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Mathematics IX ( Term -I) 1 CBSE SAMPLE QUESTION PAPER (SOLVED) Maximum Marks : 90 Maximum Time : 3 hours General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A comprises of 8 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 10 questions of 4 marks each. (iii) Question numbers 1 to 8 in Section A are multiple choice questions where you are to select one correct option out of the given four. (iv) There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and 2 questions of four marks each. You have to attempt only one of the alternatives in all such questions. (v) Use of calculators is not permitted. SECTION A (Question numbers 1 to 8 carry 1 mark each. For each question, four alternative choices have been provided of which only one is correct. You have to select the correct choice). 1. Which of the following is a rational number? (a) 2 3 (b) 15 (c) 135 (d) 2 3 Sol. (a) 2 3 2. The value of k, for which the polynomial x3 – 3x2 + 3x + k has 3 as its zero, is (a) –3 (b) 9 (c) –9 (d) 12 Sol. (c) –9 3. Which of the following is a zero of the polynomial x3 + 3x2 – 3x – 1? (a) –1 (b) –2 (c) 1 (d) 2 Sol. (c) 1 4. The factorisation of – x2 + 5x – 6 yields (a) (x – 2) (x – 3) (b) (2 + x) (3 – x) (c) – (x – 2) (3 – x) (d) – (2 – x) (3 – x) Sol. (d) – (2 – x) (3 – x) 5. In the figure, ∠DBC equals (a) 40° (b) 60° (c) 80° (d) 100° Sol. (a) 40° E D B A C 8x 4x 6x2 Mathematics IX ( Term -I) 6. In the figure, ABC is an equilateral triangle and BDC is an isosceles right triangle, right angled at D. ∠ABD equals (a) 45° (b) 60° (c) 105° (d) 120° Sol. (c) 105° 7. The sides of a triangle are 12 cm, 16 cm and 20 cm. Its area is (a) 48 cm2 (b) 96 cm2 (c) 120 cm2 (d) 160 cm2 Sol. (b) 96 cm2 8. The side of an isosceles right triangle of hypotenuse 4 2cm is (a) 8 cm (b) 6 cm (c) 4 cm (d) 4 3cm Sol. (c) SECTION B (Question numbers 9 to 14 carry 2 marks each) 9. If x = 7+ 40, find the value of x x + 1 . Sol. We have, x = 7 + 40 = 7 + 2 10 = 5 + 2 + 2 10 = 5 2 2 5 2 2 2 ( ) + ( ) + ( )( ) = 5 22 + ( ) ⇒ x= + ( ) 5 2 [Taking square root of both sides] Also, 1 15 2 x = + = 1 5 2 5 2 5 2 + × [Rationalising the denominator] = 5 2 5 2 5 2 5 2 5 2 3 2 2 ( ) ( ) = = Now, x x + = + + 1 5 2 5 2 3 = 3 5 2 5 2 3 + ( )+ ( ) = 4 5 2 2 3+ = 23 2 5 2 + ( ) 10. Factorise the polynomial : 8x3– (2x – y)3 Sol. We have, 8x3 – (2x – y)3 = (2x)3 – (2x – y)3 = [2x – (2x – y)] [(2x)2 + 2x (2x – y) + (2x – y)2] [Using the identity a3 – b3 = (a – b) (a2 + ab + b2)] = (2x – 2x + y) (4x2 + 4x2 – 2xy + 4x2 + y2 – 4xy) = y (12x2 + y2 – 6xy) A B D CMathematics IX ( Term -I) 3 11. Find the value of ‘a’ for which (x – 1) is a factor of the polynomial