297 Proofs: S.N. Statements S.N. Reasons 1. In POS and ROQ 1. ................. i. PO = RO (S) i. From given ii. POS = ROQ (A) ii. Being V.O.A. iii. SO = QO (S) iii. From given 2. POS –ROQ 2. By S.A.S axiom 3. OPS = ORQ 3. Being corresponding angles of –Triangles 4. PS//QR 4. Being equal alternate angles 5. PQ//SR 5. Proceed same as 4. 6. PQRS is parallelogram 6. Being parallel opposite sides Proved 8. (a) Prove that all the angles of a rectangle are right angles. (b) Prove that all the angles and sides of a square are equal. OR Prove that a square is an equilateral and equiangular quadrilateral. (c) Prove that diagonals of a rectangle are equal. (d) Prove that if the diagonals of a parallelogram are equal, then it is a rectangle. Solution: a. Given: In rectangle ABCD, A = 90° To-prove: B = C = D = 90° Proofs: 1. A + B = 180° (‡ being co-interior angles) or, 90° + B = 180° B = 90° 2. C = A (‡ opposite angles of ) C = 90° 3. D = B (‡ opposite angles of ) D = 90° Hence, A = B = C = D = 90° b. Given: In square ABCD, A = 90° and AB = AD To-prove: B = C = D = A = 90° AB = BC = CD = AD Proofs: 1. A + B = 180° (‡ being co-interior angles) or, 90° + B = 180° B = 90° 2. C = 90°, D = 90° (‡ proceed same as (i)) Hence, A = B = C = D = 90° 3. AB = DC, AD = BC (‡ being opposite sides of parallelogram) 4. AB = BC = CD = AD (‡ AB = AD) D C A B D C A B
298 c. Given: AC and BD are diagonals of rectangle ABCD To-prove: AC = BD Proofs: S.N. Statements S.N. Reasons 1. In ABC and BAD 1. i. BC = AD (S) i. Opposite sides of parallelogram or rectangle. ii. ABC = BAD (A) ii. being 90° iii. AB = BA (S) iii. Being common side 2. ABC –BAD 2. By S.A.S axiom 3. AC = BD 3. Being corresponding sides of –Triangles Proved d. Given: In ABCD, AC = BD To-prove: ABCD is rectangle Proofs: S.N. Statements S.N. Reasons 1. In ABC and ADB 1. i. AC = BD (S) i. From given ii. BA = BA (S) ii. Being common side iii. BC = AD (S) iii. Being opposite sides of 2. ABC –ADB 2. By S.S.S axiom 3. ABC = BAD 3. Being corresponding angles of –Triangles 4. ABC = BAD = 90° 4. Being co-interior angles ABC + BAD = 180° 5. ABCD is rectangle 5. Being parallelogram with 90° Proved 9. (a) Prove that the diagonals of a square are equal. (b) Prove that if the diagonals of a rhombus are equal, then it is a square. (c) Prove that if the adjacent sides of a rectangle are equal, it is a square. Solution: a. Given: AC and BD are diagonals of square ABCD To-prove: AC = BD Proofs: S.N. Statements S.N. Reasons 1. In ABC and BAD 1. i. BC = AD (S) i. Opposite sides of parallelogram ii. ABC = BAD (A) ii. being 90° iii. AB = BA (S) iii. Being common side 2. ABC –BAD 2. By S.A.S axiom 3. AC = BD 3. Being corresponding sides of –Triangles Proved D C A B D C A B D C A B
299 b. Given: In rhombus ABCD, AC = BD To-prove: Rhombus ABCD is square Proofs: S.N. Statements S.N. Reasons 1. In ABC and BAD 1. i. AB = AB (S) i. Common side ii. BC = AD (S) ii. Sides of rhombus iii. AC = BD (S) iii. From given 2. ABC –BAD 2. By S.S.S axiom 3. ABC = BAD 3. Being corresponding angles of –Triangles 4. ABC = BAD = 90° 4. Being co-interior angles ABC + BAD = 180° 5. Rhombus ABCD is square 5. Being rhombus with angle 90° c. Given: In rectangle ABCD, AB = BC To-prove: Rectangle ABCD is square Proofs: Here, i. AB = DC (‡ being opposite sides of rectangle) ii. BC = AD (‡ being opposite sides of rectangle) iii. AB = BC (‡ from given) iv. AB = BC = DC = AD (‡ from (i), (ii), (iii)) Hence, rectangle ABCD is square. 10. (a) ABCD is a rectangle. The mid-point E of AB is joined with D and C. Prove that ΔDEC is an isosceles triangle. (b) In the figure, AB and CD bisect each other at O. Prove that AC//DB. (c) If ABCD is a parallelogram and equal angles are marked similarly, prove that ∠AEB = 90°. Solution: a. Given: In rectangle ABCD, AE = BE To-prove: DEC is an isosceles triangle Proofs: S.N. Statements S.N. Reasons 1. In AED and BEC 1. i. AD = BC (S) i. Being opposite sides of rectangle ii. DAE = CBE (A) ii. being 90° iii. AE = BE (S) iii. From given 2. AED –BEC 2. By S.A.S axiom 3. ED = EC 3. Being corresponding sides of –Triangles 4. DEC is an isosceles triangle 4. Being triangle with 2 equal sides D C A B D C A B A D B C E A B D C A B O D C E A D B C E
300 b. Given: AO = OB, CO = OD To-prove = CA//DB Proofs: S.N. Statements S.N. Reasons 1. In AOC and BOD 1. i. AO = OB (S) i. From given ii. AOC = BOD (A) ii. Being V.O.A. iii. CO = OD (S) iii. From given 2. AOC –BOD 2. By S.A.S axiom 3. ACO = BDO 3. Being corresponding angles of –Triangles 4. AC//DB 4. Being equal alternate angles Proved c. Given: In ABCD, EAB = EAD, ABE = EBC To-prove: AEB = 90° Proofs: 1. DAB + ABC = 180° (‡ being co-interior angles) 2. (EAB + EAD) + (ABE + EBC) = 180° (‡by whole part axiom) 3. EAB + EAB + ABE + ABE = 180° (‡ EAB = EAD, ABE = EBC) or, 2(EAB + ABE) = 180° EAB + ABE = 90° 4. AEB + EAB + ABE = 180° (‡ Sum of angles of ) or, AEB + 90° = 180° AEB = 90° 11. (a) In parallelogram WXYZ, XY = 2WX and M is the mid-point of XY. Prove that ∠WMZ = 90°. (b) In the adjoining figure, O is the midpoint of MN of the square KLMN. Prove that ΔLOK is an isosceles triangle. Solution: a. Given: In WXYZ, XY = 2WX and XM = MY To-prove: WMZ = 90° Proofs: 1. XM = MY = 1 2 XY (‡ From given) 2. XM = WX, YZ = MY (‡ YZ = WX = 1 2 XY) 3. XMW = XWM, YMZ = YZM) (‡ base angles of triangles) 4. XMW = MWZ, YMZ = MZW (‡ being alternate angles) 5. XWM = MWZ, YZM = MZW (‡ from (3) & (4)) 6. XWZ + WZY = 180° (‡ being co-interior angles) A B D C O A B D C E W Z X Y M N M K L O W Z X Y M
301 7. 2MWZ + 2MZW = 180° (‡ EMWZ = 1 2 XWZ and MZW = 1 2 WZY) MWZ + MZW = 90° 8. WMZ = 90° (‡ WMZ + MWZ + MZW = 180°) b. Given: In square KLMN, O is mid-point of MN To-prove: LOK is an isosceles triangle Proofs: S.N. Statements S.N. Reasons 1. In KNO and LMO 1. i. KN = LM (S) i. Being sides of square ii. KNO = LMO (A) ii. Being 90° iii. NO = MO (S) iii. From given 2. KNO –LMO 2. By S.A.S axiom 3. OK = OL 3. Being corresponding sides of –Triangles 4. LOK is an isosceles triangle 4. Being two equal sides Proved 14.2 Application of Theorems of Parallelogram Exercise 14.2 1. (a) In ΔABC, D and E are mid-points of AB and AC respectively. Write the relation between DE and BC. (b) In ΔPQR, PS = QS and ST//QR. What is the relation between PT and RT? (c) In ΔKMN, P and Q are mid-points of KM and KN respectively. What is the length of PQ? (d) In ΔABC, AQ = 6cm. What is the length of CQ? Solution: a. If D and E are mid-points of AB and AC, then DE = 1 2 BC & DE//BC. b. If PS = QS and ST//QR, then PT = RT c. If P and Q are mid-points of KM and KN, then PQ = 1 2 MN N M K L O E A B C D T P Q R S Q K M N P Q A B C P E A B C D T P Q R S Q K M N P
302 d. Here, P is mid-point of AB and PQ//BC, then Q is mid-point of AC. So, CQ = AQ CQ = 6 cm 2. (a) In ΔABC, AB = AC and M is the midpoint of AB and MN parallel to BC. If MN = 2 cm, find BC. (b) In the ΔXYZ, YZ = 8 cm, M is the midpoint of XY and MN is drawn parallel to YZ. Find MN. (c) In the ΔPQR, ∠Q = 50°, ST = 5.7 cm and ∠STR = 110°, find ∠P and QR. Solution: a. If M is mid-point of AB and MN//BC, then N is also mid-point of AC. Finally MN = 1 2 BC or, 2 = 1 2 BC BC = 4 cm b. If M is mid-point of XY and MN//YZ, then N is mid-point of XZ. So, MN = 1 2 YZ = 1 2 × 8 cm = 16 cm c. Here, R + 110° = 180° (‡ being co-interior angles) R = 70° And, P + Q + R = 180° or, P + 50° + 70° = 180° P = 60° Also, 1 2 QR = ST or, QR = 2 × ST = 2 × 5.7 cm = 11.4 cm 3. (i) Prove that the straight line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side. (ii) In the given ΔABC, AP = BP and AQ = CQ. Prove that PQ//BC and PQ = 1 2 BC. Solution: i. Given: P is mid-point of AB and PQ//BC To-prove: Q is mid-point of AC Construction: Let draw QR//AB Q A B C P N X Y Z M 8 cm T P Q R S 50° 110° 5.7cm N A B C M 2 cm N X Y Z M 8 cm T P Q R S 50° 110° 5.7cm Q A B C P A B C P Q R
