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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions. 

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: For what values of k does the quadratic equation 4x^2 - 12x - k = 0 have no real roots ?

Answer:

Given equation: 4x^2 - 12x - k = 0

Comparing it with ax^2 + bx + c = 0, a = 4, b = -12 and c = -k

When b^2 - 4ac < 0 , there are no real roots.

\therefore (-12)^2 - 4 (4)(-k) <0

\Rightarrow 144 + 16k < 0

\Rightarrow k < -9

Therefore for k < -9 , there will be no real roots.

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Question 2: Find the distance between the points (a, b) and (- a, - b) .

Answer:

Given points (a, b) and (-a, -b)

Therefore the distance between the two points = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{4a^2 + 4b^2} = 2 \sqrt{a^2 + b^2}

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Question 3: Find a rational number between \sqrt{2} and \sqrt{7} .

Or

Write the number of zeroes in the end of a number whose prime factorization is 2^2 \times 5^3 \times 3^2 \times 17 .

Answer:

\sqrt{2} = 1.412 and \sqrt{7} = 2.645

Therefore 2.5 is between \sqrt{2} and \sqrt{7}

\therefore 2.5 = \frac{25}{10} = \frac{5}{2}  is a rational number between \sqrt{2} and \sqrt{7}

Or

The given expression:

2^2 \times 5^3 \times 3^3 \times  17 = (2 \times 5) \times (2 \times 5) \times  5 \times  3^3 \times  17

= 10 \times 10 \times 5 \times 3^3 \times 17

We know that zeros in an expression are a result of number of 10's in it.

Hence the expression we can see that there will be 2 zeros in the given expression.

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Question 4: Let \triangle ABC \sim \triangle DEF and their areas be respectively, 64 \ cm^2 and 121 \ cm^2 . If EF = 15.4 cm, find BC .

Answer:

Since \triangle ABC \sim \triangle DEF

\frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \Big( \frac{BC}{EF} \Big)^2

\Rightarrow \frac{64}{121} = \Big( \frac{BC}{EF} \Big)^2

\Rightarrow \frac{BC}{EF} = \frac{8}{11} 

\Rightarrow BC = \frac{8}{11} \times 15.4 = 11.2 cm

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Question 5: Evaluate: \frac{\tan 65^o}{\cot 25^o}

Or

Express ( \sin 67^o + \cos 75^o) in terms of trigonometric ratios of the angle between 0^o and 45^o .

Answer:

\frac{\tan 65^o}{\cot 25^o} = \frac{\tan 65^o}{\cot (90^o-65^o)} = \frac{\tan 65^o}{\tan 65^o} = 1

Or

\sin 67^o + \cos 75^o = \sin (90^o-23^o) + \cos (90^o-15^o) = \cos 23^o + \sin 15^o

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Question 6: Find the number of terms in the A.P. : 18, 15 \frac{1}{2} , 13, \cdots , - 47

Answer:

Given AP: 18, 15\frac{1}{2}, 13, \ldots , -47

a = 18, d = 15.5 - 18 = -2.5

Let -47 be the n^{th} term

T_n = a + (n-1) d

\Rightarrow -47 = 18 + (n-1) (-2.5)

\Rightarrow -65 = -2.5 (n-1)

\Rightarrow 26 = n - 1

\Rightarrow n = 27

Therefore -47 is the 27^{th} term in the given AP

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Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: A bag contains 15 balls, out of which some are white and the others are black. If the probability of drawing a black ball at random from the bag is \frac{2}{3} , then find how many white balls are there in the bag.

Answer:

Total number of balls = 15

Let the number of black balls = x

\therefore \frac{x}{15} = \frac{2}{3} \Rightarrow x = 10

Hence the number of white balls = 15 - 10 = 5

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Question 8: A card is drawn at random from a pack of 52 playing cards. Find the probability of drawing a card which is neither a spade nor a king.

