iBOOKS SAMPLE PAPER вЂ“ 01 [FOR XTH CBSE BOARD] QUESTION ANSWER SECTION вЂ“ A SECTION вЂ“ A 1. 2. [1Г—10] State the Fundamental Theorem of Arithmetic. 1. Every composite number can be expressed as the product of primes, and this factorization is unique, apart from the order in which they occur. [1] 2. -1 and 3 are the zeroes of the polynomial. 3. For infinitely many solution, In figure, the graph of the polynomial P(x) is given. Find the zeroes of the polynomial. [1] Y 4 3 2 1 X' -3 -2 -1 0 -1 1 2 3 4 X -2 a1 b1 c 1 = = a2 b2 c 2 Here a1 2 b1 3 c1 k 1 = , = , = = . a 2 k в€’ 1 b 2 k + 2 c 2 3k 3 Now 2 3 1 2 1 = = в‡’ = k в€’1 k + 2 3 k в€’1 3 -3 в‡’ k вЂ“ 1 = 6, в‡’ k = 7. -4 [1] Y' 3. For what values of k, the following pair of linear equations has infinitely many solutions ? 4. 2x + 3y = k = (k вЂ“ 1)x + (k + 2) y = 3k. 4. The Arc of length 'l' = Оё Г— 2ПЂr 360В° 30 22 Г— 2 Г— Г— 21 = 11 cm 360В° 7 в€ґ Perimeter = l + 2r = 11 + 2 Г— 21 = 53 cm. If the following figure is a sector of a circle of radius 21 cm. Find the perimeter of this sector. B 5. l [1] Total number of outcomes, throwing two dice are = 36. The favourable outcomes to the even. The total of numbers on the dice is 13 are 0. [1] 30В° A O 5. Two dice are thrown simultaneously. Find the probability that the total of the numbers on the dice is 13. 6. Here total frequency, ОЈfi = 5 + 8 + 20 + 15 + 7 + 5 = 60. в€ґ n/2 = 30 Marks obtained: 0вЂ“10 10вЂ“20 20вЂ“30 30вЂ“40 40вЂ“50 50вЂ“60 Frequency 6. : 5 8 20 15 7 5 5 13 33 48 45 60 Find the median class of the following data : Cumulative Marks obtained Frequency Frequency 0 вЂ“ 10 5 10 вЂ“ 20 8 Now 20 вЂ“ 30 is the median class since its cumulative frequency is 33 which is greater than 30. [1] 20 вЂ“ 30 20 30 вЂ“ 40 15 40 вЂ“ 50 7 50 вЂ“ 60 5 : QUESTION 7. ANSWER In figure, P and Q are points on the sides AB and AC respectively of в€† ABC such that AP = 3.5 cm, PB = 7 cm, AQ =3 cm and QC=6 cm. If PQ = 4.5 cm, Find BC. 7. In в€† ABC, we have, AP 3.5 1 AQ 3 1 = = and = = PB 7 2 QC 6 2 в€ґ PQ || BC в€ґ в€† AQP ~ в€† ABC. 2 в‡’ 2 Ar( в€†AQP) пЈ« AQ пЈ¶ пЈ« AP пЈ¶ пЈ« PQ пЈ¶ = =пЈ¬ пЈ· = пЈ¬ BC пЈ· Ar( в€†ABC) пЈ¬пЈ AC пЈ·пЈё пЈ AB пЈё пЈ пЈё 2 2 2 2 1 пЈ« 3 пЈ¶ пЈ« PQ пЈ¶ пЈ« PQ пЈ¶ в‡’пЈ¬ пЈ· = пЈ¬ BC пЈ· в‡’ пЈ¬ BC пЈ· = 3 пЈ пЈё пЈ3 + 6пЈё пЈ пЈё в‡’ BC = 3 PQ = 3 (4.5) = 13.5 cm. 8. 9. If tan A = 3/4 and A + B = 90В°, then what is the value of cot B ? Determine the 10th term from the end of the AP., 4, 9, 14, вЂ¦вЂ¦ , 254. 10. The length of the tangent from a point A at a distance of 26 cm from the centre of the circle is 10 cm. What will be the radius of the circle ? [1] 8. Given A + B = 90В° в‡’ B = 90В° - A. [1] 9. в‡’ cot B = cot (90В° - A) в‡’ cot B = tan A = 3/4. Here a = 254, d = -5 в€ґ a10 = a + 9d = 254 + 9(-5) = 254 вЂ“ 45 = 209. [1] 10. Since the tangent to a circle is perpendicular to the radius through the point of contact. в€ґ в€ OTA = 90В° 2 2 2 In rt. в€† OAT, we have, OA = OT + AT T 2 2 2 2 в‡’ 26 = OT + 10 в‡’ 676 = OT + 100 10 cm 2 в‡’ OT = 676 вЂ“ 100 =576 в‡’ OT = 24 cm. O [1] A 26 cm SECTION вЂ“ B 11. If the ten, jack, queen, king and ace of diamonds are removed, then the total of cards left is 47. Now the queens left in the deck is 3. [1] SECTION вЂ“ B [2 Г— 5] 11. The ten, jack, queen, king and ace of diamonds are removed from the deck of cards and then shuffled. One card is then picked up at random (i) What is the probability that the card is the queen ? (ii) What is the probability that the card is a face card ? OR A game consists of tossing a one rupee coin 3 times and noting its outcomes each time. Anjali wins if all the tosses give the same result and losses otherwise. Calculate the probability that Anjali will lose the game. So P (getting a queen) = 3/47. If the jack, queen and king of diamonds are removed, then the face cards left = 9. So P (getting a face card) = 9/47. [1] OR When a coin is tossed 3 times, the possible outcomes [1] are : HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. The favourable outcomes that Anjali will lose the game if all the tosses do not give the same results i.e. all heads or all tails are HHT, HTH, THH, HTT, THT, TTH. So P (lose the game) = 6/8 = 3/4. [1] 12. Any point on x вЂ“ axis is P (x, 0). Given P is at a distance of 5 units from the point A. So PA = 5 2 2 2 12. Find the points on the x вЂ“ axis, each of which is at a distance of 5 units from the point A (5, 3). 2 в‡’ PA = 25. в‡’ (x вЂ“ 5) + (0 + 3) = 25. в‡’ (x вЂ“ 5) = 25 вЂ“ 9 = 16. в‡’ x вЂ“ 5 = В± 4. [1] в‡’ x = 5 В± 4 = 9 or 1. Therefore, there are two points (1, 0) and (9, 0) which are at a distance of 5 units from the given point. [1] QUESTION ANSWER 13. PQR is a right angled triangle having 2 2 в€ Q = 90В°, If QS = SR, Show that PR = 4PS вЂ“ 2 3PQ . 13. Given PQR is a right angled triangle having в€ Q = 90В° and QS = SR. P 2 2 2 To show PR = 4PS вЂ“ 3PQ . 2 2 2 In в€† PQR, PR = PQ + QR (by Pythagoras theorem) 2 2 2 2 в‡’ PR = PQ + (2QS) ( в€µ QS = SR) 2 Q S 2 в‡’ PR = PQ + 4QS вЂ¦вЂ¦вЂ¦.. (1) 2 2 R [1] 2 Again in в€† PQS, PS = PQ + QS 2 2 2 в‡’ QS = PS вЂ“ PQ . Putting in equation (1) we get 2 2 2 2 2 2 2 PR = PQ + 4(PS вЂ“ PQ ) = PQ + 4PS вЂ“ 4PQ 2 2 = 4 PS вЂ“ 3PQ . 14. Prove the identity : tan 2 2 A вЂ“ tan 2 B =. sin 2 A в€’ sin 2 B . cos 2 A.cos 2 B 2 14. L.H.S. = tan A вЂ“ tan B = = sin 2 sin 2 A cos 2 A A.cos 2 B в€’ sin 2 B.cos 2 в€’ [1] sin 2 B cos 2 B A cos 2 A.cos 2 B = sin 2 A(1 в€’ sin 2 B) в€’ sin 2 B(1 в€’ sin 2 A) cos 2 A.cos 2 B = sin 2 A в€’ sin 2 A.sin 2 B в€’ sin 2 B + sin 2 A.sin 2 B cos 2 A.cos 2 B [1] [1] sin 2 A в€’ sin 2 B cos 2 A.cos 2 B 2 2 15. We have f(x) = abx + (b вЂ“ ac) x вЂ“ bc. 2 2 = abx + b x вЂ“ acx вЂ“ bc = bx (ax + b) вЂ“ c (ax + b) = (ax + b) (bx вЂ“ c). The value of f(x) is zero when ax + b = 0 or bx вЂ“ c = 0. в€’b c в‡’x= or x = . a b Therefore zeroes are вЂ“b/a or c/b. [1] в€’b c в€’b 2 + ac в€’(b 2 в€’ ac) Sum of the zeroes = + = = a b ab ab coefficient of x =в€’ [ВЅ] coefficient of x 2 Product of zeroes c в€’bc cons tan t term пЈ« в€’b пЈ¶ c = [ВЅ] = пЈ¬ пЈ· =в€’ = a ab coefficient of x 2 пЈ a пЈёb = 15. Find the zeroes of the quadratic polynomial f(x) 2 2 = abx + (b вЂ“ ac) x вЂ“ bc and verify the relationship between the zeroes and their coefficients. SECTION вЂ“ C [3Г—10] 16. Prove that for any prime positive integer p, в€љp is an irrational number. SECTION вЂ“ C 16. Let us assume that в€љp is rational. Then there exists positive co-primes a and b such that в€љp = a/b [1] в‡’ p = a2 /b2 в‡’ pb2 = a2 в‡’ p divides a2 в‡’ p divides a ......(1) в‡’ Let a = pc for some positive integer c 2 2 2 Now a = p c в‡’ p c = pb2 в‡’ b2 = pc2 2 2 [1] QUESTION ANSWER в‡’ p divides b2 в‡’ p divides b .......(2) 17. Solve the following system of linear equations graphically : From (1) and (2), p is a common factor of a and b This contradicts that a and b are co-prime. Hence в€љp is irrational. 5в€’x 17. x + 2y = 5, в‡’ y = 2 x y x + 2y = 5 1 2 5 0 x y Also find the points where the lines meet the x вЂ“ axis. 1 2 -2 0 (-3, 4) [1] -3 4 Again 2x вЂ“ 3y = -4, в‡’ y = 2x - 3y = -4 [1] 2x + 4 3 [1] 4 4 Y 2xвЂ“3y =- 4 (4, 4) A(1, 2) x+2y=5 X' O (5, 6)B C(-2, 0) Y' The Lines meet x вЂ“ axis at (5, 0) and (-2, 0) 18. L.H.S. = 18. Prove the identity : cos2 Оё sin3 Оё + = 1 + sin Оё.cos Оё 1 в€’ tan Оё sin Оё в€’ cos Оё OR Prove the identity : tan Оё + sec Оё в€’ 1 1 + sin Оё = tan Оё в€’ sec Оё + 1 cos Оё X [ВЅ] [ВЅ] cos2 Оё sin3 Оё + 1 в€’ tan Оё sin Оё в€’ cos Оё cos2 Оё sin3 Оё + sin Оё sin Оё в€’ cos Оё 1в€’ cos Оё [1] = cos3 Оё sin3 Оё cos3 Оё в€’ sin3 Оё в€’ = cos Оё в€’ sin Оё cos Оё в€’ sin Оё cos Оё в€’ sin Оё [1] = (cos Оё в€’ sin Оё)(cos2 Оё + sin2 Оё + cos Оё.sin Оё) (cos Оё в€’ sin Оё) [1] = = 1 + cos Оё . sin Оё = R. H. S. OR tan Оё + sec Оё в€’ 1 L.H.S. = tan Оё в€’ sec Оё + 1 (sec Оё + tan Оё) в€’ (sec 2 Оё в€’ tan2 Оё) (1 в€’ sec Оё + tan Оё) (sec Оё + tan Оё) в€’ (sec Оё + tan Оё)(sec Оё в€’ tan Оё) = (1 в€’ sec Оё + tan Оё) (sec Оё + tan Оё){1 в€’ (sec Оё в€’ tan Оё)} = (1 в€’ sec Оё + tan Оё) (sec Оё + tan Оё)(1 в€’ sec Оё + tan Оё) = (1 в€’ sec Оё + tan Оё) = sec Оё + tan Оё = [1] [1] QUESTION ANSWER 2 19. In an A.P, the sum of first n terms is 3n 5n + . 