Oasis School Mathematics 10 new

Oasis School Mathematics-10 247 Corollary The angle in a semi-circle is 900 (a right angle). Experimental Verification: Draw two circles of different radii with center 'O' and in each circle draw ∠BAC on the semi-circle taking BC as the diameter. to verify: ∠BAC = 900 Now, measure the angle BAC with the help of a protractor and tabulate them as follows; Figure ∠BAC remarks I 900 ∠BAC = 900 II 900 ∠BAC = 900 Conclusion: Hence, the angle on a semi-circle is a right angle. theoretical Proof : Given : O is the center of a circle and ∠BAC is an angle the circumference of the circle to prove : ∠BAC = 900 Proof : S.N. Statements S.N. reasons 1. ∠BAC = 1 2 ∠BOC 1. Relation between inscribed angle and central angle 2. ∠BOC = 1800 2. BOC is a straight angle 3. ∠BAC = 1 2 × 1800 = 900 3. From (1) and (2) Hence, proved. Hence, the angle on the semi-circle is a right angle. Theorem 15.6 Angles on the same segment of a circle are equal. B C A O Fig. I Fig. II C A O B B C A O

248 Oasis School Mathematics-10 Or, Inscribed angles standing on the same arc are equal. Experimental Verification: Draw two circles of different radii with center O and also draw angles BAC and BDC on the same segment BC. to Verify : ∠BAC = ∠BDC observation: Figure ∠BAC ∠BDC remarks I ....° ....° ∠BAC = ∠BDC II ....° ....° ∠BAC = ∠BDC Conclusion : Thus angles on the same segment of a circle are equal. theoretical Proof: Given : O is the center of the circle. ∠BAC and ∠BDC are the inscribed angles on the same segment BC. to Prove : ∠BAC = ∠BDC Construction : Join B and O, C and O. Proof : S.N. Statements S.N. reasons 1. ∠BAC = 1 2 ∠BOC 1. Relation between inscribed angle and central angle 2. ∠BDC = 1 2 ∠BOC 2. Same as (1) 3. ∴ ∠BAC = ∠BDC 3. From (1) and (2) Hence, proved. Hence, angles on the same segment of a circle are equal. theorem 15.7 the opposite angles of a cyclic quadrilateral are supplementary. O D A B C Fig. I Fig. II A B C D O A B D C O

Oasis School Mathematics-10 249 Experimental Verification: Draw two circles of different radii with center O and also draw a cyclic quadrilateral ABCD in each circle. to verify: (i) ∠A + ∠C = 1800 (ii) ∠B + ∠D = 1800 Observation: Figure ∠A ∠B ∠C ∠D remarks I ......0 ......0 ......0 ......0 ∠A + ∠C = 1800 II ......0 ......0 ......0 ......0 ∠B + ∠D = 1800 Conclusion: Hence, the opposite angles of a cyclic quadrilateral are supplementary. theoretical Proof: Given : O is the center of the circle. ABCD is a cyclic quadrilateral. to prove : (i) ∠A + ∠C = 1800 (ii) ∠B +∠D = 1800 Construction: Join B and O, D and O. Let the obtuse ∠BOD be x and reflex ∠BOD be y. Proof : S.N. Statements S.N. reasons 1. ∠BAD = 1 2 x 1. Relation between the central angle and inscribed angle standing on the same arc 2. ∠BCD = 1 2 y 2. Same as statement 1 3. ∠BAC + ∠BCD = 1 2 (x + y) 3. Adding statements 1 and 2 4. x + y = 3600 4. Angle in one complete rotation is 3600 5. ∴ ∠BAC + ∠BCD = 1 2 × 3600 = 1800 i.e. ∠A + ∠C = 1800 5. From 3 and 4 6. Similarly, ∠B + ∠D = 1800 6. By joining A and O, C and O and following similar process Hence, proved. Hence, the opposite angles of a cyclic quadrilateral are supplementary. If a pair of opposite angles of a quadrilateral are supplementary, then its vertices are concyclic. B A D C O Fig. I Fig. II B C D A O O A B D C y0 x

250 Oasis School Mathematics-10 Corollary If a side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the opposite interior angle. Experimental Verification: Fig. I Fig. II A B C E D A D C E B O O Draw two circles of different radii taking O as the center. Draw cyclic quadrilateral ABCD in each figure. Produce side BC to E. to verify : ∠DCE = ∠BAD observation: Figure ∠BAD ∠DCE remarks (i) ......0 ......0 ∠DCE = ∠BAD (ii) ......0 ......0 ∠DCE = ∠BAD Conclusion: Hence, if a side of a cyclic quadrilateral is produced, an exterior angle so formed is equal to the opposite interior angle. theoretical Proof: If a side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the opposite interior angles. Given : O is the center of the circle and the side BC of the cyclic quadrilateral ABCD is produced to E. to prove : ∠DCE = ∠BAD Proof: S.N. Statements S.N. reasons 1. ∠BAD +∠BCD = 1800 1. Opposite angles of a cyclic quadrilateral are supplementary 2. ∠BCD + ∠DCE = 1800 2. Sum of linear pair is 1800 3. ∠BAD + ∠BCD = ∠BCD + ∠DCE 3. From statements 1 and 2 4. ∴ ∠BAD = ∠DCE 4. From 3 Hence, proved Hence, if a side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to opposite interior angles. O C E D A B

Oasis School Mathematics-10 251 Worked out examples example 1 Find the angles marked by the letters x0 and y0 in the following figures: Solution: ∠ADB = ∠ACB = 300 (Angles on the same segment) ∴ y = 300 or, 2∠ADB = ∠AOB (Relation between inscribed angle and central angle standing on the same arc.) or, 2 × 300 = ∠AOB ∴ ∠AOB = 600 ∴ x = 600 example 2 Find the value of x in the given figure. Solution: Construction: Join A and O i. ∠BAO = ∠ABO = 400 (Base angles of isosceles ∠AOB as AO = OB) ii. ∠OAC = ∠OCA = 300 (Base angles of isosceles ∠AOB as AO = OC) iii. ∴ ∠BAO + ∠OAC = ∠BAC = 400 + 300 = 700 (whole part axiom) ∴ Obtuse ∠BOC = 2 ∠BAC (Relation between inscribed angle and central angle standing on the same arc) x = 2 × 70 x = 1400 ∴ x = 1400 example 3 In the given figure, BC is a diameter and AB = AC, find the value of x. Solution: Here, ∠BAC = 900 (Angle at the semi-circle) ∠ABC = ∠ACB (Being AB=AC) In ∆ABC, ∠BAC + ∠ABC + ∠ACB = 1800 or, 900 + ∠ACB + ∠ACB = 1800 (Base angles of an isosceles triangle, AB = AC) or, 2∠ACB = 1800 – 900 A B C D O 300 x0 y0 C 400 300 x O B A A C D B O x0 O x B A C 30 400 0

252 Oasis School Mathematics-10 ∴ ∠ACB = 900 2 = 450 iii. ∠ADB = ∠ACB = 450 (Angles on the same segment) ∴ x = 450 example 4 In the given figure, 'O' is the centre of the circle, BP⊥ AC, find the value of x. Solution: Here, 2∠CAB = ∠BOC (Relation between the inscribed angle and central angle standing on the same arc) or, 2 × 600 = ∠BOC ∴ ∠BOC = 1200 In ∆CPO, ∠CPO + ∠OCP = ∠BOC (Sum of two angles of a triangle is equal to the opposite exterior angle) 900 + ∠OCP = 1200 ∴ ∠OCP = 1200 – 900 = 300 ∴ x = 300 example 5 In the given figure, ABCD is a cyclic quadrilateral. BC is produced to E, find the value of x and y. Solution: Here, ∠BCD +∠DCE = 1800 y + 700 = 1800 (Sum of linear pair) ∴ y = 1800 – 700 = 1100 Again, x + y = 1800 (Opposite angles of a cyclic quadrilateral) x + 1100 = 1800 ∴ x = 1800 – 1100 = 700 Example 6 (a) In the given figure, O is the center of a circle in which ∠AOC = 1600 and ∠DAB = 400 . Find the value of x. Solution: Here, Obtuse ∠AOC + Ref. ∠AOC = 3600 (Angle in one complete rotation) or, 1600 + Ref. ∠AOC = 3600 A B C x O P 600 700 A D B C y E 0 x0 A B O C D 1600 x 400

