark for XZY
- 34. Paper - V Q.1. Solve the following : 3 (i) ABC ~ PQR, AB = 6 cm, BC = 8 cm, CA = 10 cm and QR = 6 cm. What is the length of side PR ? (ii) ABCD is cyclic quadrilateral such that 2m A = 3m C. What is the measure of C ? (iii) What is the point of concurrence of the medians of a triangle called ? Q.2. Solve the following : 6 (i) Ray PT is the angle bisector of QPR. Find the value of x and the perimeter of PQR. (ii) In the adjoining figure, m (arc XAZ) = m (arx YBW). Prove that XY || ZW (iii) Draw an arc such that chord ST = 5.6 cm, inscribing SVT = 80º. Q.3. Solve the following : 9 (i) Construct the circumcircle of SIM in which SI = 6.5 cm, I = 125º, IM = 4.4 cm. S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min.GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS P Q T R 4 cm 5 cm 5.6 cm x • • X Y A B Z W
- 35. Paper - V (ii) In the adjoining figure, ABCD is a square. The BCE on side BC and ACF on the diagonal AC are similar to each other. Then show that A (BCE) = 1 A ( ACF) 2 (iii) Chords AB and CD of a circle intersect in point Q in the interior of a circle as shown in figure, if m (arc AD) = 25º circle and m (arc BC) = 31º, then find BQC. Q.4. Solve the following : 12 (i) Let X be any point on side BC of ABC, XM and XN are drawn parallel to BA and CA. MN meets produced CB in T. Prove that TX2 = TB . TC. (ii) In the adjoining figure, two circles touch each other internally in a point A. The radius of the smaller circle with centre M is 5. The smaller circle passes through the centre N of the larger circle. The tangent to the smaller circle drawn through C intersects the larger circle in point D. Find CD. (iii) Construct CVX such that CX = 9.1 cm, CVX = 130º, VD CX and VD = 1.7 cm. ... 2 ... D B QA C F CD A B E A N M CXB T A D P MNC Best Of Luck
- 36. Paper - V A.1. Solve the following : (i) ABC ~ PQR [Given] AB PQ = BC QR = AC PR [c.s.s.t.] ½ 6 PQ = 8 6 = 10 PR 8 6 = 10 PR PR = 6 10 8 PR = 15 2 PR = 7.5 cm ½ (ii) 2m A = 3m C [Given] m A = 3 m C 2 .......(i) ABCD is cyclic m A + m C = 180º [Opposite angles of a cyclic ½ quadrilateral is supplementary] 3 2 m C + m C = 180º [From (i)] 3m C 2m C 2 = 180 5m C 2 = 180 m C = 180 2 5 m C = 72º ½ (iii) The point of concurrence of the medians of a triangle is called 1 Centroid. MODEL ANSWER PAPER S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min. GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS
- 37. Paper - V... 2 ... A.2. Solve the following : (i) In PQR, ray PT bisects QPR[Given] PQ PR = QT TR [Property of angle bisector of a triangle] ½ 5.6 x = 4 5 x = 5 5.6 4 x = 7 ½ PR = 7 cm QR = QT + TR [ Q - T - R] QR = 4 + 5 QR = 9 cm ½ Perimeter of PQR = PQ + QR + PR = 3.6 + 9 + 7 Perimeter of PQR = 19.6 cm ½ (ii) Construction : Draw seg XW ½ Proof : m XWZ = 1 2 m (arc XAZ) .......(i) ½ m WXY = 1 2 m (arc YBW) .......(ii) But, m (arc XAZ) = m (arc YBW) .......(iii) [Given] ½ m XWZ = m WXY [From (i), (ii) and (iii)] XY || ZW [Alternate angles test] ½ (iii) P Q T R 4 cm 5 cm 5.6 cm x • • X Y A B Z W [Inscribed angle theorem] S 80º V T5.6 cm 10º 10º160º O S 80º V T 5.6 cm (Rough Figure) ½ mark for rough figure ½ mark for drawing base angles ½ mark for drawing an arc ½ mark for drawing SVB
