Excel in Mathematics Book 10 Final_CTP (2080)_compressed (1)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 251 Vedanta Excel in Mathematics - Book 10 Geometry - Circle 4. In the given figure, O is the centre of circle and chord AB = chord DC. a) What is the relation between arcs AB and CD? b) If the length of arc AB is 5.6 cm, what is the length of arc CD? c) What is the relation between the chords BC and AD? d) If ∠ABC = 110o , what is the measure of ∠BAD? Creative section-A 5. a) In the given figure, O is the centre of circle. If ∠AOB = 60o , find: (i) the degree measure of minor arc AB. (ii) the degree measure of major arc ACB. b) In the given circle, the central angle ∠XOY = 150o , find: (i) the degree measure of minor arc XY. (ii) the degree measure of major arc XZY. 6. a) O is the centre of circle. If arc AB = arc CD and ∠COD = 80o , find: (i) the degree measure of arc AB. (ii) the measure of ∠AEB. b) In the given circle, O is the centre of circle. If arc PQ = arc QR and ∠POQ = 68o , find: (i) the degree measure of arc QR. (ii) the measure of ∠QSR. 7. Find the sizes of unknown angles in the following circles. a) b) c) d) Creative section-B 8. a) In the figure alongside, AD // BC. Prove that ∠AYC = ∠BXD. b) In the given figure, ∠ APC = ∠ BQD. Prove that AB // CD. D B C A O B C 60° A O Y Z X 150° O D C B E 80° A O R Q 68° P S O R Q S 20° x° P T U A D C x° B 84° E O A P M N Q S R 65° x° A C E D B x 41° ° X Y C A D B C D P A Q B

Vedanta Excel in Mathematics - Book 10 252 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle c) In the figure alongside, chords AB and DC of a circle intersect externally at E. Prove that ∠AED 1 2 (AD – BC ) d) In the given circle, chords DE and FG intersect within the circle at X. Prove that ∠DXF 1 2 (DF + GE ). 9. a) In the given figure, O is the centre of the circle, PQ the diameter and QR = RS . Prove that: (i) PS // OR (ii) Area of DPOR = Area of DSOR b) In the adjoining figure, AB is the diameter. OC and OD are the radii where ∠ BOC = ∠ COD. Prove that: (i) AD // OC (ii) Area of DAOD = Area of DACD c) In the given figure, O is the centre of a circle. C and D are two points on the chord AB. If AC = BD and AM = BN , prove that ∠ACM = ∠BDN. d) In the given figure, two chords AB and CD intersect at right angle at X. Prove that, AD – CA = BD – BC . Project Work and Activity Section 10. a) Take a chart paper and draw a circle as bigger as possible. Mark any four different points on it which make equal arcs. Draw the angles at the centre subtended by the equal arcs. Cut out the sectors and compare the central angles. b) Take a chart paper and draw a circle with a suitable radius. Mark any four different points on it which make equal arcs. Draw the angles at the circumference subtended by the equal arcs. Cut out the inscribed angles and compare them. 13.4 Theorems based on the angles subtended by the same arc Theorem 9 The angle at the centre of a circle is twice the angle at its circumference standing on the same arc. ‘OR’ The inscribed angle of a circle is equal to half of the central angle standing on the same arc. C A B E D D G E F X Q S R P O B D C A O M N A C D B O X C D A B

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 253 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Activity Steps: (i) Draw three circles with centre O and same radius on colourful chart paper and cut them out. (ii) Mark three points A, B and C on the circumference of the first circle. (iii) Keep all three circles of same size one on the other, fold along AC and BC and press to make the crease so that angles at the circumference i.e., ∠ACB is formed. O A B C C M N M N O A B C C P P Q Q O A B C (iv) Join A, O and B, O in each circle. (v) Cut the angles, ∠MCN and ∠PCQ from the first two circles and paste them on the central ∠AOB of the third circle. We observe that ∠AOB = ∠MCN + ∠PCQ = ∠ACB + ∠ACB = 2∠ACB [ ∠MCN = ∠PCQ = ∠ACB] Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: ∠AOB is drawn at the centre and ∠ACB is drawn at the circumference of each circle. Here, ∠AOB and ∠ACB are standing on the same arc AB. Step 3: ∠AOB and ∠ACB are measured and the results are tabulated. Figure ∠AOB ∠ACB Result (i) ∠AOB = 2 ∠ACB (ii) ∠AOB = 2 ∠ACB (iii) ∠AOB = 2 ∠ACB Conclusion: The angle at the centre of a circle is twice the angle at its circumference standing on the same arc. Theoretical proof Given: O is the centre of a circle. ∠AOB is the angle at the centre and ∠ACB is the angle at the circumference of the circle. They are standing on the same arc AB. O O (i) (ii) (iii) B B A A C C A O B C www.geogebra.org/classroom/sfftc6uy Classroom code: SFFT C6UY Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A O D B C www.geogebra.org/classroom/ennzestz Classroom code: ENNZ ESTZ Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 254 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle To prove: ∠AOB = 2 ∠ACB. Construction: C, O are joined and the line is produced to D. Proof Statements Reasons 1. In ∆OAC, ∠OAC = ∠ OCA 1. OA = OC (radii of the same circle), so base angles of isosceles ∆OAC 2. ∠AOD = ∠OAC + ∠OCA 2. Exterior and opposite interior angles of ∆OAC 3. ∠AOD = ∠OCA + ∠OCA = 2 ∠OCA 3. From statement (1) 4. Similarly, ∠BOD = 2 ∠OCB 4. Same as above facts and reasons 5. ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB 5. Adding statements (3) and (4) 6. ∠AOB = 2 (∠OCA + ∠OCB) = 2∠ACB 6. Whole part axiom Proved Thales of Miletus (c. 624 – 546 BCE) was a Greek mathematician and philosopher. Thales is often recognized as the first scientist in Western civilisation: rather than using religion or mythology, he tried to explain natural phenomena using a scientific approach. He is also the first individual in history that has a mathematical discovery named after him: Thales’ theorem. Theorem 10 Thales’ theorem: The angle in the circumference of a semi-circle is one right angle. Theoretical proof Given: O is the centre of a circle and AB is a diameter of the circle. To prove: ∠ACB = 90° Proof Statements Reasons 1. ∠AOB = 180° 1. ∠AOB is a straight angle. 2. ∠ACB = 1 2 ∠AOB 2. Inscribed angle is half of the angle at the centre of a circle standing on the same arc. 3. ∠ACB = 1 2 × 180° = 90° 3. From the statements (1) and (2) Proved Note: A diameter divides a circle into two halves and each half is called a semi-circle. Angle in the semi-circle is always a right angle (90°). A O B C

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 255 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Theorem 11 Angles on the same segment of a circle are equal. ‘OR’ The angles in the circumference of a circle (inscribed angles) standing on the same arc are equal. Activity Steps: (i) Draw a circle with centre O and a suitable radius on a chart paper and cut it out. (ii) Draw the inscribed angles ACB and ADB on the same arc AB and cut the sectors ACB and ADB. O A B C D A A A A B B C B B D C (iii) Arrange the sectors ACB and ADB and observe the size of ∠ACB and ∠ADB. Here, ∠ACB and ∠ADB are exactly fitted to each other. Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Inscribed angles ABC and ADC are drawn in the same segment ABDC and on the same arc AC. Step 3: ∠ABC and ∠ADC are measured and the results are tabulated. Figure ∠ABC ∠ADC Result (i) ∠ABC = ∠ADC (ii) ∠ABC = ∠ADC (iii) ∠ABC = ∠ADC Conclusion: Angles in the same segment of a circle are equal. Activity 1 O (i) (ii) B B A A C C D D O O (iii) B A D C www.geogebra.org/classroom/bbe75uf4 Classroom code: BBE7 5UF4 Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 256 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Theoretical proof Given: O is the centre of a circle. ∠ABC and ∠ADC are the inscribed angles standing on the same arc AC and in the same segment ABDC. To prove: ∠ABC = ∠ADC Construction: O, A and O, C are joined. So, ∠AOC is the angle at the centre of the circle. Proof Statements Reasons 1. ∠ABC = 1 2 ∠AOC 1. Inscribed angle is half of the angle at the centre of a circle standing on the same arc. 2. ∠ADC = 1 2 ∠AOC 2. Same reason as above 3. ∠ABC = ∠ADC 3. From statements (1) and (2) Proved Facts to remember 1. The angle at the centre of a circle is twice the angle at the circumference standing on the same arc. 2. The angle at the semi-circle is always a right angle. 3. The inscribed angles standing on the same arc are equal. 13.5 Theorems based on cyclic quadrilateral Theorem 12 The opposite angles of a cyclic quadrilateral are supplementary. Activity Steps: (i) Draw a circle with centre O and a suitable radius on a chart paper. (ii) Cut out the circle from the chart paper. (iii) Draw a cyclic quadrilateral ABCD on it and cut it out. B C D A 1 1 2 4 3 3 4 2 O B A D C www.geogebra.org/classroom/wknt3j8r Classroom code: WKNT 3J8R Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 257 Vedanta Excel in Mathematics - Book 10 Geometry - Circle (iv) Cut out the angles ∠A, ∠B, ∠C and ∠D of the cyclic quadrilateral ABCD and name them by ∠1, 2, ∠3 and ∠4 respectively. (v) Arrange the opposite angles ∠1 and ∠3, 2 and ∠4 and observe the sum of pair of the angles. We observe that the sum of opposite angles of a cyclic quadrilateral is 180o . Experimental verification Step 1: Three circles with centre O and different radii are drawn. Step 2: Cyclic quadrilaterals ABCD are drawn in each circle. Step 3: ∠A, ∠B, ∠C and ∠D of each cyclic quadrilateral are measured and the results are tabulated. Figure ∠A ∠C ∠A + ∠C ∠B ∠D ∠B + ∠D Result (i) ∠A + ∠C = 180°, ∠B + ∠D = 180° (ii) ∠A + ∠C = 180°, ∠B + ∠D = 180° (iii) ∠A + ∠C = 180°, ∠B + ∠D = 180° Conclusion:The opposite angles of a cyclic quadrilateral are supplementary. Theoretical proof Given: O is the centre of a circle. ABCD is the cyclic quadrilateral. To prove: ∠ABC + ∠ADC = 180° and ∠BAD + ∠BCD = 180° Construction: O, A and O, C are joined. O (i) (ii) B B A A C C D D O O (iii) C D A B www.geogebra.org/classroom/pcxgp2rb Classroom code: PCXG P2RB Vedanta ICT Corner Please! Scan this QR code or browse the link given below: O B A C D