Oasis School Mathematics-10 253 or, Ref. ∠AOC = 3600 – 1600 = 2000 ∴ ∠ABC = 1 2 × Ref. ∠AOC (Relation between inscribed angle and central angle standing on the same arc) = 1 2 × 2000 = 1000 Again, In ∆ABD ∠BAD + ∠ABD +∠ADB = 1800 (Sum of angles of a triangle) 400 + 1000 + x = 1800 ∴ x = 1800 – 1400 = 400 example 7 In the adjoining figure, O is the center of the circle. Calculate the value of x. Solution: Construction: Join B and C, i. ∠ADC + ∠ABC = 1800 (Opposite angles of the cyclic quadrilateral) or, 1150 + ∠ABC = 1800 or, ∠ABC = 1800 – 1150 = 650 ii. In ∆BOC, ∠OBC = ∠OCB = 65° (OB = OC, radii of the same circle) And ∠BOC + ∠OBC + ∠OCB = 180° (Sum of angles of the triangle) or, ∠BOC + 65° + 65° = 180° or, ∠BOC = 180° – 130° ∴ x = 50° example 8 In the given figure, AB is a diameter and ∠AEB = 500 . Find the value of ∠COD. Solution: Given AOB is a diameter ∠AEB = 500 Join AD Here, ∠ADB = 900 [Angle on the semi-circle] In ∆ADE, B C D A O x 1150 B O C D A 500 E

254 Oasis School Mathematics-10 ∠AED + ∠ DAE = ∠ADB [Sum of two angles of a triangle is equal to opposite exterior angles] or, 500 + ∠DAE = 900 or, ∠DAE = 900 – 500 or, ∠DAE = 400 Again, ∠COD = 2∠DAE (Central angle is twice the inscribed angle standing on the same arc.] or, ∠COD = 2 × 400 = 800 example 9 Points P, Q, R, S and T are on the circumference of a circle, such that PS ⊥ PQ and RT ⊥ PR. Prove that PQ = ST. Solution: Given : P, Q, R are on the circumference of a circle, such that PS ⊥ PQ and RT ⊥ PR. to prove : PQ = ST Construction : Join Q and T Proof : S.N. Statements S.N. reasons 1. ∠PRT = ∠PST = 900 1. Angles on the same segment and ∠PRT = 900 2. ∠PST + ∠PQT = 1800 2. Opposite angles of the cyclic quadrilateral 3. ∴ ∠PQT = 900 3. From 1 and 2 4. ∠SPQ + ∠QTS = 1800 4. Opposite angles of the cyclic quadrilateral. 5. ∴ ∠QTS = 900 5. ∠SPQ = 900 and from 1st no. (4) 6. ∴ PQTS is a rectangle 6. Each angle of quadrilateral being 900 7. PQ = ST 7. Opposite sides of the rectangle Hence, proved Example 10 In the given figure, O is the center of a circle. Prove that: ∠XOZ = 2(∠XZy + ∠yXZ). Solution: Given : O is the center of the circle. to prove : ∠XOZ = 2(∠XZY + ∠YXZ) Proof : P Q S T R Y Z O X

Oasis School Mathematics-10 255 S.N. Statements S.N. reasons 1. ∠XOY = 2∠XZY 1. Relation between inscribed angle and central angle standing on the same arc 2. ∠YOZ = 2 ∠YXZ 2. Same as 1. 3. ∠XOY + ∠YOZ = 2(∠XZY+∠YXZ) 3. Adding 1 and 2 4. ∠XOZ = 2(∠XZY + ∠YXZ) 4. From 3, whole part axiom Hence, proved. example 11 Prove that the quadrilateral formed by an angular bisector of a cyclic quadrilateral ABCD is also cyclic. Solution: Given : ABCD is a cyclic quadrilateral. AP, BP, CR and DR are the bisectors of ∠A, ∠B, ∠C and ∠D respectively. to prove : PQRS is a cyclic quadrilateral. Proof S.N. Statements S.N. reasons 1. ∠APB +∠PAB + ∠PBA = 1800 1. Sum of angles of a triangle 2. ∠CRD +∠RCD + ∠RDC = 1800 2. Same as 1 3. ∴ ∠APB + ∠PAB + ∠PBA + ∠CRD + ∠RCD + ∠RDC = 3600 3. Adding 1 and 2 4. ∠APB + ∠CRD + 1 2 (∠A+∠B+∠C+∠D) = 3600 4. AP, BP, CR and DR are the bisectors of ∠A, ∠B, ∠C and ∠D 5. ∠APB + ∠CRD = 3600 – 1 2 × 3600 = 1800 5. Sum of the angles of the quadrilateral 6. ∴ PQRS is a cyclic quadrilateral 6. Opposite angles of the quadrilateral being supplementary Hence, proved example 12 In the given figure, chords AB and CD are produced to meet at E. Prove that: ∠AoC – ∠BOD = 2∠AeC. Solution: Given : O is the center of the circle chord and AB and CD are produced to meet at E. to prove : ∠AOC – ∠BOD = 2∠AEC Construction : Join AD D P R S Q C B A A E D C O B A E D C O B

256 Oasis School Mathematics-10 Proof : S.N. Statements S.N. reasons 1. ∠AOC = 2∠ADC 1. Central angle is twice the inscribed angle standing on the same arc 2. ∠BOD = 2∠BAD 2. As statement 1 3. ∠AOC – ∠BOD = 2(∠ADC – ∠BAD) 3. Subtracting statement 2 from 1 4. ∠ADC = ∠BAD + ∠AED 4. Sum of two angles of a triangle is equal to the opposite exterior angle. 5. ∠AED = ∠ADC – ∠BAD 5. From 4 6. ∠AOC – ∠BOD = 2 ∠AED 6. From 3 and 5 Hence, proved. example 13 In the given figure, AB is a diameter of the circle with centre O. If AD = DE, prove that OE // BD. Solution: Given : In the given figure, AE = DE. to prove : OE // BD Construction : Join OD Proof S.N. Statements S.N. reasons 1. In ∆ADE and ∆DOE (i) AE = DE (S) (ii) OA = OD (S) (iii) OE = OE (S) 1. (i) Given (ii) Radii of same circle (iii) Common side 2. ∠AOE ≅ ∠DOE 2. By SSS axiom 3. ∠AOE = ∠DOE 3. Corresponding angles of congruent triangles 4. ∠AOE = 1 2∠AOD 4. From statement 3 5. ∠ABD = 1 2∠AOD 5. Inscribed angle is half of the centre angle standing on the same arc 6. ∠AOE = ∠ABD 6. From statements 4 and 5 7. OE // BD 7. From statement 6, being corresponding angles equal Hence, proved. A C B E O D

Oasis School Mathematics-10 257 exercise 14.2 1. (a) What is the relation between the inscribed angle and central angle standing on the same arc? (b) What is the relation between the inscribed angles standing on the same arc? (c) What is the value of an angle on a semi-circle? (d) What is the relation between the opposite angles of a cyclic quadrilateral? 2. (a) In the given figure, write the relation between ∠BAC and ∠BOC. (b) In the given figure, write two pairs of angles which are equal. (c) In the given figure, what is the value of ∠ACB? Why? (d) In the given figure, write the relation between ∠A and ∠C as well as ∠B and ∠D. (e) In the given figure, write the angle which is equal to ∠PST. 3. Find the sizes of unknown angles in the following circles with center o. A B C O A D C B C B A O A D C B T P Q S R (a) (b) (c) z 30 x y 20 x0 1000 A B C O A B C 2300 x A B C O

258 Oasis School Mathematics-10 O O E F (d) (e) (f) Z C A D O 120 x 0 0 S P o Q R x y 300 x 1400 y B O Q S R (g) (h) (i) A B C O 1000 x P Q S R x 1400 A C B E D x O y B C A O 350 y x x0 550 C B D A O 500 x0 A x Z Y D 1200 400 x A D C O B E A B C D 230 x0 D A B C 100 920 x E D A C O 350 x0 x0 O 250 500 y0 (j) (m) (p) (k) (n) (q) (l) (o) (r) D C B A x y0 3y 2 A E B C D M N E R 2300 x Q O P Q O T R S P 400 x0 (s) (t) (u) P S Q R U T (x–20)0 (v) A (w) B C D A C O B D 250 x 200 1200 x0 1100 O B

Oasis School Mathematics-10 259 4. (a) In the given figure, PQ = 6 cm, QR = 8 cm. Find the length of OQ. (b) In the given figure, AC = 6cm, OC = 5 cm. Find the length of BC. 5. (a) In the given figure, O is the center of the circle, PQ ⊥AB. If ∠BAP = 700 , find ∠ABO. (b) In the given figure, O is the center of the circle. AD⊥BC. If ∠OCD = 300 , find ∠ABC. 6. (a) OABC is a parallelogram in the given figure, where O is the centre of the given circle. Find x and y. (b) In the given figure, PQRS is a parallelogram. Find the value of ∠SPQ. 7. (a) In the given figure, O is the center of the circle. Find ∠ADB. (b) In the given figure, O is the center of the circle. C is the mid-point of BD and ∠ADB = 500 . Find ∠ABD. (c) In the given figure, ∆ABC is an isosceles triangle with AB = AC and ∠ABC = 500 . Find ∠BDC and ∠BEC. (d) A chord of a circle is equal to its radius. Prove that the angle subtended by the chord at a point on the circumference of the circle is either 1500 or 300 . Q P O 6cm 8cm R A O B C 6cm 5cm A B P O Q C B A O D x0 O C B A y0 S Q R P A B O 300 D C A O C D B B C A D E