a2x3– 4ax + 4a – 1. Sol. Let p(x) = a2 x3 – 4ax + 4a – 1 Since, (x – 1) is a factor of p(x), therefore, p(1) = 0 [∵ x – 1 = 0 ⇒ x = 1] ⇒ a2 (1)3 – 4a (1) + 4a –1 = 0 ⇒ a2 – 4a + 4a –1 = 0 ⇒ a2 = 1 ⇒ a = ± 1 12. In the figure, if AC = BD, show that AB = CD. State the Euclid’s postulate/axiom used for the same. Sol. We have, AC = BD ⇒ AC – BC = BD – BC [Subtracting BC from both sides] ⇒ AB = CD The Euclid’s axiom used here is : ‘If equals are subtracted from equals, the remainders are equals’. 13. In the figure, find the value of x. OR In the figure ABCDE is a regular pentagon. Find the relation between ‘a’, ‘b’ and ‘c’ Sol. ∠CBD = 40° + 30° = 70° [Exterior angle property] x = ∠CED = ∠CBD + ∠BDE = 70° + 45° = 115° . OR ABCD is a regular pentagon. We know that each side as well as each angle of a regular polygon are equal. So, ∠A = ∠B = ∠C = ∠D = ∠E ... (i) AB = BC = CD = DE = AE ... (ii) In ΔABE, ∠a = ∠3 [AB = AE] ... (iii) Also, ∠a + ∠3 + ∠A = 180° [Angle sum property of ΔABE] ⇒ 2∠a = 2 ∠3 = 180° – ∠A [From (iii)] ⇒ ∠a = ∠3 = 90° – ∠A 2 ... (iv) Similarly, in ΔBCD, ∠1 = ∠2 = 90° – ∠C 2 = 90° – ∠A 2 ... (v) [From (i)] ∠1 = ∠2 = ∠3 = ∠a. ... (vi) (From (iv) and (v)] Also, ∠C + ∠1 = ∠A [From (i) D B A C4 Mathematics IX ( Term -I) and ∠3 + ∠b + ∠2 = ∠A. [From (i) ⇒ ∠C + ∠1 = ∠3 + ∠b + ∠2 ⇒ ∠C + ∠1 = ∠a + ∠b + ∠1 [From (vi)] ⇒ ∠C = ∠a + ∠b Hence, c = a + b. 14. In the figure, ABC is an equilateral triangle. The coordinates of vertices B and C are (3, 0) and (– 3, 0) respectively. Find the coordinates of its vertex A. Sol. From the figure, length of side BC = 6 units Since, ABC is an equilateral triangle. ∴ AB = BC = CA = 6 units Now, AO is perpendicular bisector of BC. Therefore, AOB is a right angled triangle in which AB = 6 units and OB = 3 units ∴ AO2 = AB2 – OB2 [Using Phythagoras theorem] = (6)2 – (3)2 = 27 ⇒ AO = 3 3 units Hence, coordinates of vertex A = (0, 3 3 ). SECTION C (Question numbers 15 to 24 carry 3 marks each) 15. Evaluate : 5 2 6 8 2 15 + { }+{ }OR If a = 9 – 4 5, find the value of a a 2 2 1 + . Sol. We have, 5 2 6 3 2 2 6 + = + + = 3 2 2 3 2 3 2 3 2 2 2 2 ( ) + ( ) + = + ( ) = + Also, 8 2 15 5 3 2 15 5 3 2 5 3 2 2 = + = ( ) + ( ) = 5 3 5 3 2 ( ) = ∴ 5 2 6 8 2 15 3 2 5 3 + + = + + = 2 5 + X′C B O (–3,0) (3,0) X A YY′Mathematics IX ( Term -I) 5 OR We have, a = 9 – 4 5 ⇒ 1 1 9 4 5 a == 1 9 4 5 9 4 5 9 4 5 × ++ [Rationalising the denominator] = 9 4 5 9 4 5 9 4 5 81 80 9 4 5 2 2 + ( ) = + = + ( ) Now, a + 1a = 9 – 4 5 9 4 5 18 + + = ⇒ a a + ⎛⎝ ⎜ ⎞⎠ ⎟= 1 18 2 2 ( ) ⇒ a a 2 2 1 2 324 + + = ⇒ a a 2 2 1 324 2 322 + = = Hence, a2 + 12 a = 322. 