Vedanta Excel in Mathematics Book 9 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 251 Vedanta Excel in Mathematics - Book 9 Rectangle A rectangle is a parallelogram in which each angle is 90°. (i) The opposite sides of a rectangle are equal. (ii) The diagonals of a rectangle are equal. (iii) The diagonals of a rectangle bisect each other. (iv) The area of a rectangle = length × breadth = l × b. Square A parallelogram in which all sides are equal and each angle is 90° is called a square. (i) The diagonals of a square are equal. (ii) Each diagonal bisects the vertical angles. (iii) Diagonal bisect each other perpendicularly. (iv) The triangles formed by the diagonals are congruent. (v) The area of a square = (side)2 = l 2 or = 1 2 (diagonal)2 Rhombus A rhombus is a parallelogram in which all sides are equal. (i) The opposite angles of a rhombus are equal. (ii) Each diagonal bisects the vertical angles. (iii) The diagonals are not equal and bisect each other perpendicularly. (iv) The triangles formed by the diagonals are congruent. (v) The area of a rhombus = 1 2 × product of diagonals = 1 2 × d1 × d2 d1 d2 (i) The opposite sides of a parallelogram are equal. (ii) The opposite angles of a parallelogram are equal. (iii) The diagonals of a parallelogram bisect each other. (iv) The area of a parallelogram = base × height = b × h. Rectangle, square, and rhombus are some special types of parallelograms. Let's discuss the important properties of these special types of parallelograms. h b l Geometry - Parallelogram

Vedanta Excel in Mathematics - Book 9 252 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Theorem 9 The straight line segments that join the ends of two equal and parallel line segments towards the same sides are also equal and parallel. Given: AB = CD and AB // CD. The ends A, C and B, D are joined. To prove: AC = BD AC // BD Construction: B and C are joined. Proof Statements Reasons 1. In ∆s ABC and BCD 1. (i) (ii) (iii) (iv) AB = CD (S) ∠ABC = ∠BCD (A) BC = BC (S) ∴ ∆ABC ≅ ∆BCD (i) (ii) (iii) (iv) Given AB // CD and alternate angles Common side S. A. S. axiom 2. AC = BD 2. Corresponding sides of congruent triangles 3. ∠ACB = ∠CBD 3. Corresponding angles of congruent triangles 4. AC // BD 4. Alternate angles being equal Proved Theorem 10 The straight line segments that join the ends of two equal and parallel line segments towards the opposite sides bisect each other. Given: AB = CD and AB // CD. The opposite ends A, D and B, C are joined. Let, AD and BC intersect at O. To prove: AD and BC bisect each other at O. i.e. AO = OD and BO = OC Proof Statements Reasons 1. (i) (ii) (iii) (iv) In ∆s AOB and COD ∠ABO = ∠OCD (A) AB = CD (S) ∠BAO = ∠ODC (A) ∴ ∆AOB ≅ ∆COD 1. (i) (ii) (iii) (iv) AB // CD and alternate angles Given AB // CD and alternate angles A. S. A. axiom 2. AO = OD and BO = OC 2. Corresponding sides of congruent triangles 3. AD and BC bisect each other at O. 3. From statement (2) Proved Geometry - Parallelogram

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 253 Vedanta Excel in Mathematics - Book 9 Theorem 11 The opposite angles and sides of a parallelogram are equal. Given: ABCD is a parallelogram in which AB // DC and AD // BC. To prove: (i) ∠ABC = ∠ADC, ∠BAD = ∠BCD (ii) AB = DC, AD = BC Construction: Diagonal AC is drawn. Proof Statements Reasons 1. In ∆s ABC and ACD 1. (i) (ii) (iii) (iv) ∠BAC = ∠ACD (A) AC = AC (S) ∠ACB = ∠CAD (A) ∴ ∆ABC ≅ ∆ACD (i) (ii) (iii) (iv) AB // DC and alternate angles Common side AD // BC and alternate angles A. S. A. axiom 2. ∠ABC = ∠ADC 2. Corresponding angles of congruent triangles 3. ∠BAD = ∠BCD 3. Drawing the diagonal BD and same as above in ∆s ABD and BCD 4. AB = DC and AD = BC 4. Corresponding sides of congruent triangles Proved. Converse (I) Theorem 11 If the opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram. Given: ABCD is a quadrilateral in which AB = DC and AD =BC. To prove: ABCD is a parallelogram, i.e. AB // DC and AD // BC. Construction: Diagonal AC is drawn. Proof Statements Reasons 1. In ∆s ABC and ACD 1. (i) (ii) (iii) (iv) AB = DC (S) BC = AD (S) AC = AC (S) ∴ ∆ABC ≅ ∆ACD (i) (ii) (iii) (iv) Given Given Common side S. S. S. axiom 2. ∠BAC = ∠ACD and ∠ACB = ∠CAD 2. Corresponding angles of congruent triangles 3. AB // DC and AD // BC 3. From statement (2), alternate angles being equal 4. ABCD is a parallelogram 4. Opposite sides are parallel Proved Geometry - Parallelogram

Vedanta Excel in Mathematics - Book 9 254 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Converse (II) of Theorem 11 If the opposite angles of a quadrilateral are equal, the quadrilateral is a parallelogram. Given: ABCD is a quadrilateral in which ∠ABC = ∠ADC and ∠BCD = ∠BAD. To prove: ABCD is a parallelogram, i.e., AB // DC and AD // BC. Proof Statements Reasons 1. ∠A + ∠B + ∠C + ∠D = 360° 1. Sum of the angles of a quadrilateral 2. ∠A + ∠B + ∠A + ∠B = 360° or, 2 (∠A + ∠B) = 360° or, ∠A + ∠B = 180° 2. Given, ∠A = ∠C and ∠B = ∠D. 3. ∴ AD // BC 3. Being the sum of co-interior angles 180° 4. Similarly, ∠A + ∠D = 180° 4. Same as above 5. ∴ AB // DC 5. Same as reason (3) 6. ABCD is a parallelogram 6. Opposite sides are parallel Proved Theorem 12 The diagonals of a parallelogram bisect each other. Given: ABCD is a parallelogram. Diagonals AC and BD intersect at O. To prove: AC and BD bisect each other at O, i.e., AO = OC and BO = OD. Proof Statements Reasons 1. (i) (ii) (iii) (iv) In ∆s AOB and COD ∠OAB = ∠OCD (A) AB = DC (S) ∠OBA = ∠ODC (A) ∴ ∆AOB ≅ ∆COD 1. (i) (ii) (iii) (iv) AB // DC and alternate angles Opposite sides of a parallelogram AB // DC and alternate angles A. S. A. axiom 2. AO = OC and BO = OD 2. Corresponding sides of congruent triangle. 3. AC and BD bisect each other at O. 3. From statement (2) Proved Converse of Theorem 12 If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. Given: ABCD is a quadrilateral in which diagonals AC and BD bisect each other at O. ∴ AO = OC and BO = OD To prove: ABCD is a parallelogram, i.e., AB // DC, AD // BC, AB = DC, AD = BC. Geometry - Parallelogram

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 255 Vedanta Excel in Mathematics - Book 9 Example 1: In the given figure, PQRS is a parallelogram. If PT = SR, ∠ PSR = 4x° and ∠ QPT = x°, find the value of ∠ QRS. Solution: (i) ∠PQR = ∠PSR = 4x° [Opposite angles of parallelograms are equal] (ii) PQ = PT [ PQ = SR and PT = SR] (iii) ∠PQT + ∠PTQ = 4x° [ PQ = PT] (iv) ∠PQT + ∠PTQ + ∠QPT = 180° [Sum of angles of DPQT] or, 4x° + 4x° + x° = 180° or, 9x° = 180° ∴ x = 20° (v) ∠QRS = 180° – ∠PQR [ PQ // SR and co-interior angles] = 180° – 4 × 20° = 100° Proof Statements Reasons 1. (i) (ii) (iii) (iv) In ∆s AOB and COD AO = OC (S) ∠AOB = ∠COD (A) BO = OD (S) ∴ ∆ AOB ≅ ∆ COD 1. (i) (ii) (iii) (iv) Given Vertically opposite angles Given S. A. S. axiom 2. AB = DC 2. Corresponding sides of congruent triangle 3. ∠OAB = ∠OCD 3. Corresponding angles of congruent triangles 4. AB // DC 4. From statements (3), alternate angles are equal 5. AD = BC and AD // BC 5. AD and BC join the ends of two equal and parallel lines towards the same side. 6. ABCD is a parallelogram 6. Opposite sides are equal and parallel. Proved Example 2: In the figure alongside, ABCD is a rhombus. If ∠DAC = 35°, find the measure of ∠ABC. Solution: (i) ∠DAC = ∠ACD = 35° [AD = CD, sides of rhombus] (ii) ∠ADC + ∠DAC + ∠ACD = 180° [Sum of angles of DADC] or, ∠ADC + 35° + 35° = 180° ∴ ∠ADC = 110° (iii) ∠ABC = ∠ADC [Opposite angles of rhombus] ∴ ∠ABC = 110° Geometry - Parallelogram Worked-out examples 4x° x° P Q T R S A D 35° C B

Vedanta Excel in Mathematics - Book 9 256 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Parallelogram Example 3: In the figure alongside, WXYZ is a square. If ∠ZNX = 115°, find the sizes of ∠WZN and ∠ZMY. Solution: (i) ∠WZN + ∠ZWN = ∠ZNX [∠ZNX is the exterior angle in DZWN] or, ∠WZN + 90° = 115° ∴ ∠WZN = 25° (ii) ∠ZWY = ∠XWY=45° [Diagonal of square bisects its vertical angle] (iii) ∠ZMY = ∠WZM+∠ZWM [In DWZM, ∠ZMY is the exterior angle] ∴ ∠ZMY = 25° + 45° = 70° Z Y W X N M 115° Example 4: In the figure, ABCD is a parallelogram. The diagonal BD is produced in either sides to the points P and Q such that BP = DQ. Prove that: (i) ∠BCP = ∠DAQ (ii) AQ // PC. Solution: Given: In parallelogram ABCD, the diagonal BD is produced to the points P and Q such that BP = DQ. To prove: (i) ∠BCP = ∠DAQ (ii) AQ // PC. Proof Statements Reasons 1. ∠DBC = ∠ADB 1. AD // BC, alternate angles 2. ∠PBC = ∠ADQ 2. Supplements of equal angles, ∠DBC = ∠ADB 3. (i) (ii) (iii) (iv) In DBPC and DADQ BP = DQ (S) ∠PBC = ∠ADQ (A) BC = AD (S) DBPC ≅ DADQ 3. (i) (ii) (iii) (iv) Given From statement(2) Opposite sides of parallelogram ABCD By S.A.S, axiom 4. ∠BCP = ∠DAQ and ∠BPC = ∠AQD 4. Corresponding angles of congruent triangles 5. AQ // PC 6. From statements (4), alternate angles ∠BPC and ∠AQD are equal. Proved A D Q B P C Example 5: In the figure alongside, ABCD is a parallelogram. If 2MO = OD, prove that M is the mid-point of BC. Solution: Given: ABCD is a parallelogram in which BC // AD and AB // DC, 2 MO = OD To prove: M is the mid-point of BC, i.e. BC = 2 MC