260 Oasis School Mathematics-10 8. In the given figure, O is the center of the circle, AB is the diameter and DO⊥AB. Prove that ∠AMC = ∠ODA. 9. In the figure alongside DM⊥EF, FN⊥ED. Prove that ∠EMN = ∠EDF. 10. (a) In the given cyclic quadrilateral ABCD, DC is the bisector of ∠BDE. Prove that AC = BC. (b) In the given cyclic quadrilateral ABCD, AC = BC. Prove that DC is the bisector of ∠BDE. 11. ABCD is a parallelogram. The circle through A, B and C intersects CD produced at E. Prove that AE = AD. 12. In the figure alongside, O is the center of the circle. Then prove that 2∠APC = ∠AOC + ∠BOD. 13. In the given figure, O is the center of the circle. Prove that ∠AOB = ∠ACB, if AC = CD. 14. In the given figure, two circles intersect at M and N. PQ and RS pass through M. R, P, S and Q are joined to N. Prove that ∠PNR = ∠SNQ. 15. In the given figure, if AB = AC, prove that BE = DC. E D C B A E D C B A E C D A B B D P C A O O C N D B A M Q M N P S R B C O A D E D F E N M D B O A C M

Oasis School Mathematics-10 261 16. In the given figure, AD and BC are chords perpendicular to each other, and O is the center of the circle. Prove that ∠AOC + ∠BOD = 180°. 17. Two circles intersect each other at points A and B. PAQ and SBR are two straight lines. Prove that PS||QR. 18. In the given figure, PQ//BC, prove that AP = AQ. 19. In the given diagram, M and N are the centers of two circles, which intersect at A and C. MA and MC are produced to meet the other circle at B and D. Prove that AB = CD. 20. In the figure, AP is the radius of the bigger circle and diameter of the smaller circle. Prove that AQ = QC. 21. In the given figure, P and Q are mid–points of arcs AB and AC respectively. Prove that AX = AY. 22. In the given figure, if AB//DC and AD = BC, show that ABCD is a cyclic quadrilateral. 23. In the figure alongside, BD = DP and ∠ABD = ∠DBC. Prove that AB = CP. A O B C D P Q R B A S A P B C Q B D M N A C C A P B Q A B C x y P Q D C A B A D B C P

262 Oasis School Mathematics-10 24. (a) In the given figure, AM is the bisector of ∠BAC. Prove that EF//BC. (b) In the given figure, EF // BC. Prove that AM is the bisector of ∠BAC. 25. In ∆ABC, P, Q and R are mid-points of BC, AC and AB respectively. If AW⊥BC, prove that W, P, Q and R are concyclic points. 26. (a) In the given figure, AB is a diameter, and OE//BD, prove that AE = ED. (b) In the given figure, AB is a diameter and AE = ED, prove that OE // BD. A D C M E F B A D C M E F B A B C W P R Q A C D B A C D B Answers 1. Consult your teacher 2. Consult your teacher 3. (a) x = 500 (b) x= 650 (c) x=300 , y=200 , z=1000 (d) x=600 , y=300 (e) x = 300 (f) x = 700 , y=1100 (g) x=1350 (h) x=400 , y=700 (i) x = 550 , y=1100 (j) x = 540 , y=360 (k) x = 600 (l) x=1100 (m) x = 500 , y=550 (n) x = 400 (o) x = 440 (p) x = 800 (q) x = 980 (r) x = 1250 (s) x = 500 (t) x = 700 (u) x = 900 (v) x = 1150 (w) x = 400 4. (a) 5 cm (b) 8cm 5. (a) 500 (b) 600 6. (a) x=600 , y=1200 (b) 900 7. (a) 600 (b) 500 (c) 800 and 1000 Project Work With the help of a thin string, prepare a circle on a chart paper. With the help of a toothpick make the necessary angles and verify the above theorems. Do you know ! The geometry of a plane figure is known as Euclidean Geometry.

Oasis School Mathematics-10 263 miscellaneous exercise type – I 1. ‘O’ is the centre of the circle. If ∠BAC and ∠BOC are inscribed angles and central angle standing on the same arc BC. (a) Draw two figures of different size and verify by actual measurement that ∠BAC = 1 2 ∠BOC. (b) If the side BO is produced to P on AC such that ∠OPC = 85° and ∠BAC = 60° then find the measurement of ∠OCP. [Ans: 35°] 2. ∠QPS and ∠QRS are two angles on the circumference standing on the same arc QS. (a) Make two such figure with different measurement. Verify experimentally that ∠QPS = ∠QRS. (b) If PR > QS and PQ and RS are produced to meet at point T outside the circle and PT = RT then show that PS = QR. 3. A, B, C and D are four points on the circumference of a circle. ABCD is a cyclic quadrilateral. (a) Draw two cyclic quadrilateral and verify by measurement that ∠A + ∠C = 180° and ∠B + ∠D = 180°. (b) ‘O’ is the centre of the circle. Join OA and OC. If ∠AOC = 130°, then find the value of ∠B and ∠D. [Ans: 65° and 115°] 4. Study the given question and answer the following. (a) Draw two cyclic quadrilateral of different size and verify experimentally that opposite angles of cyclic quadrilateral are supplementary. (b) In the cyclic quadrilateral ABCD, OB and OD are joined such that OBCD is a parallelogram. Find the value of ∠BAD, ∠BCD and ∠OBC. [Ans: 60°, 120°, 60°] 5. A side BC of cyclic quadrilateral is produced to E. (a) Make two different sized cyclic quadrilateral ABCD. Produce side BC to E and verify experimentally that ∠BAD = ∠DEC. (b) ‘O’ is the centre of the circle. Join OB and OC. Join AC. If ∠DCE = 94° and ∠BOC = 144°, find the value of ∠DAC. [Ans: 22°] 6. Study the given question and answer the following. (a) Draw two cyclic quadrilateral ABCD of different measurement and verify by actual measurement that ∠A + ∠C = 180° and ∠B + ∠D = 180°. (b) Take any point E and AD and draw EF DC, where F is a point on AC, prove that ABFE is a cyclic quadrilateral.

264 Oasis School Mathematics-10 7. ABCD is a cyclic quadrilateral side AD is produced to E. (a) Draw two figures of different size and verify by actual measurement that ∠ABC = ∠CDE. (b) Join AC and BD then AC = BC, prove that ∠BDC = ∠CDE. 8. Two chords AB and CD are parallel. (a) Draw two figures of different size and verify experimentally that arc AC = arc BD. (b) Join AD and BC then prove that AD = BC. 9. PQRS is a cyclic quadrilateral, PT is a bisector of ∠SPQ. (a) Draw a figures of different size and verify experimentally that arc ST = arc TRQ. (b) Again, if RU is the bisector of ∠SRQ. Again TU is joined, then prove that TU is a diameter of the circle. 10. AOB is a diameter and C is a point on the circumference. (a) Verify experimentally that ∠ACB = 90°. (b) Join OC. If BC = 24cm, OC = 13cm, find the length of AC. [Ans: 10 cm] type II 1. ‘O’ is the centre of the circle AOB is a diameter. (a) What type of triangle is triangle ABC. (b) If OA = 5cm and BC = 6cm, what is the length of AC? [Ans: 8 cm] (c) Is it possible that AB < AC? Give reason. 2. Study the given figure and answer the questions given below. (a) In the given figure, write two pairs of angles which are equal. (b) If ∠QPS = 50°, find the value of ∠RQS. [Ans: 400 ] (c) Is ∠PQS = ∠QSR? Justify your answer. 3. In the given figure, AB = BD (a) What is the relation of AB and CD? (b) Take any point P and arc CD, if ∠APC = 25°, then what is the value of ∠BPD? [Ans: 250 ] (c) If BC is a diameter, then verify that AC = CPD. 4. ABCD is a cyclic quadrilateral. Side CD is produced to E. (a) Write the name of an angle which equal to ∠ADE. (b) If ABD is an equilateral triangle, what is the value of ∠BAC and ∠BCD? Explain with reasons. (c) If we produce AB to F, what is the relation of ∠FBC with ∠ADE? A O B C P O S R Q A C B D P C B D E A