16. Simplify the following: 7 3 5 3 5 7 3 5 3 5 ++ OR If 5 3 5 3 15 + = + a b, find the values of a and b. Sol. We have, 7 3 5 3 5 7 3 5 3 5 ++ = 7 3 5 3 5 7 3 5 3 5 3 5 3 5 + ( )( ) ( ) + ( ) + ( )( ) = 21 7 5 9 5 15 21 7 5 9 5 15 3 5 2 2 + ( ) + ( ) ( ) ( ) = 6 2 5 6 2 5 4 + ( ) = 6 2 5 6 2 5 4 4 5 4 5 + + = = OR We have, 5 3 5 3 15 + = +a b Rationalising the denominator on LHS, we get 5 3 5 3 5 3 5 3 5 3 5 3 + = + × ++6 Mathematics IX ( Term -I) = 5 3 5 3 5 3 2 15 5 3 8 2 15 2 4 15 2 2 2 + ( ) ( ) ( ) = + + = + = + Therefore, 4 + 15 15 = +a b Equating the rational and irrational parts, we get a = 4 and b = 1. 17. Factorise the following : 12(x2 + 7x)2 – 8(x2 + 7x) (2x – 1) – 15(2x – 1)2 Sol. Let x2 + 7x = p and 2x – 1 = q. Then, 12(x2 + 7x)2 – 8(x2 + 7x) (2x – 1) – 15(2x – 1)2 = 12p2 – 8pq – 15q2 = 12p2 – 18pq + 10pq –15q2 = 6p (2p – 3q) + 5q (2p – 3q) = (6p + 5q) (2p – 3q) = [6 (x2 + 7x) + 5 (2x – 1)] [2 (x2 + 7x) – 3 (2x – 1)] [Substituting the values of p and q] = (6x2 + 42x + 10x – 5) (2x2 + 14x – 6x + 3) = (6x2 + 52x – 5) (2x2 + 8x + 3) 18. Show that 2 and – 13 are the zeroes of the polynomial 3x3 – 2x2 – 7x – 2. [3] Also, find the third zero of the polynomial. Sol. Let p(x) = 3x3 – 2x2 – 7x – 2 If 2 is a zero of p(x), then p(2) = 0 So, p(2) = 3(2)3 – 2(2)2 – 7 × 2 – 2 = 3 × 8 – 2 × 4 – 7 × 2 – 2 = 24 – 8 – 14 – 2 = 0 Hence, 2 is a zero of the given polynomial. Similarly, if 1 3 is a zero of p(x), then p 1 3 ⎛⎝ ⎜ ⎞⎠ ⎟ = 0 So, p 1 3 ⎛⎝ ⎜ ⎞⎠ ⎟ = 3 1 3 2 1 3 7 1 3 2 3 2 ⎛⎝ ⎜ ⎞⎠ ⎟ ⎛⎝ ⎜ ⎞⎠ ⎟ ⎛⎝ ⎜ ⎞⎠ ⎟ = 3 1 27 2 19 73 2 ⎛⎝ ⎜ ⎞⎠ ⎟ ⎛⎝ ⎜ ⎞⎠ ⎟+ = 1 9 29 73 2 + = 1 2 21 18 9 09 0 + = = Hence, 1 3 is also a zero of the given polynomial. Since, 2 and 1 3 are the zeroes of p (x), therefore,Mathematics IX ( Term -I) 7 x – 2 and x + 13 are the factors of p (x). ⇒ (x – 2) x + ⎛⎝ ⎜ ⎞⎠ ⎟ 13 is a factor of p (x) ⇒ (x – 2) (3x + 1) is a factor of p (x) ⇒ 3x2 – 5x –2 is a factor of p (x) Now, the third zero of p(x) can be found out by dividing p(x) by 3x2 – 5x – 2. So, Thus, x + 1 is also a factor of p (x) i.e., x = –1 is the third zero of p (x). 19. In the figure, l || m || n and a ⊥ l. If ∠BEF = 55°, find the values of x, y and z.OR In the figure, l || m || n. Find the value of (y + x) : (y – x) A B lmn C E F x y 30° 20° D 100°40° Sol. We have, a ⊥ l ⇒ a ⊥ n [If a line is ⊥ to one of the given || lines, then it is also ⊥ to the other lines] ⇒ ∠AEF = 90° ⇒ z + 55° = 90° ⇒ z = 90° – 55° = 35°. Also m || n and DE is a transversal. Therefore, y + 55° = 180° [Co-interior angles on the same side of a transversal are supplementary] ⇒ y = 180° – 55° ⇒ y = 135°. Now, l || m and BD is a transversal. Therefore, x and y are corresponding angles. ⇒ x = y [Corresponding angles between two || lines are equal] ⇒ x = 135° Hence, x = 135°, y = 135° and z = 35°. 