303 Proofs: S.N. Statements S.N. Reasons 1. PBRQ is parallelogram. 1. Being parallel opposite sides 2. QR = BP 2. Opposite sides of parallelogram 3. QR = AP 3. BP = AP 4. QRC = ABC 4. Being corresponding angles 5. ADQ = ABC 5. Reason 4 6. QRC = APC 6. From statements 4 & 5 7. In APQ and QRC 7. i. AQP = QCR (A) i. Being corresponding angles ii. QRC = APC ii. From 6 iii. AP = QR (S) iii. From (3) 8. APQ –QRC 8. By A.A.S axiom 9. AQ = QC Hence, Q is mid-point of AC 9. Being corresponding sides of –Triangles Proved ii. Given: In ABC, AP = BP and AQ = CQ To-prove: PQ//BC and PQ = 1 2 BC Construction: Let extend PQ to R so that PQ = QR and, join R with C. Proofs: S.N. Statements S.N. Reasons 1. In APQ and CRQ 1. ................. i. AQ = QC (S) i. From given ii. AQP = CQR (A) ii. Being vertically opposite angles iii. PQ = QR (S) iii. From construction 2. APQ –CRQ 2. By S.A.S axiom 3. APQ = CRQ and AP = CR 3. Being corresponding parts of –Triangles 4. BP//CR and BP = CR 4. Being equal alternate angles and AP = BP 5. PR//BC, PR = BC 5. Being BP//CR and BP = CR 6. PQ//BC, PQ = 1 2 PR = 1 2 BC 6. Being PQ = QR Proved 4. (i) In the given figure, M and N are the midpoints of AB and AC respectively. Prove that AP = PD. (ii) In the parallelogram ABCD, M and N are the midpoints of AB and CD respectively. Prove that: (a) AMCN is a parallelogram. (b) BF = FE = ED. (iii) In the fig, BF = FC and AG = GC. If FG produced to meet AD at E, prove that AE = ED. Q A B C P R M A B C P N D A B D C E F N M A B D C G E F
304 (iv) From the adjoining figure, prove that: (a) BX = 1 2 AP (b) AP + CR = 2 BQ. Solution: i. Given: M and N are mid-points of AB and AC To-prove AP = PD Proofs: 1. MN//BC (‡ Line joining mid-points of two sides of parallel to third side) 2. P is mid-point of AD (‡ Line passes mid-point of one side and parallel to other side bisect remaining side) ii. Given: In ABCD, M and N are mid-points of AB and CD. To-prove: a. AMCN is a parallelogram b. BF = FE = ED Proofs: S.N. Statements S.N. Reasons 1. AM = NC & AM//NC 1. AM = 1 2 AB = 1 2 DC = NC and AB//DC 2. AN = MC & AN//MC 2. Being line segments joining equal and parallel lines on same sides 3. AMCN is a parallelogram 3. Being equal and parallel opposite sides 4. BF = FE 4. In ABE, M is mid-point of AB and MF//AE 5. ED = FE 5. In DCF, N is mid-point of DC and NE//CF 6. BF = FE = ED 6. From 4 and 5 Proved iii. Given: BF = FC, AG = CG To-prove: AE = ED Proofs: Here, 1. GF//AB (‡ In ABC, G and F are mid-points of AC and BC respectively) 2. EG//DC (‡ AB//DC) 3. AE = ED (‡ In ADC, G is mid-point of AC and EG//DC.) iv. Give: AP//BQ//CR and B is mid-point of AC To-prove: a. BX = 1 2 AP b. AP + CR = 2BQ P R A C B Q M A B C P N D A B D C E F N M A B D C G E F P R A C X B Q
305 Proofs: Here, 1. X is mid-point of CP (‡ In ACP, B is mid-point BX//AP) 2. BX = 1 2 AP (‡ Line segment joining mid-points of two sides is half of third side) 3. XQ = 1 2 CR (‡ Proceed same as (2)) 4. 1 2 AP + 1 2 CR = BX + XQ (‡ Adding (2) & (3)) or, AP + CR = 2BQ (‡ BX + XQ = BQ) 5. (i) Prove that the lines joining the midpoints of the opposite sides of a parallelogram bisect each other. (ii) Show that a quadrilateral joining the midpoints of the sides of a quadrilateral is a parallelogram. (iii) Prove that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other. Solution: i. Given: P, N, Q and M are mid-points of AD, AB, BC and CD in ABCD. To-prove: QP and MN bisect each other. Construction: Let join P and M, A and C, N and Q, P and N, M and Q. Proofs: 1. PM = 1 2 AC and PM//AC (‡ In PAC, P and M are mid-point of AD and CD) 2. NQ = 1 2 AC and NQ//AC (‡ In BAC, N and Q are mid-points of AB and BC) 3. PM = NQ and PM//NQ (‡ From (1) & (2)) 4. PN = MQ and PN//MQ (‡ Line segment joining PM = NQ and PM//NQ) 5. PNQM is parallelogram (‡ Being equal and parallel opposite sides) 6. QP and MN bisect each other (‡ Being diagonals of parallelogram) ii. Given: P, Q, R and S are mid-points of sides AB, BC, CD and AD respectively. To-prove: PQRS is parallelogram Construction: Let join A and C Proofs: 1. SR = 1 2 AC and SR/AC (‡ In ADC, S and R are mid-points of AD and CD) 2. PQ = 1 2 AC and PQ//AC (‡ In ABC, P and Q are mid-points of AB and BC) 3. SR = PQ and SR//PQ (‡ From (1) & (2)) 4. PS = QR and PS//QR (‡ Proceed same as (3)) 5. PQRS is parallelogram (‡ Being equal and parallel opposite sides) D C A B P Q N M D C B A S Q P R
306 iii. Given: P, Q, R and S are mid-points of AB, BC, CD and AD respectively To-prove: PR and QS bisect each other Construction: Let join P, Q, R and S. Also, join A and C Proofs: 1. SR = 1 2 AC, SR//AC, (‡ In ADC, S and R are mid-points of AD and CD) 2. PQ = 1 2 AC, PQ//AC (‡ Same reason as (1)) 3. SR = PQ, SR//PQ (‡ From (1) & (2)) 4. PS = QR, PS//QR (‡Proceed same as (3)) 5. PQRS is parallelogram (‡ Being equal and parallel opposite sides) 6. PR and QS bisect each other (‡ Being diagonals of ) D C A B S Q P R
307 CHAPTER CONSTRUCTION 15 15.1 Construction of Scalene Quadrilaterals Exercise 15.1 1. Construct a quadrilateral ABCD from the following data: (a) AB = 5.5 cm, BC = DA = 4.2 cm, CD = 5 cm and diagonal AC = 6.5 cm. (b) AB = 4 cm, BC = 3.2 cm, CD = 4.6 cm, DA = 3.9 cm and diagonal AC = 5.7 cm. (c) AB = 6 cm, BC = 4.4 cm, CD = 5.2 cm, DA = 4.2 cm and diagonal AC = 7 cm. Solution: a. Given: AB = 5.5 cm, BC = DA = 4.2 cm, CD = 5 cm and AC = 6.5 cm To-construct: Quadrilateral ABCD Construction: No need to mention: Step I: Draw line AB of 5.5 cm Step II: Determine point C, taking arc 4.2 cm from B and arc 6.5 cm from A Step III: Determine point D, taking arc 5 cm from C and arc 4.2 cm from A. b. Given: AB = 4 cm, BC = 3.2 cm, CD = 4.6 cm, DA = 3.9 cm, AC = 5.7 cm To-construction: Quadrilateral ABCD Construction: No need to mention: Step I: Draw line AB of 4 cm Step II: Determine point C taking arc 3.2 cm from B and 5.7 cm from A. Step III: Determine point D, taking arc 4.6 cm from C and arc 3.9 cm from A. A B D C 5 cm 5.5 cm A B D C 4 cm D C A B 5 cm 5.5 cm D C A B 4 cm
308 c. Given: AB = 6 cm, BC = 4.4 cm, CD = 5.2 cm, DA = 4.2 cm and AC = 7 cm To-construct: Quadrilateral ABCD Construction: No need to mention: Step I: Let draw AB of 6 cm Step II: Determining point C, taking arc 7 cm from A and arc 4.4 cm from B. Step III: Determining point D, taking arc 4.2 cm from A and arc 5.2 cm from C. 2. Construct a quadrilateral PQRS from the following information: (a) PQ = 5.9 cm, QR = 4.1 cm, RS = ST = 4.5 cm, ∠PQR = 60°. (b) PQ = 4.6 cm = QR = 3.6 cm, RS = 5.2 cm, PS = 4.2 cm, ∠P = 75°. (c) PQ = 5 cm = QR= 4 cm, RS = PS = 4.5 cm, ∠QRS = 75°. Solution: a. Given: PQ = 5.9 cm, QR = 4.1 cm, RS = ST = 4.5 cm, PQR = 60° To-construct: Quadrilateral PQRS Construction: No need to mention: Step I: Let draw PQ line with length 5.9 cm Step II: Constructing 60° at Q and determining point R with arc 4.1 cm Step III: Determining point S taking arc 4.5 cm from P and R. b. Given: PQ = 4.6 cm, QR = 3.6 cm, RS = 5.2 cm, PS = 4.2 cm, P = 75° To-construct: Quadrilateral PQRS Construction: No need to mention: Step I: Let draw straight line with length 4.6 cm. Step II: Constructing 75° at point P and determining point S taking arc 4.2 cm from P. Step III: Determining point R taking arc 5.2 cm from S and arc 3.6 cm from Q. A B C D 6 cm Q P S R 5.9 cm 60° D C A B 6 cm S R P Q 5.9 cm 60° S R P 4.6 cm Q 75°
