Set-1 Sec-d

CBSE X Mathematics 2009 Solution (SET 1)

CBSE X 2009 Mathematics Section D Question Number 26 to 30 carry 6 marks each.

26.

Solve the following equation for x: 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0 OR If (–5) is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the values of p and k. Solution: 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0 2a 2 5ab 2b 2 2 x ( a b) x 0 9 x2

2a b 3

a 2b 3

x

x2

2a b 3

a 2b 3

x

x2

2a b 3

a 2b 3

x

x x x

2a b 3

x

2a b 0 or x 3 2a b a 2b or 3 3

a 2b 3 a 2b 3

2a 2

4ab ab 2b 2 9

2a (a 2b) b(a 2b) 9 (a 2b)(2a b) 9

0 0

0

[Using x 2 (a b) x ab

0 0

OR It is given that 2x2 + px – 15 = 0 If (–5) is a root of this quadratic equation, then 2(–5)2 + p(–5) – 15 = 0 2(25) + p(–5) – 15 = 0 50 – 5p – 15 = 0 5p = 35 p=7 … (1) Again, it is given that p(x2 + x) + k = 0 px2 + px + k = 0

( x a )( x b)]

CBSE X Mathematics 2009 Solution (SET 1) On putting p = 7, we get: 7x2 + 7x + k = 0 If this equation has real roots, then its discriminant must be 0. (7)2 – 4 (7) (k) = 0 49 – 28k = 0 49 7 k= 28 4 7 Thus, the values of p and k are 7 and respectively. 4

27.

Prove that the lengths of the tangents drawn from an external point to a circle are equal. Using the above theorem prove that: If quadrilateral ABCD is circumscribing a circle, then AB + CD = AD + BC

Solution: The figure shows a circle with centre O. P is a point taken in the exterior of the circle. PA and PB are tangents from point P to the circle. We also construct OA, OB, and OP.

We need to prove that the lengths of the tangents drawn from an external point to a circle are equal, i.e., PA = PB. It is known that the tangent at any point of a circle is perpendicular to the radius through the point of contact. PA and PB are perpendicular to OA and OB respectively. OAP = OBP = 90 In OAP and OBP: OA = OB (Radii of the same circle) OP = OP (Common) OAP = OBP (Each 90 ) OAP PA = PB

OBP

(RHS congruence criterion) (CPCT)

CBSE X Mathematics 2009 Solution (SET 1) Thus, the lengths of the tangents drawn from an external point to a circle are equal. Hence, proved. The figure shows quadrilateral ABCD, which circumscribes a circle. Sides AB, BC, CD, and AD touch the circle at F, G, H, and E respectively.

It is known that the lengths of the tangents drawn from an external point to a circle are equal. Hence, AF = AE BF = BG CH = CG DH = DE

(Tangents from point A) (Tangents from point B) (Tangents from point C) (Tangents from point D)

Adding these four equations: AF + BF + CH + DH = AE + BG + CG + DE (AF + BF) + (CH + DH) = (AE + DE) + (BG + CG) AB + CD = AD + BC Hence, proved. 28.

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant. Solution: The given information can be represented as

CBSE X Mathematics 2009 Solution (SET 1) Let P be the point on the ground and let B and C be the two aeroplanes. Since the aeroplane at point B is flying 3125 m above the ground, AB = 3125 m The distance between the two aeroplanes at that instant is given by BC. In APB, AB tan 30 AP 3125 m 1 AP 3

AP 3125 3 m In APC, AC tan 60 AP AB BC tan 60 AP 3125 m BC 3 3125 3 m 3125 m BC 9375 m

BC 9375 m 3125 m 6250 m Thus, the distance between the two aeroplanes at that instant is BC = 6250 m.

29.

A juice seller serves his customers using a glass as shown in Figure 6. The inner diameter of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces the capacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use = 3.14)

OR A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of (i) (ii)

water displaced out of the cylindrical vessel. water left in the cylindrical vessel.

