Ap board solutions class 9 maths

AP State Syllabus AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3 Question 1. Find the remainder when x 3 + 3x 2 + 3x + 1 is divided by the following Linear polynomials i) x + 1 Each Solution: Let f(x) = x 3 + 3x 2 + 3x + 1 By remainder theorem, the remainder is f (- 1) f (- 1) = (- 1) 3 + 3(- 1) 2 + 3(- 1) + 1 = – 1 + 3 – 3 + 1 = 0 iii) x Solution: f(x) = x 3 + 3x 2 + 3x + 1 The remainder is f(0) ∴ f(0) = 0 3 + 3(0) 2 + 3(0) + 1 = 1 iv) x + π Solution: f(x) = x 3 + 3x 2 + 3x + 1 By remainder theorem, the remainder is f(- π) f(- π) = (- π) 3 + 3(-π) 2 + 3 (-π) + 1 . = – π 3 + 3π 2– 3π + 1 v) 5 + 2x f(x) = x 3 + 3x 2 + 3x + 1 The remainder is \(\mathrm\right)\) Question 2. Find the remainder when x 3– px 2 + 6x – p is divided by x – p. Solution: Let f(x) = x 3– px 2 + 6x – p (x – a) = x – p) By Remainder theorem, the remainder is f(p) ∴ f(P) = P 3– P(P) 2 + 6p – p = p 3– p 3 + 5p = 5p Question 3. Find the remainder when 2x 2– 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial ? State reason. Solution: Let f(x) = 2x 2– 3x + 5 and x – a = 2x – 3 = x – \(\frac\right)\) Question 5. If the polynomials 2x 3 + ax 2 + 3x – 5 and x 3 + x 2– 4x + a leave the same remainder, when divided by x – 2, find the value of a. Solution: Let f(x) = 2x 3 + ax 2 + 3x – 5 g(x) = x 3 + x 2– 4x + a Given that f(x) and g(x) divided x – 2 give same remainder. i e., f(2) = g(2) By Remainder theorem. But f(2) = 2(2) 3 + a(2) 2 + 3(2) – 5 ...

AP Board9th Class Maths Textbook Solutions Andhra Pradesh Board APSCERT 9th Class / SSC textbook Solutions pdf are available here for free of cost download in pdf format. These textbook are very important for all students who are studying their Class 9 in AP Board. Also if any candidates going to prepare for SSC, APPSC, UPSC, IBPS examinations can also read these standard books for their knowledge. APBoard Class 9 Maths Textbook Solutions download Free PDF AP SCERT Class 9 Maths Textbook Solutions PDF Format: Candidates who are looking for Class 9 AP School Books, syllabus, sample questions, exam pattern, and Co-Curricular Subject textbooks can refer to this entire web page. Here, we have gathered Subject Wise Andhra Pradesh Board textbook Solutionsfor Class 9th students along with the direct download links. From this Web Page, students will get to download subject-wise AP SCERT Class 9th Mathstextbook Solutionsin PDF Format for free of cost. You can also directly download them from the official site @ apscert.gov.in, scert.ap.gov.in. To make it easy for English, Telugu, Hindi, and Urdu language students, the AP board released the Class 9th Standard textbooks PDF in (English, Telugu, Hindi, and Urdu). So, we have also provided Telugu, Hindi, Urdu, and English medium Andhra Pradesh Board SCERT Textbooks & Notes PDF download links here in this Web Page.