309 c. Given: PQ = 5 cm, QR = 4 cm, RS = PS = 4.5 cm, QRS = 75° To-construct: Quadrilateral PQRS Construction: No need to mention: Step I: Let draw QR with length 4 cm Step II: Constructing 75° at point R and determining point S taking 4.5 cm arc from R. Step III: Determining point P, taking arc 4.5 cm from S and arc 5 cm from Q. 3. Construct a quadrilateral KITE from the following data: (a) KI = 5 cm, IT = 4.5 cm, KE = 4.8 cm and diagonals KT = 6 cm and IE = 6.5 cm. (b) KI = IT = 4.5 cm, TE = 5.1 cm and diagonals KT = 6.3 cm, IE = 5.8 cm. (c) KE = IT = 5 cm, TE = 4.5 cm and diagonals KT = 6.5 cm, IE = 6 cm. Solution: a. Given: KI = 5 cm, IT = 4.5 cm, KE = 4.8 cm, KT = 6 cm, IE = 6.5 cm To-construction: Kite KITE Construction: No need to mention: Step I: Let draw straight line KI with length 5 cm. Step II: Determine point T taking arc 6 cm from K and arc 4.5 cm from I. Step III: Determine point E taking arc 6.5 cm from I and arc 4.8 cm from K. P Q R S 4.6 cm 75° P S Q 4 cm R 75° K I T E 5 cm P S Q R 4 cm 75° E T K I 5 cm
310 b. Given: KI = IT = 4.5 cm, TE = 5.1 cm, KT = 6.3 cm, IE = 5.8 cm To-construct: Quadrilateral KITE Construction: No need to mention: Step I: Let draw straight line KI 4.5 cm length. Step II: Determine point T taking arc 4.5 cm from I and 6.3 cm from K. Step III: Determine point E taking 5.1 cm from I and 5.1 cm from T. c. Given: KE = IT = 5 cm, TE = 4.5 cm, KT = 6.5 cm, IE = 6 cm To-construct: Quadrilateral KITE Construction: No need to mention: Step I: Let draw st. line IT with length 5 cm. Step II: Determine the point E with arc 4.5 cm from T and arc 6 cm from I. Step III: Determine point K with arc 5 cm from E and 6.5 cm from T. 4. (a) Construct a quadrilateral KLMN with KL = 4 cm, MN = 5 cm, LM = 6 cm, ∠M = ∠L = 60°. (b) Construct a quadrilateral PQRS with PQ = QR = 3.6 cm, PS = 4.2 cm, ∠P = ∠Q = 120°. (c) Construct a quadrilateral WXYZ with YZ = XY = 5 cm, WZ = 3.6 cm, ∠Y = 60°, ∠Z = 75°. Solution: a. Given: KL = 4 cm, MN = 5 cm, LM = 6 cm, M = L = 60 To-construct: Quadrilateral KLMN Construction: No need to mention: Step I: Let draw st. line LM with length 6 cm Step II: Determine point N with length 5 cm from M where M = 60° Step III: Drawing 60° at L and determining point K 4 cm from L. T I K E 4.5 cm K E T I 5 cm E T K I 4.5 cm K E I T 5 cm K N L M 6 cm 60° 60°
311 b. Given: PQ = QR = 3.6 cm, PS = 4.2 cm, P = Q = 120° To-construction: Quadrilateral PQRS Construction: No need to mention: Step I: Let draw st. line PQ with length 3.6 cm. Step II: Drawing 120° at point P and Q. Step III: Determining point R with length 3.6 from Q and point S with length 4.2 cm from P. c. Given: YZ = XY = 5 cm, WZ = 3.6 cm, Y = 60°, Z = 75° To-construct: Quadrilateral WXYZ Construction: Note need to mention: Step I: Let draw st. line YZ with length 5 cm. Step II: Drawing 60° at Y and 75° at point Z. Step III: Determining point X with length 5 cm from Y and W with length 3.6 m from Z. 5. (a) Construct a quadrilateral ABCD in which AB = 4.3, BC = 5.5 cm, ∠A = 135°, ∠B = ∠C = 60°. (b) Construct a quadrilateral BIKE having BI = 4.3, IK = 5.5 cm, ∠B = 45°, ∠I = ∠K = 60°. (c) Construct a quadrilateral PRST with PR = RS = 4.8 cm, ∠P = 60°, ∠R = 105°, ∠S = 75°. Solution: a. Given: AB = 4.3 cm, BC = 5.5 cm, A = 135°, B = C = 60° To-construct: Quadrilateral ABCD Construction: No need to mention: Step I: Let draw straight line AB with length 4.3 cm. K N L 6 cm M 60° 60° S R P 3.6 cm Q 120° 120° 60° X W Z Y 5 cm 60° 75° S R P 3.6 cm Q 120° 120° X W Y Z 5 cm 60° 75° 60° 60° 135° D A B C 4.3 cm
312 Step II: Drawing 135° and 60° at A and B. Step III: Determining point C with length 5 cm from B. Step IV: Drawing 60° at C. b. Given: BI = 4.3 cm IK = 5.5 cm, B = 135°, I = K = 60° To-construction: Quadrilateral BIKE Construction: No need to mention: Step I: Let draw st. line IK with length 5.5 cm Step II: Drawing 60° at point I and K. Step III: Determining point B with length 4.3 cm from I. Step IV: Drawing 135° at B. c. Given: PR = RS = 4.8 cm, P = 60°, R = 105°, S = 75° To-construct: Quadrilateral PQRS Construction: No need to mention: Step I: Let draw st. line PR with length 4.8 cm. Step II: Drawing 60° and 105° at P and R. Step III: Determining point S with length 4.8 cm from R. Step IV: Drawing 75° at S 6. (a) Construct a quadrilateral ABCD with AB = BD = 4.5 cm, AD = BC = 5 cm , ∠BAC = 45°. (b) Construct a quadrilateral HERA with HE = 5 cm = EA, ER = 4.5 cm, ∠HEA = 60°, ∠EAR = 45°. Solution: a. Given: AB = BD = 4.5 cm, AD = BC = 5 cm, BAC = 45° C B A D 4.5 cm 60° 60° 135° B E I 5.5 cm K 60° 60° 135° T S P 4.8 cm R 60° 105° 75° 75° T S P R 4.8 cm 60° 105° 75° B E I K 5.5 cm 60° 60° 135°
313 To-construction: Quadrilateral ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 4.5 cm Step II: Drawing 45° at point A and determining point C with length 5 cm from B. Step III: Determining point D with arc 4.5 cm from C and 5 cm from A. b. Given: HE = 5 cm = EA, ER = 4.5 cm, HEA = 60°, EAR = 45° To-construct: Quadrilateral HERA Construction: No need to mention: Step I: Let draw st. line AE with length 5 cm. Step II: Drawing 45° at A and 60° at E. Step III: Determining point H with length 5 cm from E and point R with length 4.5 cm from E. 15.2 Construction of Trapeziums Exercise 15.2 1. Construct a trapezium ABCD in which; (a) AB = 5 cm, BC = 4.2 cm, CD = 4 cm ∠BAC = 45° and AB//CD. (b) BC = 6.5 cm, CD = 5 cm, DA = 4.5 cm ∠BCD = 60° and BC//AD. (c) BC = 4.8 cm, AB = 5.2 cm, AD = 6 cm ∠BAD = 75° and CB//DA. A B C D 45° 4.5 cm A E R H 45° 60° 5 cm D C A B 4.5 cm 45° 45° 60° R A E H 5 cm
314 Solution: a. Given: AB = 5 cm, BC = 4.2 cm, CD = 4 cm, BAC = 45°, AB//CD Construction: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 5 cm. Step II: Drawing 45° at A and determining point C with length 4.2 cm from B. Step III: Drawing 45° at C so that AB//CD and determining point D with length 4 cm from C. b. Given: BC = 6.5 cm, CD = 5 cm, DA = 4.5 cm, BCD = 60°, BC//AD To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line BC with length 6.5 cm. Step II: Drawing 60° at C and determining point D with length 5 cm from C. Step III: Drawing 120° at D so that AD//BC and determining point A with length 4.5 cm from D. c. Given: BC = 4.8 cm, AB = 5.2 cm, AD = 6 cm, BAD = 75°, CB//DA To-construction: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 5.2 cm. Step II: Drawing 75° at A and determining point D with length 6 cm from A. Step III: Drawing 105° at point B so that CB//DA and determining point C with length 4.8 cm from B. D A B C 5 cm 4 cm 45° 45° D C B A 60° 4.5 cm 6.5 cm A B C D 5.2 cm 75° 105° D C A B 5 cm 4 cm 45° D C B A 60° 4.5 cm 6.5 cm D C A 5.2 cm B 75°