2 2 th Find its 25 term. 1 sin Оё 1 + sin Оё + = [1] cos Оё cos Оё cos Оё = R.H.S. 19. Let Sn denote the sum of n terms of an AP whose nth term is an. = 3n2 5n + 2 2 3 5 в‡’ Sn в€’1 = (n в€’ 1)2 + (n в€’ 1) 2 2 We have Sn = [1] пЈ® 3n 5n пЈ№ пЈ® 3 5 пЈ№ в€ґ Sn в€’ Sn в€’1 = пЈЇ + пЈє в€’ пЈЇ (n в€’ 1)2 + (n в€’ 1)пЈє [1] 2 пЈєпЈ» пЈ° 2 2 пЈ» пЈЇпЈ° 2 3 5 в‡’ an = [n2 в€’ (n в€’ 1)2 ] + [n в€’ (n в€’ 1)] 2 2 3 5 в‡’ an = (2n в€’ 1) + 2 2 3 5 в‡’ a25 = Г— 49 + = 76 [1] 2 2 20. Given well is of diameter 3 m and 14 m deep. Hence its radius = 3/2 m 2 Volume of the earth taken out from the well = ПЂr h 3 3 = ПЂ Г— Г— Г— 14 2 2 22 3 3 = Г— Г— Г— 14 = 99 m3 [1] 7 2 2 The earth taken out of the well has been spread evenly all around it in the shape of a circular ring of width 4m to form an embakment. 2 2 Area over which the earth is spread out = ПЂ (r1 вЂ“ r2 ) 22 2 2 = 7 (5.5 в€’ 1.5 ) [1] 2 20. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment. OR Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Q = O 22 пЈ® 3025 225 пЈ№ 22 = Г— 28 = 88m2 в€’ 7 7 пЈЇпЈ° 100 100 пЈєпЈ» в€ґ Height of embankment = 99/88 = 9/8 = 1.125 m. [1] OR Q R P O R P In figure, given PQ = 24 cm, PR = 7 cm. Here, в€ P is a right angle. [в€µ Angle in a semicircle is right angle] в€ґ By Pythagoras Theorem, we have 2 2 2 2 2 2 QR = PQ + PR = (24) + (7) = 576 + 49 = 625 = (25) в‡’ QR = 25 cm. [1] Since, QR is a diameter of a circle (as passes through the centre O of the circle), Therefore, radius of a circle (r) = 25/2 cm. Now, Area of the shaded region = Area of semicircle вЂ“ Area of right в€† QPR. [1] 1 2 1 = ПЂr в€’ (Base Г— Height) 2 2 QUESTION ANSWER 2 1 22 пЈ« 25 пЈ¶ 1 Г— Г— в€’ Г— 7 Г— 24 [ в€µ Base = PR = 7 cm, 2 7 пЈ¬пЈ 2 пЈ·пЈё 2 Height = PQ = 24 cm] пЈ®11Г— 25 Г— 25 пЈ№ =пЈЇ в€’ 7 Г— 12пЈє cm 2 7 Г— 4 пЈ° пЈ» = 21. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. пЈ® 6875 2352 пЈ№ 2 пЈ® 6875 в€’ 2352 пЈ№ cm 2 =пЈЇ в€’ пЈє cm = пЈЇ пЈє 28 28 28 пЈ° пЈ» пЈ° пЈ» 4523 = cm 2 [1] 28 21. Steps of Construction : 1. Draw a line segment BC = 6 cm. 2. With B as centre and radius = 4 cm, draw an arc. 3. Again, C as centre and radius = 5 cm, draw another are cutting the arc in step 2 at A. Then, ABC is the required triangle. 4. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. 5. Locate 3 points (the greater of 2 and 3 in 2/3) B1, B2 and B3 on BX so that BB1 = B1B2 = B2B3. 6. Join B3C and draw a line through B2 parallel to B3C, to intersect BC at C'. 7. Draw a line through C' parallel to the line CA to intersect BA at A'. Then, в€† A'BC' is the required triangle. A A' 5 cm 4 cm B 6 cm C C' B1 B2 X B3 22. If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium. [3] 22. Given : A quadrilateral ABCD whose diagonals AC and AO BO BD intersect each other at O such that = . OC OD To prove : Quadrilateral ABCD is a trapezium, i.e., AB || DC. Construction : Draw OE || BA, meeting AD in E. [1] D C E O A Proof : In в€† ABD, we have OE || BA. AE BO So, = вЂ¦вЂ¦вЂ¦вЂ¦.. (1) [By BPT] ED OD B QUESTION ANSWER AO BO = вЂ¦вЂ¦вЂ¦вЂ¦.. (2) [Given] OC OD AE AO From (1) and (2), we get = [1] ED OC Thus, in в€† DCA, O and E are points on AC and AD AE AO respectively such that = ED OC Therefore, by the converse of BPT, we have EO || DC. But OE || BA в€ґ DC || BA в‡’ AB || DC. Hence, ABCD is trapezium. [1] 23. Let the speed of the steam be x km/h Speed of the boat in still water = 5 km/h But 23. Swati row her boat at a speed of 5 km/h in still water. If it takes her 1 hour more to row the boat 5.25 km upstream than to return downstream, find the speed of the stream. в€ґ Speed of the boat upstream = (5 вЂ“ x) km/h. Speed of the boat down stream = (5 + x) km/h 5.25 5.25 в€’ = 1 (given) 5в€’x 5+x [1] 1 пЈј 21 пЈ± 5 + x в€’ 5 + x пЈј пЈ± 1 в‡’ (5.25) пЈІ в€’ пЈЅ = 1, в‡’ пЈІ пЈЅ =1 4 пЈі (5 в€’ x)(5 + x) пЈѕ пЈі5 в€’ x 5 + x пЈѕ в‡’ 21 Г— 2x = 4(25 вЂ“ x2), в‡’ 21 x = 50 вЂ“ 2x2 в‡’ 2x2 + 21 x вЂ“ 50 = 0, в‡’ 2x2 + 25 x вЂ“ 4 x вЂ“ 50 = 0 [1] в‡’ (2x + 5) (x вЂ“ 2) = 0, в‡’ Either 2x + 5 = 0 or x вЂ“ 2 = 0 в‡’ x = -5/2 or x = 2, We reject x = -5/2 в‡’ x = 2 в€ґThe speed of the stream = 2 km/h. [1] 24. Find the lengths of the medians of the triangles whose vertices (1, -1), (0, 4) and (-5, 3). 24. Let the three vertices of the triangle ABC be A(1, -1), B (0, 4) and C(-5, 3), respectively. Let D, E, F be the midpoints of sides BC, CA and AB respectively. Then, coordinates of points D, E and F are пЈ« 0 в€’ 5 4 + 3 пЈ¶ пЈ« в€’5 + 1 3 в€’ 1пЈ¶ пЈ« 1 + 0 в€’1 + 4 пЈ¶ пЈ¬ 2 , 2 пЈ· , пЈ¬ 2 , 2 пЈ· and пЈ¬ 2 , 2 пЈ· пЈ пЈё пЈ пЈё пЈ пЈё пЈ« 5 7пЈ¶ пЈ« 1 3пЈ¶ i.e. пЈ¬ в€’ , пЈ· ,( в€’2,1) and пЈ¬ , пЈ· пЈ 2 2пЈё пЈ 2 2пЈё пЈ« 5 7пЈ¶ пЈ«1 3пЈ¶ i.e., D пЈ¬ в€’ , пЈ· ,E( в€’2,1) and F пЈ¬ , пЈ· , respectively. пЈ 2 2пЈё пЈ2 2пЈё A (1, -1) в€ґ Length of the median AD 2 пЈ« 5 пЈ¶ пЈ«7 пЈ¶ = пЈ¬ в€’ в€’ 1пЈ· + пЈ¬ + 1пЈ· пЈ 2 пЈё пЈ2 пЈё 2 пЈ« 7пЈ¶ пЈ«9пЈ¶ = пЈ¬в€’ пЈ· + пЈ¬ пЈ· пЈ 2пЈё пЈ 2пЈё 2 F 2 49 81 130 130 + = = 4 4 4 2 Length of the median BE = E B (0, 4) D = ( в€’2 в€’ 0) 2 + (1 в€’ 4) 2 = 4 + 9 = 13 and length of the median CF 2 2 [1] 2 пЈ«1 пЈ¶ пЈ«3 пЈ¶ пЈ« 11пЈ¶ пЈ« 3пЈ¶ = пЈ¬ + 5пЈ· + пЈ¬ в€’ 3пЈ· = пЈ¬ пЈ· + пЈ¬ в€’ пЈ· пЈ2 пЈё пЈ2 пЈё пЈ2пЈё пЈ 2пЈё C (-5, 3) [1] 2 121 9 121 + 9 130 130 + = = = 4 4 4 4 2 Hence, the lengths of the medians of the triangle whose QUESTION ANSWER 25. The coordinates of the centroid of a triangle are (1, 4) and two of its vertices are (-8, 6) and (9, 5). Find the third vertex. Also, find the coordinates of the centroid of the triangle when the third vertex is (2, 4). OR Find the point on the x вЂ“ axis which is equidistant from (2, -5) and (-2, 9). SECTION вЂ“ D vertices are (1, -1), (0, 4) and (-5, 3) are 130 130 , 13 and . [1] 2 2 25. Let the two vertices of the triangle be A (-8, 6) and B (9, 5) and let third vertex of the triangle ABP be P(x, y), the centroid of the triangle is given as G(1, 4). [1] x + x2 + x3 y + y2 + y3 в€ґ 1 =1 and 1 =4 3 3 в€’8 + 9 + x 6+5+y в‡’ =1 and =4 3 3 в‡’1+x=3 and 11 + y = 12 в‡’x=3вЂ“1 and y = 12 вЂ“ 11 в‡’x=2 and y = 1 Thus, the coordinates of the third vertex are (2, 1). [1] Let the centroid of the triangle ABC be G1 (x, y). Also, let the three vertices of the triangle ABC are A (-8, 6), B (9, 5), C (2, 4) respectively. ( в€’8) + 9 + 2 6+5+4 Then, x = and y = 3 3 в‡’ x = 3/3 and y = 15/3 в‡’x=1 and y = 5 Thus, the coordinates of the centroid are (1, 5). [1] OR Let the required point P be on the x вЂ“ axis, then its ordinate is zero. Let the abscissa of the point P be x. в€ґ The coordinates of the point P are (x, 0). Let the given points be A(2, -5) and B(-2, 9). It is given that the point P(x, 0) on x вЂ“ axis is equidistant from A(2, -5) and B (-2, 9). [1] в€ґ AP = BP (given) 2 2 в‡’ AP = BP 2 2 2 2 в‡’ (x вЂ“ 2) + (0 + 5) = (x + 2) + (0 вЂ“ 9) 2 2 в‡’ (x вЂ“ 4x + 4) + 25 = (x + 4x + 4) + 81 в‡’ -4x + 29 = 4x + 85 [1] в‡’ - 4x - 4x = 85 вЂ“ 29 в‡’ -8x = 56, в‡’ x = -7. Thus, the required point on the x вЂ“ axis is (-7, 