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 257 Vedanta Excel in Mathematics - Book 9 Geometry - Parallelogram Proof: Statements Reasons 1. (i) (ii) (iii) (iv) In ∆ AOD and ∆ MOC ∠ADO = ∠CMO ∠AOD = ∠COM ∠DAO = ∠MCO ∆AOD ~ ∆MOC 1. (i) (ii) (iii) (iv) AD // BC and alternate angles Vertically opposite angles AD // BC and alternate angles A.A.A. axiom 2. (i) (ii) (iii) AD MC = OD MO = AO CO BC MC = 2MO MO or, BC = 2 MC M is the mid-point of BC 2. (i) (ii) (iii) Corresponding sides of similar triangles AD = BC (opposite sides of parallelogram) and OD = 2MO (given) From statement (ii) Proved EXERCISE 14.1 General section 1. a) In the figure alongside, AB // DC and AB = DC. Write down the relation between AD and BC. b) In the given figure, PQ // RS and PQ = RS. What is the relation between PS and QR? c) In the figure alongside, BEST is a parallelogram. Write down the relation between BT and ES, BE, and TS. d) In the parallelogram WXYZ, diagonals, ZX and WY intersect at O. Write the relation between OW, OY and OZ, OX. 2. a) In the given figure, ABCD is a rectangle. AC and BD are diagonals. If the length of diagonal AC is 10 cm, what is the length of diagonal BD? b) In the square PQRS given alongside, what is the value of ∠QPR? c) In the given figure, READ is a rhombus. If ∠RED = 40°, what is the measure of ∠RDE? B D C A Q R S P T S E B Y O W X Z O C D B A P Q R S R E A D

Vedanta Excel in Mathematics - Book 9 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Parallelogram d) In the given rectangle WXYZ, O is the point of intersection of its diagonal WY and XZ. If ∠XOY = 120°, find the size of ∠OWZ. 3. Calculate the size of unknown angles: a) b) c) d) e) f) g) h) O Y Z X W A B C E D 3x y 2x z w P S R Q a 5y b 4y L I K E p+q p+10° 2p–50° r D 60° x 2x O B C E A x y 115° E T A D L A B D E F 110° 55° x C a a 100° b Q R S P T x y B C D 50° 60° F E A 4. a) In the given figure, ABCD is a parallelogram. If ∠BCD = 125° and AB = AE, find the size of ∠BAE. b) In the figure alongside, DBCE is a parallelogram. If ∠AFE =110° and ∠FCE = 40°, find the value of ∠DBC. 110° c) In the adjoining figure, PQRS is a parallelogram. Find the value of ∠SMR. d) In the given figure, HEAD is a parallelogram. If AM ⊥ HE, AN ⊥ HD and ∠NAD = 25°, find the measure of ∠MAE and ∠MAN. H E A D 25° N M 5. a) In the given figure, ABCD is a rhombus. If ∠ADB = 40°, find the size of ∠CDE.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 259 Vedanta Excel in Mathematics - Book 9 b) In the given figure, ABCD is a rhombus. If ∠BCD = 114°, find the measure of ∠ABD. c) In the adjoining figure, ABCD is a rhombus. If ∠OAB = 30°, find the measures of ∠ODC and ∠ABC. C O 30° B A D d) In the given figure, PQRS is a rectangle. If ∠POS = 124°, find the sizes of ∠PQO and ∠OQR. P 124° S O Q R Geometry - Parallelogram 6. a) In the given figure, ABCD is a parallelogram. If AB = 3x cm, BC = (5y – 1) cm, CD = 12 cm, and AD = 19 cm, find the values of x and y. A D 19 cm (5y – 1) 12 cm C 3x cm B e) In the figure alongside, PQRS is a square. If ∠SMR = 70°, find the measures of ∠PST and ∠STQ. f) In the given figure, FIVE is a square. If PQ // FV and ∠QPR = 75°, find the measure of ∠FIR and ∠FPI. g) In the given figure, PQRS is a rhombus and SRM is an equilateral triangle. If SN ⊥ RM and ∠PRS = 55°, find the size of ∠QSN. M 70° T S R Q P M 75° F E R Q V I P 55° P S R N M Q b) In the figure alongside PQRS is a parallelogram. If PQ = (x – y) cm, QR = 20 cm, RS = 10 cm, and PS = (x + y) cm, find the value of x and y. P R S (x + y) cm 20 cm 10 cm (x – y) cm Q

Vedanta Excel in Mathematics - Book 9 260 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) In the adjoining figure, LMNP is a rhombus. If MN = (2a – 3) cm, OM = (3b + 2) cm, OP = 5 cm, and perimeter = 28 cm, find the values of a and b. d) In the adjoining figure, PQRS is a rectangle. If OP = (2x – 3) cm and OR = (x + 1) cm, find the length of diagonal QS. e) RACE is a rectangle in which diagonal RC = 18 cm, OA = (p + q) cm and OE = 3p cm, find the values of p and q. L P 5cm (3b+2)cm (2a – 3)cm N O M S R (x+1)cm (2x–3)cm Q O P R E 3p cm (p+q)cm C O A Geometry - Parallelogram 7. a) In the adjoining figure ABCD is a square and ABE is an equilateral triangle. Find the measure of ∠ADE and ∠DCE. b) In the given figure, ABCD is a square and BEC is an equilateral triangle. Find ∠AEB and ∠DAE. D A B E C D A B C E Creative section -A 8. a) In the adjoining figure, PQ = AB and PQ // AB. Prove that (i) AP = BQ (ii) AP // BQ. b) In the given quadrilateral, AB = DC and BC = AD. Prove that the quadrilateral ABCD is a parallelogram. c) In the figure alongside, diagonals AC and BD of the quadrilateral bisect each other at O. Prove that ABCD is a parallelogram. P Q A B D C B A D C B O A

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 261 Vedanta Excel in Mathematics - Book 9 Geometry - Parallelogram 9. a) In the given figure, ABCD is a parallelogram. AP bisects ∠A. Prove that DP = BC. b) In the given figure, EXAM is a parallelogram, the bisector of ∠A meets the mid-point of EM at P. Prove that AX = 2AM. E X A M P Creative section -B c) In the adjoining figure, ABCD is a rhombus in which CD is produced to E such that CD = DE. Prove that ∠EAC = 90°. E D C B A 10. a) ABCD is a parallelogram. P and Q are two points on the diagonal BD such that DP = QB. Prove that APCQ is a parallelogram. b) ABCD is a parallelogram. DE ⊥ AC and BF ⊥ AC. Prove that BEDF is a parallelogram. c) In parallelogram PQRS, the bisectors of ∠PQR and ∠PSR meet the diagonal at M and N respectively. Prove that MQNS is a parallelogram. d) In the given figure, ABCD is a parallelogram. If P and Q are the points of trisection of diagonal BD, prove that PAQC is a parallelogram. P Q R S M N A B C D P Q 11. a) Prove that the diagonals of a rectangle are equal. b) If the diagonals of a rhombus are equal, prove that it is a square. c) Prove that the diagonals of a rhombus bisect each other perpendicularly. d) Prove that the diagonal of a parallelogram divides it into two congruent triangles.

Vedanta Excel in Mathematics - Book 9 262 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Parallelogram Project work and activity section 12. a) Take two rectangular sheets of paper of the same size. Fold one sheet through one diagonal and another sheet through other diagonal. Then, cut out each sheet of papers through diagonals. (i) Are two diagonals equal? (ii) Do these diagonals bisect each other? b) Take a square sheet of paper and fold it through both diagonals. (i) Are two diagonals equal? (ii) Do these diagonals bisect each other perpendicularly? OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. The properties of the quadrilaterals are given below. (i) The opposite sides are equal. (ii) The opposite angles are equal (iii) The diagonal bisect each other (iv) The diagonals are equal (v) The diagonal bisect each other at a right angle Which of the above are the properties of a parallelogram? (A) (i), (ii) and (iv) (B) (i), (iv) and (v) (C) (i), (ii) and (iii) (D) (i), (ii) and (v) 2. Which of the following statements is correct? (A) Every square is a rectangle. (B) Every parallelogram is a rhombus. (C) Every parallelogram is a rhombus. (D) Every rectangle is a square. 3. The line segments that join the ends of two equal and parallel line segments towards the same sides are (A) parallel (B) equal (C) parallel and equal (D) bisect each other 4. The line segments that join the ends of two equal and parallel line segments towards the opposite sides bisect each other. (A) equal (B) parallel (C) parallel and equal (D) equal and perpendicular 5. The opposite angles of a parallelogram are (A) equal (B) unequal (C) complementary (D) supplementary 6. The opposite sides of a parallelogram are (A) equal (B) parallel (C) Both (A) and (B) (D) None 7. The quadrilateral with equal opposite sides is always a (A) parallelogram (B) square (C) rectangle (D) rhombus 8. The quadrilateral with equal opposite angles is a (A) parallelogram (B) square (C) rectangle (D) rhombus 9. The diagonals of a parallelogram are (A) equal (B) unequal (C) bisect each other (D) perpendicular to each other 10. The angles made by the diagonals with a side of rhombus are (A) equal (B) complementary (C) supplementary (D) unequal https://www.geogebra.org/m/z5w6dnby Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 263 Vedanta Excel in Mathematics - Book 9 Unit 15 Geometry: Construction 15.1 Construction of quadrilaterals A quadrilateral has ten parts in all: four sides, four angles, and two diagonals. To construct a quadrilateral, we shall usually need data about five specified parts of it. However, in the case of regular quadrilaterals (square, rectangle, parallelogram, rhombus, etc.), the measurements of a few required number of parts may be sufficient. Of course, to construct any type of quadrilateral, we should know its properties and we apply these properties in its construction. Let’s study the steps of construction of different quadrilaterals under the following conditions. A. Construction of rhombus Let’s construct the rhombus under the following conditions. 1. When a side and angle made by two adjacent sides are given Example: Construct a rhombus ABCD in which AB = 4.5 cm and ∠ ABC = 45°. Steps of construction (i) Draw AB = 4.5 cm (ii) At B, construct ∠ABX = 45°. (iii) With centre at B and radius 4.5 cm, cut BX at C (iv) With centres at A and C and radius 4.5 cm, draw two arcs intersecting each other at D. (v) Join A, D and C, D. Thus, ABCD is the required rhombus. 2. When two diagonals are given Example: Construct a rhombus ABCD in which diagonals AC = 5.4 cm and BD = 4.2 cm. Steps of construction (i) Draw AC = 5.4 cm and draw its perpendicular bisector XY. Mark the mid-point of AC as O. (ii) With centre at O and radius 2.1 cm ( 1 2 of BD), draw two arcs to cut OX at D and OY at B. (iii) Join A, B; B, C; C, D and D, A. Thus, ABCD is the required rhombus. C D A B A B 45° 4.5 cm 4.5 cm 4.5 cm 4.5 cm C X D A C D B O X Y