Answer:

Total number of cards = 52

Number of space cards = 13 (includes the king of space)

No of kings other than king of spade = 3

Therefore Probability = \frac{52 - (13+3)}{52} = \frac{36}{52} = \frac{9}{13}

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Question 9: Find the solution of the pair of equations :

\frac{3}{x} + \frac{8}{y} = - 1; \frac{1}{x} - \frac{2}{y} = 2 , x, y \neq 0

Or

Find the value(s) of k for which the pair of equations kx+2y = 3, 3x+6y = 10  has a unique solution.

Answer:

Given equation:

\frac{3}{x} + \frac{8}{y} = - 1

\frac{1}{x} - \frac{2}{y} = 2

Let \frac{1}{x} = u and \frac{1}{y} = v

Therefore

3u + 8 v = -1    … … … … … iii)

u - 2v = 2    … … … … … iv)

Multiplying iv) by iii) and subtracting it from iii)

\hspace*{1cm} 3u + 8 v = -1 \\ (-1) \underline {\hspace*{0.2cm}3u - 6v = 6} \\ \hspace*{1.5cm} 14v = -7

v= \frac{-1}{2} 

Therefore u = 2v + 2 = 2 ( \frac{-1}{2} ) + 2 = -1 +2 = 1

Hence \frac{1}{x} = 1 \Rightarrow x = 1

Also \frac{1}{y} = \frac{-1}{2} \Rightarrow y = -2

Or

Given equations:

kx + 2y = 3

3x + 6y = 10

If the system of equations are ax_1 + b_1y + c_1 = 0 and ax_2 + b_2y + c_2 = 0 and they have unique solution then it satisfy the following:

\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

\Rightarrow \frac{k}{3} \neq \frac{2}{6} \neq \frac{-3}{-10} 

From first two terms

k \neq 1

From First and Third term

k \neq \frac{9}{10} 

\therefore k \neq 1 or k \neq \frac{9}{10}  for the equations to have a unique solution.

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Question 10: How many multiples of 4 lie between 10 and 205 ?

Or

Determine the A.P. whose third term is 16 and 7^{th} term exceeds the 5^{th} term by 12 .

Answer:

Since the multiples of 4 lie between 10 and 205 are 12, 16, 20, \ldots , 204

\therefore a = 12, d = 16-12 = 4

T_n = a + (n-1)d

\Rightarrow 204 = 12 + (n-1) (4)

\Rightarrow 192 = (n-1)(4)

\Rightarrow 48 = n - 1

\Rightarrow n = 49

Therefore there are 49 multiples of 4 between 10 and 205

Or

Let for the AP, first term = a and common difference = d

Given T_3 = 16

\therefore a + (3 - 1) d = 16

\Rightarrow a + 2d = 16    … … … … … i)

T_7 = a + 6d

T_5 = a + 4d

Given T_7 = T_5 + 12

\Rightarrow a + 6d = a + 4d + 12

\Rightarrow 2d = 12

\Rightarrow d = 6

Therefore form i) we get a = 16 - 2( 6) = 4

Therefore the AP is 4, 10, 16, 22, \ldots

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Question 11: Use Euclid’s division algorithm to find the HCF of 255 and 867 .

Answer:

255 ) \overline{867} ( 3 \\ \hspace*{0.7cm} \underline{765} \\ \hspace*{0.8cm} 102 ) \overline{255} ( 2 \\ \hspace*{1.5cm} \underline{204} \\ \hspace*{1.7cm}  51 ) \overline{102} ( 2 \\ \hspace*{2.25cm}  \underline{102} \\ \hspace*{2.5cm} \times

According to Eculid’s division theorem, any positive number can be expressed as a = bq+r where q is the quotient, b is the divisor and r is the remainder and 0 \leq r < b

\therefore 867 = 51 \times 17 + 0

So HCF of 867 and 255 is 51

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Question 12: The point R divides the line segment AB , where A( 4, 0) and B(0, 6) such that AR = \frac{3}{4} AB . Find the coordinates of R .