Oasis School Mathematics-10 265 5. In the given figure, AB and CD are two chords. (a) What is the relation of arc BD with ∠BAD and arc AC with ∠ADC? (b) Produce AB and CD to X, using the result of (a), express ∠AXC in terms of arc AC and arc BD. (c) If AD is an diameter and AB = BX then what type of triangle is ∆ADX? 6. In the given figure, P is the mid point of AB and Q is the mid point of AC . AP and AQ are joined. (a) Which two angels of ∆APM is equal to two angles of ∆AQN? (b) Using above relation prove that ∆APM and ∆AQN are equiangular. (c) What type of triangle is ∆AMN? Give reason. 7. In the given figure PQ// RS, and ∠PQR = ∠RQT (a) Find the two pairs of arcs which are equal. (b) Using above relation, show that RT = QS . (c) Using relation obtained in(b), show that RS = QT. 8. In the given figure, AOB is a diameter. (a) What is the value of ∠APB? What is the relation of ∠POR with ∠PBR? (b) If ∠POR = 50°, what is the value of ∠PQR? [Ans: 650 ] (c) If AP = PQ, what type of triangle is ∆ABQ? 9. In the given figure, ABCD is a parallelogram. (a) Which two angles are equal to ∠DCE? (b) Using above fact prove that AFEB is a cyclic quadrilateral. (c) Is FGD an isosceles triangle? Justify your answer. 10. In the given figure, ∆ABC is an isosceles triangle and ∆AMN is an isosceles triangle. (a) Which two pairs of arcs are equal? (b) Using the facts in (a), prove that MN BC. (c) If ∠MAB = 25°, what is the value of ∠NAC? P R Q T S A P Q R B O B A F D E G C B P C Q M N A D X A C B A N C B M

266 Oasis School Mathematics-10 Full Marks: 14 1. In the given figure, AB // DC //EF, AD // BE, AF // DE. (a) Which two parallelograms are equal i area with parallelogram ADEC? (1) (b) If area of ∆CEF = 15 cm2 , find the area of ∆ABC. (2) (c) Construct a parallelogram ABCD where AB = 6.8 cm, BC = 7 cm and AC = 8 cm. Also, construct a rectangle MBCN equal in area with parallelogram ABCD. (2) 2. 'O' is the centre of the circle. ∠BAC and ∠BDC are two inscribed angles standing on the same arc BC. (a) Draw two figures and verify by actual measurement that ∠BAC = ∠BDC. (2) (b) In the above figure, write the name of another pair of inscribed angles. If BC is produced to E and DC is the bisector of ∠ACE and ∠ACE = 1300 , find the value of ∠ABD. (2) 3. Study the given figure and answer the questions given below: (a) If arc AM = arc BN, show that arc AMN = arc MNB. (1) (b) Show that ACM BDN. (2) (c) If ∠ACM = 700 , what is the value of ∠BDN, explain with reason. (2) D B C A M N O A B C F E D

Oasis School Mathematics-10 267 STATISTICS AND PROBABIlITy Contents • Measures of central tendency • Mean • Median and Quartiles • Mode • Range • Mutually exclusive and mutually non-exclusive events • Independent event • Tree diagram expected Learning outcomes At the end of this unit, students will be able to develop the following competencies: • To convert raw data into grouped data • To find the mean from grouped data • To find the median and quartile from grouped data • To calculate mode of given data • To calculate range of given data • To calculate median class and quartile classes with their respective frequency from the given less than type ogive curve • To identify whether the given events are mutually exclusive or not • To solve the problem related to mutually exclusive and mutually non- exclusive events • To find the probability of independent events • To draw a tree diagram and find the probability of all possible outcomes materials required • A4 size paper, chart paper, flash card, chart paper, playing cards, dice, coin Estimated Teaching Hours 28

268 Oasis School Mathematics-10 15.1 Warm-up Activities Discuss the following in your class and draw a conclusion. • Identify the type of given distribution (i) 25, 27, 28, 30, 35, 40 (ii) x 10 12 14 16 18 f 5 2 7 8 5 (iii) Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 7 3 8 6 In number (ii), frequency of 12 is 2, what does it mean? In number (iii), frequency of (30-40) is 8, what does it mean? Marks obtained by 20 students in a class are given below. 25, 27, 30, 35, 20, 26, 39, 43, 42, 45, 48, 50, 38, 37, 29, 28, 33, 37, 41, 43. How do you prepare the frequency distribution table from the above data? 15.2 measures of Central tendency Statistics deals with the study of estimation. If we are working out with different variables of the same kind, there can be a single value representing all of them. The value so chosen stands for all of them and is known as the central tendency. The following are the measures of central tendency: i) Mean ii) Median i ii) Mode mean (X) The mean or average of the given distribution is denoted by X. It is calculated by using the following mathematical formulae. Unit 15 Statistics

Oasis School Mathematics-10 269 (b) For discrete distribution, Mean (x) = a + ∑fd N Where, a = assumed mean d = x –a, N = ∑f (c) For continuous distribution, Mean (x) = a + ∑fd N Where, a = assumed mean d = m – a m = mid-value In continuous distribution, we can also use the step deviation method to calculate the mean. According to this method, Mean ( X) = a + ∑fd' N × i where, a = assumed mean d' = m – a i m = mid-value i = size of the class interval. I. Direct Method: (a) For individual distribution, Mean (X) = ∑x n Where, ∑x = sum of the variables (b) For discrete distribution, Mean (X) = ∑fx N (c) For continuous distribution, (X) = ∑fm N Where, m is mid–value of variable class and N is the number of variables II. Short-cut method: (a) For individual distribution, Mean (x) = a + ∑d n where, a = assumed mean d = x – a n = number of variables

270 Oasis School Mathematics-10 Worked out examples example 1 Calculate the arithmetic mean from the given data. marks 0–20 20–40 40–60 60–80 80–100 No. of students 5 10 12 18 5 Direct method: Solution: marks mid value (m) Frequency (f) fm 0 – 20 10 5 50 20–40 30 10 300 40–60 50 12 600 60–80 70 18 1260 80–100 90 5 450 ∑f = (N) = 50 ∑fm = 2660 Now, we know, Mean ( X ) = ∑fm ∑f = 2660 50 = 53.20 Hence, the mean marks is 53.2 Short–cut method Solution: Let assumed mean (a) = 50 Class mid value (m) Frequency (f) d = m–a f × d 0 – 20 10 5 –40 –200 20 – 40 30 10 –20 –200 40 – 60 50 12 0 0 60 – 80 70 18 20 360 80 – 100 90 5 40 200 ∑f = 50 ∑fd = 160 Where, a = assumed mean = 50 d = deviation = difference between mid–value and assumed mean ∑f = frequency of the variable class

Oasis School Mathematics-10 271 We know, Mean ( X ) = a + ∑fd N = 50 + 160 50 = 50 + 3.2 = 53.2 Step–deviation method: Solution: Let, assumed mean (a) = 50 Class mid value (m) Frequency (f) d' = m – a i fd' 0 – 20 10 5 –2 –10 20 – 40 30 10 –1 –10 40 – 60 50 12 0 0 60 – 80 70 18 1 18 80 – 100 90 5 2 10 ∑f = 50 ∑fd' = 8 Where, a = assumed mean = 50 d = deviation of assumed mean from the mid-value i = size of the class interval f = frequency of the variable class Now, we know, Mean ( X ) = a + ∑fd' N × i = 50 + 8 50 × 20 = 50 + 3.2 = 53.2 exercise 15.2 1. (a) Write the formula to calculate means in the continuous data using direct method. (b) Write the formula to calculate mean in continuous data using short-cut method. and step deviation method. 2. (a) If ∑fm = 400 and N = 20, find x. (b) If ∑fm = 540 and ∑f = 27, find x. (c) If a = 20, ∑fd = 80 and N = 20, find x. (d) If a = 40 and ∑fd' = 15, i = 10, N = 30, find x. 3. (a) In a series if x = 10, ∑fm = 700 + 5m and ∑f = 40+3m, find the value of m and total number of terms. (b) If the number of terms (N) = 12 + a, ∑fm = 168 + 14a, find the mean. (c) In a continuous series, mean (x) = 32, ∑fx = K and ∑f = 20, find the value of k. (d) If ∑fm = 4000, x = 80, and N = 40 + a 10 , find the value of a. (e) If ∑fm = 2000 + 200a, ∑f = 50 and x = 80, find the value of a.