Vedanta Excel in Mathematics - Book 10 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Proof Statements Reasons 1. ∠ABC = 1 2 Ref ∠AOC 1. Inscribed angle is half of the angle at the centre of a circle standing on the same arc. 2. ∠ADC = 1 2 obt. ∠AOC 2. Same reasons as above 3. ∠ABC + ∠ADC = 1 2 (Ref. ∠AOC + obt. ∠AOC) 3. Adding the statements (1) and (2) 4. ∠ABC + ∠ADC = 1 2 × 360° i.e. ∠ABC + ∠ADC = 180° 4. The sum of reflex and obtuse angles AOC forms a complete turn. 5. Similarly, ∠BAD + ∠BCD = 180° 5. By joining O, B and O, D and following the same facts and reasons. Proved Theorem 13 If a side of a cyclic quadrilateral is extended to a point, then the exterior angle so formed is equal to its opposite interior angle. Theoretical proof Given: In a cyclic quadrilateral ABCD, side BC is extended to a point E. To prove: ∠DCE = ∠BAD Proof: S.N. Statements S.N. Reasons 1. ∠DCE + ∠BCD = 180o 1. Being linear pair 2. ∠BAD + ∠BCD = 180o 2. The sum of opposite angles of cyclic quadrilateral is 180o 3. ∠DCE + ∠BCD = ∠BAD + ∠BCD 3. From statements (1) and (2). 4. ∠DCE = ∠BAD 4. From statement (3) Proved Facts to remember 1. The opposite angles of a cyclic quadrilateral are supplementary. 2. If the opposite angles of a quadrilateral are supplementary, then the quadrilateral becomes a cyclic. 3. The exterior angle of a cyclic quadrilateral is equal to its opposite interior angle. 4. If the exterior angle and opposite interior angle of a quadrilateral are equal, it is a cyclic quadrilateral. A B C D E www.geogebra.org/classroom/q9rg782h Classroom code: Q9RG 782H Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 259 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Example 1: Find the sizes of unknown angles in the following figures: Solution: a) Ref. ∠POR = 2 ∠PQR [Angle at the centre of a circle is twice the inscribed = 2 × 110° = 220° angle standing on the same arc] Now, y = 360° – 220° = 140°. [Sum of y° and Ref. ∠POR is an angle of a complete turn] b) Let’s join B, O. (i) ∠ABO = ∠OAB = 30° [OA = OB and base angle of ∆OAB] (ii) ∠CBO = ∠OCB = 25° [OC = OB and base angle of ∆OBC] (iii) ∠ABO + ∠CBO = 30° + 25° = 55° or, ∠ABC = 55° (iv) Obt. ∠AOC = 2 ∠ABC [Angle at the centre of a circle is twice the inscribed = 2 × 55° = 110° angle, standing on the same arc] (v) Ref. ∠AOC = x = 360° – 110° = 250° c) x + 110° = 180° [sum of a linear pair] or, x = 180° – 110° = 70° Now, z + 70° = 180° [Opposite angles of cyclic quadrilateral are supplementary.] or, z = 180° – 70° = 110° Example 2: Find the sizes of unknown angles in the following figures: Solution: a) ∠BDC = ∠BAC = 40° [Inscribed angles standing on the same arc] Now,∠BDC + ∠BCD + ∠CBD = 180° [Sum of the interior angles of ∆BCD] or, 40° + x + 55° = 180° or, x = 180° – 95° = 85° b) ∠OAC = ∠OCA = 20° [OA = OC and base angles of isosceles ∆OAC] Now, x + 20° + 20° = 180° [Sum of the interior angles of ∆OAC] O y Q R P 110° x C B A 30° 25° O z° 110° a) b) c) O C B A 30° 25° O z° x 110° O C D B A 55° x 40° O C B A 20° x y D 30° O C B E x y A a) b) c)

Vedanta Excel in Mathematics - Book 10 260 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle or, x = 180° – 40° = 140° Again, y = 1 2 × 140°= 70° [Inscribed angle is half of the angle at the centre of a circle standing on the same arc] c) y = ∠CBE = 30° [Exterior angle of a cyclic quadrilateral ABCD is equal to the opposite interior angle] Now, x = 2y [Inscribed angle and angle at the centre of a = 2 × 30° = 60°. circle standing on the same base] Example 3: In the given figure, two circles intersect at A and B. Through A two straight lines PAQ and XAY are drawn terminated by the circumference. Prove that ∠PBX = ∠QBY Solution: Given: A and B are the points of intersection of two circles. Two straight lines PAQ and XAY are passing through A. To prove: ∠PBX = ∠QBY Proof Statements Reasons 1. ∠PBX = ∠PAX 1. Circumference angles standing on the same XP 2. ∠PAX = ∠QAY 2. Vertically opposite angles 3. ∠QAY = ∠QBY 3. Circumference angles standing on the same QY 4. ∠PBX= ∠QBY 4. From statements (1), (2) and (3.) Proved Example 4: In the given figure, O is the centre of the circle. If two chords AB and CD intersect at a point P, prove that 2∠APC = ∠AOC + ∠BOD Solution: Given: O is the centre of a circle. Chords AB and CD intersect at the point P. To prove: 2∠APC = ∠AOC + ∠BOD Construction: B and C are joined. Proof Statements Reasons 1 ∠ABC = 1 2 ∠AOC 1. Relation between inscribed angle and central angle standing on the same arc. 2. ∠BCD = 1 2 ∠BOD 2. Same reason as in (1) 3. ∠ABC + ∠BCD =1 2 (∠AOC + ∠BOD) 3. Adding statements (1) and (2) 4. ∠APC = 1 2 (∠AOC + ∠BOD) or, 2 ∠APC = ∠AOC + ∠BOD 4. In DBCP, ∠APC is the exterior angle. Proved A X B P Y Q D B P O A C

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 261 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Example 5: In a circle, O is the centre, AB is a diameter. D is an external point of circle such that DO ⊥ AB. AD cuts the circle at C and E is any point on the circumference. Prove that: ∠AEC = ∠ODA. Solution: Given: O is the centre of a circle, AB is a diameter. D is an external point of circle such that DO ⊥ AB. AD cut the circle at C and E is any point on the circumference. To prove: ∠AEC = ∠ODA Construction: B and C are joined. Proof: S.N. Statements S.N. Reasons 1. ∠ACB = 900 1. The angle in semi-circle is always a right angle. 2. In ∆ABC and ∆ADO (i) ∠ACB = ∠AOD (ii) ∠CAB = ∠DAO (iii) ∠ABC = ∠ODA 2. (i) Both are right angles (ii) Common angle (iii) Remaining angles 3. ∠ABC = ∠AEC 3. Both are standing on the same arc AC. 4. ∠AEC = ∠ODA 4. From statements (2), (iii) and (3) Proved Example 6: In a cyclic quadrilateral ABCD; E and F are the points on BC and AD respectively so that CD//EF. Prove that ABEF is also a cyclic quadrilateral. Solution: Given: In a cyclic quadrilateral ABCD, E and F are the points on BC and AD respectively so that CD//EF. To prove: ABEF is also a cyclic quadrilateral. Proof: S.N. Statements S.N. Reasons 1. ∠AFE = ∠ADC 1. EF//CD and corresponding angles. 2. ∠ABC + ∠ADC = 1800 2. The opposite angles of cyclic quadrilateral are supplementary. 3. ∠ABC + ∠AFE = 1800 3. From statements (1) and (2). 4. ABEF is also a cyclic quadrilateral 4. From statement (3) Proved Example 7: In a cyclic quadrilateral PQRS, RS is produced to T. If PS is an angular bisector of ∠QST, prove that ∆PQR is an isosceles triangle. C E A O D B E C A F D B

Vedanta Excel in Mathematics - Book 10 262 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Solution: Given: PQRS is a cyclic quadrilateral. RS is produced to T and ∠PSQ =∠PST. To prove: ∆PQR is an isosceles triangle Proof: S.N. Statements S.N. Reasons 1. ∠PSQ= ∠PST 1. Given 2. ∠PQR = ∠PST 2. The exterior angle of cyclic quadrilateral is equal to opposite interior angle. 3. ∠PRQ = ∠PSQ 3. The inscribed angles standing on the same arc PQ are equal. 4. ∠PQR= ∠PRQ 4. From statements (1), (2) and (3). 5. ∆PQR is an isosceles triangle 5. Base angles are equal. Proved Example 8: In the given figure, PQRS is a cyclic quadrilateral in which PQ and SR are produced to meet at T. If QT = RT, prove that (i) PS // QR (ii) PR = QS. Solution: Given: PQRS is a cyclic quadrilateral. PQ and SR are produced to meet at T and QT = RT. To prove: (i) PS // QR (ii) PR = QS Proof Statements Reasons 1. ∠TQR = ∠TRQ 1. QT = RT and base angles of isosceles ∆TQR 2. ∠TQR = ∠PSR 2. The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle. 3. ∠TRQ = ∠SPQ 3. Same reason as above 4. ∠PSR = ∠TRQ and ∠SPQ = ∠TQR 4. From statements (1), (2) and (3) 5. PS // QR 5. From statement (4), corresponding angles are equal 6. (i) (ii) (iii) (iv) In ∆s PRT and SQT ∠TPR = ∠TSQ ∠PTR = ∠STQ RT = QT ∴ ∆PRT ≅ ∆SQT 6. (i) (ii) (iii) (iv) Inscribed angles standing on the same arc Common angle Given A.S.A. axiom 7. PR = QS 7. Corresponding sides of congruent triangles Proved R T P S Q P Q R S T

