CBSE Class 10 Mathematics Previous Year Question Paper 2019 All India Set-3

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Q 1. In Figure 1, PS = 3 cm. QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, ∠PQ ⏊ RQ and RQ = 9 cm. Evaluate tan θ.

 OR

If tan ⍺ = begin mathsize 12px style 5 over 12 end style, find the value of sec ⍺.

 

Q 2. Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

 

Q 3. Find the value(s) of x, if the distance between the point A(0,0) and B(x, -4) is 5 units.

 

Q 4. Find after how many places of decimal the decimal form of the begin mathsize 12px style fraction numerator 27 over denominator 2 cubed.5 to the power of 4.3 squared end fraction end style number will terminate.

OR

Express 429 as a product of its prime factors.

 

Q 5. Write the discriminant of the quadratic equation (x + 5)2 = 2 (5x – 3).

 

Q 6. Find the sum of the first 10 multiples of 3.

 

Q 7. If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n.

OR

On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. When is the minimum distance each should walk so that each can cover the same distance in complete steps?

 

Q 8. A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number.

 

Q 9. Using completing the square method, show that the equation x2 – 8x + 18 = 0 has no solution.

 

Q 10. Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7?

 

Q 11. Solve the following pair of linear equations:

3x + 4y = 10

2x – 2y = 2

 

Q 12. Points A(3, 1), B(5, 1) C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.

OR

Point P and Q trisect the line segment joining the point A(-2, 0) and B(0, 8) such that P is near to A. Find the coordinates of point P and Q.

 

Q 13. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days:  0 - 6 6-12 12-18 18-24 24-30 30-36 36-42
Number of students: 10 11   4


Q 14. In Figure 2, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

OR

Prove that opposite side of a quadrilateral circumscribing a circle subtend supplementary angle at the centre of the circle.

 

Q 15. A, B and C are interior angles of a triangle ABC. Show that

(i) begin mathsize 12px style sin open parentheses fraction numerator straight B plus straight c over denominator 2 end fraction close parentheses equals cos straight A over 2 end style

(ii) begin mathsize 12px style If space angle straight A equals 90 degree comma space then space find space the space value space of space tan space open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses. end style

OR

begin mathsize 12px style If space tan space open parentheses straight A plus straight B close parentheses equals 1 space and space tan open parentheses straight A minus straight B close parentheses equals fraction numerator 1 over denominator square root of 3 end fraction comma space 0 degree space less than straight A space plus space straight B space less than 90 degree comma space straight A space greater than space straight B comma space then space find space the space values space of space straight A space and space straight B. end style

Q 16. Prove that  begin mathsize 12px style square root of 3 end style is an irrational number.

OR

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

 

Q 17. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the values of x and y which satisfy both the equations.

 

Q 18. Water in a cannel, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed?

 

Q 19. The perpendicular from A on side BC of a ∆ABC meets BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2.

OR

AD and PM are medians of triangles ABC and PQR respectively where ∆ ABC ∼ ∆ PQR. Prove that begin mathsize 12px style AB over PQ equals AD over PM end style.

 

Q 20. A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. begin mathsize 12px style open parentheses Use space straight pi equals 22 over 7 space and space square root of 3 equals 1.73 close parentheses end style


Q 21. Find the value of k so that area of triangle ABCD with A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

 

Q 22.  If  and -3 are the zeroes of the polynomial ax2 + 7x + b, then find the values of a and b.

 

Q 23. Change the following distribution to a ‘more than type’ distribution.

Hence draw the ‘more than type’ ogive for this distribution.

Class interval: 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Frequency: 10 8 12 24 6 25 15

 

Q 24. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower. (Given begin mathsize 12px style square root of 3 end style= 1.732)

 

Q 25. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

OR

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

 

Q 26. If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the (m + n)th term of the A.P. is zero.

OR

The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.

 

Q 27. In Figure3, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. Find begin mathsize 12px style open parentheses Take space straight pi space equals space 22 over 7 close parentheses end style

(a) the total surface area of the block.

(b) the volume of the block formed. 