0). [1] [6Г—5] 26. Prove that : The ratio of areas of similar triangle is equal to the ratio of the squares of the corresponding sides. Using the above theorem, prove that the area of equilateral triangle described on the side of a square is half the area of the equilateral triangle described its diagonal. SECTION вЂ“ D 26. Given: Triangles ABC and PQR such that в€† ABC ~ в€† PQR ( ar ( ABC ) ) = пЈ« AB пЈ¶2 = пЈ« BC пЈ¶2 = пЈ« CA пЈ¶2 пЈ¬ QR пЈ· пЈ¬ RP пЈ· To prove: ( ar (PQR ) ) пЈ¬пЈ PQ пЈ·пЈё пЈ пЈё пЈ пЈё Construction: Draw altitudes AM and PN of the triangles. P A B M C Q N 1 Proof: Now, ar(ABC) = 2 BC Г— AM 1 and ar(PQR) = 2 QR Г— PN R [1] QUESTION ANSWER пЈ«пЈ« 1пЈ¶ So, пЈ¶ ( ar ( ABC ) ) = пЈ¬пЈ пЈ¬пЈ 2 пЈ·пЈё BC Г— AM пЈ·пЈё = (BC Г— AM) ( ar (PQR ) ) пЈ«пЈ¬ пЈ« 1 пЈ¶ QR Г— PN пЈ·пЈ¶ ( QR Г— PN) (1) пЈ¬ пЈ· пЈпЈ 2 пЈё пЈё Now, in в€† ABM and в€† PQN, в€ B = в€ Q (As в€† ABC ~ в€† PQR) [1] в€ M = в€ N (Each is of 90В°) (AA similarity criterion) So, в€† ABM ~ в€† PQN AM AB Therefore, PN = PQ (2) Also, в€† ABC ~ в€† PQR (Given) AB BC CA So, PQ = QR = RP (3) ( ar ( ABC ) ) AB AM Therefore, ( ar (PQR ) ) = PQ Г— PN [From (1) and (3)] AB AB = PQ Г— PQ пЈ« AB пЈ¶ =пЈ¬ пЈ· пЈ PQ пЈё [1] [From (2)] 2 Using (3), ( ar ( ABC ) ) = пЈ« AB пЈ¶2 = пЈ« BC пЈ¶2 = пЈ« CA пЈ¶2 ( ar (PQR ) ) пЈ¬пЈ PQ пЈ·пЈё пЈ¬пЈ QR пЈ·пЈё пЈ¬пЈ RP пЈ·пЈё nd 2 Part : Given: ABCD is a square and в€†ABE is an equilateral triangle on side AB and в€†BDF is also an equilateral triangle on side BD i.e. diagonal of square ABCD. A D E C A F 1 To prove: ar (в€†ABE) = 2 ar (в€†BDF) [1] Proof: In a right в€†ABD. 2 2 2 AB + AD = BD 2 2 2 Or AB + AB = BD ( вЂ№ ABCD is a square) 2 2 в‡’ 2AB = BD or BD = в€љ2AB вЂ¦(i) в€†ABE and в€†DBF are equilateral triangles. в€ґ в€†ABE ~ в€†DBF We know that areas of two similar triangles are in the ratio of the squares of their corresponding sides. area ( в€†ABE ) AB2 в€ґ area ( в€†DBF ) = BD2 [1] From (i) BD = в€љ2AB ar ( в€†ABE ) AB2 AB2 1 = = = 2 2 в€ґ area ( в€†DBF ) 2 2AB 2AB ( ) 1 or ar (в€†ABE) = 2 ar(в€†DBF). QUESTION ANSWER 27. An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water 27. The total height of the bucket = 40 cm, which includes the height of the base. So, the height of the frustum of the cone = (40 вЂ“ 6) cm = 34 cm. Therefore, the slant height of the frustum, пЈ« 22 пЈ¶ the bucket can hold. пЈ¬ Take ПЂ = 7 пЈ· пЈ пЈё l = h2 + (r1 в€’ r2 )2 , where r1 = 22.5 cm, r2 = 12.5 cm and h = 34 cm [1] 2 2 So, l = 34 + (22.5 в€’ 12.5) cm = 342 + 102 = 35.44 cm [1] The area of metallic sheet used = curved surface area of frustum of cone + area of circular base [1] + curved surface area of cylinder 2 = [ПЂ Г— 35.44 (22.5 + 12.5) + ПЂ Г— (12.5) 2 + 2ПЂ Г— 12.5 Г— 6] cm 22 2 = 7 (1240.4 + 156.25 + 150) cm 2 28. Two poles of equal heights are standing opposite each other on either side of the road, which is 80m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60В° and 30В° respectively. Find the height of the poles and the distance of the point from the poles. OR A man standing on the deck of a ship, which is 10 m above the water level, observes the angle of elevation of the top of a hill as 60В° and angle of depression of the base of the hill as 30В°. Find the distance of the hill from the ship and height of the hill. = 4860.9 cm [1] Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket) ПЂГ—h 2 2 = 3 Г— (r1 + r2 + r1r2) 22 34 2 2 3 = Г— 7 3 Г—[(22.5) + (12.5) + 22.5 Г— 12.5] cm 22 34 3 = Г— 7 3 Г—943.75 = 33615.48 cm = 33.62 litres (approx.) [1] 28. Let AB = DE = Poles Point C is on the road such that в€ BCA = 60В°, and в€ DCE = 30В° To find AB = DE & BC, CD [1] From в€† ABC, tan 60В° = AB/BC AB в‡’ 3= BC в‡’ AB = BC в€љ3 = x в€љ3 вЂ¦вЂ¦вЂ¦. (i) [1] DE AB From в€† CDE, tan 30В° = CD = 80 в€’ x 1 AB в‡’ = [1] 3 80 в€’ x вЂ¦вЂ¦.. (ii) [1] Putting the value of AB from (i) in (ii), we get 1 x 3 = 3 80 в€’ x в‡’ 3x = 80 вЂ“ x в‡’ 4x = 80 в‡’ x = 20 в€ґ BC = 20 cm, CD = 60 cm Point C is 20 m from B and 60 m from D. [1] QUESTION ANSWER Height of poles AB = xв€љ3 = 20 Г— 1.732 = 34.64 m OR Let the man is standing on deck of a ship AB = 10 m at a point A above the water level BC. Let CD be the hill. It is given that the angle of elevation of the top D of a hill CD observed from the point A is 60В° and the angle of depression of the base C of the hill is 30В° observed from A. [1] Draw AE вЉҐ CD. Let AE = x m and DE = y m. Then, в€ DAE = 60В° and в€ EAC = в€ BCA = 30В°. In right в€† EAD, we have tan 60В° = DE/EA. y в‡’ 3 = в‡’ y = в€љ3 x вЂ¦вЂ¦вЂ¦.. (1) [1] x Again, in right в€† ABC, we have tan 30В° = AB/AC. 1 10 в‡’ = [ в€µ BC = AE = x m] 3 x 29. A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train. OR в‡’ x = 10в€љ3 вЂ¦вЂ¦вЂ¦вЂ¦вЂ¦. (2) [1] From (1) and (2), we get y = в€љ3 Г— 10в€љ3 = 30 cmвЂ¦вЂ¦вЂ¦вЂ¦вЂ¦. (3) [1] Also, CD = DE + EC в‡’ CD = y + 10 [ в€µ DE = y m and EC = 10 m] в‡’ CD = 30 + 10 = 40 m. [using (3)] Hence, the distance of the hill from the ship is AE = x = 10в€љ3 (10 Г— 1.732 m) = 17.32 m and the height of the hill CD = 40 m. [1] 29. Let original speed = x km/hr. Increased speed = (x + 5) km/hr [1] Distance traveled = 360 km The according to question, 360 360 360x + 1800 в€’ 360x в€’ = 1в‡’ =1 [2] x x+5 x(x + 5) 2 A two digit number is such that the product of its digits is 15. If 18 is added to the number, the digits interchange their places. Find the number. в‡’ 1800 = x(x + 5) = x + 5x 2 в‡’ x + 5x вЂ“ 1800 = 0 2 в‡’ x + 45 x вЂ“ 40x вЂ“ 1800 = 0 в‡’ x(x + 45) вЂ“ 40 (x + 45) = 0 [1] в‡’ (x + 45) (x вЂ“ 40) = 0 Either x + 45 = 0 в‡’ x = -45 Rejecting negative values or x вЂ“ 40 = 0 в‡’ x = 40 в€ґ Original speed = 40 km/hr. [1] OR Let the unit place digit be x and ten's place digit be y, then according to the given condition xy = 15. в‡’ y = 15/x вЂ¦вЂ¦вЂ¦вЂ¦вЂ¦вЂ¦. (1) пЈ« 15 пЈ¶ Original number = 10y + x = 10 пЈ¬ пЈ· + x [1] пЈ xпЈё пЈ« 150 пЈ¶ =пЈ¬ пЈ· + x [using (1)] пЈ x пЈё When digits interchange their places, then New number = 10x + y = 10x + 15/x [using (1)] According to the given condition, we have original number + 18 = New number QUESTION ANSWER 150 15 + x + 18 = 10x + x x пЈ« 15 150 пЈ¶ в‡’ (10x в€’ x) + пЈ¬ в€’ в€’ 18 = 0 x пЈ·пЈё пЈ x в‡’ [1] 135 в€’ 18 = 0 x 2 в‡’ 9x вЂ“ 135 вЂ“ 18x = 0 2 в‡’ 9(x вЂ“ 2x вЂ“ 15) = 0 2 в‡’ x вЂ“ 2x вЂ“ 15 = 0 [1] 2 в‡’ x вЂ“ 5x + 3x вЂ“ 15 = 0 в‡’ x(x вЂ“ 5) + 3(x вЂ“ 5) = 0 в‡’ (x вЂ“ 5) (x +3) = 0 в‡’ Either x вЂ“ 5 = 0 or x + 3 = 0 в‡’ Either x = 5 or x = -3 [1] в‡’ x = 5[ в€µ Digits can't be negative] Thus, unit place digit = 5 and ten's place digit = 15 15 = =3 x 5 Hence, the required number is 35. [1] в‡’ 9x в€’ 30. Draw the two types of cumulative frequency curves and determine the median. Wages in Rs. No. of Workers 80 - 90 9 90 - 100 17 100 - 110 19 110 вЂ“ 120 45 120 вЂ“ 130 33 130 вЂ“ 140 15 140вЂ“150 12 30. C.I.(wages) 80 вЂ“ 90 90 вЂ“ 100 100 вЂ“ 110 110 вЂ“ 120 120 вЂ“ 130 130 вЂ“ 140 140 вЂ“ 150 No. of Cum. workers Frequency < frequency type 9 9 17 26 19 45 45 90 33 123 15 138 12 150 Cum. Frequency > type 150 141 124 105 60 27 12 We plot both these types : The two curves shown in the figure give more than type and less than type. They must at P where frequency is 75 and class interval is at 117. в€ґ Median is Rs. 117. [1 QUESTION ANSWER iBOOKS SAMPLE PAPER вЂ“ 02 [FOR XTH CBSE BOARD] QUESTION ANSWER 1. SECTION вЂ“ A 2 2 3825 = 3 Г— 5 Г— 17. [1] 2. 1. [1] 3. Here SECTION вЂ“ A [1Г—10] 1. Write 3825 as a product of prime factors. 2. In figures, the graph of some polynomial p(x) is given. Find the number of zeroes of the polynomial. Y c 1 в€’5 5 = = c 2 в€’k k 4 3 For a pair of linear equations to have no solution, if 2 a1 b1 c 1 3 1 5 = в‰ в‡’ = в‰ . в‡’ k в‰ 10. 