Vedanta Excel in Mathematics - Book 9 264 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3. When the length of a diagonal and a side are given Example: Construct a rhombus ABCD in which AB = 4.3 cm and diagonal AC = 6.1 cm. Solution: Here, AB = 4.3 cm and diagonal AC = 6.1 cm Steps of construction (i) Draw AC = 6.1 cm. (ii) With the centre at A and radius 4.3 cm draw two arcs above and below the line AC. (iii) With the centre at C and radius 4.3 cm draw two arcs above and below the line AC so that the arcs above AC cut each other at D and below AC cut each other at B. (iv) Join A, D; C, D; A, B and B, C. Thus, ABCD is the required rhombus. 4. When the length of a side and angle made by the side and one of the diagonals are given Example: Construct a rhombus ABCD in which AB = 5 cm and ∠BAC = 60°. Solution: Here, AB = 5 cm and ∠BAC = 60°. Steps of construction (i) Draw AB = 5 cm. (ii) At A construct ∠BAX = 60°. (iii) With the centre at B and radius 5 cm draw an arc to cut AX at C. (iv) Join B and C. (v) From A and C, draw two arcs with radii 5 cm and the arcs intersect to each other at D. (vi) Join A, D and C, D. Thus, ABCD is the required rhombus. EXERCISE 15.1 1. Construct the rhombus ABCD in which a) AB = 5.4 cm and ∠ABC = 60o b) AB = 6 cm and ∠BAD = 45o c) BC = 4.5 cm and ∠BCD = 30o d) AD = 5.8 cm and ∠ADC = 120o 2. Construct the rhombus PQRS in which a) Diagonals PR = 6 cm and QS = 8 cm b) Diagonals PR = 6 cm and QS = 5 cm c) Diagonals PR = 4.8 cm and QS = 5.4 cm d) Diagonals PR= 6.4 cm and QS=7.2 cm 3. Construct the rhombus WXYZ in which a) XY = 5 cm and diagonal XZ = 6 cm b) YZ = 4.7 cm and diagonal WY=5.6 cm c) WZ = 5.3 cm and diagonal ZX = 4 cm d) YZ = 6.1 cm and diagonal YW=3.8 cm 4. Construct the rhombus ABCD in which a) AB = 6 cm and ∠BAC = 45o b) AB = 5.4 cm and ∠BAC = 60o c) BC = 4.6 cm and ∠ACB = 75o d) BC = 6.1 cm and ∠ABC = 45o A C D 6.1 cm 4.3 cm 4.3 cm 4.3 cm 4.3 cm B D A B C X 60° 5 cm Geometry - Construction

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 265 Vedanta Excel in Mathematics - Book 9 B. Construction of scalene quadrilateral Let’s observe the shape and size of the following quadrilateral having side lengths 3 cm, 5 cm, 7 cm and 4 cm. Do the quadrilaterals with same side lengths have exactly the same shape and size? D 4 cm 7 cm 3 cm 5 cm A B C 7 cm 4 cm 5 cm A 3 cm B C D 7 cm 4 cm A B C D 3 cm 5 cm Each interior angle of convex quadrilateral is less than 180o . However, at least one of the interior angles of concave quadrilateral is more than 180o . Thus, the shape of the quadrilaterals having same measurements may have different shapes. A 4.5 cm B D C X 60° 3 cm 3.4 cm 2.8 cm D 60° 3.4 cm A 4.5 cm B C X 3 cm 2.8 cm Concave quadrilateral Convex quadrilateral 1. When all four adjacent sides and one of the diagonals are given Example: Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 5.1 cm, CD = 4.9 cm, AD = 6.2 cm and the diagonal BD = 5.8 cm. Solution: Here, AB = 5.4 cm, BC = 5.1 cm, CD = 4.9 cm, AD = 6.2 cm and the diagonal BD = 5.8 cm Steps of construction (i) Draw AB = 5.4 cm. (ii) From A, draw an arc with radius AD = 6.2 cm and from B draw another arc with radius BD = 5.8 cm. These two arcs intersect each other at D. (iii) From B, draw an arc with radius BC = 5.1 cm and from D, draw another arc with radius DC = 4.9 cm. These two arcs intersect to each other at C. Join A, D; B, C and D, A. Thus, ABCD is the required quadrilateral. 2. When all four sides and one of the angles between adjacent sides are given Example: Construct a quadrilateral ABCD in which AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm, AD = 4 cm and ∠ABC = 60°. D C 4.9 cm 5.1 cm 6.2 cm 5.8 cm B A 5.4 cm Geometry - Construction

Vedanta Excel in Mathematics - Book 9 266 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: Here, AB = 5.2 cm, BC = 5 cm, CD = 4.2 cm, AD = 4 cm and ∠ABC = 60°. Steps of construction (i) Draw AB = 5.2 cm. (ii) At B construct ∠ABX = 60°. (iii) With the centre at B and radius 5 cm draw an arc to cut BX at C. (iv) From A, draw an arc with radius 4 cm and from C, draw another arc with radius 4.2 cm. These two arcs intersect to each other at D. Join C, D and A, D. Thus, ABCD is a required quadrilateral. 3. When any three adjacent sides and two diagonals are given Example: Construct a quadrilateral ABCD in which AB = 4 cm, BC = 5.5 cm, DA = 3.4 cm, diagonal AC = 7.2 cm and diagonal BD = 5.8 cm. Solution: Here, AB = 4 cm, BC = 5.5 cm, DA = 3.4 cm, diagonal AC = 7.2 cm and diagonal BD = 5.8 cm. Steps of construction (i) Draw AB = 4 cm. (ii) From A, draw an arc with radius 7.2 cm and from B, draw another arc with radius 5.5 cm. These two arcs intersect to each other at C. Join C to A and B (iii) Also, from A, draw an arc with radius 3.4 cm and from B, draw another arc with radius 5.8 cm. These two arcs intersect to each other at D. Join A, D; B, D and C, D. Thus, ABCD is a required quadrilateral. 4. When any three adjacent sides and two angles are given Example: Construct a quadrilateral ABCD in which AB = 5.2 cm, BC = 4.8 cm, DA = 4.1 cm, ABC = 75o and ∠BAD = 60o Solution: Here, AB=5.2 cm, BC=4.8 cm, DA=4.1 cm, ∠ABC = 75o and ∠BAD = 60o . Steps of construction (i) Draw AB = 5.2 cm. (ii) At A construct ∠BAX = 60°. (iii) With the centre at A and radius 4.1 cm draw an arc to cut AX at D. 5 cm 4 cm 4.2 cm A 5.2 cm B C D A 4 cm B C D 3.4 cm 5.8 cm 5.5 cm 7.2 cm A B C D X Y 60° 75° 5.2 cm 4.1 cm 4.8 cm Geometry - Construction

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 267 Vedanta Excel in Mathematics - Book 9 (iv) At B construct ∠ABY= 75°. (v) With the centre at B and radius 4.8 cm draw an arc to cut BY at C. (vi) Join C and D. Thus, ABCD is a required quadrilateral. 5. When any two adjacent sides and three angles are given Example: Construct a quadrilateral ABCD in which AB = 5 cm, BC = 3.9 cm, ∠BAD = 45o , ∠ABC = 120o and ∠BCD = 70o . Solution: Here, AB = 5 cm, BC = 3.9 cm, ∠BAD = 45o , ∠ABC = 120o and ∠BCD = 70o . Steps of construction (i) Draw AB = 5 cm. (ii) At A construct ∠BAX = 45°. (iii) At B construct ∠ABY = 120°. (iv) With the centre at B and radius 3.9 cm draw an arc to cut BY at C. (v) At C construct ∠BCZ= 70° so that CZ cuts AX at D. Thus, ABCD is a required quadrilateral. EXERCISE 15.2 1. Construct the quadrilateral ABCD in which a) AB = 5 cm, BC = 5.6 cm, CD = 4.5 cm, AD = 5.4 cm and, diagonal BD = 6.5 cm b) AB = 4.2 cm, BC = 5.1 cm, CD = 5.4 cm, AD = 3.8 cm and, diagonal BD = 4.6 cm c) AB = 5 cm, BC = 4.5 cm, CD = AD = 5.5 cm and, diagonal AC = 6 cm 2. Construct the quadrilateral PQRS in which a) PQ = 5.5 cm, QR = 5.7 cm, RS = 4.7 cm, SP = 4.3 cm and ∠SPQ = 60o . b) PQ= 4 cm, QR = 5 cm, RS = 5.5 cm, SP = 5.5 cm, and ∠PQR= 45° c) PQ = QR = 5.5 cm, RS = SP = 4.5 cm and ∠P = 75o . 3. Construct the quadrilateral ABCD in which a) AB= 4.5 cm, BC = 3.7 cm, AD = 4 cm, diagonals AC = 5.4 cm and BD = 7 cm. b) AB= 5.2 cm, BC = 4.3 cm, AD = 3.6 cm, diagonals AC = 6.1 cm and BD = 5.8 cm. c) AB= 4.5 cm, BC = 3.9 cm, AD = 5.1 cm and diagonals AC = BD = 6 cm. 4. Construct the quadrilateral WXYZ in which a) WX= 5.3 cm, XY = 4.6 cm, WZ = 3.9 cm, ∠W = 60o and ∠X = 75o . b) WX= 4.5 cm, XY = 5.4 cm, WZ = 4.7 cm, ∠W = 90o and ∠X = 60o . c) XY = 4.4 cm, YZ = 6.3 cm, WZ = 5.7 cm, ∠Y = 45o and ∠Z = 60o . 5. Construct the quadrilateral PQRS in which a) PQ= 5 cm, QR = 4 cm, ∠P = 60o , ∠Q = 100o and ∠R = 120o . b) PQ= 4.5 cm, SP = 3.8 cm, ∠P = 90o , ∠Q = 120o and ∠S = 70o . c) QR= 5.6 cm, RS = 4.7 cm, ∠Q = ∠R = 60o and ∠R = 90o . A 5 cm B 45° 120° 3.9 cm 70° C D X Y Z Geometry - Construction