- 38. Paper - V... 3 ... A.3. Solve the following : (i) I 4.4 cm M 125º 6.5 cm S O I 4.4 cm M 125º 6.5 cm S (Rough Figure) ½ mark for rough figure 1 mark for drawing triangle 1 mark for drawing perpendicular bisectors ½ mark for drawing circle
- 39. Paper - V (ii) BCE ~ ACF [Given] A ( BCE) A ( ACF) = 2 2 BC AC ......(i) [Areas of similar triangles] ½ ABCD is a square [Given] AB = BC = CD = AD ......(ii) [Sides of a square] ½ In ABC, m ABC= 90º [Angle of a square] AC2 = AB2 + BC2 [By Pythagoras theorem] ½ AC2 = BC2 + BC2 [From (i)] AC2 = 2BC2 .....(iii) ½ A ( BCE) A ( ACF) = BC 2BC 2 2 [From (i) and (iii)] ½ A ( BCE) A ( ACF) = 1 2 A (BCE) = 1 2 A (ACF) ½ (iii) m(arc AD) = 25º [Given] m ACD = 1 2 m(arc AD) [Inscribed ½ angle theorem] m ACD = 1 2 × 25º m ACD = 12.5º m ACQ = 12.5º [ D - Q - C] ½ m (arc BC) = 31º [Given] m BAC = 1 2 m(arc BC) [Inscribed angle theorem] ½ m BAC = 1 2 × 31º m BAC = 15.5º m QAC = 15.5º [ A - Q - B] ½ BQC is an exterior angle of AQC, m BQC = m QAC + m ACQ [Remote interior angle theorem] ½ m BQC = 15.5º + 12.5º m BQC = 28º ½ F CD A B E D B QA C ... 4 ...
- 40. Paper - V A.4. Solve the following : (i) seg AB || seg MX [Given] seg NB || seg MX [ A - N - B] On transversal TX, TBN TXM ......(i) [Converse of corresponding ½ angles test] In TBN and TXM, TBN TXM [From (i)] BTN XTM [Common angle] TBN TXM [By AA test of similarity] TB TX = TN TM .....(ii) [c.s.s.t.] 1 seg NX ll seg AC [Given] seg NX ll seg MC [ A - M - C] On transversal TC, TXN TCM ......(iii) [Converse of corresponding ½ angles test] In TXN and TCM, TXN TCM [From (iii)] NTX MTC [Common angle] TXN TCM [By AA test of similarity] TX TC = TN TM .....(iv) [c.s.s.t.] 1 TB TX = TX TC [From (ii) and (iv)] ½ TX2 = TB × TC ½ (ii) Construction : Draw seg AD. ½ Sol. NM = MA = MP = 5 units [Radii of same circle] NA = 2 NM [Diameter is twice the radius] NA = 2 × 5 NA = 10 units A N M CXB T A D P MNC ... 5 ...
- 41. Paper - V CN = NA = 10 units [Radii of same circle] CA = 2 CN [Diameter is twice the radius] CA = 2 × 10 CA = 20 units CM = CN+ NM [C - N - M] CM = 10 + 5 CM = 15 units ½ m CPM = 90º [Radius is perpendicular to tangent] m CDA = 90º [Angle subtended by a semicircle] ½ CPM CDA seg PM || seg DA ......(i) [By corresponding angles test] In CPM CPM = 90º [Radius is perpendicular to ½ the tangent] CM2 = CP2 + PM2 [By Pythagoras theorem] 152 = CP2 + 52 225 = CP2 + 25 CP2 = 225 – 25 CP2 = 200 CP = 100 × 2 [Taking square roots] CP = 10 2 units ½ In CDA, seg PM || seg DA [From (i)] CP PD = CM MA [By B.P.T.] ½ 10 2 PD = 15 5 PD = 10 2 × 5 15 PD = 10 2 3 units ½ CD = CP + PD [C - P - D] CD = 10 2 10 2 1 3 CD = 30 2 10 2 3 CD = 40 2 3 units ½ ... 6 ...
