CBSE X Mathematics 2012 Solution (SET 1) Section C

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1 CBSE X Mathematics 01 Solution (SET 1) Q19. Solve for x : 4x 4ax + (a b ) = 0 Section C The given quadratic equation is x ax a b 4x 4ax a b 0 4x 4ax a b a b x [ a a b b] x ( a b)( a b) 0 4x b a x a b x a b a b 0 4x b a x a b x a b a b 0 x[ x ( a b)] a b [ x ( a b)] 0 [ x ( a b)][ x ( a b)] 0 x ( a b) 0 or x ( a b) 0 x a b or x a b a b a b x or x Thus, the solution of the given quadratic equation is given by a b or a x x b. OR Solve for x x x :3 6 0 The given quadratic equation is3x 6 x 0. Comparing with the quadratic equation ax + bx + c = 0, we have a = 3, b 6 and c = Discriminant of the given quadratic equation, D = b 4ac

CBSE X Mathematics 01 Solution (SET 1) 6 0 b D x x 3 a x x 6 6 6 3 Thus, the solution of the given quadratic equation is x = 6 3. Q0. Prove that the parallelogram circumscribing a circle is a rhombus.

2 CBSE X Mathematics 01 Solution (SET 1) 6 0 b D x x 3 a x x Thus, the solution of the given quadratic equation is x = 6 3. Q0. Prove that the parallelogram circumscribing a circle is a rhombus. Since ABCD is a parallelogram, AB = CD (1) BC = AD () It can be observed that DR = DS (Tangents on the circle from point D) CR = CQ (Tangents on the circle from point C) BP = BQ (Tangents on the circle from point B) AP = AS (Tangents on the circle from point A) Adding all these equations, we obtain DR + CR + BP + AP = DS + CQ + BQ + AS (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) CD + AB = AD + BC On putting the values of equations (1) and () in this equation, we obtain AB = BC AB = BC (3) Comparing equations (1), (), and (3), we obtain AB = BC = CD = DA Hence, ABCD is a rhombus.

CBSE X Mathematics 01 Solution (SET 1) OR Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

3 CBSE X Mathematics 01 Solution (SET 1) OR Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle. Consider OAP and OAS, AP = AS (Tangents from the same point) OP = OS (Radii of the same circle) OA = OA (Common side) OAP OAS (SSS congruence criterion) Therefore, A A, P S, O O And thus, POA = AOS 1 = 8 Similarly, = 3 4 = 5 6 = = 360º (1 + 8) + ( + 3) + (4 + 5) + (6 + 7) = 360º = 360º (1 + ) + (5 + 6) = 360º (1 + ) + (5 + 6) = 180º AOB + COD = 180º Similarly, we can prove that BOC + DOA = 180º Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

4 CBSE X Mathematics 01 Solution (SET 1) Q1. Construct a right triangle in which the sides, (other than the hypotenuse) are of length 6 cm and 8 cm. Then construct another triangle, whose sides are 3 5 times the corresponding sides of the given triangle. Given: BC = 6 cm, C = 8 cm The triangle to be formed is to be right angled triangle. Steps of construction: 1. Draw a line segment BC = 6 cm.. Draw a ray CN making an angle of 90 at C. 3. With C as centre, taking 8 cm as the radius make an arc at CN intersecting it at A. Join AB. 4. Now, ABC is the triangle whose similar triangle is to be drawn. 5. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. 6. Locate 5 (Greater of 3 and 5 in 3 5 ) points B 1, B, B 3, B 4 and B 5 on BX so that BB 1 = B 1 B = B B 3 = B 3 B 4 = B 4 B 5 7. Join B 5 C and draw a line through B 3 (Smaller of 3 and 5 in 3 5 ) parallel to B 5C to intersect BC at C. 8. Draw a line through Cparallel to the line CA to intersect BA at A.

5 CBSE X Mathematics 01 Solution (SET 1) 9. ABCis the required similar triangle whose sides are 3 times the corresponding sides of 5 ABC. Q. In Fig. 7, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm and centre O. If POQ = 30, then find the area of the shaded region. Use 7 PQ and AB are the arcs of two concentric circles of radii 7 cm and 3.5 cm respectively. Let r 1 and r be the radii of the outer and the inner circle respectively. Suppose be the angle subtended by the arcs at the centre O. Then r 1 7 cm, r 3.5 cm and 30 Area of the shaded region = Area of sector OPQ Area of sector OAB r1 r r1 r cm 3.5 cm cm cm cm Thus, the area of the shaded region is 9.65 cm.

CBSE X Mathematics 01 Solution (SET 1) Q3. From a solid cylinder of height 7 cm and base diameter 1 cm, a conical cavity of same height and same base diameter is hollowed out.

6 CBSE X Mathematics 01 Solution (SET 1) Q3. From a solid cylinder of height 7 cm and base diameter 1 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. Use 7 It is given that, height (h) of cylindrical part = height (h) of the conical part = 7 cm Diameter of the cylindrical part = 1 cm 1 Radius (r) of the cylindrical part cm = 6cm Radius of conical part = 6 cm Slant height (l) of conical part r h cm = 6 7 cm = cm = 85 cm = 9. cm (approx.) Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Base area of the circular part = rh + rl + r 6 7 cm 6 9. cm 6 6 cm cm cm cm 551cm OR A cylindrical bucket, 3 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 4 cm, then find the radius and slant height of the heap.