Vedanta Excel in Mathematics - Book 9 268 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur C. Construction of trapezium In the case of regular quadrilaterals (square, rectangle, parallelogram, rhombus, etc.) and trapezium, the measurements of a few required number of parts may be sufficient. Of course, to construct any type of quadrilateral, we should know its properties and we apply these properties in its construction. Facts to remember 1. Trapezium is a quadrilateral in which a pair of opposite sides are parallel. 2. To construct trapezium, we need the measurements of its four independent parts. Let’s study the steps of construction of trapeziums under the following conditions. 1. When the length of three adjacent sides and one of the angles are given Example: Construct a trapezium ABCD in which AB = 5.4 cm, BC = 4.9 cm, AD = 4.1 cm, ∠BAD = 80o and AB//DC. Solution: Here, AB = 5.4 cm, BC = 4.9 cm, AD = 4.1 cm, ∠BAD = 80o and AB//DC. Steps of construction (i) Draw AB = 5.4 cm. (ii) At A, construct ∠BAD = 80o by using protractor. (iii) With the centre at A and radius 4.1 cm, draw an arc to cut AX at D. (iv) At D, construct ∠ADC = 180o – 80o = 100° by using protractor. (v) With centre at B and radius BC = 4.9 cm, draw an arc to cut DY at C. (vi) Join B and C. Thus, ABCD is the required trapezium. 2. When two adjacent sides and two angles are given Example: Construct a trapezium ABCD in which AB = 5.5 cm, BC = 4.5 cm, ∠DAB = 60°, ∠BCD = 75° and AD // BC. Steps of construction (i) Draw a line segment AB = 5.5 cm. (ii) Construct ∠BAX = 60° at A. (iii) D lies on AX and AD // BC. So, at B, construct ∠ABY = 180° – 60° = 120° (iv) With the centre at B and radius 4.5 cm draw an arc to cut BY at C. (v) At C, construct ∠BCD = 75°. The arm CD intersect AX at D. Thus, ABCD is the required trapezium. A B D C Y 4.9 cm 5.4 cm 4.1 cm X 80° 100° D X 75° 60° A B C Y Geometry - Construction

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 269 Vedanta Excel in Mathematics - Book 9 3. When two sides, a diagonal and angle made by the diagonal with one given side are given Example: Construct a trapezium ABCD in which AB = 5 cm, diagonal AC = 6.5 cm, ∠BAC = 45°, CD = 4.2 cm and AB // DC. Steps of construction (i) Draw a line segment AB = 5 cm. (ii) At A, construct ∠BAX = 45°. (iii) With centre at A and radius 6.5 cm, draw an arc to cut AX at C. (iv) Join B and C. (v) As AB // DC, alternate angles BAC and ACD are equal. So, construct ∠ACY = 45° at C. (vi) With centre at C and radius 4.2 cm, draw an arc to cut CY at D. (vii) Join D and A. Thus, ABCD is the required trapezium. 4. When three sides and diagonal are given Example: Construct a trapezium ABCD in which AB = 4.5 cm, diagonal AC = 6 cm, AD = BC = 5 cm and AB // DC. Steps of construction (i) Draw a line segment AB = 4.5 cm. (ii) With centre at A and radius 6 cm, draw an arc. (iii) With centre at B and radius 5 cm, draw another arc to intersect the previous arc at C. (iv) Join A, C and B, C. (v) At C, construct ∠ACX = ∠BAC. (vi) With centre at A and radius 5 cm, draw an arc to cut CX at D. (vii) Join A and D. Thus, ABCD is the required trapezium. 5. When the lengths of all four sides are given Example: Construct a trapezium ABCD in which AB = 6.5 cm, BC = 4.3 cm, CD = 4 cm, AD = 5 cm and DC//AB. Solution: Here, AB = 6.5 cm, BC = 4.3 cm, CD = 4 cm and AD = 5 cm. A B X Y 45° 45° D C X D C A B Geometry - Construction

Vedanta Excel in Mathematics - Book 9 270 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Steps of construction (i) Draw AB = 6.5 cm. (ii) Mark the point E on AB such that AE = DC = 4 cm. (iii) From E, draw an arc with radius AD = 5 cm and from B, draw another arc with radius 4.3 cm. These two arcs intersect to each other at C. Join A, C and E, C. (iv) Also, from A, draw an arc with radius 5 cm and from C, draw another arc with radius 4 cm. These two arcs intersect to each other at D. Join A, D and C, D. Thus, ABCD is a required trapezium. EXERCISE 15.3 1. Construct the trapezium ABCD in which a) AB = 5.8 cm, BC = 4 cm, AD = 4.5 cm, ∠BAD = 60 o and AB//DC. b) AB = 6 cm, BC = 5.1 cm, AD = 4.8 cm, ∠BAD = 75o and AB//DC. c) AB = 4.2 cm, BC = CD = 5.4 cm, ∠BCD=90 o and AD//BC. 2. Construct the trapezium PQRS in which a) PQ = 5 cm, SP = 4 cm, ∠QPS = 60o , ∠PQR = 75o and PQ // SR. b) PQ = 4.7 cm, QR = 5 cm, ∠QPS = 60 o , ∠PQR = 90o and PQ // SR. c) PQ = 4.2 cm, SP = 3.5 cm, ∠QPS = 50o , ∠PQR = 80o and PQ // SR. 3. Construct the trapezium ABCD in which a) AB = 4.8 cm, diagonal AC = 5.9 cm, ∠BAC = 60o , CD = 5 cm, and AB // DC. b) AB = 6.4 cm, diagonal AC = 7.5 cm, ∠BAC = 30o , CD = 6 cm, and AB // DC. c) AB = 5.5 cm, BC = 7 cm, diagonal BD = 6.1 cm, ∠ABD = 45o , and AB // DC. 4. Construct the trapezium WXYZ in which a) WX = 4.4 cm, diagonal WY = 6.8 cm, WZ = XY= 5.2 cm and WX//ZY. b) WX= 6 cm, diagonal WY= 7.6 cm, YZ = 6.5 cm, XY= 4.5 cm and WX//ZY. c) XY = 5 cm, diagonal XZ = 5.2 cm, YZ = WZ= 4 cm and WZ//XY. 5. Construct the trapezium ABCD in which a) AB = 6 cm, BC = 5 cm, CD = 4 cm, AD = 4.5 cm, and DC//AB. b) AB = 7 cm, BC = 4.7 cm, CD = 5 cm, AD = 4.3 cm, and DC//AB. c) AB = 5 cm, BC = CD = 3 cm, AD = 4 cm, and DC//AB. 6.5 cm A E B C D 4 cm 4.3 cm 4 cm 5 cm 5 cm Geometry - Construction

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 271 Vedanta Excel in Mathematics - Book 9 16.1 Circle and its various parts A circle is a plane curved consisting of all points that have the same distance from a fixed point, called the centre. In the figure, O is the centre of a circle. Of course, when the number of sides of a polygon is increased without a limit, the sides merge into one line and the polygon becomes a circle. Let’s study about the various parts of a circle. (i) Circumference The perimeter of a circle is called its circumference. It is the total length of the curved line of the circle. (ii) Radius It is the line segment that joins the centre of a circle and any point on its circumference. In the figure, OA is the radius. The plural form of radius is radii. All the radii of a circle are equal, i.e., OA = OB = OC = OD = ... (iii) Diameter A line segment that passes through the centre of a circle and joins any two points on its circumference is called the diameter of the circle. In the figure PQ, RS, etc., are the diameters. The length of a diameter is two times radius. So, PQ = 2OQ or 2OP and RS = 2OR or 2OS. A diameter divides a circle into two halves. (iv) Semi-circle A diameter divides a circle into two halves and each half is called the semi-circle. In the figure, PXQ and PYQ are the semi-circles. (v) Arc A part of the circumference of a circle is called an arc. It is denoted by a symbol ‘ ‘. In the figure, PQR is the minor and PSR is major arcs of the circle. A minor arc is less than half of the circumference and a major arc is greater than half of the circumference. P5 P6 P7 P8 P1 P2 P3 P4 O B A O C D P Q R S O P Q X Y O P Q X Y O Unit 16 Geometry: Circle

Vedanta Excel in Mathematics - Book 9 272 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (vi) Chord The line segment that joins any two points on the circumference of a circle is called the chord of the circle. In the figure, PQ and RS are any two chords of the circle. Diameter is the longest chord of a circle. (vii) Segment The region enclosed by a chord and corresponding arc is called the segment of a circle. In the figure, the shaded region is the minor segment and non-shaded region is the major segment of the circle. A minor segment is less than half of a circle whereas the major segment is greater than half of the circle. The minor and the major segments of a circle are known as the alternate segments to each other. (viii) Sector The region enclosed between any two radii and the corresponding arc of a circle is called a sector of the circle. In the figure, OPQ is a sector of a circle. (ix) Concentric circles Two or more circles are said to be concentric circles if they have the same centre but different radii. In the figure, circles PQR, XYZ, and ABC are three concentric circles. (x) Intersecting circles If two circles intersect each other at two points, they are said to be intersecting circles. In the figure, two circles are intersecting each other at P and Q. PQ is the common chord of these two intersecting circles. (xi) Tangent to a circle A line that intersects the circle exactly at one point is called a tangent to the circle. The point at which the tangent touches the circle is called the point of contact. In the figure, PQ is the tangent and T is the point of contact. (xii) Secant of a circle A line that intersects a circle in two distinct points is called a secant of the circle. In the figure, PQ is the secant of the circle. Q P X A O C Z R B Y Geometry - Circle P Q O P Q R S O O P T Q O P A B Q P Q O O P Q O'

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 273 Vedanta Excel in Mathematics - Book 9 14.2 Theorems related to chords of a circle Theorem 13 The perpendicular drawn from the centre of a circle to a chord, bisects the chord. Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: In each circle, a chord AB of different lengths is drawn. Step 3: OP ⊥ AB is drawn in each circle. Step 4: The lengths of AP and PB are measured and the results are tabulated. Figure AP PB Result (i) AP = PB (ii) AP = PB (iii) AP = PB Conclusion:The perpendicular drawn from the centre of a circle to a chord, bisects the chord. Theoretical proof Given: O is the centre of a circle. AB is the chord of the circle and OP ⊥ AB. To prove: AP = PB Construction: O, A and O, B are joined. Proof Statements Reasons 1. (i) (ii) (iii) (iv) In rt. ∠ed ∆s OAP and OBP ∠OPA = ∠OPB (R) OA = OB (H) OP = OP (S) ∴∆OAP ≅ ∆OBP 1. (i) (ii) (iii) (iv) Both of them are right angles. Radii of the same circle Common side R.H.S. axiom 2. AP = PB i.e. OP bisects AB 2. Corresponding sides of congruent triangle Proved https://www.geogebra.org/m/nqdktdcc Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A A B B P P A B P O O O (i) (ii) (iii) A B P O Geometry - Circle