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 263 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Example 9: WXYZ is a cycle quadrilateral. If the bisectors of ∠XWZ and ∠XYZ meet the circle at the points P and Q respectively, prove that PQ is the diameter of the circle. Solution: Given: WXYZ is a cyclic quadrilateral. WP is the bisector of ∠XWZ and YQ is the bisector of ∠XYZ. To prove: PQ is the diameter of is the diameter of the circle. Construction: W, Q and Z, Q are joined. Proof: Statements Reasons 1. ∠XWZ + ∠XYZ = 180° or, 1 2 ∠XWZ + 1 2 ∠XYZ = 90° ∠PWZ + ∠QYZ = 90° 1. Sum of opposite angles of cyclic quadrilateral 2. ∠PWZ + ∠QWZ = 90° 2. ∠QYZ = ∠QWZ , angles on the same segment 3. ∠PWQ = 90° 3. From statement 2, whole part axiom 4. ∠PWQ is in a semicircle 4. From statement 3 5. PQ is a diameter 5. From statement 4 Proved EXERCISE 13.2 General section 1. a) In the given figure, O is the centre of circle. (i) Name the arc on which ∠AOB and ∠ACB are standing. (ii) What is the relation between ∠AOB and ∠ACB? (iii) If ∠AOB = 70o , what will be the size of ∠ACB? (iv) If ∠ACB = 35o , what will be the size of ∠AOB? b) O is the centre of circle given alongside. (i) Which arc subtends the reflex ∠AOC and obtuse ∠ABC? (ii) What is the relation between reflex ∠AOC and obtuse ∠ABC? (iii) If reflex ∠AOB = 240o , what will be the size of obtuse ∠ABC? (iv) If obtuse ∠ABC = 100o , what will be the size of reflex ∠AOC? c) In the given circle, O is the centre. (i) What is the chord PQ called? (ii) What is the value of ∠PRQ? (iii) If ∠PQR = 55o , what will be the value of ∠QPR? P Q X Y W Z A C B O A C B O P Q R O

Vedanta Excel in Mathematics - Book 10 264 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle 2. a) Study the given circle and answer the following questions. (i) Name the arc on which ∠BAC and ∠BDC are standing. (ii) Write the relation between ∠BAC and ∠BDC? (iii) Which angle is equal to ∠ABD? (iv) If ∠BAC = 25o , what will be the measure of ∠BDC? (v) If ∠ACD = 40o , what will be the measure of ∠ABD? b) In the given circle with centre O, answer the following questions. (i) What is the special name for quadrilateral ABCD? (ii) Write the relation between ∠BAD and ∠BCD. (iii) Write the relation between ∠ABC and ∠ADC. (iv) Which angle is equal to ∠DCE? (v) Is OACD a cyclic quadrilateral? Give reason. 3. Find the size of unknown angles in the following figures: a) b) c) d) A D B 150° C x y O A B 110°x C O A B 140° x C O Q P R 28° x O e) f) g) h) A B C 35° x O A P B 30° x O Q P R x O 40° A B C O 30° 40° x i) j) k) l) A B C x D O 60° A B C D O x° y° 39° A B C D O x 120° A B C O 38°y° x° m) n) o) p) P R Q O x 80° A B C E D x 60° 48° y x S Q P y x R O 35° 65° C y O x B A D 4. Find the sizes of unknown angles in the following figures: a) b) c) R Q P O S x 115° C D A 40° x B O A C 32° x B O d) A B x C O A B C D A B C E D O

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 265 Vedanta Excel in Mathematics - Book 10 Geometry - Circle 5. Find the sizes of unknown angles in the following figures: a) C B A D 79° x y b) B A D x 120° C c) B A D C O 40° 30° x d) A B D E C x 72° O e) f) g) h) F E D G H O 140° C B A D 96° x S x P Q R P Q R S T x x i) j) k) l) Q R 44° b a 98° S P T C D A B E 55° O x F A D x B 72° E C K J M L N 60° 85° x y Creative section-A 6. a) In the figure alongside, O is the centre of the circle and PQRS is the cyclic quadrilateral. Find the sizes of ∠PRQ and ∠PRS. b) In the given figure, O is the centre of the circle. AB and DC are two parallel chords. If ∠BAD = 40°, find the measures of: (i) ∠BCD (ii) ∠BOD (iii) ∠ADC c) In the adjoining figure, AB is the diameter of the circle with centre O. If ∠ COD = 50°, find the size of ∠ CED. d) In the figure given alongside, O is the centre of the circle and PQ is a diameter. Chord ST is parallel to the diameter. If ∠QRS = 70°, find the measure of ∠QST. S P Q T O 65° R 35° 40° A B D C O 40° A B E D C O 50° S T Q P R 70° O

Vedanta Excel in Mathematics - Book 10 266 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle 7. a) Verify experimentally that the circumference angle NMR is half of the central angle NQR standing on same arc NR of the circle having centre Q. (Two circles with radii at least 3cm are necessary) b) In the circle of centre O, prove experimentally that inscribed angles ∠APB and ∠AQB standing on common arc AB are equal. (Two circles with radii not less than 3cm are required) c) Explore experimentally the relationship between opposite angles of a cyclic quadrilateral MNOP. (Two circles having radii at least 3 cm are necessary.) d) Explore experimentally the relationship between the exterior angle SRT formed by producing a side QR of a cyclic quadrilateral PQRS to the point T and the angle QPS of the same quadrilateral. 8. a) In the adjoining figure, O is the centre of the circle and PQ is the diameter. Show that ∠PRQ is a right angle. b) In the given figure, O is the centre of the circle. Prove that ∠POR = 2 ( ∠PRQ + ∠QPR). Creative section-B 9. a) In the figure, X is the centre of a circle. AB is a diameter of the circle. If DE ⊥ AB and AC = BD . Prove that ∠ABC = ∠BDE. b) In the given figure, O is the centre of the circle, AB the diameter and DO ⊥ AB. Prove that ∠ AEC = ∠ ODA. c) In the adjoining figure, O is the centre of the circle. If AMO ⊥ XOZ, prove that ∠OAZ = ∠XYO. 10. a) In the given figure, ABCD is a cyclic quadrilateral. Prove that ∠CBE = ∠ADC. Q P R O R P Q O X E C D A B E C D A O B X M Y A Z O A D E C B

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 267 Vedanta Excel in Mathematics - Book 10 Geometry - Circle b) D ABC is an isosceles triangle and XY // BC. If XY cuts AB at X and AC at Y, prove that the four points X, B, C and Y are concyclic. c) In the given figure, QR is the bisector of ∠ SQT and PQRS is a cyclic quadrilateral. Prove that D PSR is an isosceles triangle. d) In the adjoining figure, AC = BC and ABCD is a cyclic quadrilateral. Prove that DC bisects ∠ BDE. e) In the adjoining figure, TPQ is an isosceles triangle in which TP = TQ. A circle passing through P and Q cuts TP and TQ at R and S respectively. Prove that PQ // RS. f) ABCD is a parallelogram. The circle through A, B, and C intersects CD produced at E. Prove that AE = AD. g) In the adjoining figure, ABC is a triangle in which AB = AC. A circle passing through B and C intersects the sides AB and AC at the points D and E respectively. Prove that AD = AE. h) In the given figure, ABCD is a quadrilateral in which AB // DC and ∠ ADC = ∠ BCD. Prove that the points A, B, C, and D are concyclic. i) In the adjoining figure, PQRS is a cyclic quadrilateral. If the diagonals PR and QS are equal, prove that, QR = PS and PQ // SR. A B X Y C P Q T R S A E D B C P T Q R S E A D C B B A C D E A D C B S Q R P

Vedanta Excel in Mathematics - Book 10 268 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle j) In the figure, PQRS is a parallelogram. The inscribed circle cuts PQ at M and Rs at N. Prove that ∠ MNS = ∠ PQR. k) In the given triangle ABC, AD ⊥ BC and CE ⊥ AB. Prove that ∠BDE = ∠BAC. l) If PC is the bisector of ∠ APB, prove that XY // AB. n) The bisectors of opposite angles ∠A and ∠C of a cyclic quadrilateral ABCD intersect the corresponding circle at the points X and Y respectively. Prove that XY is the diameter of the circle. 11. a) In the given figure, PQRS is a parallelogram. Prove that UTRS is a cyclic-quadrilateral. b) In the given figure, AB and EF are parallel to each other. Prove that CDEF is a cyclic quadrilateral. c) In the given figure, NPS, MAN and RMS are straight lines. Prove that PQRS is a cyclic quadrilateral. 12. a) In the given figure, AP is the radius of circle ABC and diameter of circle APQ. Prove that AQ = QC. S N M Q R P B C D E A A B X Q Y C P Q U P S R T B C D E F A N P S M R Q A C A P B Q

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 269 Vedanta Excel in Mathematics - Book 10 Geometry - Circle b) In the adjoining figure, X and Y are the centres of the 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. c) In the adjoining figure, two circles intersect at A and B. The straight line PAQ meets the circles at P and Q and the straight line RBS meets the circles at R and S. Prove that PR // QS. d) In the given figure, two circles are intersecting at M and N. PQ and RS pass through M. If R, P, S, and Q are joined to N, prove that ∠ PNR = ∠ SNQ 13. a) Prove that the angle at the centre of a circle is twice the angle at the circumference standing on the same arc. b) The points S, T and U lie on the circumference of a circle with centre M. Prove that, ∠ TSU = 1 2 ∠ TMU. c) Prove that the angles at the circumference of a circle standing on the same arc are equal. d) In a cyclic quadrilateral PQRS, prove that ∠P + ∠R = ∠Q + ∠S = 180°. e) If ABCD is a cyclic parallelogram, prove that ABCD is a rectangle. f) If one side of a cyclic quadrilateral is produced, prove that the exterior angle is equal to the opposite interior angle of the quadrilateral. g) Prove that a cyclic trapezium is always isosceles and its diagonals are equal. h) Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. i) Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to six right angles. Project Work and Activity Section 14. By paper folding or cutting method, show the following theorems. a) The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc. b) The opposite angles of a cyclic quadrilateral are supplementary. C A B D X Y P A B S Q R P M N S Q R