315 2. (a) Construct a trapezium ABCD in which AB = 5.5 cm, BC = 4.2 cm, CD = 4.5 cm, DA = 5 cm and AB//DC. (b) Construct a trapezium PREM in which PR = 4 cm, RE = 4.5 cm, EM = 6 cm, PM = 4.6 cm and PR//EM. (c) Construct a trapezium SOVA in which SO = 6 cm, VO = 7 cm, VA = SA = 5 cm and VO//AS. Solution: a. Given: AB = 5.5 cm, BC = 4.2 cm, CD = 4.5 cm, DA = 5 cm, AB//DC To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 5.5 cm. Step II: Taking length AE = 4.5 cm on line AB and determining point C with arc 4.2 cm from B and arc 5 cm from E. Step III: Determining point D with arc 4.5 cm from C and arc 5 cm from A. b. Given: PR = 4 cm, RE = 4.5 cm, EM = 6 cm, PM = 4.6 cm, PR//EM To-construct: Trapezium PREM Construction: No need to mention: Step I: Let draw st. line EM with length 6 cm and taking length MA with length 4 cm. Step II: Determining point R with arc 4.6 cm from A and arc 4.5 cm from E. Step III: Determining point P with arc 4.6 cm from M and arc 4 cm from R. c. Given: SO = 6 cm, VO = 7 cm, VA = SA = 5 cm, VO//AS To-construct: Trapezium SOVA Construction: No need to mention: Step I: Let draw st. line VO with length 7 cm and taking VP length 5 cm of VO. Step II: Determining point S with arc PS = 5 cm and arc OS = 6 cm. Step III: Determining point A with arc 5 cm from S and 5 cm from V. D C E B A 4.5 cm 5.5 cm 4.5 cm P R M A E 4 cm 6 cm 4 cm D A B C 4.5 cm 5.5 cm P M E R 4 cm 6 cm A S V O 5 cm 7 cm
316 3. (a) Construct a trapezium ABCD, AB//DC, AB = 6.2 cm, BC = 4.8 cm, CD = 5.3 cm, a diagonal AC = 6 cm. (b) Construct a trapezium WXYZ, ZY//WX, WX = 4.4 cm, WZ = 5.6 cm, ZY = 5.5 cm, a diagonal ZX = 6.5 cm. (c) Construct a trapezium ABCD, AD//BC, AD = 4.5 cm, BC = 6.3 cm, CD = 4.3 cm, a diagonal BD = 7 cm. Solution: a. Given: AB//DC, AB = 6.2 cm, BC = 4.8 cm, CD = 5.3 cm, AC = 6 cm To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 6.2 cm. Step II: Determining point C with arc 4.8 cm from B and arc 6 cm from A. Step III: Drawing ACD = BAC so that AB//DC and determining point D with length 5.3 cm from C. b. Given: ZY//WX, WX = 4.4 cm, WZ = 5.6 cm, ZY = 5.5 cm, ZX = 6.5 cm To-construction: Trapezium WXYZ Construction: No need to mention: Step I: Let draw st. line WX with length 4.4 cm. Step II: Determining point Z with arc 5.6 cm from W and arc 6.5 cm from X. Step III: Drawing XZY = WXZ so that ZY//WX and determining point Y with arc 5.5 cm from Z. A S V P O 5 cm 7 cm 5 cm A B C D 5.3 cm 6.2 cm Z Y X W 5.5 cm 4.4 cm D C A B 5.3 cm 6.2 cm Z Y W X 5.5 cm 4.4 cm
317 c. Given: AD//BC, AD = 4.5 cm, BC = 6.3 cm, CD = 4.3 cm, BD = 7 cm To-construction: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line BC with length 6.3 cm. Step II: Determining point D with arc 7 cm from B and arc 4.3 cm from C. Step III: Drawing ADB = CBD so that AD//BC and determining point A with arc 4.5 cm from D. 4. Construct a trapezium ABCD in which; (a) AB = 5.2 cm, BC = 4.2 cm, ∠B = 60°, ∠A = 90°, AB//DC. (b) AB = 4.2 cm, BC = 5.2 cm, ∠B = 45°, CD = 5.3, AB//CD. (c) AB//DC, AB = 5 cm, AD = 3.8 cm, ∠BAD = 75°, ∠ABC = 105°. Solution: a. Given: AB = 5.2 cm, BC = 4.2 cm B = 60°, A = 90°, AB//DC To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 5.2 cm. Step II: Constructing 90° at A and 60° at B. Step III: Determining point C with arc 4.2 cm from B and drawing BCD = 120° so that AB//DC. b. Given: AB = 4.2 cm, BC = 5.2 cm, B = 45°, CD = 5.3 cm AB//CD To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 4.2 cm. Step II: Drawing 45° at B and determining point C with arc 5.2 cm from B. Step III: Drawing 135° at C so that AB//CD and determining point D with arc 5.3 cm from C. A D B C 4.5 cm 6.3 cm D C A 5.2 cm B 90° 60° 120° A B C D 4.5 cm 6.3 cm D C A B 5.2 cm 60° D A B C 5.3 cm 4.2 cm 45°
318 c. Given: AB//DC, AB = 5 cm, AD = 3.8 cm, BAD = 75°, ABC = 105° To-construction: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 5 cm. Step II: Drawing 75° at A and 105° at B. Step III: Determining point D with arc 3.8 cm from A and drawing 105° at D so that AB/DC. 5. (a) Construct a trapezium ABCD with AB = 6.2 cm, BC = 4.3 cm, AD = 4.2 cm, AB//CD and distance between the parallel sides 3.2 cm. (b) Construct a trapezium PQRS with PQ = 4.5 cm, QR = 5.2 cm, PS = 4 cm, PQ//RS and distance between the parallel sides 3 cm. (c) Construct a trapezium CDEF with CD = 4.5 cm, DE = 5.2 cm, CF = 4 cm, CD//EF and distance between the parallel sides 3.4 cm. Solution: a. Given: AB = 6.2 cm, BC = 4.3 cm, AD = 4.2 cm, AB//CD and distance between // is 3.2 cm. To-construct: Trapezium ABCD. Construct: No need to mention: Step I: Let draw st. line AB with length 6.2 cm. Step II: Drawing 90° at B and taking length of 3.2 cm to point E. Step III: Let draw 90° at point E so that AB//CD. [OR, Draw DE//AB taking equal arc on B an d E] Step IV: Determining point D with arc 4.2 cm from A and point with arc 4.3 cm from B. D C A B 5.3 cm 4.2 cm 45° 135° D C A 5 cm B 75° 105° 105° D A B C E D C A B 5 cm 75° 105° 105° D C A B 6.2 cm
319 b. Given: PQ = 4.5 cm, QR = 5.2 cm, PS = 4 cm PQ//RS and distance between // is 3 cm. To-construct: Trapezium PQRS Construction: No need to mention: Step I: Let draw st. line P with length 4.5 cm. Step II: Drawing 90° at Q and determining point T with arc 3 cm. Step III: Drawing 90° at point T. Step IV: Determining point S with arc 4 cm from P and point R with arc 5.2 cm from Q. c. Given: CD = 4.5 cm, DE = 5.2 cm, CF = 4 cm, CD//EF and distance between // lines is 3.4 cm To-construction: Trapezium CDEF Construction: No need to mention: Step I: Let draw st. line CD with length 4.5 cm. Step II: Drawing 90° at D and determining point A with arc 3.4 cm from D. Step III: Drawing 90° at point A. Step IV: Determining point F with arc 4 cm from C and point E with arc 5.2 cm from D. 15.3 Construction of Rhombus Exercise 15.3 1. Construct a rhombus ABCD having; (a) Side = 6 cm, ∠BAD = 60° (b) BC = 5.4 cm, ∠D = 30° (c) AD = 5 cm, ∠B = 120° Solution: a. Given: AB = 6 cm, BAD = 60° To-construct: Rhombus ABCD Construction: No need to mention: Step I: Let draw AB with length 6 cm. S T R Q P X Y 4.5 cm C D F A E X Y 4.5 cm F E C D 4.5 cm D C A B 60° 6 cm S R P Q 4.5 cm
320 Step II: Drawing 60° at point A and determining point D with length 6 cm. Step III: Determining point C with arc 6 cm from point B and D. b. Given: BC = 5.4 cm, D = 30° To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AD = BC = 5.4 cm Step II: Drawing 30° at point D and determining point C with arc 5.4 cm from D. Step III: Determining point B with arc 5.4 cm from A and C. c. Given: AD = 5 cm, B = 120° To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw AB = AD = 5 cm Step II: Drawing 120° at B and determining point C with arc 5 cm from B. Step III: Determining point D with arc 5 cm from point A and C. 2. Construct a rhombus PQRS having; (a) PQ = 5.5 cm, PR = 7 cm (b) RS = 6 cm, QS = 8 cm (c) PS = 5 cm, RP = 7.4 cm Solution: a. Given: PQ = 5.5 cm, PR = 7 cm To-construct: Rhombus PQRS Construction: No need to mention: Step I: Let draw st. line PQ with length 5.5 cm. Step II: Determining point with arc 7 cm from P and arc 5.5 cm from Q. D C B A 60° 6 cm A D B C 30° 5.4 cm A B C D 120° 60° B C A D 30° 5.4 cm D C A B 120° S R P Q 5.5 cm