3 –5 –2 x x 2 x + 1 3 –2 –7 –2 x x x 3 2 3 –5 –2 x x x 3 2 – + + 3 –5 –2 x x 2 3 –5 –2 x x 2 – + + 0 l m n a A E B z 55° y x F CD8 Mathematics IX ( Term -I) S T R Q P OR We have, l || m and AC is a transversal. Therefore, x + 100° = 180° [Co-interior angles on the same side of a transversal are supplementary] ⇒ x = 180° – 100° = 80° Also, sum of all angles on the same side of a line is 180°. Therefore, 30° + y + 20° = 180° ⇒ y + 50° = 180° ⇒ y = 180° – 50° = 130° Now, y + x = 130° + 50° = 210° And, y – x = 130° – 80° = 50° ∴ (y + x) : (y – x) = 210° : 50° = 21 : 5. 20. In the figure, DE || BC and MF || AB. Find (i) ∠ADE + ∠MEN (ii) ∠BDE (iii) ∠BLE Sol. (i) We have, DE || BC and BD is a transversal. Therefore, ∠ADE = ∠DBC [Corresponding angles] ⇒ ∠ADE = 40° Now, MF || AB and DE is a transversal. Therefore ∠DEL = ∠ADE [Alternate angles] ⇒ ∠DEL = 40° Also, ∠MEM = ∠DEL [Vertically opposite angles] ⇒ ∠MEN = 40° Hence, ∠ADE + ∠MEN = 40° + 40° = 80°. (ii) DE || BC and BD is a transversal. Therefore, ∠DBC + ∠BDE = 180° [Co-interior angles on the same side of a transversal are supplementary] ⇒ 40° + ∠BDE = 180° ⇒ ∠BDE = 180° – 40° = 140° Hence, ∠BDE = 140° (iii) Again MF || AB and BL is a transversal. Therefore, ∠DBL + ∠BLE = 180° [Co-interior angles on the same side of a transversal are supplementary] ⇒ 40° + ∠BLE = 180° ⇒ ∠BLE = 180° – 40° = 140° Hence, ∠BLE = 140°. 21. In the figure, PS is bisector of ∠QPR and PT ⊥ RQ. Show that ∠TPS = 12 ( ∠R – ∠Q). Sol. Since, PS is the bisector of ∠QPR. Therefore, ∠QPS = ∠SPR ........... (i) Now, in ΔPTR, we have, ∠PRT + ∠TPR + ∠PTR = 180° [Angle sum property of a triangle]Mathematics IX ( Term -I) 9 50° 20° F G P E H D B C A ⇒ ∠PRT + ∠TPR + 90° = 180° ⇒ ∠PRT + ∠TPR = 90° ⇒ ∠PRT = 90° – ∠TPR ⇒ ∠R = 90° – ∠TPR .......... (ii) Also, in ΔPQT, we have ∠PQT + ∠PTQ + ∠QPT = 180° [Angle sum property of a triangle] ⇒ ∠PQT + 90° + ∠QPT = 180° ⇒ ∠PQT + ∠QPT = 90° ⇒ ∠PQT = 90° – ∠QPT ⇒ ∠Q = 90° – ∠QPT .......... (iii) Subtracting (iii) from (ii), we get ∠R – ∠Q = (90° – ∠TPR) – (90° – ∠QPT) = ∠QPT – ∠TPR = (∠TPS + ∠QPS) – (∠SPR – ∠TPS) = ∠TPS + ∠QPS – ∠QPS + ∠TPS = 2∠TPS [∵ ∠QPS = ∠SPR using (i)] Hence, ∠R – ∠Q = 2∠TPS or, ∠TPS = 12 (∠R – ∠Q) Proved. 22. In the figure, ΔABC and ΔABD are such that AD = BC, ∠1 = ∠2 and ∠3 = ∠4. Prove that BD = AC. Sol. We have, ∠1 = ∠2 ........... (i) and ∠3 = ∠4 ........... (ii) Adding (i) and (ii), we get ∠1 + ∠3 = ∠2 + ∠4 ⇒ ∠BAD = ∠ABC Now, in ΔABD and ΔBAC AD = BC [Given] ∠BAD = ∠ABC [Proved above] AB = AB [Common] ∴ By SAS congruency ΔABD ≅ ΔBAC Hence, by CPCT, BD = AC. Proved 23. In the figure, AB || CD. If ∠BAE = 50° and ∠AEC = 20°, find ∠DCE. Sol. Draw GAF || PC. Now, GA || PE and AE is a transversal. Therefore, ∠GAE = ∠AEC [Alternate angles] ⇒ ∠GAE = 20° Also, ∠BAG = ∠BAE – ∠GAE = 50° – 20° = 30° Again, AB || DH and AH is a transversal. Therefore, ∠DHA = ∠BAG [Corresponding angles] D C B A1 3 4 2 E10 Mathematics IX ( Term -I) ⇒ ∠DHA = 30° Since, GH || PC and HC is a transversal. Therefore, ∠DCE = ∠DHA [Corresponding angles] ⇒ ∠DCE = 30°. 