Vedanta Excel in Mathematics - Book 10 270 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Geometry - Circle Objective Questions Tick the correct alternative. 1. The degree measure of an angle at the centre of a circle is ……….. the degree measure of its opposite arc. (A) equal to (B) half of (C) less than (D) greater than 2. Which of the following statement is NOT true? (A) Equal arcs of a circle subtend equal angles at the centre. (B) Equal arcs of a circle subtend equal angles at the circumference. (C) If two chords are parallel, then the arcs between them are equal. (D) The chords between equal arcs are equal. 3. Which of the following statement is NOT true? (A) The angle at the centre of a circle is twice the angle at the circumference standing on the same arc. (B) The angle at the semi-circle is always a right angle. (C) The opposite angles of a cyclic quadrilateral are equal. (D) The inscribed angles standing on the same arc are equal. 4. In the given figure, O is the centre of the circle. If OA = AB, what is the measure of ∠ACB? (A) 600 (B) 900 (C) 300 (D) 450 5. Smarika wishes to draw two angles at the circumference having equal measurements. Which of the following mathematical facts should she know? (A) The angles at the circumference standing on the same arc of a circle are equal. (B) The angles at the circumference subtended by the equal arcs of a circle are equal. (C) The angles in the circumference standing on the same segment of a circle are equal. (D) All of the above. 6. In the figure, O is the centre of the circle. If PR = QR, what is the measure of ∠QSR? (A) 300 (B) 450 (C) 600 (D) 900 7. The opposite angles of a cyclic quadrilateral are (A) Supplementary (B) Complementary (C) Equal (D)Right angles 8. Under which condition, any quadrilateral can NOT be a cyclic quadrilateral? (A) If the opposite angles are supplementary. (B) If the opposite angles are complementary. (C) If the exterior angle is equal to the opposite interior angle. (D) If a pair of angles subtended by any side at opposite vertices are equal. 9. In the semi-circle given alongside, O is the centre. If OC = 5 cm and BC = 6 cm, what is the length of AC? (A) 5 cm (B) 6 cm (C) 10 cm (D) 8 cm 10. In the given figure, arc AD = arc AC and ∠CAD = 1400 , what is the measure of ∠ABC? (A)1400 (B) 700 (C) 400 (D) 200 O A C B P O Q R S O A C B 140° B C A D

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 271 Vedanta Excel in Mathematics - Book 10 Geometry - Circle Assessment –V 1. In the adjoining figure, PQ // BC. If BX // CA and CY // BA meet the line PQ produced at X and Y respectively, answer the following questions. (a) Write the relation between the area of triangle AXB and the parallelogram XBCQ. (b) If the area of parallelogram XBCQ is 50 square metres, what is the area of parallelogram PBCY? (c) Prove that: area of ∆ABX = area of ∆ACY (d) Construct a triangle ABC having sides a = 6.4 cm, b = 6 cm and c = 5.6 cm. Also, construct another triangle having one side 7 cm and equal in area to the ∆ABC. 2. In the adjoining diagram, ABCD is a parallelogram. Answer the following questions. (a) If the area of triangle ABQ is 30 cm2 , what is the area of parallelogram ABCD? (b) Prove that: area of ∆APQ = area of ∆BPD. (c) Prove that: area of ∆APQ = area of ∆PCD. (d) Construct a parallelogram ABCD in which AB = 6 cm, BC = 4 cm and ∠BAD = 45o . Also, construct a triangle APQ having one angle 60o and equal in area with parallelogram ABCD. 3. In a circle with centre O; P, Q and R are on the circumference. (a) Draw two circles PQR of difference radii not less than 3 cm. Explore experimentally the relationship between ∠PQR and ∠POR. (b) If ∠PQR = 40o , find the value of ∠OPQ. 4. In the given figure, PQ = RT and SQ is the bisector of ∠PQR. Answer the following questions. (a) Which chord is equal to chord PS? (b) If the measurement of ∠SRT = 100o , what is the measurement of ∠QPS? (c) Prove that DSQT is an isosceles triangle. 5. In the given circle, arc AB and arc CD are equal. Chords AC and BD intersect at E. (a) Write the relation between the chords AD and BC. (b) If AC were the diameter of the circle and arc AD = arc DC, what would be the value of ∠ABD. (c) Prove that the triangles ABE and CDE are equal in area. X Y A B C P Q B C D Q A P P R S Q x T A D B C E

Vedanta Excel in Mathematics - Book 10 272 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Unit 14 Statistics Classwork - Exercise 14.1 Statistics – Looking back 1. Let’s tick (√) the correct answer from the bracket given aside. a) The numerical representation of facts is called …………….. (data, frequency) b) Each entry in the data is ………… (a data, an observation) c) The number of occurrences of a particular item in a data is ………….. (item, frequency) 2. Let’s answer the following questions. a) What is arithmetic mean? b) Write down the formula to find the arithmetic mean from a discrete data? c) Define median. d) What do you mean by mode in a set of data? e) How do you differentiate mode from range in a data? 14.2 Statistics-review Statistics is the branch of mathematics in which facts and information are collected, sorted, displayed, and analysed. Statistics are used to make decision and prediction about the future plans and policies. The word ‘statistics’ comes from the word ‘state’ largely because it was the job of the state to keep records and make decisions based on census results. Sir Ronald Aylmer Fisher was a British Statistician 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 the Linnean Society of London’s prestigious Darwin-Wallace Medal in 1958. Sir Ronald Aylmer Fisher (AD (CE) 1890 - 1962) 14.3 Measures of central tendency The measure of central tendency gives a single central value that represents the characteristics of the entire data. A single central value is the best representative of the given data towards which the values of all other data are approaching.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 273 Vedanta Excel in Mathematics - Book 10 Statistics Average of the given data is the measure of central tendency. There are three types of averages which are commonly used as the measure of central tendency. Central Tendency Mean Median Mode A systematic flow of data in a logical or specific order is called statistical series. There are mainly three types of statistical series. They are: (i) Individual series (ii) Discrete series (iii) Grouped and continuous series The process of finding averages of the given data is based on the way of presentation of the data. 14.4 Mean of grouped and continuous data We have discussed the methods of finding mean of individual and discrete series in the previous classes. So, in this chapter, we shall discuss about the methods of finding mean of grouped and continuous series. When data are grouped in class intervals and presented in the form of a frequency table, we get a frequency distribution as shown below. Age in years 0 - 4 4 - 8 8 -12 12 - 16 16 - 20 20 - 24 Number of students 5 8 13 18 10 6 The above table shows the number of students in the various age groups. For example, there are 5 students in the age group 0-4. When we form a grouped frequency table then the identity of the individual observations is lost. To represent the class intervals, we should find the mid-value (m) or class-mark of each class interval and it is written in the second column of the frequency table. The midvalue of each class interval is obtained by using the following rule: Mid-value (m) = lower limit + upper limit 2 In grouped or continuous data, we can compute the arithmetic mean by applying any one of the following methods. (i) Direct Method (ii) Assumed Mean Method (iii) Step Deviation Method Let’s learn about these methods to calculate the arithmetic mean of continuous data. (i) Direct Method The arithmetic mean of a grouped and continuous data can be calculated quickly by using the direct method. For this method, we follow the steps given below. www.geogebra.org/classroom/wb4yfxcn Classroom code: WB4Y FXCN Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 274 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics Steps: Obtain the mid-value (m) of each class Multiply those mid-values (m) by the corresponding frequency (f) of each class and obtain the sum of fm. i.e., if the corresponding frequency of each of mid-value m1 , m2 , m3 ... mn be f 1 , f 2 , f 3 , .... f n, respectively, then (i) f 1 + f 2 + f 3 + … + f n = ∑f = N (ii) f 1 m1 + f 2 m2 + f 3 m3 + … + f nmn = ∑fm Apply the formula, mean (X) = ∑fm N Example 1: The marks obtained by the students of class X of Sarawati Secondary School in a terminal examination is given below. Calculate the mean marks. Marks obtained 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 Number of students 3 8 10 7 2 Solution: Calculation of mean Marks obtained Mid-value (m) No. of students (f) fm 0 – 20 10 3 30 20 – 40 30 8 240 40 – 60 50 10 500 60 – 80 70 7 490 80 – 100 90 2 180 Total N = 30 ∑fm = 1440 Now, mean marks (X ) = ∑fm N = 1440 30 = 48 (ii) Short-Cut Method or Assumed Mean Method or Deviation Method A data in which the observation or frequency or both are in larger number, it is difficult and time consuming to find the products of the observations and corresponding frequencies, and then adding the products. However, it may have chances of errors. In such cases, we can use the Assumed Mean Method to find the arithmetic mean of grouped and continuous data. Steps: Any value of the observations is taken as the mean called assume mean and denoted by the letter ‘A’. Preferably, choose the middle value of the data. Find the difference between each mid–value m and assumed mean A. The difference is called the deviation and it is denoted by‘d’. Thus, d = m – A. Multiply each deviation ‘d’ by the corresponding frequency f then obtain ∑fd . Apply the formula, mean (X) = A + ∑fd N

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 275 Vedanta Excel in Mathematics - Book 10 Statistics Example 2: The daily wages of 100 workers of a factory are given in the following table. Find the mean wage. Daily wages (in Rs.) 450 – 550 550 – 650 650 – 750 750 – 850 850 – 950 Number of workers 16 24 30 19 11 Solution: Let, the assumed mean (A) = 700. Daily wages (in Rs.) Mid-value (m) No. of workers (f) d = m – A fd 450 – 550 500 16 – 200 – 3200 550 – 650 600 24 – 100 – 2400 650 – 750 700 30 0 0 750 – 850 800 19 100 1900 850 – 950 900 11 200 2200 Total N = 100 ∑fd = – 1500 Now, mean (X ) = A + ∑fm N = 700 + –1500 100 = 700 – 15 = 685 Hence, the daily wage of the workers is Rs. 685. (iii) Step Deviation method In this method, another deviation (d′) is found by dividing the deviation (d) by the length of class interval (c) in order to simplify the calculation. Then the mean is calculated by using the following formula: Mean (X) = A + ∑fd' N × c where, A = the assumed mean, c = the length of class internal. This method is very much useful in the case of large values of x and f. Steps: Assume a value of the observations as mean and denote it by ‘A’. Find d = m – A. Calculate d′ = m – A c and multiply each d′ by the corresponding frequency f Obtain ∑fd′ and the mean by applying the formula, X= A + ∑fd' N ×c Example 3: In a school, 250 pupils have the heights as tabulated below. Calculate the average height of pupils by using step deviation method. Height (in cm) 80 – 90 90 – 100 100 – 110 110 – 120 120 – 130 Number of pupils 32 55 80 58 25

