CBSE X Mathematics 2012 Solution (SET 1) Section B

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1 CBSE X Mathematics 01 Solution (SET 1) Section B Q11. Find the value(s) of k so that the quadratic equation x kx + k = 0 has equal roots. Given equation is x kx k 0 For the given equation to have equal roots, D = 0. b ac 0 k k k k 0 k k 1 0 k 0 or k k 0 or k Hence, for k = 0 or 1, the given equation will have equal roots. Q1. Find the sum of all three digit natural numbers, which are multiples of 11. The smallest and the largest three digit natural numbers, which are divisible by 11 are 110 and 990 respectively. So, the sequence of three digit numbers which are divisible by 11 are 110, 11, 13,, 990. Clearly, it is an A.P. with first term, a = 110 and common difference, d = 11. Let there be n terms in the sequence. So, a n = 990 a n 1 d n n n n n 891 n 81 n Now, required sum 1 a n d

CBSE X Mathematics 01 Solution (SET 1) 81 110 81 1 11 81 0 880 81 1100 550 Hence, the sum of all three digit numbers which are multiples of 11 is 550. Q13.

2 CBSE X Mathematics 01 Solution (SET 1) Hence, the sum of all three digit numbers which are multiples of 11 is 550. Q13. 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. 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 contact OAP OBP 90 On applying Pythagoras theorem in OAP, we obtain: (OP) = (OA) + (AP) (OP) = (8) + (15) (OP) = 6 + 5

3 CBSE X Mathematics 01 Solution (SET 1) (OP) = 89 OP = 89 OP = 17 Thus, the length of OP is 17 cm. On applying Pythagoras theorem in OBP, we obtain: (OP) = (OB) + (BP) (17) = (5) + (BP) 89 = 5 + (BP) (BP) = 89 5 (BP) = 6 BP = 16.5 cm (approx.) Hence, the length of BP is 16.5 cm. Q1. In Fig., an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC. 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. 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. OR

CBSE X Mathematics 01 Solution (SET 1) Given: Two concentric circles C 1 and C with centre O, and AB is the chord of C 1 touching C at C. To prove: AC = CB Construction: Join OC.

4 CBSE X Mathematics 01 Solution (SET 1) Given: Two concentric circles C 1 and C with centre O, and AB is the chord of C 1 touching C at C. To prove: AC = CB Construction: Join OC. Proof: AB is the chord of C 1 touching C at C, then AB is the tangent to C at C with OC as its radius. We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. OC AB Considering, AB as the chord of the circle C 1. So, OC AB. OC is the bisector of the chord AB. Hence, AC = CB (Perpendicular from the centre to the chord bisects the chord). Q15. The volume of a hemisphere is Given, volume of hemisphere 5 cm cm Let the radius of the hemisphere be r cm. 1 5 cm 3. Find its curved surface area. 3 3 Use 7

CBSE X Mathematics 01 Solution (SET 1) Volume of hemisphere r 3 3 851 r 3 3 851 r 3 7 3 8513 7 r 3 11 r 3 111 r 111 r 3 1 r cm.

5 CBSE X Mathematics 01 Solution (SET 1) Volume of hemisphere r r r r 3 11 r r 111 r 3 1 r cm... 1 Curved surface area of hemisphere = πr 3 1 using = 693 cm Q16. In Fig. 6, OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. Use 7 It is given that OABC is a square of side 7 cm Area of square OABC = (7) cm = 9 cm

6 CBSE X Mathematics 01 Solution (SET 1) Also, it is given that OAPC is a quadrant of circle with centre O. Radius of the quadrant of the circle = OA = 7 cm 1 Area of the quadrant of circle π r 1 π 7 cm 9π cm 9 cm 7 77 cm Area of the shaded region = Area of Square Area of Quadrant of circle cm cm 1 cm = 10.5 cm Thus, the area of the shaded region is 10.5 cm. Q17. If a point A (0, ) is equidistant from the points B (3, p) and C (p, 5), then find the value of p. The given points are A (0, ), B, (3, P) and C (P, 5). According to the question, A is equidistant from points B and C. AB = AC 3 0 p p p p 3 9 p p p 9 p p p 13 9

7 CBSE X Mathematics 01 Solution (SET 1) On squaring both sides, we obtain: p p 13 p 9 p p 1 Q18. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and. Total number of outcomes = 50 Multiples of 3 and which are less than or equal to 50 are: 1,, 36, 8 Favorable number of outcomes = Probability of the number being a multiple of 3 and Number of favourable outcomes Total number of outcomes 50 5

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1 / 24 CBSE-XII-017 EXAMINATION CBSE-X-01 EXAMINATION MATHEMATICS Paper & Solution Time: 3 Hrs. Max. Marks: 90 General Instuctions : 1. All questions are compulsory.. The question paper consists of 34 questions