OR

A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use 𝜋 = 3.14)

Q 28. Construct a triangle, the lengths of whose sides are 5 cm, 6 cm and 7 cm. Now construct another triangle whose sides begin mathsize 12px style 5 over 7 end style are  times the corresponding sides of the first triangle.

 

Q 29. Prove that :

begin mathsize 12px style fraction numerator tan cubed straight theta over denominator 1 plus tan squared straight theta end fraction plus fraction numerator cot cubed straight theta over denominator 1 plus cot squared straight theta end fraction equals secθ space space cosecθ minus 2 sinθ space cosθ. end style

 

Q 30. A motorboat whose speed in still water is 9 km/h, goes 15 km downstream and comes to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.


 Q 1. In Figure 1, PS = 3 cm. QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, ∠PQ ⏊ RQ and RQ = 9 cm. Evaluate tan θ.

OR

If tan ⍺ = begin mathsize 12px style 5 over 12 end style, find the value of sec ⍺.

Solution:

In the figure we have a right angled triangle ΔPSQ

So on applying Pythagoras theorem we get

∴ PQ2 = PS2 + QS2

begin mathsize 12px style therefore PQ equals square root of 3 squared plus 4 squared end root
space space space space space space space space space equals space square root of 25
space space space space space space space space space equals space 5 end style

In ΔPQR

begin mathsize 12px style therefore tan space straight theta equals PQ over QR
therefore tan space straight theta equals 5 over 9 end style

OR

 begin mathsize 12px style Given colon
tan space straight alpha space equals space 5 over 12
we space know comma
sec squared space straight alpha space equals 1 plus tan squared space straight alpha
sec space straight alpha equals square root of 1 plus open parentheses 5 over 12 close parentheses squared end root
sec space straight alpha equals square root of fraction numerator 144 plus 25 over denominator 144 end fraction end root equals square root of 169 over 144 end root equals 13 over 12
sec space straight alpha equals 13 over 12 end style

 

Q 2. Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:

It is given that a > b,

AB is the tangent to the smaller circle,

OB ⏊ AB…..(tangent  ⏊ radius)

So ΔABO is right angled triangle

∴ AO2 = AB2 + BO2

begin mathsize 12px style therefore space AB equals square root of straight a squared minus straight b squared end root end style

Now we know that the perpendicular drawn from the center to a chord bisects the chord so

The length of chord will be 2AB =begin mathsize 12px style 2 square root of straight a squared minus straight b squared end root end style

Q 3. Find the value(s) of x, if the distance between the point A(0,0) and B(x, -4) is 5 units.

Solution:

A(0, 0) and B(x, -4)

The distance AB is given by distance formula

begin mathsize 12px style AB equals square root of open parentheses straight x minus 0 close parentheses squared plus open parentheses negative 4 minus 0 close parentheses squared end root
AB equals square root of straight x squared plus 16 end root
Now comma
AB equals 5 space Units
therefore space 5 equals square root of straight x squared plus 16 end root
squaring
therefore 25 equals straight x squared plus 16
therefore space straight x squared space equals space 9
therefore space straight x space equals plus-or-minus space 3 end style


Q 4. Find after how many places of decimal the decimal form of the begin mathsize 12px style fraction numerator 27 over denominator 2 cubed.5 to the power of 4.3 squared end fraction end style number will terminate.

OR

Express 429 as a product of its prime factors.

Solution:

begin mathsize 12px style fraction numerator 27 over denominator 2 cubed.5 to the power of 4.3 squared end fraction
equals fraction numerator 3 cubed over denominator 2 cubed.5 to the power of 4.3 squared end fraction
equals fraction numerator 3 over denominator 2 cubed.5 to the power of 4 end fraction end style

Now power of 5 is 4 and that of 2 is 3

∴  4 > 3, so the decimal will terminate after 4 places

OR

∴ 429 = 3 × 11 × 13

 

Q 5. Write the discriminant of the quadratic equation (x + 5)2 = 2 (5x – 3).