• • • • • • AP 9th Class Textbook Pdf Download Chapter 3 The Elements of Geometry • • AP 9th Class Maths Solutions Chapter 4 Lines and Angles • • • • • 9th Class Maths TS State Syllabus Guide Pdf Chapter 5 Co-Ordinate Geometry • • • • AP 9th Class Maths Textbook State Syllabus Solutions Chapter 6 Linear Equation in Two Variables • • • • • AP Board 9th Class Maths Guide Chapter 7 Triangles • • • • • AP SCERT 9th Maths Solutions Chapter 8 Quadrilaterals • • • • • AP State 9th Class Maths Textbook Solutions Chapter 9 Statistics • • • 9th Class Maths Textbook State Syllabus Pdf Chapter 10 Surface Areas and Volumes • • • • • AP State Board 9th Class Maths Syllabus Chapter 11 Areas • • • • AP 9th Maths Solutions Chapter 12 Circles • • • • • • AP State 9th Class Maths Textbook Pdf Chapter 13 Geometrical Constructions • • • 9th Class Maths Solution Pdf Chapter 14 Probability • • 9th Class Maths Textbook Telugu Medium Chapter 15 Proofs in Mathematics • • • • • AP Board 9th Class Maths Textbook Solutions in Telugu Medium AP State 9th Class Maths Study Material Chapter 1 వాస్తవ సంఖ్యలు • • • • • AP State 9th Class Maths Solutions Chapter 2 బహుపదులు మరియు కారణాంక విభజన • • • • • • AP 9th Class Textbook Pdf Download Chapter 3 జ్యామితీయ మూలాలు • • AP 9th Class Maths Solutions Chapter 4 సరళ రేఖలు మరియు కోణములు • • • • • 9th Class Maths TS State Syllabus Guide Pdf Chapter 5 నిరూపక జ్యామితి • • • • AP 9th Class Maths Textbook State Syllabus Solutions Chapter 6 రెండు చరరాశులలో రేఖీయ సమీకరణా...

AP State Syllabus AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.3 Question 1. Draw the graph of each of the following linear equations. i) 2y = – x + 1 ii) – x + y = 6 iii) 3x + 5y = 15 iv) \(\frac\) × (-40) + 32 = -40 ii) From the graph 30°C = 86°F iii) 95°F = 35°C iv) When C = – 40 then F = – 40 Categories Post navigation

AP State Syllabus AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.4 Question 1. In the figure, ‘O’ is the centre of the circle. ∠AOB = 100°, find ∠ADB. Solution: ’O’ is the centre ∠AOB = 100° Thus ∠ACB = \(\frac\) = 5cm Question 6. In the figure, ‘O’ is the centre of the circle and OM, ON are the perpen-diculars from the centre to the chords PQ and RS. If OM = ON and PQ = 6 cm. Find RS. Solution: ‘O’ is the centre of the circle. OM = ON and 0M ⊥ PQ; ON ⊥ RS Thus the chords FQ and RS are equal. [ ∵ chords which are equidistant from the centre are equal in length] ∴ RS = PQ = 6cm Question 7. A is the centre of the circle and ABCD is a square. If BD = 4 cm then find the radius of the circle. Solution: A is the centre of the circle and ABCD is a square, then AC and BD are its diagonals. Also AC = BD = 4 cm But AC is the radius of the circle. ∴ Radius = 4 cm. Question 8. Draw a circle with any radius and then draw two chords equidistant from the centre. Solution: • Draw a circle with centre P. • Draw any two radii. • Mark off two points M and N oh these radii. Such that PM = PN. • Draw perpendicular through M and N to these radii. Question 9. In the given figure, ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of ΔOCD. Solution: ‘O’ is the centre of the circle. AB, CD are equal chords ⇒ They subtend equal angles at the centre. ∴ ∠AOB =∠COD = 70° Now in ΔOCD ∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal...

• • • • • Maths 10th Class Guide Chapter 3 Polynomials • • • • • • AP 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables • • • • • AP 10th Class Maths Guide Pdf Free Download Chapter 5 Quadratic Equations • • • • • • AP State 10th Class Maths Textbook Solutions Pdf Chapter 6 Progressions • • • • • • • SSC 10th Class Maths Guide Pdf Free Download Chapter 7 Coordinate Geometry • • • • • • AP 10th Class Maths Solutions Pdf Chapter 8 Similar Triangles • • • • • • Maths Solutions Class 10 State Board Chapter 9 Tangents and Secants to a Circle • • • • • AP 10th Class Maths Study Material Pdf 2022 Chapter 10 Mensuration • • • • • • Maths Material 10th Class Chapter 11 Trigonometry • • • • • • AP State Board Class 10 Maths Solutions Chapter 12 Applications of Trigonometry • • • • 10th Class Maths Study Material 2021-2022 Chapter 13 Probability • • • • 10th Class Maths Textbook Pdf English Medium Chapter 14 Statistics • • • • • AP 10th Class Maths Guide Telugu Medium Pdf 10th Class Maths Material 2022 Chapter 1 వాస్తవ సంఖ్యలు • • • • • • • AP SSC 10th Maths Solutions Chapter 2 సమితులు • • • • • AP State Maths Textbook Solutions Chapter 3 బహుపదులు • • • • • • C AP 10th Maths Solutions Chapter 4 రెండు చరరాశులలో రేఖీయ సమీకరణాల జత • • • • • 10th Class Maths Guide Telugu Medium Pdf Chapter 5 వర్గ సమీకరణాలు • • • • • • SCERT Maths Textbook Class 10 Solutions Chapter 6 శ్రేఢులు • • • • • • • AP 10th Class Maths Guide Pdf Download Chapter 7 నిరూపక రేఖాగ...

