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2012 Class 10 Set-3 Section-b

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  CBSE X Mathematics 2012 Solution (SET 3) Section B Q11.  If a point A (0, 2) is equidistant from the points B (3,  p ) and C (  p , 5), then find the value of  p . Solution: The given points are A (0, 2), B, (3, P) and C (P, 5). According to the question, A is equidistant from points B and C.   AB = AC                222222222222 30205232394494139  p p p p p p p p p p                       On squaring both sides, we obtain: 22 4139441  p p p p p            Q12.  A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. Solution: Total number of outcomes = 50 Multiples of 3 and 4 which are less than or equal to 50 are: 12, 24, 36, 48 Favorable number of outcomes = 4 Probability of the number being a multiple of 3 and 4  Number of favourable outcomesTotalnumberofoutcomes450225   Q13.  The volume of a hemisphere is 124252  cm 3 . Find its curved surface area. 22Use 7           CBSE X Mathematics 2012 Solution (SET 3) Solution: 33 14851Given, volume of hemisphere 2425 cm cm22     Let the radius of the hemisphere be ‘ r  ’ cm.     3333333 2Volume of hemisphere 324851322224851372485137222244121 22221212122221212122221cm ...12 r r r r r r r r                           Curved surface area of hemisphere = 2π r  2     2 22212 using 1722221212722              = 693 cm 2   Q14.  Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in Fig. 3. If AP = 15 cm, then find the length of BP. Solution:  CBSE X Mathematics 2012 Solution (SET 3) Given that: OA = 8 cm, OB = 5 cm and AP = 15 cm To find:  BP Construction:  Join OP. Tangent to a circle is perpendicular to the Now, OA AP and OB BP radius through the point of contactOAPOBP90              On applying Pythagoras theorem in  OAP, we obtain: (OP) 2  = (OA) 2  + (AP) 2   (OP) 2  = (8) 2  + (15) 2   (OP) 2  = 64 + 225   (OP) 2  = 289 OP = 289     OP = 17 Thus, the length of OP is 17 cm. On applying Pythagoras theorem in  OBP, we obtain: (OP) 2 = (OB) 2  + (BP) 2   (17) 2 = (5) 2  + (BP) 2  289 = 25 + (BP) 2   (BP) 2  = 289  –   25   (BP) 2  = 264   BP = 16.25 cm (approx.) Hence, the length of BP is 16.25 cm. Q15.  In Fig. 4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the  point of contact P bisects the base BC.  CBSE X Mathematics 2012 Solution (SET 3) Solution: Given:  An isosceles  ABC with AB = AC, circumscribing a circle. To prove:  P bisects BC Proof:  AR and AQ are the tangents drawn from an external point A to the circle.   AR = AQ (Tangents drawn from an external point to the circle are equal) Similarly, BR = BP and CP = CQ. It is given that in   ABC, AB = AC.   AR + RB = AQ + QC   BR = QC (As AR = AQ)   BP = CP (As BR = BP and CP = CQ)   P bisects BC Hence, the result is proved. OR In Fig. 5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. Solution:
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