321 Step III: Determining point S with arc 5.5 cm from point P and R. b. Given: RS = 6 cm, QS = 8 cm To-construct: Rhombus PQRS Construction: No need to mention: Step I: Let draw st. line RS with length 6 cm. Step II: Determining point Q with arc 6 cm from R and arc 8 cm from S. Step III: Determining point P with arc 6 cm from Q and S. c. Given: PS = 5 cm, RP = 7.4 cm To-construct: Rhombus PQRS Construction: No need to mention: Step I: Let draw st. line PS with length 5 cm Step II: determining point R with arc 7.4 cm from P and arc 5 cm from point P. 3. Construct a rhombus SERA having; (a) SR = 6 cm, EA = 7.4 cm (b) EA = 7 cm, RS = 8 cm (c) SR =4.6 cm, RP = 6 cm Solution: a. Given: SR = 6 cm, EA = 7.4 cm To-construct: Rhombus SERA Construction: No need to mention: Step I: Let draw st. line EA with length 7.4 cm Step II: Drawing bisector on EA, Step III: Determining point S and R with arc 3 cm from intersection point. P Q S R 5.5 cm Q P R 6 cm S Q R S P 5 cm Q P R S 6 cm Q R P S 5 cm S E A R 6 cm
322 b. Given: EA = 7 cm, RS = 8 cm To-construct: Rhombus SERA Construction: No need to mention: Step I: Let draw st. line RS with length 8 cm Step II: Drawing bisector on RS Step III: Determining points E and A with arc 3.5 cm from intersection point to bisector line. c. Given: SR = 4.6 cm, RP = 6 cm To-construct: Rhombus PQRS Construction: No need to mention: Step I: Let draw st. line RP with length 6 cm. Step II: Determining points S and Q with arc 4.6 cm from R and P. 4. Construct a rhombus RAME having; (a) ME = 4.5 cm, RM = 7.2 cm and ∠EMR = 45° (b) MA = 5 cm, EA = 6 cm and ∠AMR = 30° (c) RE = 6.2 cm, EA = 8 cm and REA = 60° S E A R 6 cm E R S A 8 cm S P R Q 6 cm E R S A 8 cm S P R Q 6 cm
323 Solution: a. Given: ME = 4.5 cm, EMR = 45° To-construct: RAME rhombus No need to mention: Step I: Let draw st. line ME with length 4.5 cm. Step II: Drawing 45° at point M. Step III: Determining points R with arc 4.5 cm from E on MR, and determining point A from M and R. b. Given: MA = 5 cm, AMR = 30° To-construct: Rhombus RAME No need to mention: Step I: Let draw st. line MA with length 5 cm. Step II: Draw 30° at point M. Step III: Determining point M with arc 5 cm from A on MR. Step IV: Determining point E with arc 5 cm from M and R. c. Given: RE = 6.2 cm, REA = 60° To-construct: Rhombus RAME Construction: No need to mention: Step I: Let draw st. line RE with length 6.2 cm. Step II: Drawing 60° at point E. Step III: Determining point A with arc 6.2 cm from R on AE. Step IV: Determining point M with arc 6.2 cm from A and E. A R M E 45° 4.5 cm E M A R 5 cm 30° R E A M 6.2 cm 60° A R M E 45° 4.5 cm E R M A 5 cm 30° A M R E 6.2 cm 60°
324 CHAPTER CIRCLE 16 16.1 Circle Exercise 16.1 1. Define the following terms. (a) Circle (b) Centre (c) Radius (d) Circumference (e) Chard (f) Diameter (g) Area (h) Semi-circle Solution: a. Circle: It is locus of moving point with certain length (radius = r) from fixed point (centre). b. Centre: It is fixed point which is equidistance from points on circumference. c. Radius: It is generating line which rotate to 360°. In another word radius is distance between centre and point at circumference. d. Circumference: It is length of outer surface or perimeter of circle. e. Chord: It is distance between one point of circumference to another point of circumference. f. Diameter: It is longest chord of circle which passes through centre. g. Area: The region covered by circle is called area which is calculated by product of square of radius with constant i.e. A = r 2 h. Semi-circle: The portion of circle divided by diameter is called semi-circle. 2. Write the relation between the following terms about the circle. (a) Diameter and Radius (b) Circumference and Diameter (c) Circumference and Radius (d) Area and Radius Solution: a. Diameter is double of radius. i.e. d = 2r b. Circumference is product of diameter and . i.e. c = d c. Circumference is product of twice radius with . i.e. c = 2r d. Area and radius: Area is product of square of radius with . i.e. A = r 2 3. (a) What do O, OP, AB, BC and represent in the adjoining figure? (b) Write the name of the radius, diameter and chord in the given figure. Solution: a. O is centre. OA is radius. AB is diameter. BC is chord. is arc. O A D B C O A D B C
325 b. Radius = OD Diameter = AB Chord = BC 4. Circle the correct answer. (a) What is the name of the shaded portion in the given circle? (i) radii (ii) segment (iii) sector (iv) semi-circle (b) What is the name of the shaded region in the given circle? (i) sector (ii) semi-circle (iii) radii (iv) segment (c) What is the length of the diameter in a circle with radius 7 cm? (i) 3.5 cm (ii) 14 cm (iii) 44 cm (iv) 21 cm (d) What is the length of circumference of circle having diameter 21 cm? (i) 66 cm (ii) 10.5 cm (iii) 132 cm (iv) 42 cm (e) In the adjoining figure, AB = 10 cm. What is the length of CD? (i) 5 cm (ii) 20 cm (iii) 7 cm (iv) 10 cm (f) What is the length of QT if RP = 6 cm? (i) 12 cm (ii) 6 cm (iii) 24 cm (iv) 3 cm Solution: a. (iii) sector b. (iv) segment c. When, radius (r) = 7 cm Diameter (d) = 2r = 2 × 7 = 14 cm d. If diameter (d) = 21 cm Circumference (d) = d = 22 7 × 21 = 66 cm e. If AB = 10 cm Then CD = 10 cm f. If RP = 6 cm, QT = 1 2 ST = 1 2 SR = RP QT = 6 cm 5. (i) What is the relation between the chord and the perpendicular drawn from the centre of a circle to the chord? (ii) Write the relation between PR and RQ in the given circle. (iii) Write the relation between OC and AB in the adjoining circle. O A B O O N M C D B A O Q R S T P O A B O O N M C D B A O Q R S T P O R P Q O C A B
326 O D A B C (iv) From which point does the line segment CD when CD ⊥ AB and AD = BD in the given circle? Solution: i. The perpendicular drawn from the centre of a circle to the chord bisect to chord. ii. PR = RQ iii. OC AB iv. If CD AB and AD = BD Then CD passes through centre O. 6. (i) The length of the chord AB of a circle with centre at O and radius 10 cm is 12 cm and ∠OAB = 45°. (a) Write the relation between AE and BE when OE ⊥ AB. (b) Write the length of AE. (c) Find the distance of the chord AB from the centre O. (ii) The radius of a circle with centre at P is 13 cm. The length of the chord AB of this circle is 24 cm. Determine the distance from P to the chord AB. (iii) Find the length of a chord which is at a distance of 4 cm from the centre of a circle with radius 5 cm. (iv) Find the length of the radius of a circle in which 32 cm long chord is 12 cm far from its origin. Solution: i. a. If OE AB, then AE = BE b. AE = 1 2 AB = 1 2 × 12 = 6 cm c. OE = OA2 – AE2 = 102 – 62 = 100 – 36 = 64 = 8 cm O R P Q O C A B O D A B C O E A B 12 cm P Q A B 24 cm O E A B 12 cm
327 ii. PQ = AP2 – AQ2 = 132 – 24 2 2 ‡ AQ = 1 2 AB = 169 – 144 = 25 = 5 cm iii. Here, Radius (r) = 5 cm distance (D) = 4 cm Length of chord (l) = ? We have, l 2 2 + D2 = r2 or, l 2 2 = r2 – D2 or, l 2 2 = 52 – 42 or, l 2 2 = 25 – 16 = 9 or, l 2 = 9 or, l = 3 × 2 l = 6 cm iv. Here, Radius (r) = ? Length of chord (l) = 32 cm distance (D) = 12 cm We have, r2 = D2 + l 2 2 = 122 + 32 2 2 = 144 + 256 = 400 or, r = 400 r = 20 cm 7. In the adjoining figure, P is the centre of the circle and its radius is 8 cm, ∠APB = 60° and PM ⊥ AB. (a) Name the types of triangle APB. (b) Calculate the length of AB. (c) Find the length of AM. (d) Justify that the length of PM is 4 3. P M A B 60°