Answer:

Given points A(-4, 0) and B(0, 6)

AR = \frac{3}{4} AB

\Rightarrow \frac{AR}{AB} = \frac{3}{4}

\Rightarrow \frac{AR}{AR + RB} = \frac{3}{4}

\Rightarrow 4AR = 3AR + 3RB

\Rightarrow \frac{AR}{RB} = \frac{3}{1}

Using sections formula, R(x, y) = \Big( \frac{3 \times 0 + 1 \times (-4)}{3+1} , \frac{3 \times 6 + 1 \times 0}{3+1}  \Big) = \Big( -1, \frac{9}{2} \Big)

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Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that : ( \sin \theta + 1 + \cos \theta) ( \sin \theta - 1 + \cos \theta) . \sec \theta \ \mathrm{cosec} \theta = 2

Or

\sqrt{\frac{\sec \theta -1}{\sec \theta +1}} + \sqrt{\frac{\sec \theta +1}{\sec \theta -1}} = 2 \ \mathrm{cosec} \theta

Answer:

LHS = ( \sin \theta + 1 + \cos \theta) ( \sin \theta - 1 + \cos \theta) . \sec \theta \ \mathrm{cosec} \theta

= [ (\sin \theta  + \cos \theta )^2 -1 ] \sec \theta \ \mathrm{cosec} \theta 

= [ \sin^2 \theta  + \cos^2 \theta  + 2 \sin \theta \cos \theta -1] \sec \theta \ \mathrm{cosec} \theta

= [ 1 + 2 \sin \theta  \cos \theta  -1 ] \sec \theta \ \mathrm{cosec} \theta

= 2 \sin \theta  \cos \theta  \sec \theta \ \mathrm{cosec} \theta

= 2\  \frac{\sin \theta \cos \theta }{\sin \theta \cos \theta } 

= 2 . Hence proved.

Or

LHS = \sqrt{\frac{\sec \theta -1}{\sec \theta +1}} + \sqrt{\frac{\sec \theta +1}{\sec \theta -1}}

= \frac{(\sqrt{\sec \theta - 1})^2 + (\sqrt{\sec \theta + 1})^2}{\sqrt{(\sec \theta + 1)(\sec \theta - 1)}}

= \frac{\sec \theta - 1 + \sec \theta + 1}{\sqrt{\sec^2 \theta -1}} 

= \frac{2}{\cos \theta } \sqrt{\frac{1}{\frac{1}{\cos^2 \theta }-1}} 

= \frac{2}{\cos \theta } \sqrt{\frac{\cos^2 \theta}{1- \cos^2 \theta}} 

= 2 \ \frac{\cos \theta}{\cos \theta } \sqrt{\frac{1}{\sin^2 \theta}} 

= \frac{2}{\sin \theta}

= 2 \ \mathrm{cosec} \theta

hence proved.

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Question 14: In what ratio does the point P(- 4, y) divide the line segment joining the points A(- 6, 10) and B(3, - 8) ? Hence find the value of y .

Or

Find the value of p for which the points (- 5, 1), (1, p) and (4, - 2) are collinear.

Answer:

Let P divides AB in the ratio of k:1

\therefore P(-4, y) = \Big(  \frac{3k-6}{k+1} , \frac{-8k + 10}{k + 1}   \Big)

\Rightarrow -4 = \frac{3k-6}{k+1} 

\Rightarrow -4k - 4 = 3k - 6

\Rightarrow 7k = 2

\Rightarrow k = \frac{2}{7} 

Therefore ratio is 2:7

\therefore y = \frac{-8k + 10}{k + 1}

= \frac{-8(\frac{2}{7}) + 10}{(\frac{2}{7}) + 1}

= \frac{-16+70}{2 + 7}

= \frac{54}{9} = 6

Or

Given point A(-5, 1), B(1, p) and C(4, -2)

If the points are collinear are of the \triangle ABC = 0

\therefore 0 = \frac{1}{2} [ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) ]

\Rightarrow -5(p+2) + 1(-2-1) + 4(1-p) = 0

\Rightarrow -5p - 10 - 3 + 4 - 4p = 0

\Rightarrow -9p - 9 = 0

\Rightarrow p = -1

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Question 15: ABC is a right triangle in which \angle B = 90^o . If AB = 8 cm and BC = 6 cm, find the diameter of the circle inscribed in the triangle.