Vedanta Excel in Mathematics - Book 10 276 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics Solution: Let, the assumed mean (A) = 105. Height (in cm) Mid-value (m) No. of pupils (f) d = m – A d′ = d 10 fd′ 80 – 90 85 32 – 20 – 2 – 64 90 – 100 95 55 – 10 –1 – 55 100 – 110 105 80 0 0 0 110 – 120 115 58 10 1 58 120 – 130 125 25 20 2 50 Total N = 250 ∑fd′ = – 11 Now, mean (X ) = A + ∑fd' N × c = 105 + –11 250 × 10 = 105 – 0.44 = 104.56 Hence, the average height of the pupil is 104.56 cm. Facts to remember Grouped and Continuous data Direct method Assumed mean method Step deviation method X = ∑fm N X = A + ∑fm N where d = m – A X = A + ∑fd' N × c where d' = m – A c Methods of finding Mean Example 3: The values of diastolic blood pressure of 80 people of age group 20–50 years are shown in the table given below. BP (in mmHg) 70 – 75 75 – 80 80 – 85 85 – 90 90 – 95 95 – 100 Number of people 10 17 23 16 9 5 Calculate the average blood pressure by using (i) direct method (ii) short-cut method (iii) step deviation method Solution: (i) Calculation of mean by direct method

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 277 Vedanta Excel in Mathematics - Book 10 Statistics BP (in mmHg) Mid-value (m) No. of people (f) fm 70 – 75 72.5 10 725 75 – 80 77.5 17 1317.5 80 – 85 82.5 23 1897.5 85 – 90 87.5 16 1400 90 – 95 92.5 9 832.5 95 – 100 97.5 5 487.5 Total N = 80 ∑fm = 6660 Now, mean marks (X ) = ∑fm N = 6660 80 = 83.25 (ii) Calculation of mean by short-cut method Let, the assumed mean (A) = 82.5 mmHg BP (in mmHg) Mid-value (m) No. of people (f) d = m – A fd 70 – 75 72.5 10 – 10 – 100 75 – 80 77.5 17 – 5 – 85 80 – 85 82.5 23 0 0 85 – 90 87.5 16 5 80 90 – 95 92.5 9 10 90 95 – 100 97.5 5 15 75 Total N = 80 ∑fd = 60 Now, mean (X ) = A + ∑fm N = 82.5 + 60 80 = 82.5 + 0.75 = 83.25 (iii) Calculation of mean by step-deviation method Let, the assumed mean (A) = 82.5 mmHg BP (in mmHg) Mid-value (m) No. of people (f) d = m – A d′ = d 5 fd 70 – 75 72.5 10 – 10 – 2 – 20 75 – 80 77.5 17 – 5 – 1 – 17 80 – 85 82.5 23 0 0 0 85 – 90 87.5 16 5 1 16 90 – 95 92.5 9 10 2 18 95 – 100 97.5 5 15 3 15 Total N = 80 ∑fd = 12

Vedanta Excel in Mathematics - Book 10 278 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics Now, mean (X ) = A + ∑fd' N × c = 82.5 + 12 80 × 5 = 82.5 + 0.75 = 83.25 Hence, the average blood pressure is 83.25 mmHg. Facts to remember 1. The value of mean calculated from all three methods is the same. 2. If the value of observations x and frequencies f are sufficiently small, we can use direct method. 3. If the value of observations x and frequencies f are larger numbers, we can use assumed mean method or step deviation method. Example 4: If the arithmetic mean of the data given below is 31, calculate the value of p. X 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 f 4 6 10 p 6 4 Solution: X Mid-value (m) f fm 0 – 10 5 4 20 10 – 20 15 6 90 20 – 30 25 10 250 30 – 40 35 p 35p 40 – 50 45 6 270 50 – 60 55 4 220 Total N = 30 + p ∑fm = 850 + 35p Now, mean ( X )= ∑fm N 31 = 850 + 35p 30 + p or, 850 + 35p = 930 + 31p or, p = 20 So, the required value of p is 20.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 279 Vedanta Excel in Mathematics - Book 10 Statistics EXERCISE 16.1 General section 1. a) If ∑fx = 450 and N = 15, find X . b) If ∑fm = 960 and n = 24, find X . 2. a) In a continuous series, if the sum of weights of 50 students is 1,850 kg, find their average weight. b) In a grouped data, the sum of heights of 44 students is 5,500 cm, find their average height. 3. a) In a continuous series, ∑f = 20p + 7 and ∑fm = 100p + 35, find the mean (X ). b) In a grouped data, ∑f = 12a2 + 3a and ∑fm = 60a2 + 15a, find the mean (X ). 4. a) If X = 35 and ∑fx = 455, find the number of terms (N). b) If the mean of a grouped data having ∑fm = 320 is 16, find the value of N. 5. a) In a continuous series, if the average marks of some students is 40 and the sum of their marks (∑fm) is 1200, find the number of students. b) In a continuous series, if the average daily wage of some workers is Rs 500 and the sum of their daily wages (∑fm) is Rs 9,000, find the number of workers. 6. a) In a continuous series, if mean (X ) = 10 + a and ∑fm = 50 + 5a, find the number of terms (N). b) In a grouped data, if mean (X ) = 6a + 9 and ∑fm = 66a + 99, find the number of terms (N). 7. a) If X = 25, N = 10 and ∑fx = p, find the value p. b) If X = 32, ∑f = 15 and ∑fm = k, find the value k. 8. a) In a continuous data, if ∑fm = 150 + 5a, X = 10 and ∑f = 10 + a, find the value of a. b) If ∑fm = 72 + 8k, X = 6 and ∑f = 16 + k, find the value of k. c) If mean (X ) = 12, ∑fm = 70 + 10a and the number of frequency (N) = 5 + a, find the value of a. 9. a) In a continuous series, if X = 20, fm = 800 + 15p and N = 10 + p, find the value of p and exact value of N. b) In a grouped data, the mean (X ) = 40, the number of terms (N) = 20 + 3n and ∑fm = 1400, find the exact value of N after finding the value of n.

Vedanta Excel in Mathematics - Book 10 280 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics Creative section 10. Find the arithmetic mean (or average) from the data given in the following tables: a) Marks 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Frequency 12 7 8 3 10 b) Wages (Rs) 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 No. of workers 6 8 12 8 6 c) Age (in years) 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 No. of workers 50 70 100 180 150 d) 11. a) Find the mean by constructing a frequency table of class interval of 10 from the data given below. 7, 47, 36, 39, 31, 19, 41, 49, 9, 51, 29, 22, 59, 17, 49, 21, 24, 12, 31, 8, 36, 18, 32, 16, 23 b) Construct a frequency table of class interval of 10 from the given data and find the mean. 23, 5, 17, 28, 39, 52, 16, 22, 69, 75 41, 33, 9, 49, 34, 59, 72, 46, 65, 58 60, 48, 64, 32, 50, 73, 57, 51, 63, 36 12. a) The mean of the data given below is 36. Determine the value of p. Age (in years) 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 No. of teachers 3 8 15 p 4 b) The mean of the given data below is 28. Find the value of k. Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Frequency 12 18 27 k 17 6 c) The mean of the data given below is 45. Find the value of a. Wages (in Rs) 10 – 20 20 – 30 30 – 40 40 – 50 50 –60 60 – 70 70 – 80 No. of workers 4 7 9 18 a 7 3 Project Work and Creative Section 13. a) Collect the marks obtained by the students of your class in mathematics in recent exam. Then construct the marks in a frequency distribution table with class interval 10 and calculate the average mark. b) Collect the ages of family members of your 10 friends. Show the ages in a frequency distribution table of class of length 10 and calculate the average age. X 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 f 5 7 8 6 4

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 281 Vedanta Excel in Mathematics - Book 10 Statistics 14.5 Median -review Let’s study the following conversation between a student and the mathematics teacher of a school through viber. Median is an average that divides the given set of data into two equal parts. For example, 10, 16, 22, 28 34, 40, 46 3 items Middle item 3 items Here, the data are arranged in ascending order. The item 28 has 3 items before it and 3 items after it. So, 28 is the middle item that divides the series into two parts. Thus, 28 is the median of the given series of data. Good morning sir! Good morning, Shashwat! How are you? I am fine. How are you sir? I am also fine. Do you have any problem? Yes of course! Then, what’s your problem? Sir! How can we get the reasonable value in such case? Ok sir. Thank you. Bye sir! Ok! See you in the class. Obviously, it is Rs 26,600. Please! Tell me it clearly. Yes sir! I finished my homework from arithmetic mean but when I was finding the average income of my family members, I got confused. The monthly incomes of 4 members of my family are Rs 15,000, Rs 18,000, Rs 20,000 and Rs 25,000. Their average income is Rs 19,500. My brother who would not work before has just joined in an INGO and earns Rs 55,000 per month. Now, the new average income of the family is Rs 26,600. Sir! Does the average income of Rs 26,600 truly represent the income status of the individuals of my family? The extreme income affects the mean and the mean may move away from the central area. Oh! In such case, we find another type of average called median. We shall discuss about it in today’s class. www.geogebra.org/classroom/rq4pa4su Classroom code: RQ4P A4SU Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 282 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics 14.6 Median of grouped and continuous data In the case of grouped and continuous series, the position of the median is obtained by using the following formula. Position of the median = N 2 th class. Then, the actual median within the median class is obtained by using the following formula. Median = N 2 – c.f. L + × c f Where, L = the lower limit of median class c.f. = c.f. of the class just preceding the median class f = frequency of median class c = length or width of size of class interval 14.7 Graphical method of determining the median We can find the median class and the value of median by using less than ogive or more than ogive or by using both the ogives of a distribution on the same graph paper.For example, Marks No. of students Upper limit Less than c.f. Coordinates (x,y) 0 - 10 4 10 4 (10, 4) 10 - 20 12 20 16 (20, 16) 20 - 30 10 30 26 (30, 26) 30 - 40 9 40 35 (40, 35) 40 - 50 5 50 40 (50, 40) Here, total number of observation (N) = 40 We know, median lies in the 50% of the total data. ∴50% of 40 students = 20 students A straight line drawn from 20 (frequency) on y-axis parallel to x-axis intersects the curve at A. A perpendicular drawn from A meets x-axis at M and its x-coordinate is 24. So, median = 24. Thus, median class is 20–30 and the median is 24. Y X Less than cumulative frequencies Upper limits 0 10 20 30 40 50 60 10 20 30 40 50 60 (10, 4) (20, 16) (30, 26) (40, 35) (50, 40) A M