Vedanta Excel in Mathematics - Book 9 274 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Converse of Theorem 13 The line joining the mid-point of a chord and the centre of a circle is perpendicular to the chord. Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: In each circle chord AB of different lengths is drawn and the mid-point of the chord is marked as P. Step 3: O and P are joined. Step 4: ∠OPA and OPB are measured and the results are tabulated. Figure ∠OPA ∠OPB Result (i) ∠OPA = ∠OPB = 90° (ii) ∠OPA = ∠OPB = 90° (iii) ∠OPA = ∠OPB = 90° Conclusion: The line joining the mid-point of a chord and the centre of a circle is perpendicular to the chord. Theoretical proof Given: O is the centre of a circle and AB is the chord. P is the midpoint of AB. O and P are joined. To prove: OP ⊥ AB Construction: O, A and O, B are joined. Proof Statements Reasons 1. (i) (ii) (iii) (iv) In rt. ∠ed ∆s OAP and OBP OA = OB (S) AP = PB (S) OP = OP (S) ∴ ∆OAP ≅ ∆OBP 1. (i) (ii) (iii) (iv) Radii of the same circle Given (P is the mid-point of AB) Common side S.S.S. axiom 2. ∠OPA = ∠OPB 2. Corresponding angles of congruent triangle 3. ∠OPA = ∠OPB = 90° 3. Adjacent angles in linear pair are equal. 4. OP ⊥ AB 4. From the statement 3. Proved A A A B B B P P P O O O (i) (ii) (iii) A B P O Geometry - Circle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 275 Vedanta Excel in Mathematics - Book 9 Theorem 14 Equal chords of a circle are equidistant from the centre. Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Two equal chords AB and CD of different lengths are drawn in each circle. Step 3: OP ⊥ AB and OQ ⊥ CD are drawn in each circle. Step 4: The lengths of OP and OQ are measured and the result is tabulated. Figure OP OQ Result (i) OP = OQ (ii) OP = OQ (iii) OP = OQ Conclusion: Equal chords of a circle are equidistant from the centre. Theoretical proof Given: O is the centre of a circle. Chords AB = CD and OP ⊥ AB, OQ ⊥ CD. To prove: OP = OQ Construction: O, A and O, C are joined. Proof Statements Reasons 1. AB = CD 1. Given 2. 2 AP = 2 CQ i.e. AP = CQ 2. OP ⊥ AB and OP bisects AB; OQ ⊥ CD and OQ bisects CD. 3. (i) (ii) (iii) (iv) In rt. ∠ed ∆s OAP = OCQ ∠OPA = ∠OQC (R) OA = OC (H) AP = CQ (S) ∴ ∆OAP ≅ ∆OCQ 3. (i) (ii) (iii) (iv) Both of them are right angles. Radii of the same circle From statements 2 R.H.S. axiom 4. OP = OQ 4. Corresponding sides of congruent triangles Proved https://www.geogebra.org/m/wueykd74 Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A A A B P Q B C D Q B C D P P D Q C O O O (i) (ii) (iii) P O Q A C B D Geometry - Circle

Vedanta Excel in Mathematics - Book 9 276 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Converse of Theorem 14 Chords which are equidistant from the centre of a circle are equal. Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Two lines OP and OQ are drawn from the centre O such that OP = OQ in each circle. Step 3: Two chords AB and CD perpendicular to OP and OQ respectively are drawn in each circle. Step 4: The lengths of chords AB and CD are measured and the results are tabulated. Figure AB CD Result (i) AB = CD (ii) AB = CD (iii) AB = CD Conclusion: Chords which are equidistant from the centre of a circle are equal. Theoretical proof Given: O is the centre of a circle. AB and CD are two chords of the circle. OP ⊥ AB, OQ ⊥ CD and OP = OQ. To prove: AB = CD Construction: O, A and O, C are joined. Proof Statements Reasons 1. (i) (ii) (iii) (iv) In rt. ∠ed ∆s OAP and OCQ ∠OPA = ∠OQC (R) OA = OC (H) OP = OQ (S) ∴ ∆OAP ≅ ∆OCQ 1. (i) (ii) (iii) (iv) Both of them are right angle Radii of the same circle Given R.H.S. axiom 2. AP = CQ 2. Corresponding sides of congruent triangles 3. 2 AB = 2 CD i.e. AB = CD 3. OP ⊥ AB and OP bisects AB; OQ ⊥ CD and OQ bisects CD. Proved A A A C B D B B D D O O O C C (i) (ii) (iii) A B D P O Q C

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 277 Vedanta Excel in Mathematics - Book 9 Geometry - Circle Example 1: The radius of a circle is 10 cm and the length of a chord of the circle is 16 cm. Find the distance of the chord from the centre of the circle. Solution: Let O be the centre of the circle. AB be the chord of the circle such that AB = 16 cm. OP be the distance between the centre and the chord. Here, OP ⊥ AB and OP bisects AB. ∴ AP = 1 2 AB = 1 2 × 16 cm = 8 cm Also, radius OA = 10 cm. Now, in rt. ∠ed ∆OAP, OP = OA2 – AP2 = 102 – 82 = 100 – 64 = 36 = 6 cm So, the required distance of the chord from the centre is 6 cm. Example 2: In the given figure O is the centre of the circle. If OA ⊥ MN, MN = 6 cm and AX = 1 cm, find the length of diameter of the circle. Solution: Here, O is the centre of the circle, OA ⊥ MN. MN = 6 cm and AX = 1 cm ∴ AM = 1 2 MN = 1 2 × 6 cm = 3 cm Let, OA = x, then OX = OM = (x + 1) cm. Now, In rt. ∠ed ∆OAM, OM2 = OA2 + AM2 or, (x + 1)2 = x2 + 32 or, x2 + 2x + 1 = x2 + 9 or, 2x = 8 ∴ x = 4 ∴ The length radius OM (r) = (x + 1) cm = (4 + 1) cm = 5 cm and diameter (d) = 2r = 2 × 5 cm = 10 cm. Example 3: In the adjoining figure AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and AB // CD. If the distance between AB and CD is 3 cm, find the radius of the circle. Solution: OP ⊥ CD is drawn and it is produced to meet AB at Q. ∴ OQ is also perpendicular to AB ( AB // CD) Let the radius of the circle OA = OC = r and OP = x cm. So, OQ = (x + 3) cm Also, CP = 1 2 of CD = 1 2 × 12 cm = 6 cm AQ = 1 2 of AB = 1 2 × 6 cm = 3 cm O A C D Q P B O A P B Worked-out examples O M X A N

Vedanta Excel in Mathematics - Book 9 278 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle In rt. ∠ed ∆OQA, OA2 = OQ2 + AQ2 = (x + 3)2 + 32 = x2 + 6x + 9 + 9 = x2 + 6x + 18 In rt. ∠ed ∆OPC, OC2 = OP2 + CP2 = x2 + 62 = x2 + 36 As, OA = OC, OA2 = OC2 ∴ x2 + 6x + 18 = x2 + 36 or, 6x = 18 or, x = 3 ∴ OC2 = r2 = x2 + 36 r2 = 32 + 36 r = 45 = 6.71 cm So, the radius of the circle is 6.71 cm. Example 4: If a line PQ intersects two concentric circles at the points A, B, C, and D as shown in the figure, prove that AB = CD. Solution: Given: O is the centre of two concentric circles. Line PQ intersects two concentric circles at A, B, C, and D. To prove: AB = CD Construction: OR ⊥ PQ is drawn. Proof: Statements Reasons 1. BR = RC 1. OR ⊥ BC and OR bisects BC. 2. AR = RD 2. OR ⊥ AD and OR bisects AD. 3. AR – BR = RD – RC 3. Subtracting (1) from (2) 4. AB = CD 4. AR = AB + BR and RD = RC + CD Proved Example 5: In the given figure, O is the centre of the circle. Two equal chords AB and CD intersect at E. Prove that OE is the bisector of ∠BED. Solution: Given: O is the centre of the circle. Chord AB = chord CD and they intersect at E. To prove: OE is the bisector of ∠BED. Construction: OP ⊥ AB and OQ ⊥ CD are drawn. P Q R A B C D O Q O B C D A P E

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 279 Vedanta Excel in Mathematics - Book 9 Geometry - Circle Proof Statements Reasons 1. OP = OQ 1. Equal chords are equidistant from the centre of circle. 2. O lies in the locus of the bisector of ∠BED. 2. O is equidistant from the arms of ∠BED. 3. OE is the bisector of ∠BED 3. From statement (2) Proved Example 6: In the given figure, equal chords AB and CD of a circle with centre O intersect each others at right angle at K. If M and N are the mid-points of AB and CD respectively, prove that MONK is a square. Solution: Given: (i) O is the centre of circle (ii) Equal chords AB and CD intersect at right angle at K (iii) M and N are the mid-points of AB and CD respectively. To prove: MONK is a square. Proof Statements Reasons 1. OM ⊥ AB i.e. ∠OMK = 90° 1. OM joins the centre O and mid-point M of chord AB. 2. ON ⊥ CD i.e. ∠ONK = 90° 2. ON joins the centre O and mid-point N of chord CD. 3. ∠MKN = 90° 3. Given 4. ∠MON = 90° 4. Remaining angle of quadrilateral MONK 5. OM = ON 5. Equal chords of a circle are equidistant from the centre. 6. MONK is a square. 6. From statements (1), (2), (3), (4) and (5). Proved Example 7: In the given figure, A and B are the centres of two intersecting circles. If CD intersects and AB perpendicularly at P. Prove that (i) CM = DN (ii) CN = DM. Solution: Given: (i) A and B are the centres of two intersecting circles. (ii) CD ⊥ AB at P To prove: (i) CM = DN (ii) CN = DM Proof: Statements Reasons 1. PM = PN 1. AP ⊥ MN and AP bisects MN 2. PC = PD 2. BP ⊥ CD and BP bisects CD 3. PC – PM = PD – PN 3. Subtracting (1) from (2) 4. CM = DN 4. Remaining parts of PC = PD 5. PC + PN = PD + PM 5. Adding PC = PD and PN = PM 6. CN = DM 6. From (5) by whole part asiom Proved O D M K C B N A A P B D M C N