Solution:

begin mathsize 12px style open parentheses straight x plus 5 close parentheses squared equals 2 open parentheses 5 straight x minus 3 close parentheses
therefore space straight x squared plus 10 straight x plus 25 space equals space 10 straight x minus 6
therefore space straight x squared plus 31 space equals 0
discriminat space equals space straight b squared minus space 4 ac
therefore space straight b squared minus space 4 ac equals 0 squared minus 4 open parentheses 1 close parentheses open parentheses 31 close parentheses equals negative 124 end style


Q 6. Find the sum of the first 10 multiples of 3.

Solution:

The multiples of 3 are

3, 6, 9, 12…

The above series is in arithmetic progression

∴ a = 3 and d = 3

We need sum of 10 multiples

begin mathsize 12px style straight S subscript straight n equals straight n over 2 open parentheses 2 straight a plus open parentheses straight n minus 1 close parentheses straight d close parentheses
therefore space straight S subscript 10 equals 10 over 2 open parentheses 2 cross times 3 plus open parentheses 10 minus 1 close parentheses 3 close parentheses
therefore space straight S subscript 10 equals 165 end style

 

Q 7. If HCF of 65 and 117 is expressible in the form 65n – 117, then find the value of n.

OR

On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. When is the minimum distance each should walk so that each can cover the same distance in complete steps?

Solution:

65 = 5 × 13

117 = 3 × 3 × 13

HCF = 13

Given that

HCF = 65n - 117

∴ 13 = 65n - 117

∴  130 = 65n

∴ n = 2

OR

We have to find the LCM of 30 cm, 36 cm and 40 cm to get the required minimum distance. Because we are asked the minimum distance

 Now,

 30 = 3 × 2 × 5,

 36 = 2 × 3 × 2 × 3 

40 = 2 × 2 × 2 × 5

∴ LCM (30, 36, 40) = 23 × 32 × 5 = 360

Minimum distance each should walk 360 cm. so that, each can cover the same distance in complete steps.

 

Q 8. A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number.

Solution:

A die is thrown,

So sample space S is given by

∴ S = {1, 2, 3, 4, 5, 6}

∴ n(S) = 6

Let A be the event such that composite number shows up

∴ A = {4, 6}

∴ n(A) = 2

(i) Probability of getting composite number

 begin mathsize 12px style therefore space straight P open parentheses straight A close parentheses equals 2 over 6 equals 1 third end style

Let B be the event such that prime number shows up

∴ B = {2, 3, 5}

∴ n(B) = 3

(ii) Probability of getting prime number

 begin mathsize 12px style therefore space straight P open parentheses straight B close parentheses equals 3 over 6 equals 1 half end style

 

Q 9. Using completing the square method, show that the equation x2 – 8x + 18 = 0 has no solution.

Solution:

The given quadratic equation is

begin mathsize 12px style straight x squared minus 8 straight x plus 18 space equals 0
The space above space equation space can space be space written space as
therefore space straight x squared minus 8 straight x plus 16 minus 16 plus 18 equals space 0
therefore space straight x squared minus 8 straight x space plus 16 equals negative 2
therefore space open parentheses straight x minus 4 close parentheses squared equals negative 2
therefore space straight x minus 4 equals square root of negative 2 end root end style

Which is not possible, so the given quadratic equation has no roots.

 

Q 10. Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7?

Solution:

Poonam selects a card in random from 7 to 40

So the sample space S is

∴ S = {7, 8, 9, 10, ……40}

∴ n(S) = 34

Let A be the event of getting multiple of 7

∴ A = {7, 14, 21, 28, 35}

∴ n(A) = 5

Probability of getting multiple of 7

begin mathsize 12px style therefore space straight P open parentheses straight A close parentheses equals 5 over 34 end style

Q 11. Solve the following pair of linear equations:

3x + 4y = 10

2x – 2y = 2

Solution:

3x + 4y = 10    …. (i)

2x – 2y = 2       …. (ii)

Multiply equation (ii) by 2, we get

4x – 4y = 4       …. (iii)

Adding equation (i) and (iii), we get

∴ 7x = 14

→ x = 2

Put x = 2 in the equation (ii), we get

∴ 2(2) – 2y = 2    

→ 2y = 4 – 2

→ y = begin mathsize 12px style 2 over 2 equals 1 end style

Hence the solutions is x = 2 and y = 1.