- 42. Paper - V (iii) C X V 130º 1.7cm D 9.1 cm (Rough Figure) V V 130º C X 40º D 1.7 cm 9.1 cm 40º O 100º A B 1.7 cm ½ mark for rough figure 1 mark for locating centre ‘O’ 1 mark for locating point V 1 mark for drawing the altitude ½ mark for drawing CVX ... 7 ...
- 43. Paper - VI Q.1. Solve the following : 3 (i) The areas of two similar triangles are 18 cm2 and 32 cm2 respectively. What is the ratio of their corresponding sides ? (ii) Two circles with centres P and Q having diameter 25 cm and 15 cm respectively touch each other externally at A, then what is the distance between P and Q ? (iii) If the circumcircle lies in the interior of the triangle, then it is ........... angled triangle. Q.2. Solve the following : 6 (i) From the information given in the adjacent figure, state whether the triangles are similar with reasons. (ii) In the adjoining figure, if m (arc APC) = 60º and m BAC = 80º Find (a) ABC (b) m (arc BQC). (iii) Construct an arc PQM such that seg PM of length 6.2 cm subtends an angle of 40º on it. Q.3. Solve the following : 9 (i) Construct the incircle of SRN, such that RN = 5.9 cm, RS = 4.9 cm, R = 95º. S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min.GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS A P Q R B C 4.6 2.3 4 8 10 5 B A C80º • Q •P
- 44. Paper - VI (ii) ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO CO = BO DO . (iii) In the adjoining figure, A and B are centers of two circles touching each other at M. Line AC and line BD are tangents. If AD = 6 cm and BC = 9 cm then find the length of seg AC and seg BD. Q.4. Solve the following : 12 (i) In the adjoining figure, XY || AC and XY divides the triangular region ABC into two equal areas. Determine AX : AB. (ii) ABC is inscribed in a circle with centre O, seg AX is a diameter of the circle with radius r. seg AD seg BC. Prove that (a) ABX ~ ADC, (b) A (ABC) = abc 4r . (a is side opposite to A, ...) (iii) Construct XYZ such that XY = 9.5 cm, XZY = 115º, ZP is median. ZP = 3.3 cm. D A B C M A C X B Y A B D C O X ... 2 ... Best Of Luck
- 45. Paper - VI A.1. Solve the following : (i) Let ABC ~ PQR A ( ABC) A ( PQR) = AB PQ 2 2 [Areas of similar triangles] ½ 18 32 = AB PQ 2 2 [Given] 9 16 = AB PQ 2 2 AB PQ = 3 4 [Taking square roots] AB : PQ = 3 : 4 ½ (ii) Diameter of the cicle with centre P = 25 cm Its radius (r1 ) = 12.5 cm Diameter of the circle with centre Q = 15 cm ½ Its radius (r2 ) = 7.5 cm Both the circles touch each other externally at point A Distance between P and Q = r1 + r2 = 12.5 + 7.5 Distance between P and Q = 20 cm ½ (iii) If the circumcircle lies in the interior of the triangle, then it is 1 acute angled triangle. A.2. Solve the following : (i) AB PQ = 4.6 2.3 AB PQ = 2 1 ......(i) ½ BC QR = 10 5 BC QR = 2 1 ......(ii) ½ MODEL ANSWER PAPER S.S.C. Test - II Batch : SB Marks : 30 Date : Time : 1 hr. 15 min. GEOMETRY – Chapter : 1, 2, 3 MAHESH TUTORIALS A P Q R B C 4.6 2.3 4 8 10 5