328 O R P Q Solution: a. APB is an equilateral triangle. b. Length of AB = radius = 8 cm c. Length of AM = 1 2 AB = 1 2 × 8 = 4 cm d. PM = AP2 – AM2 (‡ P = h2 – b2 ) = 82 – 42 = 64 – 16 = 48 = 4 3 cm 8. (i) Prove that the perpendicular drawn from the centre of a circle to a chord bisects the chord. (ii) O is the centre of a circle and R is the midpoint of PQ. The centre O and the midpoint are joined. (a) Write the relation between PQ and OR. (b) Prove the above relation by statements and reasons. (iii) Verify experimentally that the perpendicular bisector of a chord of a circle passes through the centre of the circle. Solution: i. Given: O is centre of circle where OM AB. To-prove: AM = BM Construction: Let join O with A and B. Proofs: S.N. Statements S.N. Reasons 1. OA2 = OB2 1. Being radii of same circle 2. AM2 + OM2 = BM2 + OM2 or, AM2 = BM2 AM = BM 2. h2 = p2 + b2 Proved ii. Given: R is mid-point of PQ. To-prove: OR PQ Construction: Let join O with P and Q. Proofs: S.N. Statements S.N. Reasons 1. In POR and QOR 1. ................. i. PR = QR (S) i. From given ii. OR = OR (S) ii. Being common sides iii. OP = OQ (S) iii. Being radii of same circle 2. POR –QOR 2. By S.S.S axiom 3. PRO = QRO 3. Being corresponding angles of –Triangles 4. OR PQ 4. Being equal linear pair Proved P M A B 60° O R Q P O M A B
329 iii. In above figures (i) and (ii) OM is bisectors of chord AB. Obviously OM passes through centre O. In conclusion, perpendicular bisector of a chord of a circle passes through the centre of the circle. 9. (i) In the given figure, ABC is an isosceles triangle and a circle with A as the centre cuts BC at D and E. (a) If AF BC, write the relation between DF and EF. (b) Prove that BD = CE. (ii) In the adjoining figure, OD is perpendicular to the chord AB of a circle whose centre is O. (a) Is the point D midpoint of AB ? Why ? Justify your answer with reason. (b) If BC is a diameter, show that AC = 2 OD and AC // DO. (iii) In the given figure, P and Q are the centres of two intersecting circles and AC // PQ. (a) Prove that AC = 2PQ. (b) If AC = 12 cm, what is the length of PQ? Solution: i. Given: In ABC, AB = AC and AF BC. To-prove: a. DF = EF b. BD = CE Proofs: Here, i. DF = EF (‡ drawn from centre bisect chord) ii. BF = FC (‡ drawn on base of isosceles bisect base) iii. BF – DF = FC – EF (‡ subtracting (i) from (ii)) BF = CE ii. Given: BC is diameter and OD AB To-prove: a. D is mid-point of AB b. AC = 2OD and AC//DO O M A B Fig. (i) O A M B Fig. (ii) A B C D E O A D C B P A B Q C A F B C D E O A D C B
330 Proofs: Here, i. D is mid-point of AB (‡ drawn from centre bisect chord) ii. AC = 2OD and AC//DO (‡ OD is line segment passes through mid-points of AB and BC in ABC) iii. Given: P and Q are the centres of circles and AC//PQ. To-prove: AC = 2PQ Construction: Let draw PM AB and QN BC. Proofs: i. MB = 1 2 AB, BN = 1 2 BC (‡ draw from centre bisect chord) ii. MB + BN = 1 2 (AB + BC) (‡ By addition axiom) iii. MN = 1 2 AC (‡ By whole part axiom) iv. PQ = 1 2 AC (‡ MN = PQ being opposite sides of rectangle) AC = 2PQ When, AC = 12 cm, PQ = 1 2 × 12 = 6 cm 10. (i) Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it. (ii) If a line intersects two concentric circles at the points A, B, C and D as shown in the figure, prove that AB = CD. (iii) In the figure, XY is a diameter and it is also perpendicular to AB. Prove that XA = XB. Solution: i. Given: Chord AB//CD and OM AB To-prove: ON CD and N is mid-point of CD Proofs: i. ONC = 90° (‡ ONC = OMB = 90° alternate angles) i.e. ON CD ii. N is mid-point of CD (‡ drawn from centre bisect chord) ii. Given: O is centre of two concentric circles To-prove: AB = CD Construction: Let draw OM AD Proofs: i. AM = MD (‡ drawn from centre bisect chord) ii. BM = MC (‡ drawn from center bisect chord) iii. AM – BM = MD – MC (‡ Subtracting (ii) from (i)) AB = CD P A M B Q C N A D B C O A B X Y O N C D A M B O A M D B C
331 iii. Given: XY is diameter and XM AB To-prove: XA = XB Proofs: S.N. Statements S.N. Reasons 1. In XAM and XBM 1. ................. i. AM = BM (S) i. drawn from centre bisect chord ii. AMX = BMX (A) ii. Being XM AB iii. XM = XM (S) iii. Being common side 2. XAM –XBM 2. By S.A.S axiom 3. XA = XB 3. Being S.A.S axiom Proved 11. (i) In the given figure, the chords AB and CD of a circle with centre at O intersect at X so that ∠AXO = ∠OXD. Prove that BX = CX. (ii) In the given figure, X and Y are the centres of two circles which intersect at A and C. XA and XC are produced to meet the other circle at B and D. Prove that AB = CD. Solution: i. Given: AXO = OXD To-prove: BX = CX Construction: Let draw OM AB and ON CD Proofs: S.N. Statements S.N. Reasons 1. In OXM and OXN 1. ................. i. OMX = ONX (A) i. Being 90° ii. OXM = OXN (A) ii. From given iii. OX = OX (S) iii. Being common side 2. OXM –OXN 2. By A.A.S axiom 3. OM = ON and MX = NX 3. Being corresponding sides of –Triangles 4. AB = CD 4. Equidistance chords are equal 5. BM = CN 5. BM = 1 2 AB = 1 2 CD = CN 6. BM – MX = CN – NX BX = CX 6. Subtracting 3 from 5 Proved O A M B X Y O D C B A X X C Y B A D O N D C M B A X
332 X C Y B A D ii. Given: Circles with centre X and Y intersect at A and C. To-prove: AB = CD Construction: Let join Y with A, B, C and D Proofs: S.N. Statements S.N. Reasons 1. In AXY and CXY 1. ................. i. AX = CX (S) i. Being radii of same circle ii. AY = CY (S) ii. Being radii of same circle iii. XY = XY (S) iii. Being common side 2. AXY –CXY 2. By S.S.S axiom 3. XAY = XCY 3. Being corresponding angles of –Triangles 4. BAY = DCY (A) 4. Being linear pairs of XAY = XCY 5. ABY = CDY (A) 5. Being ABY = BAY = DCY = CDY base angles of isosceles triangles. 6. BY = DY (S) 6. Radii of same circle 7. ABY – CDY 7. By A.A.S. axiom 8. AB = CD 8. Being corresponding sides of –Triangles Proved Addition Practice – V 1. In triangle ABC, a side BC is produced to D. (a) What would be the value of ∠ABC when ∠ACD = 120° and ∠BAC = 50°. (b) Experimentally verify that the relation of exterior angle of the triangle with sum of two non-adjacent interior angles of the triangle ABC by making different size of two triangles. (c) If ABC is an equilateral triangle, calculate and mention the ratio between the ∠ABC and ∠CAD formed by joining the vertex A and the point D in such a way that AC = CD of ΔACD. Solution: a. Here, ACD = 120°, BAC = 50°, ABC = ? We have, ABC + BAC = ACD [⸪ The exterior angle property of triangle] or, ABC + 50° = 120° ABC = 70° D C B A D C B A
333 b. Given: ACD is the exterior angle of different types of triangle ABC To-verify: ACD = ABC + BAC Experiments: Fig. ACD ABC BAC Result i. 130° 80° 50° ACD = ABC + BAC ii. 90° 50° 40° iii. 60° 40° 20° Conclusion: An exterior angle equals to sum of two non-adjacent interior angles of the triangle. c. Given: AB = BC = AC = CD, ABC CAD = ? Here, CDA = CAD (‡ Being base angles of ) ABC = ACB = 60° (‡ Being angles of ) And, ACB = CAD + ADC (‡ An exterior angle equals to sum of non-adjacent interior angles of ) or, ABC = CAD + CAD or, ABC = 2CAD or, ABC CAD = 2 1 ABC : CAD = 2:1 2. In the figure of an isosceles triangle ABC, the straight line AD is drawn perpendicular from the vertex A to the base BC. (a) If ∠BAC = 68°, and BD = CD, what is the value of ∠BAD? (b) Experimentally verify that the straight line drawn perpendicular from the vertex to the base of an isosceles triangle ABC bisects the vertical angle and the base by making two triangles of different size. (c) In which axiom, ΔABD is congruent to ΔADC? Solution: a. If ABC is isosceles , AD BC and BD = CD Then, AD is bisector of BAC. So, BAD = 1 2 BAC = 1 2 × 68° = 34° b. Given: In ABC, AB = BC and BD AC To- verify: ABD = CBD D C B A Fig. (i) D C B A Fig. (ii) D C B A Fig. (iii) A B C D A B C D A B C D A B C D