CBSE X Mathematics 2009 Solution (SET 1) [Take

22 ] 7

Solution:

Apparent capacity of the glass will be the same as the capacity of the cylindrical portion having its base diameter as 5 cm and height as 10 cm. 5 2.5 cm Base radius = 2 Apparent capacity = πr2h = 3.14 × (2.5)2 × 10 = 196.25 cm3 Actual capacity of the glass will be the difference between the cylindrical portion and the hemispherical portion. From the figure, it is clear that the base radius of the hemispherical portion is 5 also cm or 2.5 cm. 2 2 3 πr Actual capacity of the glass = πr 2 h 3 2 196.25 cm 3 3.14 (2.5)3 cm 3 3 = 196.25 cm3 – 32.71 cm3 = 163.54 cm3 Thus, the apparent capacity of the glass is 196.25 cm3 and the actual capacity of the glass is 163.54 cm3. OR

CBSE X Mathematics 2009 Solution (SET 1) (I)

When the solid cone is completely immersed in water, some water will be displaced out of the cylindrical vessel.

The volume of water thus displaced out will be the same as the volume of the solid cone having its height as 6 cm and base diameter as 7 cm. 7 3.5 cm Base radius 2 1 2 πr h Volume of the water displaced out 3 1 22 (3.5) 2 6 3 7 22 3.5 3.5 2 7 3 77 cm Thus, the water displaced out of the cylindrical vessel is 77 cm3 . (II)

In the beginning, the volume of water in the completely-filled cylindrical vessel will be the same as the capacity of this cylindrical vessel. 10 5 cm Base radius 2 Volume of water in the cylindrical vessel = πr2h 22 (5) 2 10.5 7 33 25

825 cm3 Water left in the cylindrical vessel = Total water – Water displaced out = 825 cm3 – 77 cm3 = 748 cm3 Thus, the water left in the cylindrical vessel is 748 cm3 .

CBSE X Mathematics 2009 Solution (SET 1) 30.

During the medical check-up of 35 students of a class their weights were recorded as follows: Weight (in kg) 38 – 40

Number of students 3

40 – 42

2

42 – 44

4

44 – 46

5

46 – 48

14

48 – 50

4

50 – 52

3

Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph. Solution: For the given data, the cumulative frequency distribution of the less than type can be computed as follows. Weight (in kg) Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52

Number of students (Cumulative frequency) 3 3+2=5 5+4=9 9 + 5 = 14 14 + 14 = 28 28 + 4 = 32 32 + 3 = 35

To draw a less than ogive, we mark the upper class limits of the class intervals on the x-axis and their corresponding cumulative frequencies on the y-axis by taking a convenient scale. Now, plot the points corresponding to the ordered pairs [(upper class limit, cumulative frequency) – i.e., (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)] on the graph paper as follows:

CBSE X Mathematics 2009 Solution (SET 1)

Similarly, we can compute the cumulative frequency distribution of the more than type as follows: Weight (in kg) More than 38 More than 40 More than 42 More than 44 More than 46 More than 48 More than 50

Number of students (Cumulative frequency) 35 35 – 3 = 32 32 – 2 = 30 30 – 4 = 26 26 – 5 = 21 21 – 14 = 7 7–4=3

Now, to draw a more than ogive, we mark the lower class limits of the class intervals on the x-axis and their corresponding cumulative frequencies on the y-axis by taking a convenient scale. Now, plot the points corresponding to the ordered pairs [(lower class limit, cumulative frequency) – i.e., (38, 35), (40, 32), (42, 30), (44, 26), (46, 21), (48, 7), (50, 3)] on the graph paper as follows:

CBSE X Mathematics 2009 Solution (SET 1) Now, to obtain the median weight from the graph, we draw both ogives on the same graph paper. They intersect at (46.5, 17.5). 46.5 kg is the median weight of the given data.