 

Q 12. Points A(3, 1), B(5, 1) C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.

OR

Point P and Q trisect the line segment joining the point A(-2, 0) and B(0, 8) such that P is near to A. Find the coordinates of point P and Q.

Solution:

Points A (3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of parallelogram

We know that diagonals of parallelogram bisect each other

The M be the midpoint of both AC and BD

So by midpoint formula

begin mathsize 12px style therefore space straight M equals open parentheses fraction numerator 3 plus straight a over denominator 2 end fraction plus fraction numerator 1 plus straight b over denominator 2 end fraction close parentheses
also comma
straight M equals open parentheses fraction numerator 5 plus 4 over denominator 2 end fraction plus fraction numerator 1 plus 3 over denominator 2 end fraction close parentheses equals open parentheses 9 over 2 comma space 2 close parentheses
so comma
therefore open parentheses fraction numerator 3 plus straight a over denominator 2 end fraction plus fraction numerator 1 plus straight b over denominator 2 end fraction close parentheses equals open parentheses 9 over 2 comma space 2 close parentheses
therefore fraction numerator 3 plus straight a over denominator 2 end fraction equals 9 over 2
therefore straight a equals 6
therefore fraction numerator 1 plus straight b over denominator 2 end fraction equals 2
straight b equals 3 end style

OR

P and Q trisect AB, given that A (-2, 0) and B (0, 8)

Now P will divide AB in ratio 1: 2

So by using section formula we get

begin mathsize 12px style straight P open parentheses straight x comma space straight y close parentheses equals open parentheses fraction numerator 2 straight x minus 2 plus 1 cross times 0 over denominator 1 plus 2 end fraction comma fraction numerator 2 cross times 0 plus 1 cross times 8 over denominator 1 plus 2 end fraction close parentheses
therefore straight P open parentheses straight x comma space straight y close parentheses equals open parentheses fraction numerator negative 4 over denominator 3 end fraction comma 8 over 3 close parentheses end style

Now P is the midpoint of AQ

Let coordinates of Q be (a, b)

So by using midpoint theorem

begin mathsize 12px style straight P open parentheses fraction numerator negative 4 over denominator 3 end fraction comma 8 over 3 close parentheses equals open parentheses fraction numerator negative 2 plus straight a over denominator 2 end fraction comma fraction numerator 0 plus straight b over denominator 2 end fraction close parentheses
fraction numerator negative 4 over denominator 3 end fraction equals fraction numerator negative 2 plus straight a over denominator 2 end fraction
therefore space straight a equals fraction numerator negative 2 over denominator 3 end fraction
8 over 3 equals straight b over 2
therefore space straight b equals 16 over 3
therefore The space cordinates space of space straight Q space are space open parentheses fraction numerator negative 2 over denominator 3 end fraction comma 16 over 3 close parentheses end style

 

Q 13. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

 

Number of days:  0 - 6 6-12 12-18 18-24 24-30 30-36 36-42
Number of students: 10 11   4

 Solution:

We will use direct method to find the mean:

No. of days No. of students Class marks(xi)     fixi
0-6 10  begin mathsize 12px style fraction numerator 0 plus 6 over denominator 2 end fraction equals 3 end style  30
6-12 11  begin mathsize 12px style fraction numerator 6 plus 12 over denominator 2 end fraction equals 9 end style 99 
12-18 7 begin mathsize 12px style fraction numerator 12 plus 18 over denominator 2 end fraction equals 15 end style 105 
18-24 4 begin mathsize 12px style fraction numerator 18 plus 24 over denominator 2 end fraction equals 21 end style  84 
24-30 4 begin mathsize 12px style fraction numerator 24 plus 30 over denominator 2 end fraction equals 27 end style  108 
30-36 3 begin mathsize 12px style fraction numerator 30 plus 36 over denominator 2 end fraction equals 33 end style  99 
36-42 1 begin mathsize 12px style fraction numerator 36 plus 42 over denominator 2 end fraction equals 39 end style  39 
   Σf1 = 40    Σf1x1 = 564

begin mathsize 12px style Mean space open parentheses straight x with bar on top close parentheses equals fraction numerator sum for blank of straight f subscript straight i straight x subscript i over denominator sum for blank of straight f subscript i end fraction
space space space space space space space space space space space space space space space equals 564 over 40
space space space space space space space space space space space space space space equals space 14.1 end style

Hence, the number of days a student was absent is 14.1

 

Q 14. In Figure 2, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

OR

Prove that opposite side of a quadrilateral circumscribing a circle subtend supplementary angle at the centre of the circle.