272 Oasis School Mathematics-10 4. Find the value of mean by using: (i) Direct method (ii) Short-cut method (iii) Step deviation method (a) marks 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 Frequency 20 10 30 50 40 (b) Classes 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequency 3 5 8 4 2 (c) Size: 0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 Frequency 5 7 2 3 4 5. (a) Find the value of 'a' if the mean of the given data is 28. Class 17.5–22.5 22.5–27.5 27.5–32.5 32.5–37.5 Frequency 12 18 12 a (b) Find the value of f if x is 54. Class 0–20 20–40 40–60 60–80 80–100 Frequency 7 f 10 9 13 (c) If the mean of given data is 25, find the value of x. marks obtained 0–10 10–20 20–30 30–40 40–50 Number of students 3 5 4 5 x Answers 1. Consult your teacher 2. (a) 20 (b) 20 (c) 24 (d) 45 3. (a) 12, 76 (b) 14 (c) 640 (d) 100 (e) 10 4. (a) 20.17 (b) 23.64 (c) 4.43 5. (a) 18 (b) f = 11 (c) 3 15.3 median In the given set of variables when arranged either in ascending or in descending order of magnitude, the value which divides the whole arrangement into two equal halves is called the "median". It is also called positional average. It is denoted by Md. e.g. median of 1, 5, 10, 11, 20 is 10. Calculation for median: a) Individual distribution: The positional value of median for individual distribution is given by n+1 2 th ( ( item, where n is No. of observations.

Oasis School Mathematics-10 273 b) Discrete distribution: Let x1 , x2 , x3 , ............ xn be the values of the variables and their respective frequencies be f1 , f2 , f3 .......... fn. Then the value of the median is given by N+1 2 th ( ( position, where N is the number of variables, i.e., total number of frequencies. c) Continuous series, or grouped frequency distribution: Median position for grouped data is obtained by the formula N 2 th ( ( items, where N is number of variables. But here we will be able to find the class variable in which median falls. Now, in order to calculate the exact value of the median from its class, we use the following formula. Median = l + Where, l = lower limit of median class N = number of variables i = size of the median class c.f. = cumulative frequency preceding the median class f = frequency of the median class f – c.f. N 2 × i Worked out examples example 1 Find the median from the given distribution: x: 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 f: 2 5 7 12 4 Solution: Computation for the median: x f c.f. 0 – 10 2 2 10 – 20 5 2 + 5 = 7 20 – 30 7 7 + 7 = 14 30 – 40 12 14 + 12 = 26 40 – 50 4 26 + 4 = 30 ∑f = 30 Here, number of variables (N) = 30

274 Oasis School Mathematics-10 For median N 2 30 2 = = 15 The c.f. which is just greater than 15 is 26, whose corresponding class interval is (30 – 40). Hence, the median that lies in the class is (30 – 40). example 2 Find the value of x from the given data, if the median (md) = 52. Class interval 15–30 30–45 45–60 60–75 75–90 No. of students 10 8 15 p 10 Solution: Computation for the median: class No. of students (f) c.f. 15–30 10 2 30–45 8 2 + 5 = 7 45–60 15 7 + 7 = 14 60–75 p 14 + 12 = 26 75–90 10 26 + 4 = 30 N = 43+p Since, Md = 52, it lies as the class (45 – 60). ∴ l = 45, f = 15, c.f = 18, i = 15 We have, Md = l + f – c.f. N 2 × i 52 = 45 + 15 – 18 43+p 2 × 15 52–45 = 43 + 8 – 36 2 × 1 15 × 15 7 = 7 + 8 2 or, 14 = 7 + 8 or, 14 = 14 – 7 = 7 ∴ p = 7 l

Oasis School Mathematics-10 275 exercise 15.2 1. (a) What is the position of median in continuous series? Write the formula to calculate exact median in continuous series. (b) In a continuous series, the median lies in the class (20-30), whose corresponding frequency is 10. If N = 50, c.f. proceeding the median class is 20, find the exact median. (c) In a continuous series, N = 60, the median lies in the class (30–40), whose corresponding frequency is 10, c.f. proceeding the median class is 27, find the exact median. 2. Compute the median from the following data: (a) x: 0–5 5–10 10–15 15–20 20– 25 f: 10 15 12 18 5 (b) marks 0–10 10–20 20–30 30–40 40–50 50–60 No. of students 1 5 8 6 8 2 (c) Age (in years) 0–2 2–4 4–6 6–8 8–10 No. of persons 8 5 7 3 4 (d) Class 0 – 30 0 – 60 0 – 90 0 – 120 0 – 150 0 – 180 Frequency 5 15 37 62 76 80 (e) marks 0–10 0–20 0–30 0–40 0–50 No. of Students 4 12 24 44 62 3. (a) The following are the marks obtained by 35 students of class X in Mathematics. 28, 47, 22, 56, 69, 70, 82, 36, 40, 50, 55, 62, 31, 37, 42, 57 60, 72, 81, 90, 91, 56, 34, 57, 49, 70, 90, 61, 65, 80, 81, 87, 63, 28, 34 (i) make a frequency table of class interval 10. (ii) find the median marks. (b) The following are the daily wages of 30 workers in a factory. 95, 80, 76, 91, 70, 52, 50, 45, 67, 70, 85, 96, 98, 72, 65, 55, 48, 73 86, 82, 49, 55, 72, 79, 78, 62, 63, 82, 89, 92

276 Oasis School Mathematics-10 (i) make the frequency table of class interval 10. (ii) find the median. 4. (a) If the median of the given distribution is 17.5, find the value of f. x : 5–10 10–15 15–20 20–25 25–30 f : 20 30 40 f 10 (b) If the median of the given distribution is 25.625, find the value of p. x : 0–10 10–20 20–30 30–40 40–50 f : 3 p 8 6 3 (c) If the median of the given data is 24, calculate the value of x. marks 0–10 10–20 20–30 30–40 40–50 No. of students 9 21 x 15 10 5. If the median of the given distribution is 30, find the values of 'x' and 'y'. marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 –60 Total Number of students 10 x 25 30 y 10 100 Answers 1. (a) Consult your teacher (b) 25 (c) 33 2. (a) 12.08 (b) 31.67 (c) 4.14 (d) 93.6 (e) 33.5 3. (a) 59.16 (b) 75 4. (a) f = 40 (b) p = 5 (c) x = 25 5. x = 15, y = 10 15.4 Quartiles Quartiles are the values which divide the whole arrangement of data into four equal parts. Hence, there are three quartiles in a distribution, which are denoted by Q1 , Q2 and Q3 . Q2 represents the value of median. The lower quartile is denoted by Q1 and upper quartile is denoted by Q3 . Calculations for Q1 and Q3 are as follows: a) Positional value of Q1 (i) In individual distribution Q N th 1 1 4 = + ( ) n item.

Oasis School Mathematics-10 277 (ii) In discrete distribution Q N th 1 1 4 = + ( ) item. (iii) In continuous distribution, position of Q1 = N 4 th ( ( . Where, l = lower limit of Q1 class f = frequency of Q1 class c.f. = c.f. preceding the Q1 class i = size of the class interval b) Positional value of Q3 (i) In individual distribution Q item n item th 3 3 1 4 = ( ) + (ii) In discrete distribution, (iii) In continuous distribution, Q item N item th 3 3 4 = ( ) Q l N c f f i 3 3 4 = + − × . . Where, l = lower limit of Q3 class f = frequency of Q3 class c.f. = c.f. preceding the Q3 class i = size of the class interval Worked out examples example 1 Compute the lower and upper quartiles: x: 50 – 60 60 – 60 70 – 80 80 – 90 90 – 100 f: 5 7 8 10 3 Q1 = l + f – c.f. N 4 × i Q item item th 3 3 1 4 = (N+ )

278 Oasis School Mathematics-10 Solution: We have, Class (x) Frequency (f) c.f. 50 – 60 5 5 60 – 70 7 5 + 7 = 12 70 – 80 8 12 + 8 = 20 80 – 90 10 20 + 10 = 30 90 – 100 3 30 + 3 = 33 ∑f = 33 For lower quartile (Q1 ) c.f. just greater than 8.25 is 12, which belongs to the class (60 – 70). Hence, Q1 class is (60 – 70). Here, l = 60, i = 70 – 60 = 10 c.f. = 5, f = 7 We have, For upper quartile (Q3 ): 3 4 3 33 4 24 75 N = × = . . c.f. just greater than 24.75 is 30, which corresponds to the variable class (80 – 90) Q3 class is (80 – 90). Now, Lower limit (l) of Q3 class = 80 Class height (i) = (90 – 80) = 10 c.f. = 20 f = 10 N 4 33 4 = =8 2. . 5 ∴ = + −           × = + − × = + × = + Q N c f f i 1 4 60 8 25 5 7 10 60 3 25 7 10 60 4 l . . . . .64 = 64.64 ∴ = + − ( ) × = + − × = + = Q l N c f f i 3 3 4 80 24 75 20 10 10 80 4 75 84 75 . . . . . l

Oasis School Mathematics-10 279 example 2 Calculate the value of 'a' if Q1 = 33. marks obtained 0-20 20-40 40-60 60-80 80-100 Frequency 3 5 a 2 10 Solution: marks f c.f. 0 - 20 3 3 20 - 40 5 8 40 - 60 a 8 + a 60 - 80 2 10 + a 80 - 100 10 20 + a N = 20 + a Given, Q1 = 33 Since Q1 = 33, it lies on the class (20 - 40) Where, l = 20, f = 5, c.f. = 3, i = 20 We have, Q1 = l l + N c f f + i − × 4 . . or a or a or a or , , , 33 20 20 4 3 5 20 33 20 20 12 4 5 20 13 8 20 20 = + + − × − = + − × × = + × , , 13 8 13 8 5 = + = − ∴ = a or a a exercise 15.3 1. (a) What is the position of Q1 in a continuous series? Write the formula to calculate the exact value of Q1 . (b) What is the position of Q3 in a continuous series. Write the formula to calculate the exact Q3 .