334 Experiments: Fig. ABD CBD Result i. 45° 45° ABD = CBD ii. 65° 65° Conclusion: The straight line drawn perpendicular from the vertex to the base of an isosceles triangle ABC bisects the vertical angle. c. ABD – ADC by R.H.S. axiom 3. Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. (a) Prove that: Δ QPT ~ Δ RST. (b) Prove that PT × TR = ST × TQ. Solution: Given: Right angled triangles PQR and QSR are standing on QR To-prove: a. QPT ~ RST b. PT × TR = ST × TQ Proofs: 1. QPT = RST (A) (‡ Being 90°) 2. PTQ = STR (A) (‡ Being vertically opposite angles) 3. PQT = SRT (A) (‡ Being remaining angles) 4. QPT ~ RST (‡ By A.A.A. axiom) 5. PT ST = TQ TR (‡ Being corresponding sides of similar triangles) PT × TR = ST × TQ 4. In the given figure, a girl looks the reflection of the top of the lamppost on the mirror which is 6.6 m away from the foot of the lamppost. The girl whose height is 1.25 m is standing 2.5 m away from the mirror. (Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line.) (a) Prove that: ΔABD ~ ΔECD. (b) Find the height of the lamppost. Solution: In figure alongside, 1. ABD = ECD (‡ Being 90°) 2. BAD = CED (‡ Being corresponding angles) 3. ADB = CDE (‡ Being common angle) 4. ABD ~ ECD (‡ By A.A.A. axiom) 5. AB EC = BD CD (‡ Being corresponding sides of similar triangles) A B C D T Q R P S T Q R P S A B D C E 1.25m 2.5m 6.6m x
335 or, x 1.25 = 6.6 2.5 x = 3.3 Hence, height of lamppost is 3.3 m. 5. The figure of rhombus MNOP is given below. (a) Write the relationship between the diagonals of rhombus. (b) If MN = (4m – 7 ) cm and MP = (m + 2) cm, find the value of m. (c) Construct a rhombus MNOP having diagonal MO = 6 cm and diagonal PN = 8 cm. Solution: a. Diagonals of rhombus bisect each other and mutually perpendicular. b. MN = MP (‡ Being sides of rhombus) or, 4m – 7 = m + 2 or, 3m = 9 m = 3 c. Given: Diagonal MO = 6 cm, diagonal PN = 8 cm To-construct: Rhombus MNOP Construction: No need to mention: Step I: Let draw st. line PN with length 8 cm. Step II: Drawing bisector on PN. Step III: Determining point M and O with arc 3 cm from intersection point to bisector. 6. Parallelograms BCDE and ABDE are given in the trapezium ACDE as shown in the figure with measurement. (a) Write the relationship between opposite angles of parallelogram. (b) Find the value of x, y and z. (c) Construct a trapezium in which AB = 6 cm, BC = CD = 4 cm and DA = 5 cm. Also, AB || CD. Solution: Given: In BCDE and ABDE a. Opposite angles of parallelogram are equal. b. x + (55° + 65°) = 180° (‡ Being co-interior angles) x = 60° O P N M 8 cm P M O N P M O N O P N M 8 cm D E C A B 55° 65° x z y D E C A B 55° 65° x z y
336 y = 55° (‡ Being alternate angles) z = y (‡ Being opp. angles of ) z = 55° c. Given: AB = 6 cm, BC = CD = 4 cm, DA = 5 cm, AB//CD To-construct: Trapezium ABCD Construction: No need to mention: Step I: Let draw st. line AB with length 6 cm and AE with length 4 cm on AB. Step II: Determining point C with arc 5 cm from E and arc 4 cm from B. Step III: Determining point D with arc 5 cm from A and arc 4 cm from C. 7. In the given figure, O is the center of the circle. OP is the perpendicular distance from the center to the chord. (a) What type of triangle is OAB? (b) Find the radius of circle having chord AB = 16 cm and OP = 6 cm. (c) Can we draw a circle through the points A and B having radius 5 cm? Write with reason. Solution: a. OAB is an isosceles triangle being OA = OB as radii of same circle. b. AB = 16 cm, OP = 6 cm Then, radius (r) = OA = OP2 + AP2 = OP2 + AB 2 2 ‡ AP = 1 2 AB = 62 + 82 = 36 + 64 = 100 = 10 cm c. No, we cannot draw a circle through the points A and B having radius 5 cm because. To include A and B the radius of circle should be greater than half length of AB. i.e., Radius (r) > 1 2 AB or, r > 1 2 × 16 or, r > 8 cm 8. In a circle, O is the center of the circle. OA = 10 cm and AB = 12 cm. (a) Write the length of AP. (b) Prove that: OP = 8 cm (c) Prove that: ΔOAP ≅ ΔOBP. D C E B A 4 cm 6 cm 4 cm D A B C 4 cm 6 cm O P A B O P A B O P A B
337 Solution: Given: OA = 10 cm, AB = 12 cm a. AP = 1 2 AB = 1 2 × 12 = 6 cm b. OP = OA2 – AP2 = 102 – 62 = 100 – 36 = 64 = 8 cm c. Given: OP AB To-prove: OAP – OBP Proofs: S.N. Statements S.N. Reasons 1. In OAP and OBP 1. i. OPA = OPB (R) i. Being OP AB ii. OA = OB (H) ii. Being radii of same circle iii. OP = OP (S) iii. Being common side 2. OAP –OBP 2. By R.H.S axiom Proved 9. In the given circle with center O, AS = AK, OS ⊥ AH and OK ⊥ AD. (a) Prove that: ΔOAS ≅ ΔOAK (b) Are AH equal to AD? Justify with reason. (c) If AH = 12 cm, what should be the value of AS. Solution: Given: AS = AK, OS AH and OK AD To-prove: a. OAS –OAK b. AH = AD Proofs: S.N. Statements S.N. Reasons 1. In OAS and OAK 1. i. ASO = AKO (R) i. Being 90° ii. OA = OA (H) ii. Being common side iii. AS = AK (S) iii. From given 2. OAS –OAK 2. By R.H.S axiom 3. OS = OK 3. Being corresponding sides of –Triangles 4. AH = AD 4. Being equidistance chords Proved c. If AH = 12 cm, then AS = 1 2 AH = 1 2 × 12 = 6 cm O P A B O P A B O S H D A K O S H D A K
338 10. In the given figure, the length of two parallel chords AB and CD are 12 cm and 6 cm respectively. Both chords are lying in the same side from the center O of the circle. (a) Write the relation of chords which are equidistant from the center of a circle. (b) If the distance between two chords AB and CD is 3 cm, find the radius of the circle. Solution: a. If chords are equidistance from centre, they are equal. b. In figure alongside, AM = 1 2 AB and, CN = 1 2 CD = 1 2 × 12 = 1 2 × 6 = 6 cm = 3 cm Now, OM = OA2 – AM2 and, ON = OC2 – CN2 = r 2 – 62 = r 2 – 32 = r 2 – 36 = r 2 – 9 We have, MN = 3 cm or, ON – OM = 3 or, r 2 – 9 – r 2 – 36 = 3 or, ( r 2 – 9 – r 2 – 36 )2 = 32 or, r2 – 9 – 2 (r2 – 9) (r2 – 36) + r2 – 36 = 9 or, 2r2 – 54 = 2 r 4 – 45r2 + 324 or, (r2 – 27)2 = ( r 4 – 45r2 + 324 )2 or, r2 – 54r2 + 729 = r4 – 45r2 + 324 or, 405 = 9r2 or, 45 = r2 r = 3 5 = 6.70 cm O 6 cm A B 12 cm C D O 6 cm A B 12 cm N M C D
339 UNIT VI STATISTICS AND PROBABILITY CHAPTER CLASSIFICATION AND REPRESENTATION 17 OF DATA 17.1 Collection of Data and Frequency Table PRACTICE 17.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (i) Given the length of some pencils (in cm) : 12, 8, 6, 11, 9, 10, 12, 14, 12, 8. (a) Define frequency of an item. (b) Make the frequency table. (c) Find the sum of frequencies. (d) Which length of the pencil is equal to the sum of frequencies? (ii) Given the marks of 20 students out of 25 : 21, 23, 19, 17, 12, 15, 15, 17, 17, 19, 23, 23, 21, 23, 25, 25, 21, 19, 19,19 (a) What value corresponds to tally marks | | | | ? (b) Construct the frequency table by using tally bars. (c) Find the sum of frequencies. (d) Why the sum of frequencies is not equal to 25? Give reason. Solution: (i) (a) Frequency of an item: The number of repeating of an item is called frequency of an item. (b) Frequency table, x Tally marks f 6 | 1 8 || 2 9 | 1 10 | 1 11 | 1 12 ||| 3 14 | 1 N = 10 (c) Sum of frequencies (N) = 10 (d) The 10 cm length of pencil is equal to the sum of frequency.