Answer:

AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = 10 cm2019-07-07_15-28-35

Area of  \triangle ABC = Area of \triangle AOB + Area of  \triangle BOC + Area of \triangle AOC

\frac{1}{2} \times BC \times AB =  \frac{1}{2} \times AB \times r  + \frac{1}{2} \times BC \times r = \frac{1}{2} \times AC \times r

\Rightarrow 6 \times 8 = 8r + 6r + 10r

\Rightarrow 48 = 24r

\Rightarrow r = 2 cm

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Question 16: In Figure 1, BL and CM are medians of a \triangle ABC right-angled at A . Prove that 4 (BL^2 + CM^2) = 5 BC^2 .

2019-05-24_9-17-16
Figure 1

Or

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer:

Given BL and CM are medians

\Rightarrow BM = MA \Rightarrow AM = \frac{1}{2} AB

\Rightarrow CL = AL \Rightarrow AL = \frac{1}{2} AC

From

\triangle ABC, BC^2 = AC^2 + AB^2    … … … … … i)

\triangle BAL, BL^2 = AL^2 + AB^2    … … … … … ii)

\triangle ACM, CM^2 = AC^2 + AM^2    … … … … … iii)

Adding ii) and iii)

BL^2 + CM^2 = AL^2 + AB^2 + AC^2 + AM^2

BL^2 + CM^2 = \Big( \frac{1}{2} AC \Big)^2 + AB^2 + AC^2 + \Big( \frac{1}{2} AB \Big)^2

BL^2 + CM^2 = \frac{1}{4} AC^2 + AB^2 + AC^2 + \frac{1}{4} AB^2

BL^2 + CM^2 = \frac{1}{4} (AC^2 + AB^2) + AB^2 + AC^2

4 (BL^2 + CM^2) = 5 BC^2

Hence Proved.

Or

Given: ABCD is a rhombus2019-07-07_15-22-52

To prove: AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2

We know that the diagonals of a rhombus bisect at right angles

Therefore from \triangle AOB, AB^2 = AO^2 + OB^2

\Rightarrow AB^2 = \Big( \frac{AC}{2} \Big)^2 + \Big( \frac{BD}{2} \Big)^2

\Rightarrow 4 AB^2 = AC^2 + BD^2

Since AB = BC = CD = DA

\Rightarrow AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2

Hence proved.

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Question 17: In Figure 2, two concentric circles with centre O , have radii 21 cm and 42 cm. If \angle  AOB = 60^o , find the area of the shaded region.

2019-05-24_9-00-10
Figure 2

Answer:

Area of shaded area = \frac{300}{360} \pi \times (42)^2 - \frac{300}{360} \pi \times (21)^2

= \frac{5}{6} \pi (42^2 - 21^2)

= \frac{5}{6} \times \frac{22}{7} \times 1323

= 3465 \ cm^2

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Question 18: Calculate the mode of the following distribution :

Class: 10-15 15-20 20-25 25-30 30-35
Frequency: 4 7 20 8 1

Answer:

2019-07-07_11-07-27.png

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Question 19: A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere and hence find the surface area of this sphere.

Or

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/hr, in how much time will the tank be filled ?