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 283 Vedanta Excel in Mathematics - Book 10 Statistics Facts to remember 1. Median is the position value which divides the arranged data in to two equal halves. 2. Frequency of median class (f) = c.f. of median class – c.f. of pre-median class. 10.8 Derivation of formula of median of grouped and continuous data Step 1: Draw a less than ogive from a grouped data having N observations. Step 2: We know, median = size of N 2 th observation. Thus, mark the point 0, N 2 on y-axis and draw a line parallel to x-axis meeting the curve at A. Step 3: From A, draw a line AM perpendicular to x-axis at M. Then, median = OM = x-co-ordinate of point M Step 4: Suppose, L and U are the lower limit and upper limit of the median class. i.e., OP = L and OQ = U, PQ = median class and PM = BE = x. Then, median, OM = L + x. Step 5: Assume that BAC is a straight line. From B draw a horizontal line that cuts CQ at D and AM at E. Then, AM = N/2 Now, CD = CQ – DQ = CQ – BP = c.f. of median class – c.f. of pre-median class = frequency of median class = f Also, AE = AM – EM = N/2 – BP = N/2 – c.f. of pre-median class = N/2 – c.f. Again, BD = PQ = OQ – OP = Upper limit of median class (U) – Lower limit of median class (L) = Width of median class = c Since, ∆ ABE and ∆ CBD are similar. Y X N A B E O P M Q N 2 C D

Vedanta Excel in Mathematics - Book 10 284 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics ∴ BE BD = AE CD [The corresponding sides of similar triangles are proportional] or, x c = N 2 – c.f. f or, x = N 2 – c.f. f × c Step 6: Median = OM = L + x = L + N 2 – c.f. f × c ∴ Median = L + N 2 – c.f. f × c Worked-out Examples Example 1: The table given below shows the marks obtained by the students of class 10 of a school. Marks obtained 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Number of students 3 5 8 18 10 6 (i) Construct the cumulative frequency table. (ii) Find the class in which the median lies. (iii) Compute the median. Solution: (i) Constructing the cumulative frequency table Marks obtained Number of students (f) c.f. 20 - 30 3 3 30 - 40 5 8 40 - 50 8 16 50 - 60 18 34 60 - 70 10 44 70 - 80 6 50 Total N = 50 (ii) Now, the position of median class = N 2 th class = 50 2 th class = 25th class In c.f. column, the c.f. just greater than 25 is 34 and its corresponding class is 50 - 60 ∴ Median class is (50 - 60) (iii) Again, L = 50, N 2 = 25, c.f. = 16, f = 18 and c = 10 We know, Median (Md) = L + N 2 – c.f. f × c = 50 + 25 – 16 18 × 10 = 55 Hence, the median mark is 55.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 285 Vedanta Excel in Mathematics - Book 10 Statistics Example 2: The speed of vehicles (in km/hr.) in a road was recorded by traffic police this morning. 51, 23, 60, 34, 48, 58, 61, 79, 44, 50, 25, 65, 39, 52, 43, 55, 46, 37, 40, 67, 76, 55, 71, 54, 33 (i) Construct the frequency table of class interval 10. (ii) In which class does the median lie? (iii) Compute the median speed. Solution: Here, Minimum speed = 23 km/hr, maximum speed = 79 km/hr, class size = 10 km/hr Class intervals are 20 - 30, 30 - 40, ... , 70 - 80 (i) Constructing frequency and cumulative frequency table Speed Tally bars Number of vehicles (f) c.f. 20 - 30 | | 2 2 30 - 40 |||| 4 6 40 - 50 |||| 5 11 50 - 60 |||| || 7 18 60 - 70 |||| 4 22 70 - 80 ||| 3 25 Total N = 25 (ii) Now, the position of median class = N 2 th class = 25 2 th class = 12.5th class In c.f. column, the c.f. just greater than 12.5 is 18 and its corresponding class is 50-60 Median class is (50 - 60) (iii) Again, L = 50, N 2 = 12.5, c.f. = 11, f = 7 and c = 10 We know, median (Md) = L + N 2 – c.f. f × c = 50 + 12.5 – 11 7 × 10 = 52.14 Hence, the median speed of the vehicles is 52.14 km/hr. Example 3: The people of a community organized the ‘Afforestation Program’ on March 21, the International Day of Forests. The heights of plants planted on the program are given below. Height of plants (in cm) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Number of plants 5 12 18 x 15 11 14 (i) If the median height of the plants is 36 cm, find the class which includes the median height of the plants.

Vedanta Excel in Mathematics - Book 10 286 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics (ii) Draw the cumulative frequency table. (iii) Find the value of x. (iv) What percent of plants of height less than 40 cm were planted? Solution: (i) Since, the median height of the plants is 36 cm; it lies in the class 30-40. So, the median class is 30 - 40. (ii) Cumulative frequency table. Height (in cm) Number of plants (f) c.f. 0 - 10 5 5 10 - 20 12 17 20 - 30 18 35 30 - 40 x 35 + x 40 - 50 15 50 + x 50 - 60 11 61 + x 60 - 70 14 75 + x Total N = 75 + x (iii) Also, L = 30, N 2 = 75 + x 2 , c.f. = 35, f = x and c = 10 We know, median (Md) = L + N 2 – c.f. f × c or, 36 = 30 + 75 + x 2 – 35 x × 10 or, 36 – 30 = 75 + x – 70 2 x × 10 or, 6 = x + 5 2x × 10 or, 6 = x + 5 x × 5 or, 6x = 5x + 25 or, x = 25 Hence, the required value of x is 25. (iv) Total number of plants (N) = 75 + x = 75 + 25 = 100 Number of plants with height less than 40 cm = 35 + x = 35 + 25 = 60 ∴ Percentage of plants of height less than 40 cm = 60 100 × 100% = 60%

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 287 Vedanta Excel in Mathematics - Book 10 Statistics 14.9 Quartiles Quartiles are the values that divide the arrayed data into four equal parts. A distribution is divided into four equal parts by three quartiles. (i) The first or lower quartile (Q1 ) is the point below which 25% of the items lie and above which 75 % of the items lie. (ii) The second quartile (Q2 ) is the point below which 50% of the items lie and above which 50% of the items lie. So, the second quartile is the median of a distribution. (iii) The third or upper quartile (Q3 ) is the point below which 75% of the items lie and above which 25% of the items lie. Q1 25% 25% Q 25% 25% 2 Q3 14.10 Quartiles of grouped and continuous series In this case, we obtain the position of the first quartile and the third quartile by using the following formulae. The position of the first quartile, Q1 = N 4 th class. The position of the third quartile, Q3 = 3N 4 th class Then, the actual first and third quartiles are obtained by using the following formulae. The first quartile, Q1 = N 4 – c.f. L + × c f and the third quartile, Q3 = 3N 4 – c.f. L + × c f Facts to remember Name of partition value/Quartiles Position of class Exact value Median Median class = N 2 th class Md = N 2 – c.f. L + × c f Lower/First Quartile (Q1 ) Q1 class = N 4 th class Q1 = N 4 – c.f. L + × c f Upper/Third Quartile (Q3 ) Q3 class = 3N 4 th class Q3 = 3N 4 – c.f. L + × c f

Vedanta Excel in Mathematics - Book 10 288 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics Example 4: In a continuous series, the first quartile lies in the class interval 20-30 and cumulative frequency of its preceding class is 7. If the frequency of class interval 20-30 is 12 and the sum of frequencies of the series is 40, find the first quartile. Solution: Here, lower limit of Q1 class (L) = 20, frequency (f) = 12, c.f. of pre-Q1 class = 7, Sum of frequencies (N) = 40 and width of class (c) = 10 Now, the first quartile (Q1 ) = N 4 – c.f. L + × c f = 20 + 40 4 – 7 12 × 10 = 20 + 10 – 7 12 × 10= 22.5 Hence, the first quartile of the series is 22.5 Example 5: The age of the students of a school is given in the table. Age (in years) 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 Number of students 15 40 95 84 16 (i) Calculate the lower quartile. (ii) Calculate the upper quartile. Solution: Cumulative frequency table Age (in years) Number of students (f) c.f. 0-4 15 15 4-8 40 55 8-12 95 150 12-16 84 234 16-20 16 250 Total N = 250 (i) Now, the position of lower quartile (Q1 ) class = N 4 th class = 250 4 th class = 62.5th class In c.f. column, the c.f. just greater than 62.5 is 150 and its corresponding class is 8 - 12 ∴Q1 class is (8 -12) Here, L = 8, N 4 = 62.5, c.f. = 55, f = 95 and c = 4 We know, Lower quartile (Q1 ) = N 4 – c.f. L + × c f = 8 + 62.5 – 55 95 × 4 = 8.32 Hence, the lower quartile age is 8.32 years.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 289 Vedanta Excel in Mathematics - Book 10 Statistics (ii) Again, The position of upper quartile (Q3 ) class = 3N 4 th class = 3 × 250 4 th class = 187.5th class In c.f. column, the c.f. just greater than 187.5 is 234 and its corresponding class is 12-16 Q3 class is (12 - 16) Here, L = 12, 3N 4 = 187.5, c.f. = 150, f = 84 and c = 4 We have, Upper quartile (Q3 ) = 3N 4 – c.f. L + × c f = 12 + 187.5 – 150 84 × 4 = 13.79 Hence, the upper quartile age is 13.79 years. EXERCISE 14.2 General section 1. a) Name the average which divides the given set of data into two equal parts. b) Write the name of quartile below which 25% of items lie. c) Write the name of quartile above which 75% of items lie. 2. a) Write the formula to calculate the median class of a data having N items. b) In a continuous series, if the lower limit of median class = L, frequency of median class = f, cumulative frequency of pre-median class = c.f., width of class = h and the sum of frequencies = N, write the formula for finding the median. c) Write the formula for finding the lower quartile of a grouped data and mention the meanings of each letter used in it. d) Write the formula for finding the upper quartile of a continuous series and mention the usual meanings of its each symbol. 3. a) The median of a series lies in the class interval of 20 – 30. If the necessary variables in computing median are given as N 2 = 15, c.f. = 12, f = 5 and c = 10, compute the median. b) The median of the given data lies in class interval of 40 – 50. If N 2 = 18, c.f. = 16, f = 8 and c = 10, find the value of median. c) If the lower quartile class of a series is 40 – 60, N 4 = 9, c.f. = 8, f = 12 and c = 20, compute the first quartile. d) If the Q3 of a series lies in 30 – 40, 3N 4 = 25, c.f. = 20, f = 5 and c = 10, find Q3 . 4. a) Find the median class from the following table. Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 No. of students 2 5 7 6 3 2