CBSE X Mathematics 01 Solution (SET 1) Height (h 1 ) of cylindrical bucket = 3 cm Radius (r 1 ) of circular end of bucket = 18 cm Height (h ) of conical heap = 4 cm Let the radius of the circular end

Volume of sand in the cylindrical bucket = Volume of sand in conical heap 1 πr1 h1 π r h 3 1 π18 3 πr 4 3 1 π18 3 π r 4 3 318 3 r 18 4 4 r = 18 = 36 cm Slant height = 36 4 1 (3 ) 1 13 cm Therefore,

7 CBSE X Mathematics 01 Solution (SET 1) Height (h 1 ) of cylindrical bucket = 3 cm Radius (r 1 ) of circular end of bucket = 18 cm Height (h ) of conical heap = 4 cm Let the radius of the circular end of conical heap be r. The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap. Volume of sand in the cylindrical bucket = Volume of sand in conical heap 1 πr1 h1 π r h 3 1 π18 3 πr π18 3 π r r r = 18 = 36 cm Slant height = (3 ) 1 13 cm Therefore, the radius and slant height of the conical heap are 36 cm and 1 13 cm respectively. Q4. The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45 and 30. If the ships are 00 m apart, find the height of the light house. The given situation can be represented diagrammatically as,

8 CBSE X Mathematics 01 Solution (SET 1) Here, AB is the light house and the ships are at the points C and D; h is the height of the light house and BC = x. In right angled ABC: AB tan 45 AC h tan 45 x h 1 x x h In right angled ABD, AB tan 30 BD h tan 30 x 00 h tan 30 x h h 00 1 h 3 h 00

9 CBSE X Mathematics 01 Solution (SET 1) h 00 h 3 h 3 h 00 h h 00 m h h 31 h Hence, the height of the light house is m. Q5. A point P divides the line segment joining the points A (3, 5) and B ( 4, 8) such that AP K. If P lies on the line x + y = 0, then find the value of K. PB 1 The given points are A (3, 5) and B ( 4, 8). Here, x 3, y 5, x 4 and y Since AP K, the point P divides the line segment joining the points A and B in the ratio K:1. PB 1 mx nx1 my ny1 The co-ordinates of P can be found using the section formula, m n m n. Here, m = K and n = 1 K K K+3 8K 5 Co-ordinates of P,, K + 1 K + 1 K+1 K+1 It is given that, P lies on the line x + y = 0.

10 CBSE X Mathematics 01 Solution (SET 1) 4K+3 8K 5 0 K+1 K+1 4K 3 8K 5 0 K +1 4K 0 4K 1 K Thus, the required value of K is 1. Q6. If the vertices of a triangle are (1, 3), (4, p) and ( 9, 7) and its area is 15 sq. units, find the value(s) of p. Given, vertices of a triangle are (1, 3), (4, p) and ( 9, 7). x 1, y x 4, y p x 9, y Area of given triangle 1 x y y x y y x y y p p 1 p p 1 10 p 60 5 p 6 = Here, the obtained expression may be positive or negative. 5 p 6 15 or 5 p 6 15 p 6 3 or p 6 3 p 3 or p 9

11 CBSE X Mathematics 01 Solution (SET 1) Q7. A box contains 100 red cards, 00 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be (i) a blue card (ii) not a yellow card (iii) neither yellow nor a blue card. Number of red cards = 100 Number of yellow cards = 00 Number of blue cards = 50 Total number of cards = = 350 Number of blue cards i P a blue card Total number of cards Number of cards other than yellow ii P not a yellow card Total number of cards Number of red cards + Number of blue cards Total number of cards Number of cards which are neither yellow nor blue iii P Neither yellow nor a blue card Total number of cards Number of red cards Total number of cards Q8. The 17 th term of an AP is 5 more than twice its 8 th term. If the 11 th term of the AP is 43, then find its n th term.

12 CBSE X Mathematics 01 Solution (SET 1) Let a be the first term and d be the common difference of the given A.P. According to the given question, 17 th term = 8 th term + 5 i.e., a a a 17 1 d a 8 1 d 5 (as an a n 1 d) a 16d a 7d 5 a 16d a 14d 5 a d Also, 11 th term, a 11 = 43 a 111 d 43 a10d 43 d 5 10d 43 Using 1 1d 48 d 4 a d 5 (4) Thus, n th term of the AP, 1 a a n d On putting the respective values of a and d, we get a 3 n n 4 4n 1 n Hence, n th term of the given AP is 4n 1. n

Time: 3 hour Total Marks: 90

Time: 3 hour Total Marks: 90 General Instructions: 1. All questions are compulsory. 2. The question paper consists of 34 questions divided into four sections A, B, C, and D. 3. Section A contains of 8