24. Find the area of a triangle whose perimeter is 180 cm and two of its sides are 80 cm and 18 cm . Also calculate the altitude of the triangle corresponding to the shortest side. Sol. Let a = 80 cm, and b = 18 cm be two of the given sides of the triangle. Also, let c be its third side. Then, Perimeter of the triangle = a + b + c ⇒ 180 cm = 80 cm + 18 cm + c ⇒ c = 180 cm – 98 cm = 82 cm Now, a2 + b2 = (80)2 + (18)2 = 6400 + 324 = 6724 = (82)2 Therefore, a2 + b2 = c2 Hence, the given triangle is a right angled triangle ∴ Area of ΔABC = 12 × base × corresponding altitude = 12 × 80 × 18 cm2 = 720 cm2 Also, AC is the shortest side, therefore its corresponding altitude is BC. Thus, the length of the altitude corresponding to the shortest side is 80 cm. SECTION D (Question numbers 25 to 34 carry 4 marks each) 25. If x = 1 2 3, find the value of x3– 2x2 – 7x + 5. OR Simplify : 1 1 2 1 2 3 1 3 4 1 8 9 + + + + + + + + ......... Sol. We have, x = 1 2 3 Rationalising the denominator on RHS, we get x = 1 2 3 2 3 2 3 2 3 2 3 2 3 4 3 2 3 2 2 ( ) × ++ = + ( ) = + = + A B C 18 cm 82 cm 80 cmMathematics IX ( Term -I) 11 ⇒ x – 2 = 3 ⇒ (x – 2)2 = 3 2 ( ) ⇒ x2 – 4x + 4 = 3 ⇒ x2 – 4x + 1 = 0 ......... (i) ∴ x3 – 2x2 – 7x + 5 = x (x2 – 4x + 1) + 2 (x2 – 4x + 1) + 3 = x × 0 + 2 × 0 + 3 = 3 [Using (i)] OR Rationalising the denominator of first term, we get 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 2 2 2 + = + × = ( )= = + ( ) Also, 1 2 3 1 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 2 + = + × =( ) ( ) = = + Similarly, 1 3 4 3 4 + = + And, 1 8 9 8 9 + = + Now, 1 1 2 1 2 3 1 3 4 1 8 9 + + + + + + + + .............. = ... 1 2 2 3 3 4 8 9 + + + + + = 1 9 + = – 1 + 3 = 2. 26. If x = p q p q p q p q + + +2 2 2 2 , then show that qx2 – px + q = 0. OR If x = 2 1 2 1 1 2 1 + + and y = 2 , find the value of x2 + y2 + xy . Sol. We have, x = p q p q p q p q + + +2 2 2 2 Rationalising the denominator on RHS, we get x = p q p q p q p q p q p q p q p q + + + × + + + + 2 2 2 2 2 2 2 2 = p q p q p q p q + + ( ) + ( ) ( ) 2 2 2 22 2 2 = p q p q p q p q p q p q p p q q + + + + + + = + 2 2 2 2 2 2 2 2 2 4 42 2 ( )( ) = p p q q + 2 2 4 212 Mathematics IX ( Term -I) Now, x = p p q q + 2 2 4 2 ⇒ 2qx = p + p q 2 2 4 ⇒ 2qx – p = p q 2 2 4 ⇒ (2qx – p)2 = p q 2 22 4 ( ) ⇒ 4q2x2 + p2 – 4pqx = p2 – 4q2 ⇒ 4q2x2 – 4pqx + 4q2 = 0 ⇒ 4q (qx2 – px + q) = 0 ⇒ qx2 – px + q = 0. Proved. OR We have, x = 2 1 2 1 1 2 1 + + and = 2 y Rationalsing the denominators of x and y, we get x = 2 1 2 1 1 2 1 2 1 2 1 2 1 2 2 2 1 3 2 2 2 2 2 + × ++ = + ( ) ( ) ( ) = + + = + 2 And, y = 2 1 2 1 1 2 1 2 1 2 1 2 1 2 2 2 1 3 2 2 2 2 2 +× =( ) ( ) ( ) = + = 2 Now, x + y = 3 + 2 2 3 2 2 6 + = Also, xy = 3 2 2 3 2 2 3 2 2 9 8 1 2 2 + ( )( )= ( )= = ( ) So, x2 + y2 + xy = (x + y)2 – xy = (6)2 – 1 = 36 – 1 = 35. 27. If x3 + mx2 – x + 6 has (x – 2) as a factor, and leaves a remainder n when divided by (x – 3), find the values of m and n. Sol. Let p (x) = x3 + mx2 – x + 6 Now, (x – 2) is a factor of p (x) ⇒ p (2) = 0 [∵ x – 2 = 0 ⇒ x = 2] ⇒ (2)3 + m (2)2 – 2 + 6 = 0 ⇒ 8 + 4 m + 4 = 0 ⇒ 4 m + 12 = 0 ⇒ 4 m = – 12 ⇒ m = – 3. Also, p(x) leaves a remainder n when divided by (x – 3). Then, p (3) = n [∵ x – 3 = 0 ⇒ x = 3] ⇒ (3)3 + (–3) (3)2 – 3 + 6 = n [Substituting m = –3] ⇒ 27 – 27 + 3 = n ⇒ n = 3. 28. Prove that (x + y)3 + (y + z)3 + (z + x)3 – 3 (x + y) (y + z) (z + x) = 2 (x3 + y3 + z3 – 3xyz)Mathematics IX ( Term -I) 13 1 2 3 4 5 12345 –1 –2 –4 –5 –1 –2 –3 –4 –6 –3 Y X Y′ X′ C 6 6 –6 E F B –5 A D Sol. Let x + y = a, y + z = b and z + x = c. Then, (x + y)3 + (y + z)3 + (z + x)3 – 3(x + y) (y + z) (z + x) = a3 + b3 + c3 – 3 abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = (x + y + y + z + z + x) {(x + y)2 + (y + z)2 + (z + x)2 – (x + y) (y + z) – (y + z) (z + x) – (z + x) (x + y)} = 2 (x + y + z) {x2 + y2 + 2xy + y2 + z2 + 2yz + z2 + x2 + 2zx – (xy + xz + y2 + yz) – (yz + xy + z2 + zx) – (zx + yz + x2 + xy)} = 2(x + y + z) [2x2 + 2y2 + 2z2 + 2xy + 2yz + 2zx – xy – zx – y2 – yz – yz – xy – z2 – zx – zx – yz – x2 – xy] = 2(x + y + z) [2x2 + 2y2 + 2x2 + 2xy + 2yz + 2zx – 3xy – 3zx – 3yz – x2 – y2 – z2] = 2(x + y + z) (x2 + y2 + z2 – xy – yz – zx) = 2(x3 + y3 + z3 – 3xyz) Proved. 29. If A and B be the remainders when the polynomials x3 + 2x2 – 5ax – 7 and x3 + ax2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively and 2A + B = 6, find the value of ‘a’. Sol. Let p (x) = x3 + 2x2 – 5ax – 7 and q (x) = x3 + ax2 – 12x + 6. Now, remainder when p (x) is divided by (x + 1) = A ⇒ p (–1) = A ⇒ (–1)3 + 2 (–1)2 – 5a × (–1) – 7 = A ⇒ –1 + 2 + 5a –7 = A ⇒ 5a – 6 = A .......... (i) And, remainder when q (x) is divided by (x – 2) = B ⇒ q (2) = B ⇒ (2)3 + a (2)2 – 12 × 2 + 6 = B ⇒ 8 + 4a – 24 + 6 = B ⇒ 4a – 10 = B ........ (ii) Substituting the values of A and B in 2A + B = 6, we get 2 (5a – 6) + 4a – 10 = 6 ⇒ 10a – 12 + 4a – 10 = 6 ⇒ 14a – 22 = 6 ⇒ 14a = 28 ⇒ a = 2. 