6 2 k a2 b2 c 2 1 X' -4 -3 -2 -1 O 1 1 2 3 4 5 X 2 4. 3 Y' 4. 5. Find the value of k, so that the following system of equations has no solution : 3x вЂ“ y вЂ“ 5 = 0; 6x вЂ“ 2y = k. 5. X P Y Given, 5 tan О± = 4. в‡’ tan О± = 4/5. = 5 sin О± в€’ 3 cos О± If 5 tan О± = 4, Find the value of . 5 sin О± + 2 cos О± In the given figure, PQ || YZ. Find the length of QX. 4 5 в€’3 4в€’3 1 4 = = . 4 4+2 6 5 +2 5 XY XZ = XP XQ Now in в€† XYZ and в€†XPQ, we have 6 cm Q XY XZ = and в€ X = в€ X. XP XQ Z 9 cm [1] In в€† XYZ, we have PQ || YZ. в‡’ cm 33 cm [1] 5 sin О± 5 sin О± в€’ 3 cos О± cos О± в€’3 5 tan О± в€’ 3 Now = = 5 sin О± + 2cos О± 5 sin О± 5 tan О± + 2 cos О± +2 4 3. a1 3 1 b1 в€’1 1 = = , = = . a 2 6 2 b 2 в€’2 2 6. For what values of k, the quadratic equation 2 12 x + 4kx + 3 = 0 has equal roots ? 7. In the given figure, PQ and PR are tangents to the circle drawn from an external point P. LM is a third tangent touching the circle at M. If PQ = 15 cm, KM = 4 cm, Find the length PL. [1] в€ґ в€† XYZ ~ в€† XPQ в‡’ 6. YZ XZ 9 6 = в‡’ = в‡’ XQ = 2 cm. PQ XQ 3 XQ [1] 2 For equal roots D = b вЂ“ 4ac = 0 2 в‡’ (4k) вЂ“ 4.12.3 = 0 [1] 2 в‡’ 16k вЂ“ 144 = 0 L P k M Q 2 в‡’ k = 144/16, в‡’ k = В± 3. O R 7. [1] We have PQ = PR = 15 cm. and KL = KM = 4 cm. Now length of PL = PQ вЂ“ LQ = 15 вЂ“ 4 = 11 cm. [1] QUESTION Cumulative frequency (No. of students) 8. 9. ANSWER A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the student of the class. Y 40 30 20 10 O X 10 20 30 40 50 60 70 80 Upper limit of class intervals (Marks) Two dice are thrown at the same time. What is the probability that the sum of two numbers appearing on the top of the dice is 13 ? 10. A cone, a cylinder and a hemisphere are of equal base and have the same height. What is the ratio of their volumes ? SECTION вЂ“ B [2Г—5] 11. Without drawing the graphs, state whether the following pair of linear equations will represent intersecting lines, coincident lines or parallel lines : 6x вЂ“ 3y = - 10 2x вЂ“ y + 9 = 0 OR For which value of k will the following system of linear equations have no solution ? 3x + y = 1 (2k вЂ“ 1) x + (k вЂ“ 1) y = 2k + 1. 12. If A (-2, -1), B (x, 0), C (4, y) and D (1, 2) are the vertices of a parallelogram taken in order, find x and y. n 40 = = 20 2 2 Locate 20 on the y вЂ“ axis. From this pint, draw a line parallel to the x вЂ“ axis cutting the curve at a point. From this point, draw a perpendicular to the x вЂ“ axis. The point of intersection of this perpendicular with the x вЂ“ axis gives the median of the data. Hence, the median marks obtained by the student of the class = 54 marks. [1] 9. 0. [1] 10. Let r be the radius of cone, cylinder and the hemisphere respectively and 'h' be the height of the cone. Cylinder and the hemisphere respectively. Then w = r. 2 V1 = Volume of the cone = 1/3 ПЂr h. 2 V2 = Volume of the cylinder = ПЂr h. 3 V3 = Volume of the hemisphere = 2/3 ПЂr . 2 2 3 в€ґ V1 : V2 : V3 = 1/3 ПЂr h : ПЂr h : 2/3 ПЂr = h : 3h : 2r = 1 : 3 : 2. [1] SECTION вЂ“ B a 6 11. Here 1 = = 3 . a2 2 8. Here n = 40 в‡’ b 1 в€’3 = =3 b 2 в€’1 [1] c 1 в€’10 10 = = c2 в€’9 9 a1 b1 c 1 = в‰ , the given pair of linear equations a2 b2 c 2 represents parallel lines. [1] OR a b c The system of equation has no solution, if 1 = 1 в‰ 1 a2 b2 c 2 Since 3 1 1 = в‰ 2k в€’ 1 k в€’ 1 2k + 1 3 1 1 1 в‡’ = and в‰ [1] 2k в€’ 1 k в€’ 1 k в€’ 1 2k + 1 3 1 Now = 2k в€’ 1 k в€’ 1 в‡’ 3k вЂ“ 3 = 2k вЂ“ 1 в‡’ k = 2. Hence, the given system of linear equations will have no solution, if k = 2. [1] 12. We know that the diagonals of a parallelogram bisect each other. Therefore, the coordinates of the mid-point of the diagonal AC = the coordinates of the mid-point of the diagonal BD. пЈ« в€’2 + 4 в€’1 + y пЈ¶ пЈ« x + 1 0 + 2 пЈ¶ в‡’пЈ¬ , = , 2 пЈ·пЈё пЈ¬пЈ 2 2 пЈ·пЈё пЈ 2 в‡’ пЈ« y в€’ 1пЈ¶ пЈ« x + 1 пЈ¶ в‡’ пЈ¬ 1, ,1пЈ· пЈ·=пЈ¬ пЈ 2 пЈё пЈ 2 пЈё x +1 y в€’1 в‡’ = 1and =1 2 2 в‡’ x + 1 = 2 and y вЂ“ 1 = 2 в‡’ x = 1 and y = 3 Hence, x = 1 and y = 3. [1] [1] QUESTION 13. Prove that (1 + cot A вЂ“ cosec A) (1+tan A + sec A) = 2. ANSWER 13. (1 + cot A вЂ“ cosec A) (1 + tan A + sec A) 1 пЈ¶пЈ« sin A 1 пЈ¶ пЈ« cos A 1+ = пЈ¬1 + в€’ + sin A sin A пЈ·пЈё пЈ¬пЈ cos A cos A пЈ·пЈё пЈ пЈ« sin A + cos A в€’ 1пЈ¶ пЈ« cos A + sin A + 1пЈ¶ = пЈ¬ пЈ·пЈ¬ пЈ· sin A cos A пЈ пЈёпЈ пЈё (sin A + cos A в€’ 1)(cos A + sin A + 1) = sin A.cos A [1] (sin A + cos A) 2 в€’ 1 1 + 2 sin A.cos A в€’ 1 = sin A.cos A sin A.cos A 2 sin A.cos A = = 2. [1] sin A.cos A 14. Given : в€† ABC ~ DEF and AP and DQ are the medians. = 14. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. To prove: area ( в€†ABC ) area ( в€†DEF ) = AP2 DQ2 D A B P C E Proof: вЂ№ в€†ABC ~ в€†DEF AB BC AC в‡’ = = DE EF DF AB BC в‡’ = вЂ¦вЂ¦вЂ¦.. (1) DE EF AB BP + PC 2 BP BP = в‡’ = = DE EQ + QF 2 EQ EQ Q F [1] Now, in в€†ABP and в€†DEQ AB BP = DE EQ в€ ABP = в€ DEQ в‡’ в€†ABP ~ в€†DEQ Their corresponding sides must be proportional AB BP AP в€ґ = = вЂ¦вЂ¦вЂ¦.. (2) [1] DE EQ DQ AP BC From (1) and (2), = DQ EF We know that the areas of two similar triangles are in the ratio of the square of their corresponding sides. area ( в€†ABC ) BC2 AP2 в€ґ = = . area ( в€†DEF ) EF2 DQ2 15. One card is drawn from a well shuffled deck of 52 playing cards. Find the probability of getting (i) a non-face card. (ii) a black king or a red queen. 15. No. of face cards in a deck of playing cards = 12. No. of non-face cards = 40. 40 10 в€ґ Probability of getting a non-face card = = . [1] 52 13 Since there are 2 black kings and 2 red queens in 52 playing cards в‡’ P (getting a black king or a red queen) 4 1 = = . [1] 52 13 QUESTION ANSWER SECTION вЂ“ C [3Г—10] 16. Prove that 5 + в€љ2 is irrational. 17. For what values of a and b, the following system of linear equations have an infinite number of solution ? 2x + 3y вЂ“ 7 = 0. (a вЂ“ b) x + (a + b) y = 3a + b вЂ“ 2. 18. Represent the following pair of equations graphically and write the coordinates of points where the lines intersect y вЂ“ axis : x + 3y = 6 2x вЂ“ 3y = 12 SECTION вЂ“ C 16. Let 5 + в€љ2 be rational. a в‡’ 5 + 2 = , [a, b are coprimes and b в‰ 0] [1] b a в‡’ 2 = в€’5 b a в€’ 5b в‡’ 2= . b a Since a and b are integers, we get в€’ 5 is rational. b So в€љ2 is rational. [1] But this contradicts the fact that в€љ2 is irrational. This contradiction has arisen due to our wrong assumption. So 5 + в€љ2 is irrational. [1] 17. Given 2x + 3y вЂ“ 7 = 0 (a вЂ“ b) x + (a + b) y = 3a + b вЂ“ 2 2 3 7 For infinitely many solutions = = [1] a в€’ b a + b 3a + b в€’ 2 2 3 3 7 = Again = a в€’b a +b a + b 3a + b в€’ 2 в‡’ a = 5b вЂ¦вЂ¦ (1) в‡’ 2a вЂ“ 4b = 6 вЂ¦вЂ¦вЂ¦..(2) [1] Solving (1) and (2), we get b=1 в‡’a=5 [1] 18. Two solutions of each of the equations : x + 3y = 6 вЂ¦вЂ¦вЂ¦вЂ¦. (1) 2x вЂ“ 3y = 12 вЂ¦вЂ¦вЂ¦. (2) are given in tables below : [1] x + 3y = 6 2x вЂ“ 3y = 12 x 6 0 x 6 0 y 0 2 y 0 -4 A B A C 4 Y 3 X' B(0,2) 1 -6 -5 -4 -3 -2 -1 O -1 x+3y=6 X 1 -2 -3 B(0,-4) 19. Determine an AP whose 3rd term is 16 and when 5th term is subtracted from 7th term, we get 12. 