- 46. Paper - VI... 2 ... AC PR = 8 4 AC PR = 2 1 ......(iii) ½ In ABC and PQR, AB PQ = BC QR = AC PR [From (i), (ii) and (iii)] ABC PQR [By SSS test of similarity] ½ (ii) (a) m ABC = 1 2 m(arc APC) [Inscribed angle theorem] ½ m ABC = 1 2 × 60 m ABC = 30º ½ (b) m BAC = 1 2 m(arc BQC) [Inscribed angle theorem] ½ 80 = 1 2 m(arc BQC) m(arc BQC) = 80 × 2 m(arc BQC) = 160º ½ (iii) B A C80º • Q •P MP 6.2 cm 40º Q (Rough Figure) MP 6.2 cm 50º 50º 40º Q O 80º ½ mark for rough figure ½ mark for drawing base angles ½ mark for drawing an arc ½ mark for drawing PQM
- 47. Paper - VI... 3 ... A.3. Solve the following : (i) (ii) ABCD is a trapezium side AB || side DC [Given] On transversal AC, BAC DCA [Converse ½ of alternate angles test] BAO DCO ......(i) [ A - O - C] ½ In AOB and COD, BAO DCO [From (i)] AOB COD [Vertically opposite angles] AOB ~ COD [By AA test of similarity] 1 AO CO = BO DO [c.s.s.t.] ½ AO CO = BO DO [By Alternendo] ½ (iii) A - M - B [If two circles are touching ½ circles then the common point lies on the line joining their centres] 5.9 cm NR 4.9 cm S 95º × ×•• O 5.9 cm NR 4.9 cm S 95º (Rough Figure) A B O D C D A B C M ½ mark for rough figure ½ mark for drawing SRN 1 mark for drawing the angle bisectors 1 mark for drawing the incircle
- 48. Paper - VI AM = AD = 6 cm ........(i) [Radii of the same circle] BM = BC = 9 cm ........(ii) AB = AM + MB [ A - M - B] AB = 6 + 9 [From (i) and (ii)] ½ AB = 15 cm ........(iii) In ABC, m ACB = 90º [Radius is perpendicular to the tangent] ½ AB² = AC² + BC² [By Pythagoras theorem] 15² = AC² + 9² [From (ii) and (iii)] 225 = AC² + 81 AC² = 225 – 81 AC² = 144 AC = 12 cm [Taking square roots] ½ In ADB, m ADB = 90º [Radius is perpendicular to the tangent] ½ AB2 = AD2 + BD2 [By Pythagoras theorem] 152 = 62 + BD2 [From (i) and (iii)] 225 = 36 + BD2 BD2 = 225 – 36 BD2 = 189 BD = 9 21 BD = 3 21 cm. [Taking square roots] ½ The lengths of seg AC and seg BD are 12 cm and 3 21cm respectively. A.4. Solve the following : (i) seg XY divides ABC in two parts of equal areas A(XYB) = 1 2 A(ACB) A ( XYB) A ( ACB) = 1 2 .......(i) ½ seg XY || side AC [Given] On transversal BC, XYB ACB .....(ii) [Converse of corresponding ½ angles test] In XYB and ACB, XYB ACB [From (ii)] XBY ABC [Common angle] XYB ~ ACB [By AA test of similarity] ½ A C X B Y ... 4 ...
- 49. Paper - VI A ( XYB) A ( ACB) = XB AB 2 2 [Areas of similar triangles] ½ 1 2 = 2 2 XB AB [From (i)] 1 2 = XB AB [Taking square roots] ½ XB AB = 1 2 AB – AX AB = 1 2 [ A - X - B] ½ AB AB – AX AB = 1 2 ½ AX 1 – AB = 1 2 1 1 – 2 = AX AB AX AB = 2 – 1 2 ½ (ii) seg AX is a diameter [Given] arc ACX is a semi circle. m ABX = 90º [Angle subtended by ½ a semicircle] In ABX and ADC ABX ADC [Each is 90º] AXB ACD [Angles inscribed in the same arc and C - D - B] ABX ~ ADC [By AA test of similarity] 1 BC = a, AB = c, AC = b [Given] AB AD = AX AC [c.s.s.t.] ½ c AD = 2r b AD = bc 2r .......(i) ½ A B D C O X ... 5 ...
- 50. Paper - VI A (ABC) = 1 2 × base × height ½ = 1 2 × BC × AD = 1 2 × a × bc 2r [From (i)] ½ A (ABC) = abc 4r ½ (iii) X Y Z 115º 3.3cm P 9.5 cm (Rough Figure) Z Z Y 9.5 cm 25ºX P 3.3 cm 115º 25º O 130º ½ mark for rough figure ½ mark for drawing centre O 1 mark for drawing the perpendicular bisector 1 mark for locating point Z 1 mark for drawing XZY ... 6 ...

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