Vedanta Excel in Mathematics - Book 10 290 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics b) Find the median class. x 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 f 3 8 17 20 22 c) Find the first quartile class from the table given below. Wages (in Rs) 50 – 65 65 – 80 80 – 95 95 – 110 110 – 125 No. of workers 7 3 6 9 5 d) Find the upper quartile class from the following table. Marks obtained 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 No. of students 5 8 10 15 7 5 Creative section 5. From each of the following data, answer the questions. (i) Construct the cumulative frequency table. (ii) Find the class in which the median lies. (iii) Compute the median. a) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 No. of students 4 6 10 7 3 b) Wages (in Rs) 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 No. of workers 8 11 10 6 3 c) Class 0 – 30 30 – 60 60 – 90 90 – 120 120 – 150 150 – 180 Frequency 5 10 22 25 14 4 d) Marks obtained 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 No. of students 5 12 30 10 8 5 6. Re-write the data given below in continuous series and work out the following question. (i) Draw the cumulative frequency table. (ii) Find the median class (iii) Compute the median. a) Marks obtained 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50 No. of students 4 12 24 44 62 b) X 5 – 15 5 – 25 5 – 35 5 – 45 5 –55 5 – 65 f 5 13 23 32 43 44

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 291 Vedanta Excel in Mathematics - Book 10 Statistics 7. a) The following table shows the marks obtained by the students in a Unit Test of mathematics. Marks obtained 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Number of students 4 12 x 9 5 (i) Draw the cumulative frequency table. (ii) If the median mark is 24, find the median class. (iii) Find the value of x. (iv) What percent of students got less than 30 marks? b) The weight of a few students are recorded in the following table. Weight (in kg) 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Number of students 5 7 p 4 8 (i) Draw the cumulative frequency table. (ii) If the median is 35 kg, find the median class. (iii) Find the value of p. (iv) What percent of students have less than 40 kg weight? 8. Based on the data given below, answer the following questions. (i) Construct the cumulative frequency table. (ii) Find the first or lower quartile (Q1 ) class. (iii) Compute the first or lower quartile (Q1 ). a) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 No. of students 2 5 7 6 3 2 b) Wages (in Rs) 50 – 65 65 – 80 80 – 95 95 – 110 110 – 125 No. of workers 7 3 6 9 5 c) Wages (in Rs) 100–200 200–300 300–400 400–500 500–600 600–700 No. of workers 4 9 15 11 8 3 d) Marks obtained 20 – 30 20 – 40 20 – 50 20 – 60 20 – 70 20 – 80 No. of students 5 9 14 18 21 24 9. From the data given in the following tables below, answer the following questions. (i) Construct the cumulative frequency table. (ii) Find the third or upper quartile (Q3 ) class (iii) Find the third or upper quartile (Q3 ). a) Marks obtained 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 No. of students 2 5 7 6 3 2

Vedanta Excel in Mathematics - Book 10 292 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics b) Marks obtained 0 – 15 15 – 30 30 – 45 45 – 60 60 – 75 No. of students 8 6 12 15 7 c) Weight (in kg) 40 – 44 44 – 48 48 – 52 52 – 56 56 – 60 60 – 64 No. of students 8 10 14 16 3 1 d) Marks obtained 0 – 20 0 – 40 0 – 60 0 – 80 0 – 100 No. of students 12 15 26 34 40 10. a) The daily expenses ( in Rs) of 30 students of a school are given below. 7, 15, 24, 10, 5, 12, 20, 8, 27, 18, 25, 16, 9, 13, 14, 22, 28, 11, 4, 17, 12, 19, 14, 15, 21, 12, 12, 14, 10, 11 (i) Display the data in c.f. table with class interval of length 5. (ii) Compute the median expense. b) The following are the marks obtained by students in mathematics in an examination: 51, 22, 63, 35, 46, 57, 79, 21, 39, 51, 32, 43, 52, 59, 38, 45, 40, 32, 60, 63 (i) Make a frequency table of class interval 10. (ii) Compute the first quartile. c) The marks obtained by 20 students in an examination are given below. 15, 12, 23, 35, 46, 57, 18, 12, 39, 51, 32, 43, 25, 59, 18, 38, 45, 40, 32, 33 (i) Make a frequency table of class interval of 10 (ii) Find the third quartile. 11. a) The following table shows the marks obtained by the students of class 10 in an exam. Marks obtained 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Number of students 9 11 p 20 30 16 (i) Draw the cumulative frequency table. (ii) If the first quartile (Q1 ) is 25, find the Q1 class. (iii) Find the value of p. (iv) If the pass mark is 20, what percent of students passed the exam? b) A health officer organized a medical camp in a remote village. During the medical check-up of a few children of the village of age group 15 years, their weights were recorded as follows. Weight (in kg) 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Number of students 7 k 10 6 4 (i) Draw the cumulative frequency table. (ii) If the first quartile (Q1 ) is 22 kg, find the Q1 class. (iii) Find the value of k. (iv) If the children of weight group (30 – 50) kg are considered as good in health, what percent of children may have health issue?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 293 Vedanta Excel in Mathematics - Book 10 Statistics Project Work and Activity Section 12. a) Collect the marks obtained by the students of your class in mathematics in recent exam. Then, write the marks in a frequency distribution table with class interval 10 and calculate: (i) the median (ii) Lower quartile (Q1 ) (iii) Upper quartile (Q3 ) b) Collect the weights of family members of your 20 friends. Show the weights in a frequency distribution table of class of length 10 and calculate the quartiles. 14.11 Mode of grouped and continuous data Let’s take a set of ordered observations. 2, 3, 5, 4, 3, 6, 6, 5, 5, 2, 5, 2, 5, 3, 5 Here, the observation ‘5’ appears maximum number of times. Therefore, 5 is the mode of the given observations. Thus, the mode of a set of data is the value with the highest frequency. A distribution that has two modes is called bimodal. The mode of a set of data is denoted by Mo . (i) Mode of individual data: In the case of individual data, the item which is mostly repeated in the given set of data is the required mode. Example 1: Find the mode for the following distribution, 25, 20, 28, 25, 20, 25, 28, 25, 20 Solution: Arranging the data in ascending order 20, 20, 20, 25, 25, 25, 25, 28, 28 Here, 25 has the highest frequency. ∴ Mode = 25 (ii) Mode of discrete data: In the case of discrete data, mode can be found just by inspection, i.e., just by taking an item with highest frequency. Example 2: The following table shows the shoes number against the number of its demands. Find the modal size of shoes. Shoes size number 4 5 6 7 8 Number of demands 500 1200 1400 2500 1700 Solution: Here, the shoes number 7 has the highest number of demands. ∴ Modal size of shoes is 7. (iii) Mode of grouped and continuous data In the case of grouped and continuous data, the class with highest frequency is observed and it is taken as the model class. Then, by using the following formula, mode can be computed. Mode (M0 ) = L + f 1 – f 0 2f 1 – f 0 – f 2 × c Where, L = the lower limit of the model class f 0 = the frequency of the class preceding the model class f 1 = the frequency of the model class f 2 = the frequency of the class succeeding the model class c = the width of the class interval

Vedanta Excel in Mathematics - Book 10 294 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics 14.12Derivation of formula of mode in grouped data Step 1: Draw a histogram as shown in the figure given below from a grouped data. Step 2: Suppose, f 1 = the frequency of the modal class, f 0 = the frequency of the class preceding the modal class, f 2 = the frequency of the class succeeding the modal class, Step 3: Join the upper corners E and A of modal class to the right top B of pre-modal class and left top D of post-modal class respectively. Then, AB = AP – BP = f 1 – f 0 and ED = EQ – DQ = f 1 – f 2 Step 4: Suppose, L and U are the lower limit and upper limit of the modal class. i.e., OP = L and OQ = U and PQ = modal class. Step 5: Draw a perpendicular line from the point C (point of intersection of lines AD and EB) to x-axis at M. Step 6: Suppose, size of modal class = AE = PQ = c and PM = FC = x. Then, Mode = OM = OP + PM = L + x Since, ∆ AFC and ∆ AED are similar. ∴ FC AE = AC AD [The corresponding sides of similar triangles are proportional] or, x c = AC AD ∴x = AC AD × c … (i) Also, ∆ CDE and ∆ ABC are similar. CD AC = ED AB [The corresponding sides of similar triangles are proportional] or, CD AC = f 1 – f 2 f 1 – f 0 or, CD AC + 1 = f 1 – f 2 f 1 – f 0 + 1 [Adding 1 on both sides] or, CD + AC AC = f 1 – f 2 + f 1 – f 0 f 1 – f 0 ∴ AD AC = 2f1 – f 0 – f 2 f 1 – f 0 … (ii) From (i) and (ii), we get x = f 1 – f 0 2f 1 – f 0 – f 2 × c Step 7: Mode = OM = L +x = L + f 1 – f 0 2f 1 – f 0 – f 2 × c