Answer:

Volume of cone = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 (24) = 288 \pi \ cm^3

Let the radius of the sphere be R

\therefore \frac{4}{3} \pi R^3 = 288 \pi

\Rightarrow R^3 = 3 \times 72 = 216

\Rightarrow R = 6 cm

Surface are of the sphere = 4 \pi R^2 = 4 \times \frac{22}{7} \times 36 = 452.57 \ cm^2

Or

Rate of flow of water = \frac{3 \times 1000}{60}  \frac{m}{min} = 50 \frac{m}{min}

Volume of tank = \pi r^2 h = \pi (5)^2 \times 2 = 50 \pi m^3

Volume of water flowing per minute = \pi (0.1)^2 \times 50 \frac{m^3}{min} = 0.5 \pi \frac{m^3}{min}

Therefore the time taken to fill the tank = \frac{50 \pi m^3}{0.5 \pi \frac{m^3}{min} } = 100 minutes

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Question 20: Prove that 2 + 3 \sqrt{3} is an irrational number when it is given that \sqrt{3} is an irrational number.

Answer:

Let us assume that 2 +3 \sqrt{3} is a rational number.

The rational number is in the form of \frac{p}{q}

\therefore 2 + 3\sqrt{3} = \frac{p}{q}

3\sqrt{3} = \frac{p}{q} - 2

3\sqrt{3} = \frac{p-2q}{q}

\sqrt{3} = \frac{p-2q}{3q}

\therefore \sqrt{3} is a rational no as \frac{p-2q}{3q} is a rational number

But it is given that \sqrt{3} is an irrational number which contradicts our initial assumption.

Hence 2 + 3 \sqrt{3} is an irrational number.

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Question 21: Sum of the areas of two squares is 157 \ m^2 . If the sum of their perimeters is 68 m, find the sides of the two squares.

Answer:

Let the sides of the two squares be a and b respectively

\therefore a^2 + b^2 = 157    … … … … … i)

Also 4a + 4 b = 68

\Rightarrow a + b = 17    … … … … … ii)

From i) (a+b)^2 - 2ab = 157

17^2 - 2ab = 157

289 - 157 = 2ab

ab = 66    … … … … … iii)

From ii) b = 17 - a

\therefore a (17-a) = 66

17a - a^2 -= 66

a^2 - 17a + 66 = 0

(a-11) (a-6) =0

a =11 or a = 6

When a = 11, b = 17 - 11 = 6

When a = 6, b = 17 - 6 = 11

Hence the sides of the two squares are 6 units and 11 units

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Question 22: Find the quadratic polynomial, sum and product of whose zeroes are -1  and - 20 respectively. Also find the zeroes of the polynomial so obtained.

Answer:

Sum of zeros: \alpha + \beta = 1

Product of zeros: \alpha \beta = -12

The quadratic polynomial is of the form:

x^2 - (sum of zeros) x + (product of zeros) = 0

x^2 - x - 12 = 0

\Rightarrow \alpha + \beta = 1 \Rightarrow \beta = (1-\alpha)

\alpha \beta = -12

\Rightarrow \alpha (1 - \alpha ) = -12

\Rightarrow \alpha - \alpha^2 = -12

\Rightarrow \alpha ^2 - \alpha  - 12 = 0

\Rightarrow (\alpha -4)(\alpha +3) = 0

\Rightarrow \alpha  = 4 or -3

When \alpha  = 4, \beta = 1-4 = -3

When \alpha  = -3, \beta = 1 - (-3) = 4

Therefore zeros are (4, -3)

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Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time, it has to increase its speed by 250 km/hr from its usual speed. Find the usual speed of the plane.

Or

Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 \ m^2 .