Vedanta Excel in Mathematics - Book 9 280 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur B O D X A M C N Example 8: In the adjoining figure, two equal chords AB and CD of a circle with centre O meet at an external point X. Prove that: (i) BX = DX and (ii) AX =CX. Solution: Given: (i) O is the centre of the circle. (ii) Equal chords AB and CD meet at an external point X. To prove: (i) BX = DX (ii) AX = CX Construction OM ⊥ AB, ON ⊥ CD are drawn and OX is joined. Proof: Statements Reasons 1. (i) (ii) (iii) In DMOX and DNOX ∠OMX = ∠ONX (R) OX = OX (H) OM = ON (S) 1. (i) (ii) (iii) Both are right angles. Common side. Equal chords are equidistant from the centre. 2. DMOX ≅ DNOX 2. By R.H.S. axiom 3. MX = NX 3. Corresponding sides of congruent triangles 4. MB = 1 2 AB and ND = 1 2 CD 4. OM ⊥ AB and ON ⊥ CD 5. MB = ND 5. AB = CD (Given) and from statement (4) 6. MX – MB = NX – ND 6. Subtracting (5) from (3) 7. BX = DX 7. Remaining parts of MX and NX. 8. BX + AB = DX + CD 8. Adding equal chords AB and CD on both sides of (7) 9. AX = CX 9. By whole part axiom Proved Example 9: In the figure alongside, two circles with centres A and B intersect at M and N. The common chord MN intersects AB at P. Prove that AB is perpendicular bisector of MN. Solution: Given: (i) A and B are centres of two intersecting circles. (ii) The common chord MN intersecting AB at P. To prove: AB is perpendicular bisector of MN i.e. MP = NP and AB ⊥ MN. Construction A and B are joined to M and N. A B P M N Geometry - Circle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 281 Vedanta Excel in Mathematics - Book 9 Geometry - Circle Proof: Statements Reasons 1. (i) (ii) (iii) In DAMB and DANB AM = AN (S) BM = BN (S) AB = AB (S) 1. (i) (ii) (iii) Both are right angles. Common side. Common side. 2. DAMB ≅ DANB 2. By S.S.S. axiom 3. ∠MAB = ∠NAB i.e. ∠MAP = ∠NAP 3. Corresponding angles of congruent triangles 4. (i) (ii) (iii) DAMP and DANP AM = AN (S) ∠MAP = ∠NAP (A) AP = AP (S) 4. (i) (ii) (iii) Radii of the same circle From statement (3) Common side 5. DAMP ≅ DANP 5. S.A.S. axiom 6. MP = NP 5. Corresponding sides of congruent triangles 7. ∠APM = ∠APN 6. Corresponding angles of congruent triangles 8. ∠APM = ∠APN = 90° 7. Equal angles in linear pair 9. AP ⊥ MN i.e. AB ⊥ MN 9. From the statement (8) Proved Example 10: Anjali tied her three pets cat, dog and rabbit to a stake by three ropes of 5 feet each. They were moving in a circular path always keeping the ropes tight. Once, the dog was equidistant from cat and rabbit and the distance between the dog and cat was 6 feet. Find the distance between the cat and rabbit at the same time. Solution: Let C, D and R be the position of cat, dog and rabbit respectively and O be the centre of circular path. ∴ OC = OD = OR = 5 ft, CD = RD = 6 ft. Since OC = OR and CD = RD. So, OCDR is a kite in which the diagonal OD bisects the diagonal CR at M at a right angle. Let OM = x ft. then MD = OD – OM = (5 – x) ft. Now, In rt. ∠ed DOCM, CM2 = OC2 – OM2 = 52 – x2 = 25 – x2 ... (i) In rt. ∠ed DCMD, CM2 = CD2 – MD2 = 62 – (5 – x)2 = 36 – (25 – 10x + x2 ) = 11 + 10x – x2 ... (ii) O C R 5 ft. 6 ft. 6 ft. D M

Vedanta Excel in Mathematics - Book 9 282 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur From (i) and (ii), we get, 25 – x2 = 11 + 10x – x2 or, 14 = 10x ∴ x = 1.4 Also from (i); CM2 = 25 – (1.4)2 = 23.04∴ CM = 4.8 ft. ∴ CR = 2 × CM = 2 × 4.8 ft = 9.6 ft Hence, the distance between the cat and the rat was 9.6 ft. EXERCISE 16.1 General section 1. Name the shaded portion in each figure. a) b) 2. Write down the special type of DAOB in each figure with appropriate reasons. a) b) c) O C A B O P Q 3. a) In the given figure, O is the centre of the circle and AB is the chord. If OP ⊥ AB, write the relation between AP and PB. B O P A b) In the adjoining figure, O is the centre of the circle and PQ is its chord. If PX = XQ, write the relation between OX and PQ. c) O is the centre of the given circle. If OP ⊥ AB, OQ ⊥ CD and OP = OQ, write the relation between AB and CD. O B A O A B Q O X P A O B D C P Q O A C B Geometry - Circle

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 283 Vedanta Excel in Mathematics - Book 9 4. a) In the adjoining figure, the radius of the circle OX = 13 cm and the length of a chord XY = 10 cm. (i) Find the length of XP. (ii) Find the distance of the chord from the centre of the circle. b) In the adjoining figure, O is the centre of a circle with radius 5 cm. If OP ⊥ MN and OP = 3 cm, answer the following questions. (i) What is the length of MP? (ii) What is the length of chord MN? c) Find the length of a chord which is at a distance of 9 cm from the centre of the circle of diameter 30 cm. d) In the given figure, O is the centre of a circle and CD is a diameter. If OE ⊥ AB, CD = 20 cm and AB = 16 cm, answer the following questions. (i) What is the length of OA. (ii) What is the length of AE. (iii) Find the length of ED. 5. a) In the figure alongside, O is the centre of two concentric circles. If AP = 5 cm and CD = 6 cm, answer the following questions. (i) Find the length of CP. (ii) Find the length of AC. b) In the adjoining figure, O is the centre of both circle. If PX = 3 cm and AQ = 8 cm, answer the following questions. (i) What is the length of PA? (ii) What is the length of XA? (iii) What is the length of XY? O X Y P O M N P O C A B D E 6. a) In the figure alongside, O is the centre of a circle. AB = 20 cm, CD = 16 cm and AB // CD. Find the distance between AB and CD. b) In the given figure, O is the centre of a circle. AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively. If the radius of the circle is 10 cm, find the distance between the chords. A B O C D d) O is the centre of the given circle. If OX ⊥ PQ, OY ⊥ RS and PQ = RS, write the relation between OX and OY. Y O X Q R S P Geometry - Circle Creative section - A O A C P D B O P X A Y Q A B C D O

Vedanta Excel in Mathematics - Book 9 284 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) In the adjoining figure, O is the centre of a circle, PQ and RS are two equal and parallel chords. If the radius of the circle is 5 cm and the distance between the chords is 8 cm, find the length of the chords. 7. a) In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords, if they are: (i) on the same side of the centre (ii) on the opposite side of the centre. b) Two chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre of the circle. If AB = 1.4 cm and CD = 4 cm , find the radius of the circle. c) AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle. 8. a) In the adjoining figure, O is the centre of the circle and AB is a chord. If OP ⊥ AB, prove that AP = BP. b) In the figure given alongside, O is the centre of the circle and M is the mid-point of the chord XY, prove that OM ⊥ XY. c) In the given figure, O is the centre of the circle and PQ and RS are two chords. If PQ = RS, OX ⊥ PQ and OY ⊥ RS, show that OX = OY. d) In the given figure, O is the centre of the circle, KL and MN are two chords. If OA ⊥ KL, OB ⊥ MN and OA = OB, prove that KL = MN. Q S O P R P B O A Y M O X Q X Y O P R S K A B O L N M Geometry - Circle Creative section - B 9. a) In the given figure, O is the centre of the circle, chords MN and RS are intersecting at P. If OP is the bisector of ∠MPR, prove that MN = RS. O M N R P S

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 285 Vedanta Excel in Mathematics - Book 9 Geometry - Circle b) In the figure alongside, PQ and RS are two chords intersecting at T in a circle with centre O. If OT is the bisector of ∠PTR, prove that (i) PT = RT (ii) ST = TQ c) In the given figure, O is the centre of the circle. Two equal chords AB and CD intersect each other at E. Prove that (i) AE = CE (ii) BE = DE d) In the figure, L and M are the mid-points of two equal chords AB and CD of a circle with centre O. Prove that (i) ∠OLM = ∠OML (ii) ∠ALM = ∠CML O P R Q T S O D C B E A O L C Q B D A M 10. a) In the adjoining figure, AB is the diameter of a circle with centre O. If chord CD // AB, prove that ∠AOC = ∠BOD. O A B C D b) In the given figure, equal chords PQ and RS of a circle with centre O intersect each other at right angle at A. If M and N are the mid-points of PQ and RS respectively, prove that OMAN is a square. c) In the adjoining figure, two chords WX and WY are equally inclined to the diameter at their point of intersection. Prove that the chords are equal. S P Q A N M R O W Z X Y O 11. a) Two equal chords AB and CD of a circle with centre O are produced to meet at E, as shown in the given figure. Prove that BE = DE and AE = CE. D A B E C O D A B P C X Y O b) In the given figure, O is the centre of circle ABCD. If OY ⊥ PC, OX ⊥ PB and OX = OY, prove that PB = PC.

Vedanta Excel in Mathematics - Book 9 286 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) In the given figure, ABC is a triangle in which AB = AC. Also a circle passing through B and C intersects the sides AB and AC at the points D and E respectively. Prove that AD = AE. O D E B C A Geometry - Circle 12. a) In the given figure, two circles with centres P and Q intersect at A and B. Prove that the line joining the two centres of the circles is the perpendicular bisector of the common chord. b) In the given figure, P and Q be the centres of two intersecting circles and AB // PQ. Prove that AB = 2 PQ. c) In the given figure, two circles with centres P and Q are intersecting at A and B. If MN is parallel to common chord AB, prove that (i) MC = ND (ii) MD = NC 13. a) Five students Amrit, Bibika, Chandani, Dipesh, and Elina are playing in a circular meadow. Amrit is at the centre, Bibika and Dipesh are inside the boundary line. Similarly, Chandani and Elina are on the boundary of the meadow. If Amrit, Bibika, Chandani, and Dipesh form a rectangle and the distance between Bibika and Dipesh is 10 m, find the distance between Amrit and Elina. b) The diameter of a circular ground with centre at O is 200 m. Two vertical poles P and Q are fixed at the two points in the circumference of the ground. Find the length of a rope required to tie the poles tightly at a distance of 60 m from the centre of the ground. c) Three students i.e., Pooja, Shaswat, and Triptee are playing a game by standing on the circumference of a circle of radius 25 feet drawn in a park. Pooja throws a ball to Shaswat and Shaswat to Triptee and Triptee to Pooja. What is the distance between Pooja and Triptee when the distance between Pooja and Shaswat and the distance between Shaswat and Triptee is 30 feet each? C A E D 10m B P O T S A B P Q A B X P Q A C B M N P Q D

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 287 Vedanta Excel in Mathematics - Book 9 Project work and activity section 14. a) Draw a large circle in a chart paper. Label its parts and paste in ‘Math Corner’ of your classroom. b) Draw three circles and draw the chords AB and PQ of your own choice. By drawing the perpendicular bisectors of the chords, find the centre of circle in each circle. c) A ground is in the shape of a circle. You need to fix a pole at it’s centre. How do you find the centre of this ground? Write a report to show your process by steps and discuss in the class. Is your process accepted by your friends and your teacher? d) Without taking any measurement, draw two equal intersecting circles with a pencil compass. Join the point of intersection of these two circles and also join their centres. Prove that the common chord is the perpendicular bisector of the line joining the centres of the circle. Now, verify it by the actual measurements. OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. Which of the following statements is NOT true in any circle? (A) Every chord is a diameter. (B) Every diameter is a chord. (C) All radii are equal. (D) All diameters are equal. 2. The radius of the circle is …… the diameter. (A) equal to B) two times (C) three times (D) half of 3. The region of circle enclosed between two radii and the corresponding arc is (A) segment of circle (B) sector of circle (C) semi-circle (D) circle itself 4. The line joining the mid-point of a chord and the centre of circle is … to the chord. (A) equal (B) perpendicular (C) parallel (D) inclined 5. The perpendicular drawn from the centre of a circle to a chord …. the chord. (A) intersects (B) bisects (C) trisects (D) none 6. The equal chords of a circle are (A) equidistance from the centre. (B) equidistance from the circumference. (C) not equidistance from the centre. (D) parallel to each other. 7. The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is (A) 23 cm (B) 24 cm (C) 12 cm (D) 3 cm 8. The length of a chord which is at a distance of 6 cm from the centre of the circle of diameter 20 cm is (A) 8 cm (B) 16 cm (C) 4 cm (D) 14 cm 9. In the circle with centre O, the length of each chord AB and chord CD is 16 cm, the chord AB is at a distance of 6 cm, what is the distance of the chord CD from the centre O? (A) 22 cm (B) 10 cm (C) 12 cm (D) 6 cm 10. The chords PQ and RS of a circle are equidistance from the centre. If the length of the chord PQ is 12 cm, what is the length of the chord RS? (A) 3 cm (B) 6 cm (C) 12 cm (D) 24 cm Geometry - Circle