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 295 Vedanta Excel in Mathematics - Book 10 Statistics ∴ Mode = L + f1 – f0 2f1 – f0 – f2 × c Where, L = lower limit of modal class, f0 = the frequency of the class preceding the modal class, f1 = frequency of modal class, f2 = the frequency of the class succeeding the modal class, c = width of class Example 3: The monthly consumption of electricity of 185 consumers is given in the table below. Monthly consumption (in units) 0 - 30 30 - 60 60 - 90 90 - 120 120 - 150 Number of consumers 20 15 25 75 50 (i) Find the modal class. (ii) Compute the mode. Solution: (i) Here, the highest frequency is 75 and its corresponding class is (90-120). ∴ Modal class is (90-120) (ii) L = 90, f1 = 75, f0 = 25, f2 = 50 and c = 30 We have, mode (Mo ) = L + f 1 – f 0 2f 1 – f 0 – f 2 × c = 90 + 75 – 25 2 × 75 – 25 – 50 × 30 = 90 + 50 75 × 30 = 90 + 20 = 110 Hence, the required mode is 110 units. Example 4: The following table represents the age distribution of cases of a certain viral disease admitted in a hospital. Age (in years) 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Number of cases 3 7 10 x 12 6 8 (i) If the modal age is 36 year, find the modal class. (ii) Find the missing frequency x. (iii) Find the percentage of cases found in highest age group. Solution: (i) Since, the modal age is 36 years, it lies in the class (30-40). ∴ Modal class is (30-40) (ii) Again, L = 30, f 1 = x, f 0 = 10, f 2 = 12 and c = 10 We have, mode (Mo ) = L + f 1 – f 0 2f 1 – f 0 – f 2 × c or, 36 = 30 + x – 10 2 × x – 10 – 12 × 10

Vedanta Excel in Mathematics - Book 10 296 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics or, 36 – 30 = x – 10 2x – 22 × 10 or, 6 = x – 10 2x – 22 × 10 or, 12x – 132 = 10x – 100 or, 2x = 32 or, x = 16 Hence, the missing frequency is 16. (iii) Total number of cases = 3 + 7 + 10 + x + 12 + 6 + 8 = 46 + x = 46 + 16 = 62 Number of cases found in highest age group = 8 ∴ Percentage of cases found in highest age group = 8 62 × 100% = 12.9% 14.13Relationship between Mean, Median and Mode An empirical relationship is a relationship that is supported by experiment and observation but not necessarily supported by theory. According to Karl Pearson, there exists an empirical relationship between three measures of central tendency: Mean, Median and Mode in a moderately skewed distribution. Mode can be computed by the following empirical relation between: Mode = 3 Median – 2 Mean Example 5: In a continuous data, if the mean and median are 20.5 and 20 respectively, find the approximate value of mode. Solution: Here, Mean = 20.5 and Median = 20 Mode =? By using Karl Pearson’s relation, Mode = 3 Median – 2 Mean = 3 × 20 – 2 × 20.5 = 19 ∴ The required mode is 19. Example 6: In a continuous data, the median is 35 and mode is 36, find the mean of the data. Solution: Here, Median = 35 and Mode = 36 Mean =? By using Karl Pearson’s relation, Mode = 3 Median – 2 Mean or, 36 = 3 × 35 – 2 Mean or, 2 Mean = 69 ∴ Mean = 34.5 ∴ The required mean is 34.5 14.14 Range of grouped and continuous data The difference between the largest and the smallest score is called range. ∴ Range = Largest score – Smallest score. It is the simplest measure of dispersion. It gives a comprehensive view of the total spread of the observations. (i) Range of ungrouped data The formula to find the range of ungrouped data or discrete distribution of data is given as: ∴Range = Largest value (L) – Lowest value (S)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 297 Vedanta Excel in Mathematics - Book 10 Statistics (ii) Range of Grouped Data In the case of continuous frequency distribution or grouped data, the range is defined as the difference between the upper limit of the highest interval and the lower limit of the lowest interval. Thus, the formula to calculate the range of a grouped data is given below ∴Range = Upper limit of highest interval (ULH) – Lower limit of lowest interval (LLL). Example 1: The weights of 7 boy and 7 girl students of class 10 are given below. Gender Weights (in kg) Boys 40 55 60 35 48 54 45 Girls 35 41 45 30 50 47 36 (i) Find the range of each group. (ii) Find the range of combined group. Solution: Here, (i) For boys’ group: minimum weight = 35 kg and maximum weight = 60 kg Range = Maximum weight – Minimum weight = 60 kg – 35 kg = 25 kg For girls’ group: minimum weight = 30 kg and maximum weight = 50 kg ∴ Range = Maximum weight – Minimum weight = 50 kg – 30 kg = 20 kg (ii) For combined group: minimum weight = 30 kg and maximum weight = 60 kg Range = Maximum weight – Minimum weight = 60 kg – 30 kg = 30 kg Example 2: The table shows the marks obtained by the students in a Mathematics test. Find the range. Marks obtained 20 30 40 50 60 70 80 Number of students 4 6 11 15 12 10 7 Solution: Here, highest mark = 80 and lowest mark = 20 ∴Range = Highest mark – Lowest mark = 80 – 20 = 60 Example 3: Find the range from the following data. Monthly salary (in Rs 1000) 8 - 10 10 - 12 12 - 14 14 - 16 16 - 18 18 - 20 Number of workers 5 10 15 12 7 4 Solution: Here, Upper limit of highest class interval = Rs 20,000 Lower limit of lowest class interval = Rs 8000 ∴Range = Upper limit of highest class interval – Lower limit of lowest class interval = Rs 20,000 – Rs 8,000 = Rs 12,000

Vedanta Excel in Mathematics - Book 10 298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics EXERCISE 14.3 General section 1. a) Define mode. Write the formula to find the mode of a grouped and continuous data. b) Define range. Write the formula to find the range of a grouped and continuous data. 2. Find the modal class from the given data. Marks obtained 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 Number of students 7 10 18 20 12 6 a) Age (in years) 0 - 15 15 - 30 30 - 45 45 - 60 60 - 75 Number of people 150 140 200 90 50 b) Height (in cm) 145 - 150 150 - 155 155 - 160 160 - 165 165 - 170 Number of persons 8 10 9 15 12 c) 3. From the data given in the tables below, find the range. Marks obtained 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Number of students 3 6 9 12 7 4 a) Weight (in kg) 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 Number of people 16 18 24 17 11 b) G.P.A. 2.00-2.40 2.40-2.80 2.80-3.20 3.20-3.60 3.60-4.00 Number of students 2 5 12 9 15 4. Using the empirical relationship between mean, median and mode (Karl Pearson’s Formula), find the approximate value of mean or median or mode from the table given below. No. Mean Median Mode a) 40 39 …… b) 15 14.5 …… c) …… 27 28 d) …… 34 33 e) 26 …… 27 f) 45.5 …… 45 c)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 299 Vedanta Excel in Mathematics - Book 10 Statistics Creative section 5. From each of the data given below, (i) find the modal class (i) compute the mode. Age (in years) 6 - 8 8 - 10 10 - 12 12 - 14 14 - 16 Number of pupils 10 35 70 35 12 a) Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Number of students 20 24 40 36 15 b) Wage (in Rs.) 0 - 200 200 - 400 400 - 600 600 - 800 800 - 1000 Number of workers 15 6 18 10 7 c) 6. Find the mode of the following data Class 0 - 10 0 - 20 0 - 30 0 - 40 0 - 50 Frequency 24 54 104 124 140 a) Marks 35 - 40 35 - 45 35 - 50 35 - 55 35 - 60 Number of students 12 48 88 112 130 b) 7. a) The following table shows the marks obtained by the students in Mathematics in an examination. Marks 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Number of students 8 10 p 6 4 (i) If the modal mark is 45, find the modal class. (ii) Find the missing frequency p. (iii) How many students are passed in the examination, if the pass mark is 40? b) The table given below represents the age distribution of cases of a certain viral disease. Age (in years) 0 - 15 15 - 30 30 - 45 45 - 60 60 - 75 75 - 90 Number of cases 3 7 12 m 13 6 (i) If the modal age is 54 year, find the modal class. (ii) Find the missing frequency m. (iii) Find the percentage of cases found in lowest age group. Project Work and Activity Section 9. a) Collect the SEE result of last year in GPA. Write the GPAs in a frequency distribution table with class interval 0.4 and calculate: (i) mean (ii) median (iii) mode

Vedanta Excel in Mathematics - Book 10 300 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Statistics b) Collect the ages of family members of your 15 friends. Display the weights in a frequency distribution table of class of length 10 and calculate the central tendencies and compare them. OBJECTIVE QUESTIONS Tick the correct alternative. 1. The average which divides the arranged data into two equal parts is…. (A) mean (B) median (C) mode (D) first quartile 2. What does c.f. represent in the formula, Md = N 2 – c.f. L + × c f ? (A) Cumulative frequency of median class (B) Cumulative frequency of class preceding the median class (C) Cumulative frequency of class succeeding the median class (D) Cumulative frequency of any one of the class of the given data 3. The formula for finding the lower quartile of a continuous data is (A) N 2 – c.f. L + × i f (B) N 4 – c.f. L + × i f (C) 3N 4 – c.f. L + × i f (D) N 2 – c.f. L – × i f 4. The upper quartile of a data is (A) the maximum item of below 25% items of the arranged data. (B) the maximum item of below 75% items of the arranged data. (C) the minimum item of above 50% items of the arranged data. (D) the maximum item of above 25% items of the arranged data. 5. The first quartile divides the data in the ratio of (A) 1:2 (B) 1:3 (C) 1:4 (D) 2:3 6. The observation having highest frequency is called (A) mean (B) mode (C) median (D) range 7. The formula for finding the mode of a grouped and continuous data is (A) L + f 1 – f 0 2f 1 – f 0 – f 2 × c (B) L + f 1 – f 0 2f 1 – f 0 – f 2 × c (C) L + f 1 – f 0 f 1 – f 0 – f 2 × c (D) L + f 1 – f 0 2f 2 – f 0 – f 2 × c 8. In the formula of mode, Mo = L + f 1 – f0 2f1 – f0 – f2 × c , what does f1 represent for? (A) Frequency of modal class (B) Frequency of class preceding the modal class (C) Frequency of class succeeding the modal class (D) Frequency of the greatest class interval 9. The empirical relationship between mean, median and mode is: (A) Mean>Median>Mode (B) Mean=Median=Mode S (C) Mode = 3Median – 2 Mean (D) Mode = 3Mean – 2 Median 10. Which of the following is not the measure of central tendency? (A) mean (B) median (C) mode (D) range