Solution:

Let TP = x.

In ΔPRT and ΔQRT

PT = TQ …(tangents drawn from external point are equal in length)

RT ….common side

PTR = QTR…(the line joining the center of circle and the external point bisects the angle between the tangents, which are drawn from that external point)

ΔPRT approximately equal to ΔQRT…( SAS test )

PRT = QRT…( C.P.C.T.)

Also,

PRT = QRT = 90o…(linear pair)
We know that the perpendicular drawn from the centre of the circle to a chord bisects it.
Therefore, PR = RQ = 4 cm and OR perpendicular PQ
Using Pythagoras theorem in right angled triangle PRO, we have
begin mathsize 12px style open parentheses OP close parentheses squared equals open parentheses PR close parentheses squared plus open parentheses OR close parentheses squared
rightwards double arrow 5 squared equals 4 squared plus open parentheses OP close parentheses squared
rightwards double arrow open parentheses OR close parentheses squared equals 5 squared minus 4 squared
rightwards double arrow open parentheses OR close parentheses squared equals 25 minus 16 equals 9
rightwards double arrow OR space equals space 3 space cm
end style
Now, again using Pythagoras theorem in triangle PRT we have

(PT)2 = (PR)2 + (RT)2 

→ x2 = 42 + RT2 ....(i)

Since we know that the radius drawn to tangent is perpendicular to the tangent so, we have
In right triangle OPT,

begin mathsize 12px style open parentheses OT close parentheses squared equals open parentheses OP close parentheses squared plus open parentheses PT close parentheses squared
rightwards double arrow open parentheses OT close parentheses squared equals 5 squared plus straight x squared
rightwards double arrow open parentheses OR plus RT close parentheses squared equals 5 squared plus straight x squared
rightwards double arrow open parentheses 3 plus RT close parentheses squared equals 5 squared plus 4 squared plus RT squared space.... Using space left parenthesis straight i right parenthesis
rightwards double arrow 9 plus RT squared plus 6 RT space equals space 25 plus 16 plus RT squared
rightwards double arrow 9 plus 6 RT space equals space 41
rightwards double arrow 6 RT space equals space 32
rightwards double arrow space RT space equals space 16 over 3 end style

Again using (i), we have

Error converting from MathML to accessible text.

OR

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S.
Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒ ∠7 = ∠8
Similarly we get,
∠4 = ∠3

∠5 = ∠6

∠1 = ∠2

Adding all these angles,

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360º
⇒ ∠1 + ∠1 + ∠4 + ∠4 + ∠5 + ∠5 + ∠8 + ∠8 = 360º
⇒ 2 ∠1 + 2 ∠8 + 2 ∠4 + 2 ∠5 = 360º
⇒ 2(∠1 + ∠8) + 2(∠4 + ∠5) = 360º
⇒ (∠1 + ∠8) + (∠4 + ∠5) = 180º
⇒ ∠AOD + ∠COB = 180º
Similarly, we can prove that ∠AOB + ∠COD = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.


Q 15. A, B and C are interior angles of a triangle ABC. Show that

(i) begin mathsize 12px style sin open parentheses fraction numerator straight B plus straight c over denominator 2 end fraction close parentheses equals cos straight A over 2 end style

(ii) begin mathsize 12px style If space angle straight A equals 90 degree comma space then space find space the space value space of space tan space open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses. end style

OR

begin mathsize 12px style If space tan space open parentheses straight A plus straight B close parentheses equals 1 space and space tan open parentheses straight A minus straight B close parentheses equals fraction numerator 1 over denominator square root of 3 end fraction comma space 0 degree space less than straight A space plus space straight B space less than 90 degree comma space straight A space greater than space straight B comma space then space find space the space values space of space straight A space and space straight B. end style