280 Oasis School Mathematics-10 (b) In a continuous series, N = 60, Q3 lies in the class (20–25), whose corresponding frequency is 8, c.f. proceeding Q3 class is 40, find the exact Q3 . 3. Calculate the lower and upper quartile from the following data: (a) marks 10 – 20 20 – 30 30 – 40 40 – 50 No. of students 20 12 8 30 (b) Age in years 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 No. of people 5 4 3 6 7 (c) marks 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49 No. of students 5 4 3 2 1 (d) marks obtained 5–10 5–15 5–20 5–25 5–30 No. of students 20 50 100 140 150 4. (a) The following are the marks obtained by students in Maths in the SEE Test examination. 50, 52, 53, 29, 46, 85, 90, 84, 64, 39, 32, 77, 65, 43, 80, 95, 85, 09, 12, 01. (i) Make a frequency table of class interval 10. (ii) Calculate the first and third quartiles. (b) The following are the marks obtained by students in English in an examination. 51, 22, 63, 46, 57, 79, 21, 39, 51, 32, 43, 52, 60, 38, 45, 40, 32, 60, 63. Make a frequency table of class interval 10 and find its first quartile (Q1 ). (c) 15, 12, 23, 35, 46, 57, 18, 12, 39, 51, 32, 43, 25, 59, 18, 38, 45, 40, 32, 33. Make a frequency table of class interval 10 and find its third quartile (Q3 ) 5. (a) If first quartile of the given data is 31, find the value of 'k'. Class 10–20 20–30 30–40 40–50 50–60 60–70 Frequency 4 5 k 8 7 6 (b) If Q3 of the given data is 60, find the value of f1 Class 10–20 20–30 30–40 40–50 50–60 60–70 70–80 Frequency 3 5 4 5 4 f1 3

Oasis School Mathematics-10 281 (c) If Q3 of the given data is 22.5, find the value of m. Class 0-6 6-12 12-18 18-24 24-30 30-36 Frequency 5 6 4 5 m 2 (d) If Q1 of the given data is 27.5, find the value of 'p'. Class 10–20 20–30 30–40 40–50 50–60 60–70 Frequency 5 7 8 p 8 6 6. (a) If Q1 of the given data is 15.625, find the value of x and y. Class 0-10 10-20 20-30 30-40 40-50 50-60 Total Frequency 6 x 12 10 y 2 42 (b) If the Q3 of the given data is 56.25, find the value of 'a' and 'b'. marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total No. of students 3 5 a 5 4 b 3 26 Answers 1. Consult your teacher 2. (a) 14 (b) 23.12 3. (a) Q1 = 18.75, Q3 = 44.17 (b) Q1 = 11.56, Q3 = 25.54 (c) Q1 = 7.5, Q3 = 27 (d) Q1 = 12.92, Q3 = 21.56 4. (a) (i) Consult your teacher. (ii) Q1 = 35, Q3 = 82.5 (b) Q1 = 36.875 (c) Q3 = 45 5. (a) 10 (b) 4 (c) 3 (d) 7 6. (a) x = 8, y = 4 (b) a= 4, b = 2 15.5 mode Mode is that variate value which repeats maximum number of times. That is it is the value of the set f observations which has the highest frequency. For example, the following are the marks obtained by 15 students in mathematics. 20, 25, 15, 10, 20, 20, 15, 15, 10, 20, 20, 25, 10, 20, 15. Here, the frequency of 10 is 3, the frequency of 15 is 4, the frequency of 20 is 6 and the frequency of 25 is 2. Since 20 has the highest frequency i.e. 6. Thus, 20 is the mode of the above data. It is denoted by (Mo). mode of the continuous data In the case of the continuous data, the following steps are followed for the calculation of mode.

282 Oasis School Mathematics-10 • Find the modal class. It is the class having the highest frequency. • Find the frequency of the modal class and the frequencies of the classes preceding and succeeding to the modal class. • Use the following formula to find the mode of the data Mo = L + f1 – f0 2f1 – f0 – f2 × h Where, L = lower limit of the modal class f 1 = frequency of the modal class. f 0 = frequency of the class preceding to the modal class f 2 = frequency of the class succeeding to the modal class remember ! The relation between mean, median and mode is mode = 3 Median – 2 Mean i.e. Mo = 3Md – 2x Worked out examples 1. Find the mode (Mo) of the following data. Ages of year 0–10 10–20 20–30 30–40 40–50 No. of people 5 12 18 10 3 Solution: Here, the maximum frequency is 18. So, just by inspection, 20 – 30 is the modal class. Here, L = 20, f1 = 18, f0 = 12, f2 = 10 and h = 10 We know that, Mo = L + f1 – f0 2f1 – f0 – f2 × h = 20 + 18 – 12 2×18 – 12 – 10 × 10 = 20 + 10 3 = 70 3 = 23.33. ∴ The mode of the above data is 23.33.

Oasis School Mathematics-10 283 2. The following are the marks obtained by 50 students in a certain subject. Mark obtained 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 –89 90 – 99 no of students 3 8 15 12 6 4 2 Solution: Here, the given data is an inclusive data. So, it should be converted into the exclusive (continuous) data as follows. Mark obtained Continuous class No. of students (f) 30039 40–49 50–59 60–69 70–79 80–89 90–99 29.5–39.5 39.5–49.5 49.5–59.5 59.5–69.5 69.5–79.5 79.5–89.5 89.5–99.5 3 8 15 12 6 4 2 N = 50 Now, correction factor (c.f) = 40 – 39 2 = 1 2 = 0.5. Then subtract 0.5 from the lower limits and add 0.5 to the upper limit of the intervals. Since the highest frequency is 15, the modal class is 49.5 – 59.5. Here, L = 49.5, f1 = 15, f0 = 8, f2 = 12 and h = 10 we know that, Mo = L + f1 – f0 2f1 – f0 – f2 × h = 49.5 + 15 – 8 2×15 – 8 – 12 × 10 = 49.5 + 7 10 × 10 = 56.7 ∴ The modal mark of the given data is 56.7. 3. If the mode for the following distribution is 24, find value of x. Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequency 14 23 27 x 15 Solution:

284 Oasis School Mathematics-10 Since Mo = 24, the modal class is 20 – 30. Here, L = 20, f1 = 27, f0 = 23, f2 = x and h = 10 we know that Mo = L + f1 – f0 2f1 – f0 – f2 × h or, 24 = 20 + 27 – 23 2×27 – 23 – x × 10 or, 4 = 20 + 4 31 – x × 10 or, 1 = 10 31 – x or, 31– x = 10 or, x = 21 ∴ The required value of x is 21. exercise 15.4 1. a. What do you mean by mode of data? b. What is the modal letter of the ward 'STATISTICS'? c. Write a relation among x , Md and Mo. d. If man = 30, and mode = 24 then find median. 2. Find the modal class from the following data. a. Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequency 10 18 25 30 7 b. Wages in Rs. 0 – 500 500 – 1000 1000 – 1500 1500 – 2000 2000 – 2500 No of workers 50 46 37 22 11 3. a. Find the mode from the following frequency distributions. Mark 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 No. of student 4 10 12 14 5 b. Determine the mode from the following continuous data. Age (in year) 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 –65 No. of student 8 12 16 25 19 10

Oasis School Mathematics-10 285 4. Find the mode from the following data. a. Marks 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50 No. of students 2 10 37 37 40 b. Wage (in Rs.) 40 – 100 50 – 100 60 – 100 70 – 100 80 – 100 90 –100 No. of workers 30 25 19 14 9 6 c. Mid value 10 20 30 40 50 60 Frequency 5 14 12 18 15 8 d. Marks obtained 10 ≤ x ≤ 19 20 ≤ x ≤ 29 30 ≤ x ≤ 39 40 ≤ x ≤ 49 50 ≤ x ≤ 59 No. of students 5 12 30 10 8 5. a. If the mode of the given data is 24, find the following frequency. Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 No. of students 3 ? 8 5 2 b. If the mode of the following frequency distribution is 25, find the value of p. Mark 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 no of students 5 16 20 p 12 range Marks obtained by 20 students in class X, out of 100 full marks is given below. 72, 76, 92, 85, 70, 52, 50, 45, 62, 82 27, 65, 83, 19, 16, 98, 59, 67, 72, 93 • What is the highest marks of the class? • What is the lowest marks of the class? • What is the difference between the highest marks and the lowest marks? In the given data, The highest marks is 98. The lowest marks is 16. Their difference = 98 – 16 = 72. Hence, 72 is the range of given data. Hence, the rage of the given data is the difference between the largest item and the smallest item.