Question 2. In the given figure AS // BT; ∠4 = ∠5, \(\overline\) = 35° Also BC // DE ∴∠D = 105° (∵ alternate interior angles) Now in ΔCDE 24° + 105° + y = 180° (∵ angle sum property) ∴ y = 180° – 129° = 51° Question 4. In the given figure BE ⊥ DA and CD ⊥ DA then prove that m∠1 = m∠3. Solution: Given that CD ⊥ DA and BE ⊥ DA. ⇒ Two lines CD and BE are perpendicular to the same line DA. ⇒ CD // BE (or) ∠D =∠E ⇒ CD // BE (∵ corresponding angles for CD and BE and DA are transversal) Now m∠1 = m∠3 (∵alternate interior angles for the lines CD // BE ; DB are transversal) Hence proved. Question 5. Find the values of x, y for which lines AD and BC become parallel. Solution: For the lines AD and BC to be parallel x – y = 30° (corresponding angles) ……… (1) 2x = 5y ………….(2) (∵ alternate interior angles) Solving (1) & (2) y = \(\frac\) ∠ACB = 72° iii) From the figure ∠ABD + ∠ABC = 180° ∠ABD = 180° – 72° = 1086 In ΔABD ∠DAB + ∠ABD + ∠D = 180° ∠DAB + 108° + 45° = 180° ∠DAB = 180° – 153° = 27° iv) In ΔADE ∠D + ∠A + ∠E = 180° 45° + ∠A + 40° = 180° ⇒ ∠A = 180° -85° = 95° But ∠A = ∠DAB + 36° + ∠EAC 95° = 27°, + 36° + ∠EAC ∴ ∠EAC = 95° – 63° = 32° Question 17. Using information given in the figure, calculate the values of x and y. Solution: From the figure In ∆ACB 34° + 62° + ∠ACB = 180° (∵ angle sum property) ∴ ∠ACB = 180° – 96° = 84° . And x + ∠ACB = 180° (∵ linear pair of angles) . ∴ x + 84° = 180° x = 180°-84° = 96° (OR) x = 34° + 62° = 96° ( ∵ x is exterior angle, ∆ABC) y = 24° + x° = 2...

Question 1. Weights of parcels in a transport office are given below. Find the mean weight of the parcels. Solution: Weight in kg x i No. of parcels f i x 1f i 50 25 1250 65 34 2210 75 38 2850 90 40 3600 110 47 5170 120 16 1920 Σf i = 200 Σf ix i = 17000 \(\begin\) Mean = 85 Question 2. Number of familles In a village in correspondence with the number of children are given below. Find the mean number of children per family. Solution: No. of childrens x i No. of families f i x 1f i 0 11 0 1 25 25 2 32 64 3 10 30 4 5 20 6 1 5 Σf i = 84 Σf ix i = 144 \(\overline\) = 10 Question 4. Number of villages with respect to their population as per India census 2011 are given below. Find the average population in each village. Solution: Population (in thousands x i) Villages f i x 1f i 12 20 240 5 15 75 30 32 960 20 35 700 15 36 540 8 7 56 Σf i = 145 Σf ix i = 2571 thousands \(\overline\) Given that, Mean of 4 numbers = 15 ⇒ Sum of the 4 numbers = 4 x 15 = 60 Mean of the first 3 numbers = 9 ⇒ Sum of the first 3 numbers = 3 x 9 = 27 Mean of the first 2 numbers = 4 ⇒ Sum of the first 2 numbers = 2 x 4 = 8 Fourth number = sum of 4 numbers – sum of 3 numbers = 60 – 27 = 33 Third number = sum of 3 numbers – sum of 2 numbers = 27 – 8 = 19 Second number = Sum of 2 numbers – given number = 8-2 = 6 ∴ The other three numbers are 6, 19, 33.