2 3 4 5 A(6,0) 2x-3y=12 Y' [1.5] The first line x + 3y = 6 intersects the y вЂ“ axis at B (0, 2). The second line 2x вЂ“ 3y = 12 intersects the y вЂ“ axis at C (0, -4). [0.5] 19. Let the first term and common difference of AP be a and d, respectively. Let a3, a5 and a7 denote the 3rd term, 5th term and 7th term of an AP, then a3 = a + (3 вЂ“ 1)d [в€µ n = 3] в‡’ 16 = a + 2d вЂ¦вЂ¦вЂ¦. (1) [в€µ a3 = 16 (given)] It is given that the difference of 5th term from the 7th term is 12. [1] QUESTION 20. Prove that : 1 + cos A sin A + = 2 cos ec A . sin A 1 + cos A OR Prove that : (cosec A вЂ“ sin A) (sec A вЂ“ cos A) = 1 . tan A + cot A ANSWER в€ґ a7 вЂ“ a5 = 12 (given) в‡’ [a + (7 вЂ“ 1)d] вЂ“ [a+(5 вЂ“ 1)d] = 12 в‡’ (a + 6d) вЂ“ (a + 4d) = 12 [1] в‡’ 2d = 12 в‡’ d = 6 вЂ¦вЂ¦вЂ¦. (4) From (1) and (4), we get a + 2 Г— 6 = 16. в‡’ a = 16 вЂ“ 12 = 4 Thus, the AP is 4, 10, 16, 22, 28, 34, 40, вЂ¦вЂ¦.. [1] 1 + cos A sin A 20. L.H.S. = + sin A 1 + cos A = (1 + cos A) 2 + sin 2 A sin A(1 + cos A) [1] = 1 + 2 cos A + cos 2 A + sin 2 A 1 + 2 cos A + 1 = sin A(1 + cos A) sin A(1 + cos A) [1] 2 + 2cos A 2 + (1 + cos A) 2 = = sin A(1 + cos A) sin A(1 + cos A) sin A = 2 cosec A = R.H.S. OR We have L.H.S. = (cosec A вЂ“ sin A) (sec A вЂ“ cos A) пЈ« 1 пЈ¶пЈ« 1 пЈ¶ =пЈ¬ в€’ sin A пЈ· пЈ¬ в€’ cos A пЈ· пЈ sin A пЈё пЈ cos A пЈё = [1] [1] пЈ« 1 в€’ sin 2 A пЈ¶ пЈ« 1 в€’ cos 2 A пЈ¶ =пЈ¬ пЈ·пЈ¬ пЈ· пЈ sin A пЈё пЈ cos A пЈё 21. Find the area of the quadrilateral whose vertices taken in order are A(-5, -3), B(-4, -6), C(2, -1) and D(1, 2). пЈ« cos 2 A пЈ¶ пЈ« sin 2 A пЈ¶ 2 2 =пЈ¬ [1] пЈ· пЈ¬ cos A пЈ· [ в€µ sin A + cos A = 1] sin A пЈ пЈёпЈ пЈё = cos A sin A 1 Now, R.H.S. = tan A + cot A 1 1 = = sin A cos A sin 2 A + cos 2 A + cos A sin A sin A cos A 2 2 = sin A cos A [в€µ sin A + cos A = 1] Hence, L.H.S. = R.H.S. [1] 21. By joining A to C, we will get two triangles ABC and ACD. Now, the area of в€† ABC 1 = [x1(y2 вЂ“ y3) + x2(y3 вЂ“ y1) + x3 (y1 вЂ“ y2)] 2 1 = [(-5)(-6 + 1) + (-4) (-1 + 3) + 2 (-3 + 6)] [1] 2 D(1,2) 1 C(2,-1) = [25 вЂ“ 8 + 6] 2 23 = sq. units. 2 Also, the area of в€† ACD A(-5,-3) B(-4,-6) 1 = [x1(y2 вЂ“ y3) + x2(y3 вЂ“ y1) + x3(y1 вЂ“ y2)] [1] 2 1 = [(-5)(-1 -2) + 2(2 + 3) + 1 (-3 + 1)] 2 1 = [15 + 10 вЂ“ 2] 2 23 = sq. units. 2 QUESTION 22. Show that the points A(2, -2), B (14, 10), C(11, 13) and D(-1, 1) are the vertices of a rectangle. OR Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. ANSWER So, the area of the quadrilateral ABCD = area of в€† ABC + area of в€† ACD 23 23 = + sq. units = 23 sq. units. [1] 2 2 22. Coordinates of the given points are A(2, -2), B (14, 10), C(11, 13) and D(-1, 1) respectively. (14 в€’ 2) 2 + (10 + 2) 2 Then AB = = (12) 2 + (12) 2 = 144 + 144 = 288 = 144 Г— 2 = 12 2 BC = (11 в€’ 14) 2 + (13 в€’ 10) 2 [1] = ( в€’3) 2 + (3) 2 = 9 + 9 = 9 + 9 = 18 = 9 Г— 2 = 3 2 CD = ( в€’1 в€’ 11) 2 + (1 в€’ 13) 2 = ( в€’12) 2 + ( в€’12) 2 D(-1, 1) C(11, 13) A(2, -2) B(14, 10) = 144 + 144 = 288 = 144 Г— 2 = 12 2 DA = (2 + 1) 2 + ( в€’2 в€’ 1) 2 = (3) 2 + ( в€’3) 2 = 9 + 9 = 18 = 9 Г— 2 = 3 2 Thus, AB = CD and BC = DA. в‡’ Opposite sides are equal in length. Now, AC = [1] (11 в€’ 2) 2 + (13 + 2) 2 = (9) 2 + (15) 2 = 81 + 225 = 306 BD = ( в€’1 в€’ 14) 2 + (1 в€’ 10) 2 = ( в€’15) 2 + ( в€’9) 2 = 225 + 81 = 306 Thus, Length of diagonal AC = Length of diagonal BD. Since, opposite sides are equal and diagonals are also equal. Hence, the points A(2, -2), B (14, 10), C(11, 13) and D (-1, 1) are the vertices of a rectangle. [1] OR Vertices of в€†ABC are (0, -1), (2, 1) and (0, 3) respectively. A (0, -1) F B (2, 1) E D C (0, 3) [1] Their mid-points of sides are : пЈ« 2 + 0 1+ 3 пЈ¶ пЈ« 0 + 0 3 в€’ 1пЈ¶ D=пЈ¬ , , пЈ· = (1,2) , E = пЈ¬ пЈ· = (0,1) 2 пЈё 2 пЈё пЈ 2 пЈ 2 пЈ« 0 + 2 в€’1 + 1 пЈ¶ F=пЈ¬ , пЈ· = (1,0) 2 пЈё пЈ 2 Area of в€† DEF = ВЅ [1(1-0)+0(0-2)+1(2-1)] [1] QUESTION ANSWER 23. O is any point inside a rectangle ABCD. Prove 2 2 2 2 that OB + OD = OA + OC . = ВЅ [1+1] = 1 sq. units. Area of в€†ABC = ВЅ [0(1 -3)+2(3 +1) +0(-1-1)] = ВЅ (0 + 8+ 0) = 8/2 = 4 sq. units. Ratio of areas = 1 : 4. [1] 23. Through O, draw a line segment PQ || BC so that P lies on AB and Q lies on DC. D A D A P Q O O B B C Now, PQ || BC Therefore, PQ вЉҐ AB and PQ вЉҐ DC [в€ B = 90В° and в€ C = [1] 90В°] So, в€ BPQ = 90В° and в€ CQP = 90В° Therefore, BPQC and APQD are both rectangles. 2 2 2 In в€† OPA, we have OA = OP + PA вЂ¦вЂ¦ (1) [By Pythagoras Theorem] [1] C 2 2 2 In в€† OCQ, we have OC = OQ + CQ Adding (1) and (2), we get 2 2 2 2 2 2 OA + OC = OP + PA + OQ + CQ 2 2 2 2 = OP + DQ + OQ + PB [в€µ AP = DQ and QC = PB] 2 2 2 2 = (OP + PB ) + (OQ + DQ ) 2 2 = OB + OD 2 [In в€† OPB and в€† OQD by Pythagoras Theorem, OP + 2 2 2 2 2 PB = OB and OQ + DQ = OD ] 2 2 2 2 Hence, OA + OC = OB + OD . [1] 24. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60В°. 24. [1] Steps of construction : 1. Take a point O on the plane of the paper and draw a circle of radius OA = 5 cm. 2. Extend OA to B such that OA = AB = 5 cm. 3. With A as centre draw a circle of radius OA = AB = 5 cm. Suppose it intersect the circle drawn in step 1 at the points P and Q. 4. Join BP and BQ. [1] Then BP and BQ are the required tangents which are inclined to each other at an angle of 60В° (see figure). For justification of the construction : In в€† OAP, we have QUESTION ANSWER OA = OP = 5 cm (= Radius) Also, AP = 5 cm (= Radius of circle with centre A). в€† OAP is equilateral в‡’ в€ PAO = 60В° в‡’ в€ BAP = 120В° In в€† BAP, we have AB = AP and в€ BAP = 120В° в€ґ в€ ABP = в€ APB = 30В° Similarly we can prove that в€ ABQ = в€ AQB =30В° в‡’ PBQ = 60В°. 25. Find the area of the shaded region from figure, if the diameter of the circle with centre O is 28 cm and AQ = 1 AB. 4 [1] 25. We have AB = Diameter of a circle = 28 cm в€ґ OA = OB = 14 cm [Each = radius of a circle] 1 It is given that AQ = AB 4 1 в‡’ AQ = Г— 28 = 7 cm 4 в€ґ AO = AQ + QO в‡’ 14 cm = 7 cm + QO в‡’ QO = (14 вЂ“ 7) cm = 7 cm [1] Now, QB = QO + OB в‡’ QB = 7cm + 14 cm = 21 cm Area of the shaded region = Area of the semi-circle on diameter AQ (= 7 cm) + Area of the semi-circle on diameter QB (= 21 cm) SECTION вЂ“ D [6Г—5] 26. Some students arranged a picnic. The budget for food was Rs. 240. Because four students of the group failed to go, the cost of food to each student got increased by Rs. 5. How many students went for the picnic ? OR A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed пЈ® 1 пЈ« AQ пЈ¶ 2 1 пЈ« QB пЈ¶ 2 пЈ№ пЈ® ПЂ Г— AQ 2 ПЂ Г— QB 2 пЈ№ = пЈЇ ПЂГ—пЈ¬ + пЈє пЈ· + 2ПЂГ—пЈ¬ 2 пЈ· пЈє = пЈЇ 8 8 пЈЇпЈ° 2 пЈ 2 пЈё пЈ пЈё пЈєпЈ» пЈ° пЈ» ПЂ 2 2 2 = [(7) + (21) ] cm [1] 8 ПЂ 2 = [49 + 441] cm 8 22 1 11 = Г— Г— 490cm 2 = Г— 70cm 2 7 8 4 11 385 2 2 = Г— 35cm = cm 2 = 192.5 cm . [1] 2 2 SECTION вЂ“ D 26. Let the original number of students be x. Total cost of food for x students = Rs. 240. [1] в€ґ The cost of food for each student = Rs. 240/x. It is given that 4 students of the group failed to go. в€ґ The number of students go for a picnic = (x вЂ“ 4) [1] 240 Now, the cost of food for each student = Rs. xв€’4 It is given that the four students of a group failed to go, the cost of food to each student got increased by Rs. 5. QUESTION by 250 km/h from the usual speed. Find its usual speed. ANSWER 240 240 в€’ = 5(given) [1] xв€’4 x 240x в€’ 240x + 960 в‡’ =5 x(x в€’ 4) 2 в‡’ 960 = 5(x вЂ“ 4x) [1] 2 в‡’ 192 = x вЂ“ 4x 2 в‡’ x вЂ“ 4x вЂ“ 192 =0 [1] 2 в‡’ x вЂ“ 16x + 12x вЂ“ 192 = 0 в‡’ x(x вЂ“ 16) + 12(x -16) = 0 в‡’ (x вЂ“ 16) (x + 12) = 0 [1] в‡’ Either x = 16 or x = -12 в‡’ x = 16, as x cannot be negative. Hence, the number of students go for the picnic = x вЂ“ 4, i.e., 16 вЂ“ 4 = 12. [1] OR Let the usual speed of the plane be x km/h. Then, time taken to cover 1500 km with usual speed = 1500 h [1] x When the speed of a plane is increased by 250 km/h, then the time taken to cover 1500 km with the speed of (x 1500 h. [1] + 250) km/h = x + 250 1500 1500 1 в€ґ в€’ = hour (or 30 minutes) x x + 250 2 1500 + 1500 Г— 250 в€’ 1500x 1 в‡’ = [1] x(x + 250) 2 в€ґ 1500 Г— 250 1 = x 2 + 250x 2 2 