286 Oasis School Mathematics-10 Therefore, Range = Largest item – Smallest item = L – S As the co-efficient of range relative measure of dispersion. So co–efficient of range = L – S L + S Again, lets take an example of discrete series. Marks obtained by 50 students of class X is given below. Marks 20 30 40 50 60 70 No. of students 6 7 12 10 8 7 In the above data, The highest marks of the class (L) = 70 The lowest marks of the class (S) = 20 Hence, Range = L – S = 70 – 20 = 50. Co-efficient of Range = L – S L + S = 70 – 20 70 + 20 = 50 90 = 5 9 = 0.56 Worked out examples example 1 Find the range of its co-efficient of: Class 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 6 9 8 7 14 3 2 Solution: Here, The largest item (L) = 80. The smallest item (S) = 20 We have, Range = L – S = 80 – 20 = 60 Co-efficient of Range = L – S L + S = 80 – 20 80 + 20 = 60 100 = 0.6. exercise 15.5 1. Find the range and its co-efficient of: (a) 16, 18, 7, 25, 32, 44, 66, 78, 27 (b) 53, 66, 77, 182, 190, 177, 152, 67 (c) 92, 165, 97,82, 200, 315, 612, 145

Oasis School Mathematics-10 287 2. Find the range and its co-efficient of: (a) x 50 54 57 58 60 65 f 6 5 7 3 2 4 (b) Wages (in Rs.) 45 55 65 75 85 95 105 115 Number of workers 6 3 4 7 5 4 6 2 (c) Marks 20 40 60 70 90 100 Number of Students 2 8 4 5 7 1 3. Find the range and its co-efficient of: (a) Class 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 7 7 9 5 4 6 2 (b) Wages (In Rs.) 100 – 150 150 – 200 200 – 250 250 – 300 300 – 350 350 – 400 Number of workers 7 4 5 7 8 6 (c) Marks 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 Number of students 4 5 5 10 5 8 6 Answers 1. (a) 71, 0.86 (b) 124,0.48 (c)520, 0.74 2. (a) 15, 0.13 (b) 70, 0.44 (c) 80, 0.87 3. (a) 70, 0.78 (b) 300, 0.6 (c) 35, 0.78 Project Work 1. Collect the marks obtained by the students of your class in Mathematics in an examination. • Convert the rate into suitable series (Individual, discrete of continuous) • Calculate mean (x), median (Md), Mode (M0), Quartiles and Range. • Present the result in your classroom. 2. Collect the marks obtained by the students of your class in your terminal examination. Prepare the necessary class and the suitable continuous data and find subject-wise mean, median and quartiles. Present the results in your class.

288 Oasis School Mathematics-10 miscellaneous exercise Statistics 1. The marks obtained by 52 students of class X is given in the table. Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Number of students 7 8 5 9 12 6 5 Answer the questions given below. (a) In which series the given data is presented? (b) Make the cumulative frequency distribution table of above data. (c) Calculate the median marks obtained by the students. (d) If 3 more, students were there whose marks is between 60 to 70, in which class does Md lie? [Ans: (c) 46.44 (d) (40 – 50)] 2. Daily wages of 20 workers working in a company is given below. Daily wages (in Rs.) 500 – 600 600 – 700 700 – 800 800 – 900 900 – 1000 1000 – 1100 Number of workers 3 2 6 3 2 4 Answer the questions given below. (a) In which series the data is presented? (b) Make the cumulative frequency distribution table of the above data. (c) Calculate the third quartile in the above data. (d) What is the average wages of the workers whose wage is more than Rs. 700? [Ans: (c) 950, (d) 876.67] 3. The marks obtained by 30 students of in class in given below. Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 Number of students 4 3 5 6 2 3 3 3 1 Observe the table answer the questions given below. (a) In which series the data is presented? (b) Make the frequency distribution table of given data. (c) Calculate the lower quartile from the given data. (d) If the marks is less than 40, students have to give re-exam, find the average marks obtained by passed students. [Ans: (c) 31, (d) 63.89] 4. The ages of the people in the community is given below. 9, 46, 37, 38, 32, 18, 42, 48, 8, 53, 27, 21, 57, 16, 47, 25, 26, 14, 33, 6, 32, 16, 31, 15, 26. (a) What is the ages of youngest and oldest people? (b) Make the class interval of 10 and make frequency distribution table.

Oasis School Mathematics-10 289 (c) Make the cumulative frequency distribution table and calculate Q1 . (d) Find the average age of the people in the community. [Ans: (a) 6 yrs and 57 yrs, (c) 16.5 yrs, (d) 28.6yrs] 5. The average marks of some students in the class is 41.5. If the marks of 3 students is in between 10 – 20, 4 students in between 20 – 30, 15 students in between 40 – 50, 3 students in between 50 – 60 and 5 students in between 60 – 70. (a) Show this information in frequency distribution table. (b) Find marks of how many students is in between 30 – 40? (c) Find the Md of the data. (d) Find the modal class of this distribution. Give reason. [Ans: (b) 10, (c) 43, (d) 40 to 50] 6. If Md of the given distribution is 23.75. Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequency 2 3 4 p 3 (a) On which class median of the given distribution lie? (b) Prepare the cumulative frequency distribution table. (c) Find the value of ‘p’? (d) Find the average of the data greater than 20. [Ans: (c) 1, (d) 33.75] 7. The Q1 of given data is 6. Marks obtained 0 – 4 8 – 12 12 – 16 4 – 8 16 – 20 20 – 24 No. of students 3 2 3 6 a 3 (a) Make the frequency distribution table of above data. (b) In which class Q1 of above data lie? (c) Find the value of ‘a’. (d) Is the mode and the median lie on the same class? Give reason. [Ans: (c) 7, (d) Yes] 8. The third quartile of the given data is 45. Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Frequency 3 2 m 3 4 3 (a) On which class Q3 lie? (b) Prepare the cumulative frequency distribution table. (c) Find the value of m. (d) On which class mode and Q3 lie? [Ans: (c) 5, (d) 40 to 50] 9. Given table shows the wages of 50 workers. Wages 100 – 200 100 – 300 100 – 400 100 – 500 100 – 600 100 – 700 Number of workers 5 15 24 36 45 50

290 Oasis School Mathematics-10 (a) Convert the above data into frequency distribution table. (b) Find the class on which Q1 and Q3 lie? (c) Calculate the value of Q1 . (d) Find the modal class. [Ans: (b) (200-300) and (500-600), (c) 275, (d) (400-500)] 10. The median of the given distribution is 93.6 Class 0 – 30 30 – 60 60 – 90 90 – 120 150 – 150 150 –180 Total Frequency 5 x 22 25 y 4 80 (a) Prepare the cumulative, frequency distribution table. (b) Write the relation of ‘x’ and ‘y’. (c) On which class does the median lie? (d) Find the values of ‘x’ and ‘y’. [Ans:(c) (90-120), (d) x = 10, y = 14] Full Marks: 13 1. Study the given data and answer the questions given below. Class Interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Frequency 12 8 27 a 17 6 (a) If the mean of the given data is 28, find the value of 'a'. (2) (b) Find the mode. (2) (c) What do you mean by the modal value of given data. (1) 2. Study the given data and answer the questions given below. Class Interval 20 - 24 24 - 28 28 - 32 32 - 36 36 - 40 Frequency 2 a 6 3 1 (a) Present the above data in cumulative frequency table. (1) (b) If the lower quartile of the data 25 1 3 , find the value of 'a'. (2) (c) What is the average of the above data? (1) 3. Marks obtained by 25 students in a class out of 100 full marks is given below. 56, 62, 92, 41, 35, 46, 59, 61, 70, 84, 96, 87, 82, 74, 60, 59, 48, 36, 35, 31, 48, 49, 57, 68, 67 (a) Present the above data in frequency distribution table. (1) (b) Calculate the median of the data. (2) (c) What is the mode of given data? (1) (d) If one more student having score 95 is there, is there any changes in the modal value? Give reason. (1)