Solution:

(i) In triangle ABC, sum of all interior angles is 180°

begin mathsize 12px style rightwards double arrow straight A plus straight B plus straight C equals 180 degree
rightwards double arrow straight B plus straight C equals 180 degree minus straight A space space... left parenthesis straight a right parenthesis
rightwards double arrow fraction numerator straight B plus straight C over denominator 2 end fraction equals fraction numerator 180 degree minus straight A over denominator 2 end fraction
rightwards double arrow sin open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses equals sin open parentheses fraction numerator 180 degree minus straight A over denominator 2 end fraction close parentheses
rightwards double arrow sin open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses equals sin open parentheses 90 degree minus straight A over 2 close parentheses
Since comma space sin space open parentheses 90 degree minus straight theta close parentheses equals cos space straight theta
rightwards double arrow sin open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses equals cos open parentheses straight A over 2 close parentheses end style

(ii) Using equation (a), we have

begin mathsize 12px style straight B plus straight C space equals space 180 degree minus straight A
Since comma space angle straight A space equals 90 degree
rightwards double arrow straight B plus straight C space equals 180 degree minus 90 degree
rightwards double arrow straight B plus straight C space equals space 90 degree
rightwards double arrow fraction numerator straight B plus straight C over denominator 2 end fraction equals 45 degree
rightwards double arrow tan open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses space equals space tan space 45 degree
rightwards double arrow tan open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses space equals space 1 end style

OR

begin mathsize 12px style Given colon space tan space open parentheses straight A plus straight B close parentheses equals 1 space... left parenthesis straight i right parenthesis
And space tan space open parentheses straight A minus straight B close parentheses equals fraction numerator 1 over denominator square root of 3 end fraction space... left parenthesis ii right parenthesis
Using space first space equation comma space we space have
tan space open parentheses straight A plus straight B close parentheses equals 1 equals tan space 45 degree
rightwards double arrow straight A plus straight B space equals 45 degree space... left parenthesis iii right parenthesis
Since space straight A greater than straight B comma space so space straight A minus straight B greater than 0
Now comma space using space left parenthesis ii right parenthesis
tan space open parentheses straight A minus straight B close parentheses equals fraction numerator 1 over denominator square root of 3 end fraction equals tan space 30 degree
rightwards double arrow straight A minus straight B equals 30 degree space... left parenthesis iv right parenthesis
Solving space left parenthesis iii right parenthesis space and space left parenthesis iv right parenthesis comma space we space have
2 straight A equals 75 degree rightwards double arrow straight A equals 37.5 degree
Thus comma space we space have space straight B equals 7.5 degree end style


Q 16. Prove that  begin mathsize 12px style square root of 3 end style is an irrational number.

OR

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

Solution:

Prove that begin mathsize 12px style square root of 3 end style is an irrational number.
Let us assume that begin mathsize 12px style square root of 3 end style is a rational number.
That is, we can find integers a and b (b ≠ 0) such that begin mathsize 12px style square root of 3 equals straight a over straight b end style
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
begin mathsize 12px style square root of 3 straight b equals straight a space rightwards double arrow 3 straight b squared space equals straight a squared end style(Squaring on both the sides) …. (i)
Therefore, a2 is divisible by 3
Hence ‘a’ is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
begin mathsize 12px style 3 straight b squared equals open parentheses 3 straight c close parentheses squared
rightwards double arrow 3 straight b squared equals 9 straight c squared
therefore straight b squared equals 3 straight c squared end style
This means that b2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that begin mathsize 12px style square root of 3 end style is rational.
So, we conclude that begin mathsize 12px style square root of 3 end style is irrational.

OR

Let the largest number which divides the given numbers be x.

Since, it leaves the remainders 1, 2 and 3 while dividing 1251, 9377 and 15628
So, the numbers which are divisible by x are:
1251 – 1 =1250,  9377 – 2 =9375  and  15628 – 3 = 15625
Here, x is the largest number dividing these 3 numbers so x becomes the GCD or HCF
Now, 1250 = 54 × 2
9375 = 55 × 3 and 15625 = 56
Therefore x = HCF (1250, 9375, 15625) = 5= 625.
Hence, the required number is 625.