Answer:

Let the speed of the plane = x \frac{km}{hr}

New speed of the plane = (x + 250) \frac{km}{hr}

\therefore \frac{1500}{x} - \frac{1500}{x+250} = \frac{30}{60}

\frac{1500x + 1500 \times 250 - 1500 x}{x(x+250)} = \frac{1}{2}

1500 \times 250 \times 2 = x(x+250)

x^2 + 250 x = 750000

x^2 + 250x - 750000=0

x^2 +1000x - 750x - 750000 = 0

x(x+1000) - 750(x+1000) = 0

(x+1000)(x-750)=0

x = 750 or -1000 ( not possible as speed cannot be negative)

\therefore x = 750 \frac{km}{hr}

Or

Let length = l and breadth = b

\therefore lb = 200    … … … … … i)

2(l+b) = 60

l+b = 30    … … … … … ii)

From ii) b = (30-l)

Substituting in i)

l(30-l) = 200

30l - l^2 = 200

l^2 - 30l + 200 = 0

l^2 - 20l - 10l +200 = 0

l(l-20) - 10(l-20) = 0

(l-10)(l-20) = 0

When l = 10 m, b = 30 - 10 = 20 m

When l = 20 m, b = 30 -20 = 10 m

Hence the dimensions are 10 m by 20 m

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Question 24: Find the value of x , when in the A.P. given below 2 + 6 + 10 + \cdots + x = 1800 .

Answer:

First term of AP = a = 2

Common difference = d = 6-2 = 4

Sum of n terms = \frac{n}{2} [ 2a + (n-1)d]

\frac{n}{2} [ 2 \times 2 + (n-1)(4)] = 1800

n [ 2 + (n-1) 2 ] = 1800

n [ 2n + 2 - 2] = 1800

2n^2 = 1800

n^2 = 900

\Rightarrow n = 30 ( total number of terms)

x is the n^{th} term

\therefore x = a + (n-1) d

x = 2 + (30-1) (4)

x = 118

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Question 25: If \sec \theta + \tan \theta = m , show that \frac{m^2 -1 }{m^2 + 1} = \sin \theta

Answer:

\sec \theta  + \tan \theta  = m

\Rightarrow \sec^2 \theta + \tan^2 \theta  + 2 \sec \theta \ \tan \theta  = m^2

\Rightarrow \sec^2 \theta + \tan^2 \theta  + 2 \sec \theta \ \tan \theta - 1  = m^2 - 1

\because \tan^2 \theta  = \sec^2 \theta  - 1

\Rightarrow 2 \tan^2 \theta  + 2 \sec \theta \ \tan \theta  = m^2 - 1

\Rightarrow 2 \tan \theta  (\tan \theta  + \sec \theta ) = m^2 -1    … … … … … i)

Similarly,

\sec^2 \theta + \tan^2 \theta  + 2 \sec \theta \ \tan \theta + 1  = m^2 + 1

\because \sec^2 \theta  = \tan^2 \theta + 1

\Rightarrow 2 \sec^2 \theta  + 2 \sec \theta \ \tan \theta  = m^2 + 1

\Rightarrow 2 \sec \theta  (\tan \theta  + \sec \theta ) = m^2 +1  … … … … … ii)

Dividing i) by ii) we get

\frac{m^2-1}{m^2+1} = \frac{2 \tan \theta (\tan \theta  + \sec \theta) }{2 \sec \theta (\tan \theta  + \sec \theta)} 

\Rightarrow \frac{m^2-1}{m^2+1} = \sin \theta

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Question 26: In \triangle ABC (Figure 3), AD \perp BC . Prove that AC^2 = AB^2 + BC^2 - 2BC \times BD

2019-05-23_9-02-05
Figure 3

Answer:

Given AD \perp BC

In \triangle ABD \ \ \ \ \ AB^2 = AD^2 + BD^2 \Rightarrow AD^2 = AB^2 -BD^2    … … … … … i)

In \triangle ADC \ \ \ \ \ AC^2 = AD^2 + DC^2 \Rightarrow AD^2 = AC^2 -DC^2    … … … … … ii)

Frim i) and ii)

AB^2 - BD^2 = AC^2 -DC^2

\Rightarrow AC^2 = AB^2 + DC^2 - BD^2

\Rightarrow AC^2 = AB^2 + (DC - BD)(DC + BD)

\Rightarrow AC^2 = AB^2 + BC(DC-BD)

\Rightarrow AC^2 = AB^2 + BC [(BC-BD)-BD]

\Rightarrow AC^2 = AB^2 + BC[BC-2BD]

\Rightarrow AC^2=AB^2 + BC^2 - 2 BC \times BD

Hence Proved.