Vedanta Excel in Mathematics - Book 9 288 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Assessment -V 1. DABC is a given triangle. Answer the following questions. (a) Write down the relationship between the sum of any two sides of the triangle and its third side. (b) Verify experimentally that the relationship between the sum of any two sides and the third side. (Two triangles ABC of different shapes and sizes are necessary) (c) Can a triangle be constructed with the sides 5.5 cm, 8.4 cm and 4.6 cm? Give reason 2. In an isosceles triangle ABC; bisector of vertical ∠BAC is drawn to a point D on its base BC. Answer the following questions. (a) What is the relation between AD and BC? (b) If the length of BD is 6 cm, what is the length of CD? (c) Draw two triangles ABC and verify experimentally that the bisector of the vertical angle of an isosceles triangle is perpendicular bisector of the base. 3. The figure alongside is a parallelogram PQRS. Answer the following questions. (a) Which of the following is not the property of parallelogram? (i) The opposite sides are equal and parallel. (ii) The opposite angles are equal. (iii) The diagonals bisect to each other. (iv) The diagonals are equal. (b) If PQ = (3x – 4) cm and RS = (x + 6) cm, find the value of x. (c) Construct a rhombus PQRS having diagonal QS = 6.5 cm and AB = 4.5 cm. 4. In the circle given alongside; O is the centre, PQ is a chord and OA ⊥ PQ. (a) Write down the relation between PA and AQ. (b) If OA = 6 cm and PQ = 16 cm, find the length of the radius OP. (c) If B is a point lying in the circumference of the circle, what is the relation between OB and OP? Write with reason. 5. In the circle given alongside; O is the centre, A is the middle point of the chord PQ. If OB = 10 cm and AB = 4 cm, answer the following questions. (a) Write down the relation between OA and PQ. (b) Find the length of the chord PQ. (c) If another chord RS which is at a distance of 6 cm from the centre O is drawn, what is the length of the chord RS? P Q R S A Q O P Q B A O P