 

Q 17. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Using this graph, find the values of x and y which satisfy both the equations.

Solution:

The given equations are:
x - y + 1 = 0…… (i)
3x + 2y - 12 = 0…… (ii)
To draw the graph of these equations, let us find some points lying on them:
For equation (i)

x 0 1 -1 2
y 1 2 0 3

Plot these points i.e. (0, 1), (1, 2), (-1, 0) and (2, 3) on the graph
Now, draw a straight line connecting them and extend the line.

For equation (ii)

x 0 1 4 2
y 6 begin mathsize 12px style 9 over 2 end style 0 3

Plot these points i.e. (0, 6), begin mathsize 12px style open parentheses 1 comma 9 over 2 close parentheses end style , (4, 0) and (2, 3) on the graph
Now, draw a straight line connecting them and extend the line.

Clearly, the two lines intersect at a point (2, 3).
Thus, x = 2 and y = 3 satisfy both the equations.

Q 18. Water in a cannel, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed?

Solution:

Speed of flowing of water from canal is 10 km/h

This means the length of water flows in 1 hour = 10 km
→ length of water flows in 30 minutes = 5 km

Now,
volume of water flowing from canal = length of water flows in 30 minutes × breadth of canal × depth of canal

= 5000 m × 6 m × 1.5 m = 45000 m³

Let the area of field = x m²

Then, volume of water irrigates into the field = area of field × height of water during irrigation

= x m² × 8× 10⁻² m = 0.08x m³

Now, volume of water flowing from canal = volume of water in field

45000 m³ = 0.08x m³

x = 562500 m²

We know, 1 hectare = 10000 m²

So, area of field in hectare = 562500/10000 = 56.25 hectares.


Q 19. The perpendicular from A on side BC of a ∆ABC meets BC at D such that DB = 3CD. Prove that 2AB2 = 2AC2 + BC2.

OR

AD and PM are medians of triangles ABC and PQR respectively where ∆ ABC ∼ ∆ PQR. Prove that begin mathsize 12px style AB over PQ equals AD over PM end style.

Solution:

Given that in ΔABC, we have

AD ⊥ BC and BD = 3CD
Now, using Pythagoras theorem in right angle triangles ADB and ADC, we have

begin mathsize 12px style AB squared equals space AD squared plus BD squared space... left parenthesis straight i right parenthesis
AC squared equals space AD squared plus DC squared space... left parenthesis ii right parenthesis end style

Subtracting equation (ii) from equation (i), we get

begin mathsize 12px style AB squared space minus space AC squared space equals space BD squared space minus space DC squared
equals open parentheses 3 CD close parentheses squared space minus space DC squared
equals 9 space DC squared minus DC squared
equals 8 space DC squared
Since comma space BC equals BD plus DC
rightwards double arrow BC equals 3 DC plus DC
rightwards double arrow BC equals 4 DC
rightwards double arrow DC equals BC over 4
Then comma space we space have
AB squared minus AC squared equals 8 open parentheses BC over 4 close parentheses squared
rightwards double arrow AB squared minus AC squared space equals BC squared over 2 space
rightwards double arrow 2 AB squared minus 2 AC squared equals BC squared
rightwards double arrow 2 AB squared equals 2 AC squared plus BC squared
Hence comma space proved. end style

OR

Given: AD and PM are medians of triangle ABC and PQR respectively.

begin mathsize 12px style rightwards double arrow BD equals DC equals 1 half BC space... left parenthesis straight i right parenthesis
QM equals MR equals 1 half QR space... left parenthesis ii right parenthesis
Since space triangle ABC space tilde triangle PQR comma space using space similarity space criterion
angle straight A equals angle straight P comma space angle straight B equals angle straight Q space and space angle straight C equals angle straight R
Also comma space AB over PQ equals BC over QR equals AC over PR
rightwards double arrow AB over PQ equals BC over QR equals fraction numerator 2 BD over denominator 2 QM end fraction equals BD over QM space.... space using space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
rightwards double arrow PQ over AB equals QM over BD
Therefore space in space tringles space ABD space and space PQM comma
AB over PQ equals BD over QM and space angle straight B space equals space angle straight Q
therefore triangle ABD space tilde triangle PQM space... By space SAS
therefore space AB over PQ equals AD over PM space.... open parentheses Since comma space sides space of space similar space triangles space are space in space proportion close parentheses
Hence comma space proved.
end style