в‡’ 3000 Г— 250 = x + 250x [1] 2 в‡’ x + 250x вЂ“ 750000 = 0 2 в‡’ x + 1000x вЂ“ 750x вЂ“ 750000 = 0 в‡’ x(x + 1000) вЂ“ 750(x + 1000) = 0 [1] в‡’ (x + 1000) (x вЂ“ 750) = 0 в‡’ Either x = 750 or x = -1000 [1] в‡’ x = 750, as x cannot be negative. [1] Hence, the usual speed of the plane is 750 km/h. 27. Let AB be the height of lighthouse = 75 m. Let C and D be the position of the two ships approaching to a light house such that the distance CD between the two ships = x m. [1] It is given that as observed from the top A of the light house the angles of depression of the two ships at C and D be 30В° and respectively. [1] в€ґ в€ BCA = angle of depression = 30В° and в€ BDA = angle of depression = 45В°. в‡’ 27. As observed from the top of a 75 m high lighthouse (from sea-level), the angles of depression of two ships are 30В° and 45В°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [1] In right triangle ADB, we have tan 45В° = AB DB QUESTION ANSWER 75 в‡’ DB = 75 m. DB в‡’ DB = 75 m вЂ¦вЂ¦вЂ¦вЂ¦. (1) в‡’1= In right triangle ACB, we have tan 30В° = Using the above theorem, do the following : If ABC is an equilateral triangle with AD вЉҐ BC, 2 2 then AD = 3DC . AB CB 75 = [ в€µ CB = CD + DB] 3 CD + DB в‡’ CD + DB = 75 Г— в€љ3 [Using (1)] [1] в‡’ x = [75в€љ3 вЂ“ 75]m в‡’ x = 75(в€љ3 вЂ“ 1) m в‡’ x = 75(1.73 вЂ“ 1) m [1] в‡’ x = 75 Г— 0.73 m [в€µ в€љ3 = 1.73] в‡’ x = 54.75 m Thus, the distance between the two ships = 54.75 m. [1] 28. Given : A right triangle ABC, right angled at B. 2 2 2 To prove : (Hypotenuse) = (Base) + (Perpendicular) 2 2 2 i.e., AC = AB + BC [1] Construction : Draw BD вЉҐ AC в‡’ 28. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [1] 1 [1] Proof : в€† ADB ~ в€† ABC. [If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other]. [1] AD AB So, = [Sides are proportional] AB AC 2 в‡’ AD. AC = AB . вЂ¦вЂ¦вЂ¦вЂ¦вЂ¦. (1) [1] Also, в€† BDC ~ в€† ABC CD BC So, = [Sides are proportional] BC AC 2 в‡’ CD. AC = BC вЂ¦вЂ¦вЂ¦вЂ¦.. (2) [1] Adding (1) and (2), we have 2 2 AD. AC + CD. AC = AB + BC 2 2 в‡’ (AD + CD). AC = AB + BC 2 2 в‡’ AC . AC = AB + BC 2 2 2 Hence, AC = AB + BC [1] For the Second Part : Let ABC be an equilateral triangle and AD вЉҐ BC. In в€† ADB and в€† ADC, we have AB = AC [Given] QUESTION 29. An iron pillar has lower part in the form of a right circular cylinder and the upper part in the form of a right circular cone. The radius of the base of each of the cone and a cylinder is 8 cm. The cylindrical part is 240 cm high and conical part is 36 cm high. Find the weight of 3 the pillar if 1cm of iron weighs 7.5 grams. [Take ПЂ = ANSWER в€ B = в€ C [Each = 60В°] [Each 90В°] and в€ ADB = в€ ADC в€ґ в€† ADB в‰… в€† ADC в‡’ BD = DC вЂ¦вЂ¦вЂ¦.. (1) в€ґ BC = BD + DC = DC + DC = 2. DC вЂ¦(2) [Using (1)] In right angled в€† ADC, we have 2 2 2 AC = AD + DC 2 2 2 в‡’ BC =AD + DC [ в€µ AC=BC, sides of an equilateral в€†] 2 2 2 в‡’ (2DC) = AD + DC [using (2)] 2 2 2 в‡’ AD = 4DC вЂ“ DC 2 2 в‡’ AD = 3DC . 29. Let r1 and r2 be the radii of the cone and cylinder respectively, then r1 and r2 = 8 cm Let h1 and h2 be the heights of the cone and cylinder [1] respectively, then h1 = 36 cm and h2 = 240 cm. Total volume of the iron pillar = Volume of the cone + Volume of the cylinder. 22 ] 7 30. Find the mean, median and mode of the following data Class Frequency 0 вЂ“ 20 6 20 вЂ“ 40 8 40 вЂ“ 60 10 60 вЂ“ 80 12 80 вЂ“ 100 6 100 вЂ“ 120 5 [1] 1 2 2 = ПЂr1 h1 + ПЂr2 h2 3 ПЂ 2 2 Г— (8) Г— 36 + ПЂ Г— (8) Г— 240 [1] = 3 = ПЂ Г— 64 Г— 12 + ПЂ Г— 64 Г— 240 = 64ПЂ [12 + 240] [1] 3 3 22 = 64 Г— Г— 252 = 64 Г— 22 Г— 36 cm = 50688 cm 7 Hence, total weight of the iron pillar = 50688 Г— 7.5 g 3 [ в€µ Weight of 1 cm = 7.5 grams (given)] = 50688 Г— 15 g [1] 2 = 25344 Г— 15 g = 380160 g = 380160 = 380.16 kg. [1] 1000 30. The cumulative frequency distribution table with the given frequency becomes : Class Frequency Cumulative Class di=xi-70 x в€’ 70 fiui u = i (fi) 0 вЂ“ 20 6 20вЂ“ 40 8 40 вЂ“ 60 10f0 60 вЂ“ 80 12 f1 80 вЂ“ 100 6 f2 100вЂ“ 120 5 120вЂ“ 140 3 Total N=ОЈfi=50 frequency mark (cf) (xi) 6 10 14 30 24cf 50 36 70=a 42 90 47 110 50 130 i -60 -40 -20 0 20 40 60 20 -3 -2 -1 0 1 2 3 -18 -16 -10 0 6 10 9 ОЈfiui=-19 From the table, n = ОЈfi = 50 в‡’ n/2 = 25, a = 70, h = 20 Using the formula for calculating the mean : [1] QUESTION 120 вЂ“ 140 3 OR The mean of the following frequency table is 53. But the frequencies f1 and f2 in the classes 20 вЂ“ 40 and 60 вЂ“ 80 are missing. Find the missing frequencies. Age (in years) No. of People 0 вЂ“ 20 15 20 вЂ“ 40 f1 40 вЂ“ 60 21 60 вЂ“ 80 f2 80 вЂ“ 100 17 Total 100 ANSWER Mean = a + в€‘ f iu i Г— h = 70 + ( в€’19) Г— 20 в€‘ fi 50 38 = 70 вЂ“ 7.6 вЂ“ 62.4, [1] = 70 в€’ 5 Now, 60 вЂ“ 80 is the class whose cumulative frequency 36 is greater than n/2 = 25. Therefore, 60 вЂ“ 80 is the median class. Thus, the lower [1] limit (l) of the median class is 60. Using the formula for calculating the median : пЈ«n пЈ¶ пЈ¬ в€’ cf пЈ· 25 в€’ 24 Median = l + пЈ¬ 2 Г— h = 60 + Г— 20 f пЈ· 12 пЈ¬ пЈ· пЈ пЈё 20 5 = 60 + = 60 + = 60 + 1.67 = 61.67. [1] 12 3 Since the maximum frequency is 12, therefore, the modal class is 60 вЂ“ 80. Thus, the lower limit (l) of the modal class = 60. Using the formula for calculating the mode : пЈ« fi в€’ f0 пЈ¶ Mode = l + пЈ¬ пЈ·Г—h пЈ 2f i в€’ f 0 в€’ f 2 пЈё 12 в€’ 10 Г— 20 [1] 2 Г— 12 в€’ 10 в€’ 6 2 Г— 20 40 = 60 + = 6+ = 60 + 5 = 65. [1] 24 в€’ 16 8 OR The missing frequencies are f1 and f2. Calculation of Mean fixi Age Mid-point Frequency (xi) (fi) (in years) 0 вЂ“ 20 10 15 150 20 вЂ“ 40 30 f1 30 f1 40 вЂ“ 60 50 21 1050 60 вЂ“ 80 70 F2 70 f2 80 вЂ“ 100 90 17 1530 Total n=53+f1+f2 ОЈfixi=2730+30f1+70f2 We have, n = 100 = Total frequency = ОЈfi [2] в‡’ 53 + f1 + f2 = 100 в‡’ f1 + f2 = 100 вЂ“ 53 в‡’ f1 + f2 = 47 в‡’ f1 = 47 вЂ“ f2 [1] Also, Mean = 53 (given) 2730 + 30f1 + 70f 2 в‡’ 53 = 100 273 + 3f1 + 7f 2 в‡’ 53 = 10 в‡’ 530 = 273 + 3f1 + 7f2 [1] в‡’ 530 вЂ“ 273 вЂ“ 7f2 = 3f1 в‡’ 257 вЂ“ 7f2 = 3f1 в‡’ 257 вЂ“ 7f2 = 3(47 вЂ“ f2) в‡’ 257 вЂ“ 7f2 = 141 вЂ“ 3f2 [1] в‡’ 257 вЂ“ 141 = 7f2 вЂ“ 3f2 в‡’ 116 = 4f2 в‡’ 29 = f2 Now, from (1), f1 = 47 вЂ“ f2 = 47 вЂ“ 29 = 18. Hence, f1 = 18 and f2 = 29. [1 = 60 + QUESTION ANSWER iBOOKS SAMPLE PAPER вЂ“ 03 [FOR XTH CBSE BOARD] QUESTION ANSWER SECTION вЂ“ A [1Г—10] 1. What is the maximum no. of factors of a prime number? 2. In the figure, the graph of some polynomial p(x) is given. Find the number of zeroes of the polynomial. 1. 2. 3. 2 3 4 5 Y 1 y = f(x) X' -5 -4 -3 -2 -1 O X -11 -2 -3 -4 -5 2 3 4 5 Y' 3. If the co-ordinates of the midpoints of the sides of a triangle are (1, 1), (2, -3) and (3, 4). Find its centroid. 4. Which measures of central tendency is given by the x-co-ordinate of the point of intersection of the 'more than' ogive and 'less than' ogive ? 5. 6. 7. 8. 9. 4. 5. Cards bearing of numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out from the bag at random. What is the probability that the number on the card taken out is an even number ? 6. A chord of a circle of radius 14 cm subtends an angle 60В° at the centre. Find the area of its major sector. 7. From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm. Then what is the radius of the circle? For what value of k, are the roots of the 2 quadratic equation 3x + 2kx + 27 = 0 real and equal ? Given that tan Оё = 2 2 cos ec Оё в€’ sec Оё cos ec 2Оё + sec 2 Оё ? 1 3 2 2 x1 + x 2 =1 2 в‡’ x1 + x2 = 2 вЂ¦вЂ¦.. (1) x 2 + x3 = 2 в‡’x2 + x3 = 4 вЂ¦вЂ¦.. (2) 2 x1 + x3 = 6 вЂ¦вЂ¦.. (3) Again y1 + y1 =1 2 в‡’y1 + y2 = 2 вЂ¦вЂ¦.. (4) вЂ¦вЂ¦.. (5) y2 + y3 = -6 вЂ¦вЂ¦..