30. From the figure, find the coordinates of the points A, B, C, D, E and F. Which of the points are mirror images in (i) x-axis (ii) y-axis. Sol. Coordinates of A = (2, 2) Coordinates of B = (1, 4) Coordinates of C = (5, 5) Coordinates of D = (–1, 4) Coordinates of E = (2, – 2) Coordinates of F = (–5, –5) From the figure, it is clear that point E is the mirror image of point A in the x-axis.14 Mathematics IX ( Term -I) And, point D is the mirror image of point B in the y-axis. 31. In the figure, QT ⊥ PR, ∠TQR = 40° and ∠SPR = 30°. Find the values of x, y and z. Sol. In ΔQTR, we have ∠QTR = 90° [∵ QT ⊥ PR] ∴ By angle sum property of a triangle ∠TQR + ∠QTR + ∠QRT = 180° ⇒ 40° + 90° + x = 180° ⇒ 130° + x = 180° ⇒ x = 180° – 130° = 50° Now, in ΔPSR ∠PSQ = ∠SPR + ∠PRS [By exterior angle property] ⇒ y = 30° + 50° [∵ ∠PRS = x = 50°] ⇒ y = 80° Similarly, z = 40° + y ⇒ z = 40° + 80° = 120° Therefore, x = 50°, y = 80° and z = 120°. 32. In the figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that (i) DF = BE (ii) AM bisects ∠BAD. Sol. In ΔBCD (i) BC = CD [Sides of a square] ⇒ ∠2 = ∠4 [Angles opp. to equal sides are also equal]......... (i) Also, ∠1 = ∠2 [Corresponding angles, since EF || BD] ......... (ii) And, ∠3 = ∠4 [Same reason] ......... (iii) From (i), (ii) and (iii) we get, ∠1 = ∠3 Now, in ΔCEF ∠1 = ∠3 ⇒ CE = CF [Sides opp. to equal angles are also equal] ⇒ BC – BE = CD – DF ⇒ BE = DF (∵ BC = CD) Proved. (ii) In ΔADF and ΔABE AD = AB [Sides of a square] ∠ADF = ∠ABE [Each = 90°] DF = BE [Proved above] ∴ By SAS congruency, ΔADF ≅ ΔABE Hence, by CPCT, AF = AE and ∠5 = ∠6 Again, in ΔAMF and ΔAME AF = AE [Proved above] FM = EM [Given] 40° 30° z T x y P R Q S D A F M C E B D A F M C E B 2 578 6 1 3 4Mathematics IX ( Term -I) 15 AM = AM [Common] ∴ By SSS congruency, ΔAMF ≅ ΔAME Hence by CPCT, ∠7 = ∠8 ⇒ ∠7 + ∠5 = ∠8 + ∠6 [∵ ∠5 = ∠6] ⇒ ∠MAD = ∠MAB ⇒ AM bisects ∠BAD Proved. 33. In the figure, AB = BC, ∠A = ∠C and ∠ABD = ∠CBE. Prove that CD = AE. Sol. We have, ∠ABD = ∠CBE [Given] Adding ∠DBE both sides, we get ∠ABD + ∠DBE = ∠CBE + ∠DBE ⇒ ∠ABE = ∠CBD Now, in ΔABE and ΔCBD ∠BAE = ∠BCD [Given] AB = BC [Given] ∠ABE = ∠CBD [Proved above] ∴ By ASA congruency, ΔABE ≅ ΔCBD Hence, by CPCT, CD = AE Proved. 34. In the figure, AB = AC, D is a point in the interior of ΔABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ΔABC. Sol. In ΔBCD, we have, ∠DCB = ∠DBC [Given] ⇒ BD = CD [Sides opp. to equal angles are also equal] Now, in ΔABD and ΔACD, AB = AC [Given] AD = AD [Common] BD = CD [Proved above] ∴ By SSS congruency, ΔABD ≅ ΔACD Hence, by CPCT, ∠BAD = ∠CAD ⇒ AD bisects ∠BAC. Proved. B C AD D E A B C

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If H and K two mid points of sides EF & EG and EPG. Then FGKH be a tropryium then prov that are FGKH =3/4 (EFG)

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