340 (ii) (a) The tally marks | | | | represents 5. (b) Frequency table, x Tally marks f 12 | 1 15 || 2 17 ||| 3 19 |||| 5 21 ||| 3 23 |||| 4 25 || 2 N = 20 (c) Sum of frequencies (N) = 20 (d) The sum of frequencies is not equal to 25 because there were only 20 items. 2. (i) Weight of school' bag of 25 students (in kg) are given below : 5.3, 6.2, 5.6, 6.2, 5.2, 4.8, 6.3, 5.2, 5.3, 5.6, 5.8, 5.3, 5.6, 6.2, 6.3, 5.3, 6.4, 6.2, 5.2, 5.8, 6.3, 5.7, 5.6, 6.3, 5.4. (a) Define cumulative frequency. (b) Construct the cumulative frequency table of the data. (c) Of which weight of bag is found most? (d) Find the difference of number of the heaviest bag and number of the lightest bag. (ii) The marks obtained by 30 students in class test of statistics are given as below: 34, 44, 54, 25, 56, 65, 45, 34, 29, 56, 67, 76, 47, 78, 67, 52, 71, 68, 59, 20, 45, 37, 73, 45, 25, 28, 56, 34, 56, 73. (a) Define grouped data. (b) Make the frequency table taking class interval 10 of the following data. (c) Which group has the highest number of students? (d) Find the difference between total number of students and total frequency. Why the answer is like that? Solution: (i) (a) The accumulated frequency with each previous frequency is called cumulative frequency. (b) cumulative frequency table, x Tally marks Frequency (f) 4.8 | 1 5.2 ||| 3 5.3 |||| 4 5.4 | 1 5.6 |||| 4 5.7 | 1 5.8 || 2 6.2 |||| 4 6.3 |||| 4 6.4 | 1 N = 25 (c) The bags of weights 5.3 kg, 5.6 kg, 5.6 kg, 6.2 kg and 6.3 kg are found equally as most number. (d) The difference between heaviest number of bag and number of lightest bag = 1 – 1 = 0
341 (ii) (a) The data presented in the collective form and their frequencies with in some fixed interval is called grouped data. (b) The frequency table, Marks (x) Tally marks No. of students (f) 20 - 30 |||| 5 30 - 40 |||| 4 40 - 50 |||| 5 50 - 60 |||| || 7 60 - 70 |||| 4 70 - 80 |||| 5 N = 30 (c) The group of 50 - 60 has highest number of students. (d) the difference between total number of students and sum of frequency is 0 as every individual student contributes 1 frequency, so that the total number of students will be equal to sum of frequency. 3. (i) The ages of teachers in a government school are as follows: 45, 44, 56, 45, 45, 56, 45, 46, 53, 46, 56, 54, 54, 57, 54, 49, 59, 48, 47, 59, 53, 51, 50, 40, 55, 52, 41, 59, 49, 47 (a) Define the class interval in frequency distribution table. (b) Construct the cumulative frequency distribution table for the data taking the class interval of 4. (c) How many teachers are there in the youngest group? (d) If the retirement age limit is 60, how many teachers will be retired in coming 5 years? (ii) Here the number of family members are given which was asked by team Sagarmatha of grade nine in project work of Chapter "Sets": 3, 4, 5, 4, 3, 6, 4, 6, 7, 3, 5, 4, 7, 5, 2, 4, 2, 5, 4, 3, 6, 2, 4, 6, 7, 4, 3, 5, 6. (a) Define cumulative frequency. (b) Construct the cumulative frequency distribution table for the data. (c) How much families did the team visited? (d) How many families has most number of family members? (iii) The score secured by some students in IQ test. 75, 85, 84, 83, 88, 90, 89, 70, 98, 97, 90, 86, 79, 70, 78, 93, 82, 81, 71. (a) Determine the best class interval for the cumulative frequency distribution table. (b) Construct the cumulative frequency distribution table with the same class interval as you determined. (c) Which group has less IQ ? (d) By how much percent the highest IQ is more than the least IQ? Solution: (i) (a) Class interval of: The range or length of group or the number of possible items in a group is called class interval of frequency distribution. (b) Cumulative frequency table, Ages (x) Tally marks Frequency (f) c.f. 40 - 44 || 2 2 44 - 48 |||| |||| 9 11
342 48 - 52 |||| 5 16 52 - 56 |||| || 7 23 56 - 60 |||| || 7 30 N = 30 (c) There are only 2 teachers in youngest group. (d) 7 teacher of age group 56 - 60 will be retired in upcoming 5 years. (ii) (a) Cumulative frequency: The accumulated frequency of with each previous frequencies is called cumulative frequency. (b) Cumulative frequency table Family members (x) Tally marks No. of families (f) c.f. 2 ||| 3 3 3 |||| 5 8 4 |||| ||| 8 16 5 |||| 5 21 6 |||| 5 26 7 ||| 3 29 N = 29 (c) The team visited 29 families. (d) 3 families has most number of family member (7). (iii) (a) The largest value = 98 The smallest value = 70 Range = 98 – 70 = 28 To make about 5 classes = 28 5 = 5.6 i.e. if we make the class interval of 5 there would be 6 classes Hence, the best class interval = 5. (b) Cumulative frequency table, IQ scones (x) Tally marks Frequency (f) c.f. 70 - 75 ||| 3 3 75 - 80 ||| 3 6 80 - 85 |||| 4 10 85 - 90 |||| 4 14 90 - 95 ||| 3 17 95 - 100 || 2 19 N = 19 (c) The group 70 - 75 has less IQ. (d) The percent of difference = Highest – Lowest Lowest × 100% = 98 – 70 70 × 10% = 40% 4. (i) The following table represents the frequency distribution of marks obtained by some students of class 9 in the First terminal examination, 2080 Obtained Marks 25 36 45 48 53 55 67 78 89 No. of Students 1 3 2 7 6 8 4 5 3 (a) What is the type of data series given above? (b) Construct the cumulative frequency distribution table for the data.
343 (c) How many students scored the highest marks? (d) Find the difference between the marks obtained by highest number of students and least number of students. (ii) A shopkeeper weighed some of available books in his store and found the result as follows: Weight of Books (in gm) 245 432 342 356 432 453 453 342 No. of Students 5 7 6 5 4 2 9 1 (a) Which alphabet is used to present the number of students? (b) Construct the cumulative frequency distribution table for the given data. (c) How many books did the shopkeeper weigh altogether? (d) What is the difference between the heaviest and the lightest book? Solution: (i) (a) The given data resents a discrete data. (b) Cumulative frequency table Marks (x) No. of students (f) c.f 25 1 1 36 3 4 45 2 6 48 7 13 53 6 19 55 8 27 67 4 31 78 5 36 89 3 39 N = 39 (c) 3 students scored highest marks. (d) Marks obtained by highest number of students = 55 Marks obtained by least number of students = 25 Difference = 55 – 25 = 30. (ii) (a) f (Frequency) is used to present the number of students. (b) Cumulative frequency table. (re arranged) Weight of books (kgm) x No. of student (f) c.f 245 5 5 342 7 12 356 5 17 432 11 28 453 11 39 N = 39 (c) He weighed 39 books altogether. (d) The heaviest book's weight = 453 gm The lightest book's weight = 245 gm. difference = 453 gm – 245 gm = 208 gm. 5. (i) The health post is distributing the nutrition supplements "Baal Vita" to the babies under 15 months. The preliminary reports show the number of children by their ages months is as follows:
344 Age of Children (in month.) 2-4 4-6 6-8 8-10 10-12 12-14 No. of Children 16 15 18 17 14 12 (a) What type of data presentation is it? (b) Construct the cumulative frequency table from the following data: (c) If each children are to be given 5 packets of Bal Vita, find the total number of packets needed. (d) How many more packets need are if the children above 10 months are given 10 packets each? (ii) 19 students of taekwondo class are found to be the weight of as the following: Weight of Students (in kg) 25-30 30-35 35-40 40-45 45-50 No. of Students 5 3 4 2 5 (a) What is the class interval in the given data presentation? (b) Construct the cumulative frequency distribution table for the data. (c) How many students are of more than 40 kgs? (d) How many students need to increase their weight if they want to select in 40kg weight championship? Solution: (i) (a) Grouped data/continuous data/series. (b) Cumulative frequency table, Age (x) (yrs) No. of children (f) c.f 2 - 4 16 36 4 - 6 15 31 6 - 8 18 49 8 - 10 17 66 10 - 12 14 80 12 - 14 12 92 N = 92 (c) Total number of children = 92 Hence to give 5 packets to each = 92 × 5 = 460 packets are needed. (d) Packets needed for up to 10 years = 5 × 66 = 330 Packets needed for above 10 years = 10 × 26 = 260 Total packets = 330 + 260 = 590 Hence, 130 more packets are needed. (ii) (a) The class interval is 5 in given data. (b) Cumulative frequency table Weight (kg) (x) No. of students (f) c.f 25 - 30 5 5 30 - 35 3 8 35 - 40 4 12 40 - 45 2 14 45 - 50 5 19 N = 19 (c) There were 7 students of more than 40 kgs. (d) 12 students need to increase their weight if they want to be selected in 40 kg weight championship.
345 17.2 Histogram PRACTICE 17.2 Read / Understand / Think / Do Keeping Skill Sharp 1. Read the given histogram and answer the questions given below it. (a) Which axis should the marks be shown to draw a histogram? (b) How many students are there in all? (c) Which age group has maximum students? (d) Find the difference between the number of students above 6 years and the number of students below 7 years. Solution: (a) x-axis (b) 30 + 50 + 60 + 40 + 20 + 10 = 210 students. (c) Age group of 6 - 7 has maximum numbers of students. (d) The number of student above 6 yrs. = 60 + 40 + 20 + 10 = 130 The number of student below 7 yrs. = 30 + 50 + 60 = 140 Difference = 140 – 130 = 10. 2. A veterinarian recorded the weights of animals in a histogram. Answer the following questions: (a) Which axis should the frequencies be shown to draw a histogram? (b) Find the total number of animals recorded? (c) How many animals weight between 5 kg and 10 kg? (d) Find the difference between the heaviest group and the lightest group. Solution: (a) y-axis. (b) 8 + 18 + 10 + 17 + 15 = 68 animals. (c) 18 (d) Heaviest group has 15 animals Lightest group has 8 animals Difference = 15 – 8 = 7 animals. 3. A group of students in an entrance exam of Engineering college scored the following marks out of 75 full marks. Marks 20 30 40 50 60 70 Frequency 5 18 32 45 20 3 (a) Which axes should the marks and frequencies be shown to draw a histogram? (b) Construct the histogram for the data given above. (c) What is the marks secured by most of the students? (d) If the pass marks is 35, how many students will have to be back from engineering college?
346 Solution: (a) The x-axis should the marks and frequencies be shown to draw a histogram. (b) Constructing the histogram for the given data, (c) The marks obtained by most students is 50. (d) 5 + 18 = 23 students have to be back from engineering collage if pass marks is 35. 4. The Birthing Centre recorded the height of mothers as follows: Height (cm) 120 125 130 140 145 150 Frequency 8 15 27 30 25 10 (a) Define histogram. (b) Construct the histogram for the given data. (c) What is the total number of mothers? (d) If above 135 cm is the ideal height of mother then how many ideal mothers are there? Solution: (a) Histogram: The graphical presentation of statistical data into rectangular bars attaching each order especially when the data is in continuous series. (b) Constructing the histogram for the given data, (c) The total number of mothers is 115. (d) 30 + 25 + 10 = 65 ideal mothers are there. 5. The marks scored by standardized test in mathematics grade nine is as below: Marks 0-20 20-40 40-60 60-80 80-100 Frequency 50 150 300 200 80 (a) What data series type is given above? (b) Construct histograms to represent the data given below. (c) Find the number of students who attended the test. (d) If below 40 marks is considered as Non Graded then how many of them have not achieved any grade? 50 45 40 35 30 25 20 15 10 5 0 20 30 40 50 60 70 Marks 35 30 25 20 15 10 5 0 120 125 130 140 145 150 Heights (cm)