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 289 Vedanta Excel in Mathematics - Book 9 17.1 Statistics –Looking back Classwork-Exercise 1. The bar graph given below shows the monthly expenditure of a family in the last 6 months. Answer the following questions. Months 0 2000 2500 3000 3500 4000 4500 Jan Feb Mar Apr May Jun Expenditure (in Rs) (i) In which month the expenditure was minimum? (ii) In which month the expenditure was maximum? (iii) In how many months was the expenditure more than Rs 3,000? (iv) If the income of the family is Rs 10,000 every month, express the expenditures of every month in percent. 2. The pie chart alongside shows the percentage of types of transportation used by 400 students to come to school. (i) What is the total number of students in the school? (ii) How many students come to school by bus? (iii) How many students do not come to school by van? 17.2 Statistics The term ‘statistics’ was derived from the Latin word ‘Status’ or Italian word ‘Statista’ or German word ‘Statistiks’ all of which means the political state. In those days, statistics was used to collect the information relating to the population, military strength and income of the state etc. In the age of information technology, statics has a wide range of applications. Different departments and authorities require various facts and figures to frame policies and guidelines in order to function smoothly. Walking 40% Van 10% Bus 30% Bicycle Unit 17 Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 290 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistical information helps to understand the economic problems and formulation of economic policies. In social science, statistics is used in the field of demography for studying mortality, fertility, population growth rate, and so on. In science and technology, there is a regular use of statistical tools for collecting, presenting and analysing the observed data for various researches. The present time is the time of information and communication. In social, economical and technical area, we frequently require information in the form of numerical figures. The information collected in the form of numerical figures are called data. Facts to remember Statistics is a branch of mathematics dealing with collection, organisation, analysis, interpretation, and presentation of data. Statistics in plural form refer to data, whereas in the singular form it refers a subject. Sir Ronald Aylmer Fisher (17 February 1890 – 29 July 1962) was a British statistician, mathematician and biologist. Also, he was known as the Father of Modern Statistics and Experimental Design. Fisher did experimental agricultural research, which saved millions from starvation. He was awarded by the Linnean Society of London’s prestigious Darwin-Wallace Medal in 1958. 17.3 Types of data (i) Primary data The data collected by the investigator him/herself for definite purposes are called primary data. These data are highly reliable and relevant. (ii) Secondary data The data collected by someone other than the user oneself are called secondary data. (iii) Raw data The data obtained in original form are called raw data. (iv) Array data The data arranged in ascending or descending order are called array data. 17.4 Frequency tables Let’s consider the following marks obtained by 10 students in a unit test in mathematics. 18, 10, 15, 15, 12, 12, 10, 15, 12, 12 → Raw data 10, 10, 12, 12, 12, 12, 15, 15, 15, 18 → Array data Here, the marks 10 are repeated 2 times. So, 2 is the frequency of 10. The marks 12 are repeated 4 times. So, 4 is the frequency of 12. The marks 15 are repeated 3 times. So, 3 is the frequency of 15. The mark 18 is repeated only one time. So, 1 is the frequency of 18. Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 291 Vedanta Excel in Mathematics - Book 9 Thus, a frequency is the number of times a value occurs. Data and their frequencies can be presented in a table called frequency table. Marks Tally marks Frequency 10 12 15 18 || |||| ||| | 2 4 3 1 Tallying is a system of recording and counting results using diagonal lines grouped in fives. Each time five is reached, a horizontal line is drawn through the tally marks to make a group of five. The next line starts a new group. For example, the tally marks of frequency 5 is ||||, frequency 6 is |||| |, 7 is |||| ||, and so on. 17.5 Grouped and continuous data Let’s consider the following marks obtained by 20 students in a test of mathematics. 25, 38, 21, 18, 7, 28, 49, 17, 27, 36 26, 45, 32, 16, 24, 39, 20, 33, 40, 35 The above mentioned data are called individual or discrete data. Another way of organising data is to present them in a grouped form. For grouping the given data, we should first see the smallest value and the largest value. We have to divide the data into an appropriate class–interval. The numbers of values falling within each class– interval give the frequency. For example, Marks Tally marks Frequency 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 | ||| |||| || |||| | ||| 1 3 7 6 3 Total 20 In the above series, 7 is the smallest value and 49 is the largest value. So, the data are grouped into the interval of 0 – 10, 10 – 20, ... 40 – 50, so that the smallest and the largest values should fall in the lowest and the highest class–interval respectively. Let’s consider a class–interval 10 – 20. Here, 10 is called the lower limit and 20 is the upper limit of the class–interval. The difference between the two limit is called the length or height of each class–interval. For example, in 10 – 20 the length of the class–interval is 10. Again, let’s take class–intervals, 0 – 10, 10 – 20, 20 – 30, … Here, the upper limit of a pervious class–interval has repeated as the lower limit of the consecutive next class–interval. Such an arrangement of data is known as grouped and continuous data. Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 292 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Facts to remember 1. In individual series the items are listed singly after observation. 2. In discrete series, the discrete variables along with the corresponding frequencies are tabulated. 3. In continuous series, the continuous variables (class intervals) along with the corresponding frequencies are tabulated. 17.6 Cumulative frequency table The word ‘cumulative’ is related to the word ‘accumulated’, which means to ‘pile up’. The frequency distribution in which the frequencies are accumulated is called cumulative frequency distribution. For example, (i) The table given below shows the heights (in cm) of 40 students of class 9 and the corresponding cumulative frequency table. Height (in cm) Number of students (f) Cumulative frequency (c.f.) 110 118 120 124 130 143 7 4 6 10 8 5 7 7 + 4 = 11 11 + 6 = 17 17 + 10 = 27 27 + 8 = 35 35 + 5 = 40 Total 40 (ii) The table given below shows the marks obtained by 20 students in a mathematics test of full marks 50 and the corresponding cumulative frequency of each class– interval. Marks Frequency (f) Cumulative frequency (c.f) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 1 3 7 6 3 1 1 + 3 = 4 4 + 7 = 11 11 + 6 = 17 17 + 3 = 20 Total 20 There are two types of cumulative frequency distribution. a) Less than cumulative frequency distribution. b) More than cumulative frequency distribution. 1 student obtained marks 0 to less than 10. 4 students obtained marks 0 to less than 20. 11 students obtained marks 0 to less than 30. 17 students obtained marks 0 to less than 40. 20 students obtained marks 0 to less than 50. Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 293 Vedanta Excel in Mathematics - Book 9 Less than cumulative frequency distribution Less than cumulative frequency is obtained by adding successively the frequencies of all the previous classes including the class against which it is written. The cumulate starts from the lowest to the highest-class intervals. For example: Simple frequency distribution Less than cumulative frequency distribution Ages (in years) Number of students (f) Ages (in years) Less than c.f. 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 25 60 48 75 62 Less than 4 Less than 8 Less than 12 Less than 16 Less than 20 25 25 + 60 = 85 85 + 48 = 133 133 + 75 = 208 208 + 62 = 270 Total 270 More than cumulative frequency distribution More than cumulative frequency is obtained by adding successively the frequencies starting from the highest to lowest class intervals. In more than, cumulative frequency distribution, each value of c.f. refers to the total number of observations of more than or equal to the corresponding value in discrete series or lower limit of corresponding class in continuous series. For example. Simple frequency distribution More than cumulative frequency distribution Ages (in years) Number of students (f) Ages (in years) More than c.f. 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 25 60 48 75 62 0 or more than 0 4 or more than 4 8 or more than 8 12 or more than 12 16 or more than 16 245 + 25 = 270 185 + 60 = 245 137 + 48 = 185 62 + 75 = 137 62 Total 270 Worked-out Examples Example 1: The monthly salary (in Rs) of 30 workers of a company is given below. 15000, 20000, 8000, 25000, 15000, 10000, 30000, 20000, 20000, 25000, 20000, 30000, 15000, 25000, 30000, 10000, 15000, 25000, 20000, 15000, 8000, 25000, 15000, 10000, 30000, 20000, 25000, 15000, 20000, 15000 a) Present this individual series into a discrete series with tally bars. Also, construct a cumulative frequency table. b) How many workers have monthly salary Rs 25,000 or less than Rs 25,000? Solution: Here, minimum salary = Rs 8,000 and the maximum salary = Rs 30,000 a) Presenting the given data into discrete series with tally bars. Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 294 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Monthly salary (in Rs) Tally bars Number of workers (f) Cumulative frequency (c.f.) 8000 10000 15000 20000 25000 30000 || ||| ||||||| |||||| ||||| |||| 2 3 8 7 6 4 2 2 + 3 = 5 5 + 8 = 13 13 + 7 = 20 20 + 6 = 26 26 + 4 = 30 Total 30 b) From the above cumulative frequency table, 26 workers have monthly salary Rs 25,000 or less than Rs 25,000. Example 3: During the medical check-up of 50 patients, their weights were recorded as follows: Weight in kg Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Number of patients 6 14 25 33 38 50 Form the above table, construct the more than cumulative frequency distribution table. Solution: Frequency distribution table More than cumulative frequency table Weight in kg Number of patients (f) Weight in kg More than c.f. 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 6 14 - 6 = 8 25 - 14=11 33 - 25=8 38 - 33=5 50 - 38=12 0 or more than 0 (≥ 0) 10 or more than 10 (≥ 10) 20 or more than 20 (≥ 20) 30 or more than 30 (≥ 30) 40 or more than 40 (≥ 40) 50 or more than 50 (≥ 50) 44 + 6=50 36 + 8=44 25 + 11=36 17 + 8=25 12 + 5=17 12 EXERCISE 17.1 General section 1. a) Write a few paragraphs about the application of statistics. b) Define primary, secondary, raw, and array data. c) What do you mean by frequency of an observation in data? d) Define cumulative frequency. 2. a) The marks obtained by 50 students of class IX in Mathematics are shown in the table given below. Construct a cumulative frequency table to represent the data. Marks obtained 30 40 50 60 70 80 90 Number of students 4 7 10 12 8 6 3 Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 295 Vedanta Excel in Mathematics - Book 9 b) The daily wages of 45 workers are given in the table below. Construct a cumulative frequency table to represent the data. Daily wages (in Rs) 450 500 550 600 650 700 750 Number of workers 6 4 8 10 9 3 5 3. a) The daily sales of 20 shops are recorded in the following table. Sales (in Rs) 0-1000 1000-2000 2000-3000 3000-4000 4000-5000 Number of shops 2 5 6 4 3 (i) Construct a less than cumulative frequency distribution table. (ii) Construct a more than cumulative frequency distribution table. b) The values of diastolic blood pressure of 90 people of age group 20 – 40 are shown in the table below. Blood pressure (mmHg) 70-75 75-80 80-85 85-90 90-95 Number of people 20 25 18 17 10 (i) Construct a less than cumulative frequency distribution table. (ii) Construct a more than cumulative frequency distribution table. 4. a) From the following frequency and less than cumulative frequency distribution table, find the values of p, q and r. Class interval 0-10 10-20 20-30 30-40 40-50 Frequency 5 p 10 6 r Cumulative frequency 5 9 19 q 30 b) Find the values of a, b and c from the given frequency and less than cumulative frequency distribution table. Ages (in years) 4-8 8-12 12-16 16-20 20-24 Number of students 10 15 b 25 30 Cumulative frequency 10 a 45 70 c Creative section 5. a) The marks obtained by 20 students in mathematics in a class test is given below. 15, 18, 12, 16, 18, 10, 15, 16, 15, 12, 10, 12, 15, 12, 16, 18, 12, 15, 12, 16 (i) Present this individual data in a frequency distribution table with tally marks. (ii) Construct a cumulative frequency distribution table. (iii) From the cumulative frequency table, find the number of students who obtained 15 or less than 15 marks. b) The amount of money (in Rs) brought by 30 students of class IX for tiffin is given below. 30, 20, 50, 40, 25, 50, 40, 35, 20, 50 40, 50, 35, 25, 30, 40, 50, 25, 40, 30 35, 25, 40, 50, 40, 30, 25, 50, 35, 40 (i) Draw a frequency distribution table with tally bars to represent the data. (ii) Construct the cumulative frequency distribution table. (iii) How many students brought less than or equal to Rs 40? Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 296 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 6. a) The marks obtained by 40 students in mathematics in the final examination of class 9 are given below. 42, 68, 80, 45, 92, 36, 8, 17, 49, 30, 5, 26, 98, 74, 53, 65, 72, 28, 55, 46, 86, 70, 62, 27, 16, 44, 85, 59, 51, 73 66, 78, 38, 81, 97, 77, 69, 45, 33, 67 (i) Make a frequency distribution table with class interval of length 10. (ii) Construct the less than cumulative frequency distribution table. (iii) Construct the more than cumulative frequency distribution table. b) The weights (in kg) of 30 teachers of a school are given below. 71, 53, 60, 50, 64, 74, 59, 67, 61, 58, 65, 70, 73, 51, 63, 66, 70, 50, 60, 70, 56, 62, 52, 67, 59, 77, 64, 78, 57, 57 (i) Make a frequency distribution table with class interval of width 5. (ii) Construct the less than cumulative frequency distribution table. (iii) Construct the more than cumulative frequency distribution table. 7. a) The weight (in kg) of 50 wrestlers are recorded as follows: Weight in kg Below 100 Below 110 Below 120 Below 130 Below 140 Below 150 Number of wrestlers 6 14 25 33 38 50 (i) Construct the frequency distribution table. (ii) Construct the more than cumulative frequency distribution table. b) The following cumulative frequency table shows the marks obtained by 60 students of class 9 in mathematics. Marks obtained ≥ 40 ≥ 50 ≥ 60 ≥ 70 ≥ 80 ≥ 90 Number of students 35 30 24 16 7 4 (i) Construct the frequency distribution table. (ii) Construct the less than cumulative frequency distribution table. Project work and activity section 8. a) Measure the weights of each of your classmates. Show the weights in a frequency distribution table of continuous class with length 5 kg. Then, construct the less than and more than cumulative frequency table. b) Show the marks obtained by at least 20 friends of your class in mathematics in recently conducted exam in a frequency distribution table of continuous class of length 10. Then, construct the less than and more than cumulative frequency table. c) Visit a website and write the value of p up to 50 decimal places. Make the frequency distribution table of the digits on the value from 0 to 9 after decimal places. Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 297 Vedanta Excel in Mathematics - Book 9 17.7 Graphical representation of data We have already discussed to present data in frequency distribution tables. Alternatively, we can also present data graphically. Different types of diagrams are used for this purpose. Here, we shall discuss three types of diagrams: histogram, frequency polygon and ogive. Graphical representation Ungrouped data Grouped data line graph bar graph pictograph pie chart histogram frequency polygon frequency curve cummulative frequency curve 1. Histogram Statistical data can be represented by various types of diagrams, such as, bar diagram, pie-chart, line graph, etc. A histogram is a graphical representation of a continuous frequency distribution. For this purpose, we draw rectangular bars with class intervals bases and corresponding frequencies as heights. Study the following steps to construct a histogram. (i) If the given frequency distribution has inclusive classes, change it into exclusive. (i.e. continuous) classes. (ii) Choose a suitable scale and mark the class intervals on the x-axis and the corresponding frequencies on the Y-axis. (iii) Draw adjacent rectangles on x-axis with class interval as base and the corresponding frequencies as height. Facts to remember 1. Bar graph is used for ungrouped data whereas histogram is used for grouped data. 2. There are gaps between the bars in bar graph but there is no gap between the bars in histogram. Worked-out Examples Example 1: The table given below shows the marks obtained by 36 students in mathematics. Represent the data by drawing a histogram. Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 No. of students 2 3 5 8 10 5 3 Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 10 20 30 40 50 60 70 80 X Y Histogram of marks obtained by 45 students No. of Students Marks In the case of mid-points of class intervals are given, we should first find the class intervals of each mid-point. Example 2: Construct a histogram from the data given in the table below. Solution: The class intervals of each mid-point are given in the table below. Class interval 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 Frequency (f) 4 7 10 8 13 6 2 In the case of inclusive frequency distribution, we should first change it into exclusive (or continuous) class intervals. Mid-point 7.5 12.5 17.5 22.5 27.5 32.5 37.5 Frequency (f) 4 7 10 8 13 6 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 5 10 15 20 25 30 35 40 X Y Marks Histogram of the given data Frequency Statistics (I): Classification and Graphical Representation of Data

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 299 Vedanta Excel in Mathematics - Book 9 Example 3: The table given below shows the heights of 75 plants of a garden. Construct a histogram to represent the data. Solution: The given frequency distribution is inclusive type. So, it should be changed into exclusive (or continuous) class interval with the help of a correction factor. Correction factor = Lower limit of a class interval – Upper limit of previous class interval 2 = 10 – 9 2 = 1 2 = 0.5 Then, the correction factor 0.5 is subtracted from the lower limit and added to the upper limit of each class interval. Height in inch (Inclusive class) Height in inch (Exclusive class) No. of plants 1 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 - 69 0.5 – 9.5 9.5 – 19.5 19.5 – 29.5 29.5 – 39.5 39.5 – 49.5 49.5 – 59.5 59.5 – 69.5 15 6 18 11 14 9 2 Heights (in inch) 1 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 No. of plants 15 6 18 11 14 9 2 No. of plants Heights (in cm) 0.5 9.5 19.5 29.5 39.5 49.5 59.5 69.5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Histogram of heights of 75 plants Statistics (I): Classification and Graphical Representation of Data

Vedanta Excel in Mathematics - Book 9 300 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 17.2 General section 1. a) What is histogram? b) Write the difference between the bar graph and histogram. 2. a) The given histogram shows the distribution of marks obtained by the students of class - IX in mock test of mathematics. Answer the following questions. (i) Find the number of students who obtained more than 70 marks. (ii) Find the number of students who obtained between 50 and 80 marks. (iii) Find the number of students who obtained at most 60 marks. b) The adjoining histogram is drawn against the number of students and their weights. Study it and answer the following question. (i) How many students are there who have the weight less than 30 kg? (ii) In which weight group is there maximum number of students and what is the number of students? (iii) Find the number of students who have the weight at least 30 kg. Creative section 3. a) Draw a histogram to represent the data given below in the table. Age (in years) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 No. of people 6 10 16 20 12 7 3 b) The marks obtained by 70 students in mathematics are given below. Construct a histogram to represent the marks. Marks 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 No. of Students 4 7 12 18 10 8 9 2 c) The daily wages of 60 workers are shown in the table given below. Wages (in Rs) 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 No. of people 10 15 6 17 8 4 Construct a histogram to represent the wages of the workers. d) The values of diastolic blood pressure of 80 people of 20-50 years are shown in the table given below. Blood pressure (mmHg) 70 – 75 75 – 80 80 – 85 85 – 90 90 – 95 95 – 100 No. of people 10 17 23 16 9 5 Construct a histogram to show the above data. No. of students Weight (in kg) 10 20 30 20 25 30 35 40 45 40 50 60 No. of students Marks obtained 5 10 15 40 50 60 70 80 90 20 25 30 Statistics (I): Classification and Graphical Representation of Data