Oasis School Mathematics-10 291 16.1 Warm-up Activities Discuss the following in your class and draw a conclusion. • Write the set of all possible outcomes in the given random experiment. (i) A coin is tossed once. (ii) A dice is rolled once. (iii) A coin is tossed twice. • On whose range lies the probability of an event? • What is the probability of a sure event? • What is the probability of an impossible event? 16.2 Probability random experiment: If an experiment is repeated under homogeneous conditions, the result of which is not the same but may be one of the various possible outcomes, then such an experiment is called a random experiment. In a random experiment, the outcome of each trial depends on chance, which cannot be predicted with certainty. For example, if we toss a coin, we may get either a head or a tail. If we throw a dice which has six faces marked with 1, 2, 3, 4, 5 and 6, then any one of the faces may come up. It is not possible to say which face will come up. These are examples of a random experiment. trial and event: Performing a random experiment is called a trial, and the outcome or combination of outcomes is called events. For example, throwing a dice is a trial and getting any one of the faces 1, 2, …, 6 is an event. Sample space: The set of all possible outcomes of a random experiment is called sample space. The sample space is usually denoted by the letter 'S'. Some examples of sample space are given below. For example: i. In tossing a single coin, there are two possible outcomes, namely a head (H) or a tail (T). So, the sample space for a coin is given by S = {H, T}. ii. If two coins are tossed simultaneously, then the sample space is given by: Unit 16 Probability

292 Oasis School Mathematics-10 S = {H, T} which means head on the first coin and tail on the second coin. Similarly, {T, H} means tail on the first coin and head on the second coin and so on. exhaustive outcomes: The total number of all possible outcomes of a random experiment is called the exhaustive outcome for the experiment. For examples: In the case of a fair coin, the total number of exhaustive outcomes = 2. i.e., {H, T}. If two coins are tossed simultaneously, the possible outcomes are {HH, HT, TH, TT}, i.e., • heads on both tosses • head on the first toss and tail on the second toss • tail on the first toss and head on the second toss • tails on the second toss ∴ Total number of exhaustive events = 22 = 4. Similarly, if three coins are tossed, then the total number of exhaustive events = 23 = 8. If a dice is rolled and a coin is tossed, then the possible results are- (1, H), (2, H), (3, H), (4, H), (5, H), (6, H) (1, T), (2, T), (3, T), (4, T), (5, T), (6, T). ∴ Total number of exhaustive events = 2 × 6 = 12. Coin → Dice ↓ H t 1 (1, H) (1, T) 2 (2, H) (2, T) 3 (3, H) (3, T) 4 (4, H) (4, T) 5 (5, H) (5, T) 6 (6, H) (6, T) H H HH HT T TH TT Coin 1 Coin 2 T

Oasis School Mathematics-10 293 Probability of an event Let S be the sample space of a random experiment and E the event which is a subset of the sample space, then probability of happening an event E is denoted by P(E). Symbolically, it is defined as, P (E) = n(E) n(S) Where, n (E) = total number of favorable cases for event E, n(S) = total number of all possible outcomes of equally likely cases P (E) = probability of not getting an event E = 1– P(E) Nature of probability (i) The value of probability always lies between 0 and 1, i.e., 0 ≤ P(E) ≤ 1 (ii) If E is an impossible event, then P(E) = 0. (iii) If E is a sure event, then P(E) = 1 Playing Cards Club (13 cards) Diamond (13 cards) Heart (13 cards) Spade (13 cards) These are known as 4 suits of playing cards. Red Cards ⇒ All the cards of Diamond and Heart (26 cards) Black Cards ⇒ All the cards of Club and Spade (26 cards) ♣ ◊ ♥ ♠

294 Oasis School Mathematics-10 Face cards ⇒ Jack, Queen and King. There are 12 face cards in a pack of playing cards. mutually exclusive events: Two or more events are said to be mutually exclusive events, if both of them cannot occur together in the same trial. If A and B are two mutually exclusive events, then n(A∩B) = 0 and P (A∩B) = 0. example : If a card is drawn from 52 cards, (a) getting a king and an ace is a mutually exclusive event. (b) getting a diamond and a heart is a mutually exclusive event. Additive law for mutually exclusive events: If 'A' and 'B' are mutually exclusive events, then the probability of occurrence of at least one of the events A and B is P (A∪B) = P(A) + P(B). mutually non-exclusive events: Two events 'A' and 'B' are said to be mutually non-exclusive if both of them can occur in the same trial. For example, if a card is drawn from a pack of 52 cards, the probability of getting a face card and a red card is a mutually non-exclusive event. Similarly, getting a king and a diamond is a mutually non-exclusive event. In this type of event, P(A∪B) = P(A) + P(B)–P(A∩B) P(A∪B) = 1–P (A∪B) Worked out examples example 1 'A' and 'B' are mutually exclusive events if P(A) = 1 5 , P(B) = 2 3 . Find (i) P(A∪B) (ii) P(A∩B). Solution: Here, given P(A) = 1 5 , P(B) = 2 3 Since 'A' and 'B' are two mutually exclusive events, then we have

Oasis School Mathematics-10 295 P(A∪B) = P(A) + P(B) = 1 5 + 2 3 = 3+10 15 = 13 15 Again, P(A∪B) = 1–P (A∪B) = 1– 13 15 = 2 15 example 2 A card is drawn at random from a well shuffled pack of 52 cards. Find the probability of the the card is an ace or a queen. Solution: Let 'A' and 'Q' be the event of getting an ace or a queen respectively. Number of possible cases n(S) = 52 Then, Probability of getting an ace P(A) = n(A) n(S) 4 52 1 13 = = Probability of getting a queen P(Q) = n(Q) n(S) = 4 52 1 13 = P (A or Q) = P(A) + P(Q) = 1 13 1 13 2 13 + = ∴ The probability that the card is an ace or a queen is 2 13. example 3 From a number of cards numbered 1 to 20, a card is drawn randomly. Find the probability of getting a card whose number is divisible by 3 or 7. Solution: Here, Number of possible cases n(S) = 20 Let A be a set of numbers divisible by 3, then, A = {3, 6, 9, 12, 15, 18} then, n(A) = 6 Let B be a set of numbers which are divisible by 7 i.e. B = {7, 14} then, n(B) = 2 Now, finding the possibility of event A P(A) = n(A) n(S) 6 20 3 10 = = Probability of event B P(B) = n(B) n(S) 2 20 1 10 = = Since, A and B are mutually exclusive events,

296 Oasis School Mathematics-10 P(A ∪ B) = P(A) + P(B) = 3 10 1 10 4 10 2 5 + ( ) = = ∴ The probability of getting a card whose number is divisible by 3 or 7 having an ace or queen is 2 5 . example 4 From a pack of 52 cards, a card is chosen at random. Find the probability of getting a red card or a face card. Solution: Here, n(S) = 52 Let, 'A' and 'B' be the events of getting a red card and a face card respectively. Then, ∴ The probability of getting a red card or a face card is 8 13. example 5 From a number of cards numbered from 1 to 21, a card is drawn at random. Find the probability of getting a card whose number is divisible by 3 or 7. Solution: Here, Number of possible cases n(S) = 21 Let A be a set of numbers which are divisible by 3 i.e. A = {3, 6, 9, 12, 15, 18, 21} then, n(A) = 7 Let B be a set of numbers which are divisible by 7, i.e., B = {7, 14, 21} then, n(B) = 3 Also, A ∩ B = {21} n A n B n A B n A n S P B n n ( ) ( ) ( ) Wehave, P(A) ( ) ( ) ( ) (B) ( = = ∩ = = = = 26 12 6 26 52 S P A B n A B n S Since AandB aremutaully non exc ) ( ) ( ) ( ) , = ∩ = ∩ = − 12 52 6 52 lusive events Wehave P A B P A P B P A B , , ( ) ∪ = ( ) + − ( ) ( ) ∩ = + − 26 52 12 52 6 52 = = ∴ = 32 52 8 13 8 13 P d (Re ) orface card . Since, A and B are mutually non-exclusive events, n A n B n A B n A n S P B n n ( ) ( ) ( ) Wehave, P(A) ( ) ( ) ( ) (B) ( = = ∩ = = = = 26 12 6 26 52 S P A B n A B n S Since AandB aremutaully non exc ) ( ) ( ) ( ) , = ∩ = ∩ = − 12 52 6 52 lusive events Wehave P A B P A P B P A B , , ( ) ∪ = ( ) + − ( ) ( ) ∩ = + − 26 52 12 52 6 52 = = ∴ = 32 52 8 13 8 13 P d (Re ) orface card .