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Question 27: A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60^o to 45^o in 2 minutes. Find the speed of the boat in m/min.

Or

There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30^o and 60^o  respectively. Find the width of the river and height of the other pole.

Answer:

\frac{AB}{DB} = \tan 45^o = 1 \Rightarrow DB = AB = 150 m

\frac{AB}{BC} = \tan 60^o = \sqrt{3} \Rightarrow BC = \frac{150}{\sqrt{3}} = 50 \sqrt{3} 2019-07-07_15-12-27

\therefore DC = DB - CB = (150 - 50\sqrt{3} ) m

Therefore speed = \frac{distance}{time} = \frac{150 - 50\sqrt{3}}{2} \frac{m}{min}

= 75 - 25(1.732) \frac{m}{min}

= 31.7 \frac{m}{min}

Or

\frac{AB}{CB} = \tan 60^o = \sqrt{3} 2019-07-07_15-18-28

\Rightarrow CB = \frac{60}{\sqrt{3}} = 20\sqrt{3} m (width of the river)

\frac{AE}{20\sqrt{3}} = \tan 30^o \Rightarrow AE = \frac{20\sqrt{3}}{\sqrt{3}} = 20 m

Hence the height of the pole is 60-20 = 40 m

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Question 28: Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \frac{3}{5} of the corresponding sides of the first triangle.

Answer:

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Question 29: Calculate the mean of the following frequency distribution :

Class: 10-30 30-50 50-70 70-90 90-110 110-130
Frequency: 5 8 12 20 3 2

Or

The following table gives production yield in kg per hectare of wheat of 100 farms of a village :

Production Yield

(kg/hectare):

40-45 45-50 50-55 55-60 60-65 65-70
Number of Farms: 4 6 16 20 30 24

Change the distribution to a ‘more than type’ distribution, and draw its ogive.

Answer:

Class Interval Frequency ( f_i ) x_i f_ix_i
10-30 5 20 100
30-50 8 40 320
50-70 12 60 720
70-90 20 80 1600
90-110 3 100 300
110-130 2 120 2400
\Sigma f_i = 50 \Sigma f_ix_i = 3280

Mean = \frac{\Sigma f_ix_i}{\Sigma f_i} = \frac{3280}{50} = 65.6

Or

Production Yield Cumulative Frequency
More than 40 100
More than 45 96
More than 50 90
More than 55 74
More than 60 54
More than 65 24

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Question 30: A container opened at the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16   cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container, at the rate of Rs. 50 per liter. Also find the cost of metal sheet used to make the container, if it costs Rs. 10 per 100 \ cm^2 . (Take \pi = 3.14 )

Answer:2019-07-07_15-09-15

h = 16 cm, r_1 = 8 cm, r_2 = 20 cm

l = \sqrt{h^2 + (r_2-r_1)^2} = \sqrt{254 + 144} = 20 cm

Volume of frustum = \frac{1}{3} \pi h ({r_1}^2 + {r_2}^2 + r_1 r_2 ) 

= \frac{1}{3} \times 3.14 \times 16 (64+400+160)

= 10449.94 \ cm^3

= 10.4492 liters

Cost of milk = 50 \times 10.45 = 522.5 Rs

Surface Area of the container = \pi l (r_1+r_2) + \pi r_1^2

= 3.14 \times 20 (8 + 20+ 3.14 (8)^2

= 1758.4 + 200.96

= 1959.36 \ cm^2

Cost of metal sheet = \frac{10 Rs}{100 cm^2} = 0.1 \frac{Rs}{cm^2} 

Cost of sheet metal = 1959.36 \ cm^2 \times 0.1 \frac{Rs}{cm^2} = 195.936 Rs

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