(6) y1 + y3 = 8 Adding (1) (2) and (3) в‡’ (x1 + x2 + x3) = 12 (x1, y1) в‡’ x1 + x2 + x3 = 6 Now x3 = 6 вЂ“ 2 = 4 x2 = 0 (3,4) (1,1) x1 = 2 Adding (4), (5) and (6) в‡’ (y1 + y2 + y3) = 4 y1 + y2 + y3 = 2 (x2, y2) (2,-3) Now y3 = 0 y2 = -6 y1 = 8 y1 + y 2 + y 3 пЈ¶ пЈ« 2 пЈ¶ пЈ« x + x 2 + x3 So G = пЈ¬ 1 пЈ· = пЈ¬ 2, пЈ· 3 3 пЈ пЈё пЈ 3пЈё Median 9 1 or 18 2 Here Оё = 300В° O Оё So Area = Г— ПЂr 2 60В° 14 360В° Here = 300В° 22 Г— Г— 14 Г— 14 = 513.3 cm2 360В° 7 A Here OA = OP2 в€’ AP2 15 cm = 289 в€’ 225 17 cm P = 64 = 8 cm For real and equal roots D = 0 2 в‡’ (2k) вЂ“ 4Г—3Г—27 = 0 2 2 2 в‡’ 4k вЂ“ 324 = 0 в‡’ 4k = 81в‡’ k = 81в‡’ k = В± 9 9. tan Оё = 1 3 2 Now cos ec Оё в€’ sec 2 Оё cos ec 2Оё + sec 2 Оё 2 Now multiply sin Оё , what is the value of = 1 в€’ tan2 Оё 1 + tan2 Оё в€’ B A = 172 в€’ 152 8. (x3, y3) 1 в€’ 31 1 + 31 = 2 3 4 3 = 2 1 = 4 2 O QUESTION ANSWER 10. In the given figure, determine whether LM || QR when PL =5.2 cm, LQ = 7.8 cm, PM = 6 cm and PR = 15 cm. 10. 11. P Product of zeroes = 4(-3) = -12 = в€’12 cons tan t term = 1 coefficient of x 2 M L 12. Q R SECTION вЂ“ B [2Г—5] 11. Find a quadratic polynomial whose zeroes are 4 and -3. Verify the relationship between the coefficient and zeroes of the polynomial. 13. 12. Without using trigonometric tables, find the value of cos 70В° + cos 57В°. cosec 33В° - 2 cos sin 20В° 60В°. 13. A card is drawn at random from a well shuffled deck of playing cards. Find the probability of drawing a (i) face card (ii) card which is neither a king nor a red card. 14. OR Five cards вЂ“ ten, jack, queen, king and ace, are well shuffled with their face downwards. One card is then picked up random. (i) What is the probability that the card is a queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace (b) a queen. 15. 14. Determine the ratio in which the line 2x + y вЂ“ 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7). Also find the co-ordinates of the point of division. 15. In the figure, DE || BC. If AD = 2.4 cm, DB = 3.6 cm and AC = 5 cm find AE. A D B Yes Given the zeroes are 4 and -3 2 So the quadratic equation is x вЂ“ x вЂ“ 12 в€’1 в€’coefficient of x So sum of zeroes = 4+(-3) = 1 = = 1 coefficient of x 2 16. E C SECTION вЂ“ C [3Г—10] 16. Show that 3 + 5в€љ2 is an irrational number. cos 70В° + cos 57В° В· cosec 33В° - 2cos60В° sin 20В° cos(90В° в€’ 20В°) = + cos 57В°icos ec(90В° в€’ 57В°) в€’ 2 cos 60В° sin 20В° sin 20В° = + cos 57В°isin 57В° в€’ 2 cos 60В° sin 20В° 1 = 1 + 1 вЂ“ 2 В· = 2- 1 = 1 2 12 3 (i) P(face card) = = 52 13 24 6 (ii) P(neither king nor a red card) = = 52 13 OR 1 (i) P(a queen) = 5 1 (ii) (a) P(ace) = (b) P(queen) = 0 4 Suppose the line 2x + y вЂ“ 4 = 0 divides the line segment joining A(2, -2) and B(3, 7) in the ratio k : f at point C. Then the coordinates of C are пЈ« 3k + 2 7k в€’ 2 пЈ¶ , пЈ¬ пЈ·. пЈ k +1 k +1 пЈё But C lies on the line 2x + ky вЂ“ 4 = 0 пЈ« 3k + 2 пЈ¶ 7k в€’ 2 в‡’ 2пЈ¬ в€’ 4 = 0 в‡’ 6k+4+7k вЂ“ 2вЂ“4k вЂ“ 4 = 0 пЈ·+ пЈ k +1 пЈё k +1 в‡’ 9k вЂ“ 2 = 0 в‡’ k = 2/9 So the required ratio is 2 : 9 4пЈ¶ пЈ« 24 Now the coordinate of C are пЈ¬ , в€’ пЈ· пЈ 11 11 пЈё Given DE || BC в‡’ в€ ADE = в€ ABC (corresponding) A In в€† ADE and в€†ABC в€ A = в€ A (common) в€ ADE = в€ ABC D E в‡’ в€† ADE = в€†ABC AD AE 2.4 AE B C в‡’ = в‡’ = AB AC 6 AC в‡’ 6AE = 12 в‡’ AE = 2 cm. Let 3 + 5в€љ2 is an irrational number. a в‡’ 3 + 5в€љ2 = , [a, b are co-primes and bв‰ 0] b a a в€’ 3b a в€’ 3b в‡’ 5 2 = в€’3 в‡’ 5 2 = в‡’ 2= b b 5b a в€’ 3b Since a and b are integers в‡’ is rational 5b a в€’ 3b But в€љ2 is an irrational number в‡’ is irrational. 5b QUESTION ANSWER 17. Solve the following system of linear equations graphically : 2x вЂ“ y = 4, x вЂ“ y = 1. 17. Also shade the region bounded by the lines and x-axis. Which is a contradiction. This contradiction has arisen due to our wrong assumption. A 3 + 5в€љ2 is an irrational number 2x вЂ“ y = 4 y = 2x вЂ“ 4 x = 0 в‡’ y = -4 Y x = 1 в‡’ y = -2 6 x=2в‡’y=0 x 0 1 2 y -4 -2 0 x вЂ“ y = 1в‡’ y = x вЂ“ 1 x = 0 в‡’ y = -1 x=1в‡’y=0 x=2в‡’y=1 18. Find the roots of the equation 1 1 11 , в€’ = x + 4 x в€’ 7 30 18. x в‰ -4, 7. OR Find the value of k, so that the quadratic equation 2 2x вЂ“ (k вЂ“ 2)x + 1 = 0 has equal roots. 19. The first and the last term of an AP are 4 and 19. 81 respectively. If the common difference is 7, how many terms are there in the A. P. and what is their sum ? 20. Prove that : sinОё(1 + tan Оё)+cos Оё(1 + cot Оё) = secОё + cosec Оё 20. x 0 1 2 y -1 0 1 5 4 3 2 1 X' -5 -4 -3 -2 -1 O 1 2 3 4 5 -1 -2 -3 -4 -5 Y' Here x = 4, y = 4 1 1 11 x в€’ 7 в€’ x в€’ 4 11 в€’ = в‡’ = x + 4 x в€’ 7 30 (x + 4)(x в€’ 7) 30 11 11 2 2 в‡’ = в‡’ x вЂ“3x вЂ“ 28=-30 x вЂ“ 3x + 2 = 0 2 x в€’ 3x в€’ 28 30 2 в‡’ x вЂ“ 2x вЂ“ x + 2 = 0в‡’ x(x вЂ“ 2) вЂ“ 1 (x вЂ“ 2) = 0 в‡’ (x вЂ“ 2) (x вЂ“ 1) = 0в‡’ x вЂ“2 = 0 x вЂ“ 1 = 0в‡’ x=2 or x = 1 OR 2 2x вЂ“ (k вЂ“ 2) x + 1 = 0 The quadratic equation has equal roots when D = 0 2 2 в‡’ [-(k вЂ“ 2)] вЂ“ 4 Г— 2 Г— 1 = 0в‡’ k + 4 вЂ“ 4k вЂ“ 8 = 0 2 2 2 в‡’ k вЂ“ 4k вЂ“ 4 = 0в‡’ k вЂ“ 2.k.(2) + 4 = 8в‡’ (k вЂ“ 2) = 8 в‡’ k . 2 = В± 2в€љ2в‡’ k = 2 В±2в€љ2 So k = 2 + 2в€љ2 or 2 вЂ“ 2в€љ2 st Let вЂ�aвЂ™ be the 1 term of the given A.P. Given a=4, d=7 Let an be the last term в‡’ a +(n-1)d = 81в‡’ 4 + (n-1)7 = 81 в‡’ 7n = 84в‡’ n = 12 So there are 12 term in the given AP n st Now S12 = (1 term + last term) 2 12 = ( 4+81)=6(85)=510 2 L.H.S = sinОё(1+tanОё) + cosОё(1+cotОё) 1 пЈ¶ пЈ« = sinОё(1+tanОё) + cosОё пЈ¬ 1 + пЈ· tan Оё пЈё пЈ пЈ« tan Оё + 1 пЈ¶ = sinОё(1+tanОё) + cosОё пЈ¬ пЈ· пЈ tan Оё пЈё OR 2 Prove that : (sin Оё + cosec Оё) + (cos Оё + sec 2 2 2 Оё) = 7 + tan Оё + cot Оё. пЈ« 1 пЈ¶ cos2 Оё пЈ¶ пЈ« = (1+tanОё) пЈ¬ sin Оё + cos Оё пЈ· = (1+tanОё) пЈ¬ sin Оё + sin Оё пЈ· tan Оё пЈё пЈ пЈ пЈё sin Оё 1 пЈ¶ пЈ« 1 пЈ¶ пЈ« 1 = (1+tanОё) пЈ¬ + . пЈ· =пЈ¬ пЈ· пЈ sin Оё пЈё пЈ sin Оё cos Оё sin Оё пЈё =cosecОё + secОё = R.H.S 2 2 LHS = (sinОё+cosecОё) + (cosОё+secОё) 2 2 2 2 =sin Оё + cosec Оё + 2 + cos Оё+sec Оё+2 2 2 2 2 = sin Оё + cos Оё + cosec Оё + sec Оё+4 2 2 2 2 = 1 + cosec Оё +sec Оё+4= 5 + cosec Оё+ sec Оё 2 2 2 2 =5 + 1+cot Оё+1 + tan Оё=7+cot Оё+ tan Оё 6 X QUESTION 21. ANSWER D and E are points on the sides CA and CB respectively of в€† ABC right angled at C. Prove 2 2 2 2 that AE +BD =AB + DE . 21. OR In the given figure, DB вЉҐ BC, DE вЉҐ AB and AC вЉҐ BC. Prove that BE = AC . DE BC A D E B C 22. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and в€ ABC = 60В°. Construct a в€† AB'C' similar to в€† ABC such that sides of в€† AB'C' are th 3/4 of the corresponding sides of в€† ABC. 22. 23. 23. If A(4, -8), B(3, 6) and C(5, -4) are the vertices of a в€† ABC, D is the midpoint of BC and P is a point on AD joined such that AP = 2 , find the PD co-ordinate of P. Given in в€† ABC, в€ C = 90В°. D and E are points on the sides CA and CB respectively. 2 2 2 In в€† ACE, AE = AC + CE вЂ¦вЂ¦вЂ¦..(1) 2 2 2 In в€† BCD, BD = BC + CD вЂ¦вЂ¦вЂ¦..(2) B 2 2 Adding (1) and (2), We get AE + BD 2 2 2 2 = AC + CE + BC + CD 2 2 2 2 E =AC + BC + CE + CD 2 2. =AB + DE D C OR In the given figure, DB вЉҐ BC, DE вЉҐ AB and AC вЉҐ BC D BE AC To prove : = DE BC Proof : Since DB вЉҐ BC в‡’ в€ DBC = 90В° E в‡’в€ DBE + в€ CBE = 90В° вЂ¦вЂ¦.(1) Again in в€† BDE, DE вЉҐ AB E в‡’ в€ DEB = 90В° в‡’в€ BDE + в€ DBE = 90В° вЂ¦вЂ¦вЂ¦(2) From (1) and (2), в€ DBE + в€ CBE = в€ BDE + в€ DBE в‡’ в€ CBE = в€ BDE Now in в€† BDE and в€†ABC в€ DEB = в€ ACB (=90В°) в€ BDE = в€ CBE в‡’ в€†BDE ~ в€† ABC (By A вЂ“ A criterion) BE DE BE AC в‡’ = в‡’ = AC BC DE AC Try yourself Given A (4, -8), B(3, 6) and C(5, -4) are the vertices of a в€† ABC D is the midpoint of BC пЈ« 3 + 5 6 + ( в€’4) пЈ¶ So co-ordinate of D is пЈ¬ , пЈ· 2 пЈ 2 пЈё = (4, 1) AP P is a point on AD. Such that =2 PD Let the co-ordinates of P be (x, y) 2 Г— 4 + 1Г— 4 2 Г— 1 + 1Г— ( в€’8) So, x = =4, y= = в€’2 3 3 Hence the co-ordinates of P are (4, 2) A 2 P 1 24. If the point P(x, y) is equidistant from the points 24. A(3, 6) and B(-3, 4), prove that 3x + y вЂ“ 5 = 0. 25. OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region. 25. B (3,6) (5,4) C D Since the point (x, y) is equidistant from the points A(3, 6) and B(-3, 4) в‡’ PA = PB 2 2 2 2 2 2 в‡’ PA = PB в‡’ (x вЂ“ 3) + (y вЂ“ 6) = (x + 3) + (y вЂ“ 4) 2 2 2 2 в‡’ x + 9-6x + y + 36 вЂ“ 12 y =x + 9 + 6x + y +16 вЂ“ 8y в‡’ 12x + 4y вЂ“ 20 = 0в‡’ 3x + y вЂ“ 5 = 0 Here radius = 3.5 cm в‡’ OA = OB = 3.5 cm OD = 2cm 2 Area of the quadrant OACB = Вј ПЂr 1 7 1 22 35 35 77 2 2 (3.5 Г— 2) = = Г— Г— = cm = 9.625 cm 2 2 4 7 10 10 8 Area of в€† BOD = ВЅ Base Г— Height A A C QUESTION ANSWER 1 7 2 (3.5 Г— 2) = = 3.5 cm 2 2 Area of the shaded region = 9.625 вЂ“ 3.5 2 = 6.125 cm A C D = ВЅ OB Г— OD = SECTION вЂ“ D [6Г—5] 26. Prove that the ratio of areas of two similar triangles is equal to the ratio of the squares on their corresponding sides. Using this do the following : The diagonals of a trapezium ABCD, with AB || DC, intersect each other at the point O. If AB = 2 CD, find the ratio of the area of the в€† AOB to the area of the в€† COD. OR Prove that the lengths of tangents drawn from an external point to a circle are equal. Using this do the following : P OвЂў Q A R T B In the figure TP and TQ are tangents from T to the circle with centre O and R is any point on the circle. If AB is a tangent to the circle at R, prove that TA + AR = TB + BR. 26. B Given - в€† ABC ~ в€† DEF A B C D To prove - O D E M ar ( ABC ) 2 F 2 пЈ« AB пЈ¶ пЈ« BC пЈ¶ пЈ« AC пЈ¶ = =пЈ¬ пЈ· = пЈ¬ DF пЈ· ar (DEF ) пЈ¬пЈ DE пЈ·пЈё пЈ EP пЈё пЈ пЈё 2 Construction вЂ“ Draw AD вЉҐ BC and DM вЉҐ EF 1 Proof вЂ“ Here area of в€† ABC = Г— BC Г— AD 2 1 area of в€† DEF = Г— EF Г— DM 2 в€ґ ar ABC = ar DEF 1 2 Г— BC Г— AD 1 2 Г— EF Г— DM = BC Г— AD вЂ¦вЂ¦..(1) EF Г— DM In в€† ABD and в€† DEM LB = LE в€µ (в€† ABC ~ в€† DEF) в€ ADB = в€ DME (Each 90В°) в€ґ By AA condition в€† ABD ~ в€† DEM AB AD в€ґ = вЂ¦вЂ¦вЂ¦(2) DE DM в€ґ (sides are proportional) Since в€† ABC ~ в€† DEF AB BC AC в€ґ = = вЂ¦вЂ¦..(3) DE EF DF AB BC в€ґ = вЂ¦вЂ¦вЂ¦(4) DE EF ar ABC BC Г— AD пЈ« AB AB пЈ¶ пЈ« AB пЈ¶ = = Г— = ar DEF EF Г— DM пЈ¬пЈ DE DE пЈ·пЈё пЈ¬пЈ DF пЈ·пЈё using (2) and (4) 2 From (1) 2 2 2 ar ABC пЈ« AB пЈ¶ пЈ« BC пЈ¶ пЈ« AC пЈ¶ =пЈ¬ =пЈ¬ =пЈ¬ пЈ· пЈ· пЈ· using (3) ar DEF пЈ DE пЈё пЈ EP пЈё пЈ DF пЈё Hence, proved 2nd Part A B 4 2 Now O 1 D Since AB || CD в€ґ в€ 1 = в€ 2 3 C QUESTION ANSWER в€ 3 = в€ 4 (all interior angles are equal) вЂ¦вЂ¦вЂ¦.(1) In в€† ABO and в€† COD в€ 1 = в€ 2 в€ 3 = в€ 4 Using (1) в€ґ By AA condition в€† ABO ~ в€†COD As we know that ratio of the area of 2 similar triangles are equal to ratio of the square of their corresponding sides ar AOB пЈ« AB пЈ¶ = ar COD пЈ¬пЈ CD пЈ·пЈё в€ґ 2 Since AB = 2CD в€ґ ar AOB пЈ« 2 CD =пЈ¬ ar COD пЈ¬пЈ CD 2 пЈ¶ 4 пЈ·пЈ· = Hence 4 : 1 1 пЈё OR A B O C Given : AB and BC are tangent. To prove : AB = BC Construction : O to A, O to C and O to B Proof : Since tangent at the point of contact with circle makes right angle with radius. в€ґ в€ OAB = в€ OCB вЂ¦вЂ¦ (1) In в€† OAB acid в€† OBC OA = OC (each radius) в€ OAB = в€ OCB (each 90В°) flow Hence by RHS configuration condition в€†OAB в‰… в€† OBC в€ґ by CPCT AB = AC. 2 nd part :P A R O Q T B given : TP = TQ to prove : TA + AR = TB + BR since tangent drawn flows are external point to a circle are equal in length в€ґ TP = TQ = ... (1) AP = AR = вЂ¦. (2) BQ = BR вЂ¦.. (3) Using (1) twice TP = TQ в€ґ AT + AP = BT + BQ AT + AR = BT + BR [using (k) and 3] Hence TA + AR = TB + BR Hence verified. 2 в‡’ 8x вЂ“ 70x вЂ“ 750 = 0 70 В± 4900 + 24000 в‡’ x= 16 70 В± 170 240 100 = = = rejected= 15 hr. 16 16 10 QUESTION ANSWER 27. A motor boat whose speed is 18 km/h in still 27. Let the speed of the stream be x km/hr Given speed of the boat in still water is 18 km/hr в€ґ speed in up stream = (18-x) km/hr в€ґ speed in down stream = (18+x) km/hr Time taken to travel 24 Km up steam пЈ« 24 пЈ¶ =пЈ¬ пЈ· Km / hr пЈ 18 в€’ x пЈё Time taken to travel 24 Km down пЈ« 24 пЈ¶ stream = пЈ¬ пЈ· Km / hr пЈ 18 + x пЈё According to the questions 24 24 432 + 24x в€’ 432 + 24x в€’ =1 в‡’ =1 18 в€’ x 18 + x 324 в€’ x 2 2 2 в‡’ 48x = 324 вЂ“ x в‡’ x + 48x вЂ“ 324 = 0 в€’48 В± 2304 + 1296 в€’48 В± 60 в‡’x= = 2 2 12 108 = or (rejected) = 6 Km/hr 2 2 Hence speed of the stream be 6 Km/hr OR Let the tap of smaller diameter takes x hours to fill the tank в‡’ Tap of larger diameter will take (x + 10) hrs to fill the tank. In 1 hour the small tap can fill in 1/x of the tank and the пЈ« 1 пЈ¶ larger tap can fill пЈ¬ пЈ· of the tank. пЈ x + 10 пЈё 1 1 8 According to the question + = x x + 10 75 2x + 10 8 2 в‡’ = в‡’ 150x + 750 = 8x + 10x 2 75 x + 10x 28. Here height of the flight from the ground AB = CD E be the point of observation. water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. OR Two water taps together can fill a tank in 9 3 8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. 28. The angle of elevation of a jet fighter from a point A on the ground is 60В°. After a flight of 10 seconds, the angle of elevation changes to 30В°. If the jet is flying at a speed of 648 km/hour, find the constant height at which the jet is flying. [Use в€љ3 = 1.732]. A E 60В° 30В° B D C в€ AEB = 60В° and в€ CED = 30В°. Given speed = 648 km/hr = 180 m/s After a flight 10 seconds, the distance traveled BC = 180 Г— 10 = 1800 m. AB Now in в€† AEB, tan 60В° = BE в‡’ AB = BE в€љ3 вЂ¦.. (i) CD Again in в€† CDE, tan 30В° = CE BR 3 = в‡’ CE = 3BEв‡’ BE + BC = 3BE CE 3 в‡’ BC = 2BE в‡’ BE = 900 m. Hence AB = 900 в€љ3 m. (the height of the flight from ground). в‡’ 1 QUESTION ANSWER 29. A tent consists of a frustrum of a cone, 29. surmounted by a cone. If the diameters of the upper and lower circular ends of the frustrum be 14 m and 26 m respectively, the height of the frustrum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of the canvas required to make the tent. (Assume that the radii of the upper circular end of the frustrum and the base of surmounted conical portion are equal). Given height of the frustrum = 8m. Upper radius of the frustrum = r1 = 14 m. Lower radius of the frustrum = r2 = 26m A X E D в€ґ l = h2 + (r2 в€’ r1)2 8 = 64 + 144 = 208 = 14.4 m. B C Y Slant height of the conical portion=12m. and its radius=14 m. So total canvas required = C.S.A of frustrum + C.S.A of cone. 144 = ПЂ(r1 + r2)l + ПЂr1l = ПЂ(40) + ПЂ Г— 14 Г— 12 10 22 2 = 576 ПЂ + 168 ПЂ= 744 ПЂ= 744 Г— = 2338.28 m 7 30. Find the mean, mode and median of the following data: Classes Frequency 0 вЂ“ 10 3 10 вЂ“ 20 4 20 вЂ“ 30 7 30 вЂ“ 40 15 40 вЂ“ 50 10 50 вЂ“ 60 7 60 вЂ“ 70 4 30. C.I 0-10 10-20 20-30 30-40 40-50 50-60 60-70 fi 3 4 7 15 10 7 4 xi 5 15 25 35 45 55 65 Here, mean = A + di -30 -20 -10 0 10 20 30 ui -3 -2 -1 0 1 2 3 uifi -9 -8 -7 0 10 14 12 в€‘ uifi = 12 в€‘ uifi Г— h в€‘ fi 12 = 35 + Г— 10 = 35+ 2.4 = 37. 4 50 N Again = 25 2 The median class is 30-40 So l = 30, c.f = 14, f = 15, h = 10 N в€’ cf 25 в€’ 14 Median = l + 2 Г— h = 30 + Г— 10 = f 15 11 30 + Г— 10 15 22 = 30 + = 30 + 7.3= 37. 3 3 Modal class = 30 -40 So l = 30, f1 = 15, f0 = 7, f2= 10 and h = 40 f1 в€’ f0 So mode = l + Г—h 2f1 в€’ f0 в€’ f1 = 30 + 8 80 Г— 10 = 30 + = 30 + 6.15 = 36.15 13 13 cf 3 7 14 29 39 46 50

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