Q 20. A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. begin mathsize 12px style open parentheses Use space straight pi equals 22 over 7 space and space square root of 3 equals 1.73 close parentheses end style

Solution:


Radius of the circle = OA = OB = 14 cm

Let θ be the angle at the centre = 60°
Now in  ΔAOB,

begin mathsize 12px style angle straight A equals angle straight B space.... left parenthesis Angles space opposite space to space equal space sides space are space equal right parenthesis end style

By the angle sum property of triangle,

begin mathsize 12px style angle straight A plus angle straight B plus angle straight O equals 180 degree
straight x space plus straight x space plus space 60 degree space equals space 180 degree
rightwards double arrow space 2 straight x space equals space 120 degree
rightwards double arrow space straight x space equals 60 degree end style

Therefore, all the angles of a triangle AOB is  60°
So, ΔAOB is equilateral

→ OA = OB = AB

Area of minor segment = Area of sector – Area of triangle AOB

begin mathsize 12px style equals fraction numerator theta over denominator 360 degree end fraction cross times pi r squared minus fraction numerator square root of 3 over denominator 4 end fraction open parentheses A B close parentheses squared
equals fraction numerator 60 degree over denominator 360 degree end fraction cross times pi open parentheses 14 close parentheses squared minus fraction numerator square root of 3 over denominator 4 end fraction open parentheses 14 close parentheses squared
equals 1 over 6 cross times 22 over 7 cross times 14 cross times 14 minus fraction numerator square root of 3 over denominator 4 end fraction cross times 14 cross times 14
equals 308 over 3 minus 49 square root of 3
equals fraction numerator 308 minus 49 square root of 3 over denominator 3 end fraction space c m cubed end style

 

Q 21. Find the value of k so that area of triangle ABCD with A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

Solution: 

Given: A(k + 1, 1), B(4, -3) and C(7, - k)
Area of triangle ABD is 6 square units
We know that the area of triangle is given by

begin mathsize 12px style straight A equals 1 half open square brackets straight x subscript 1 open parentheses straight y subscript 2 minus straight y subscript 3 close parentheses plus straight x subscript 2 open parentheses straight y subscript 3 minus straight y subscript 1 close parentheses plus straight x subscript 3 open parentheses straight y subscript 1 minus straight y subscript 2 close parentheses close square brackets
rightwards double arrow 6 equals 1 half open square brackets open parentheses straight k plus 1 close parentheses open parentheses negative 3 plus straight k close parentheses plus 4 open parentheses negative straight k minus 1 close parentheses plus 7 open parentheses 1 plus 3 close parentheses close square brackets
rightwards double arrow 12 equals open parentheses straight k plus 1 close parentheses open parentheses negative 3 plus straight k close parentheses plus 4 open parentheses negative straight k minus 1 close parentheses plus 7 open parentheses 1 plus 3 close parentheses
rightwards double arrow 12 equals negative 3 straight k plus straight k squared minus 3 plus straight k minus 4 straight k minus 4 plus 28
rightwards double arrow 12 equals straight k squared minus 6 straight k plus 21
rightwards double arrow straight k squared minus 6 straight k plus 9 equals 0
rightwards double arrow open parentheses straight k minus 3 close parentheses squared equals 0
rightwards double arrow straight k equals 3 end style

 

Q 22.  If  and -3 are the zeroes of the polynomial ax2 + 7x + b, then find the values of a and b.

Solution:

begin mathsize 12px style Given space polynomial space is space straight p open parentheses straight x close parentheses equals ax squared plus 7 straight x plus straight b
2 over 3 space and space minus 3 space are space the space zeroes space of space straight p open parentheses straight x close parentheses
rightwards double arrow straight p open parentheses 2 over 3 close parentheses equals 0 space and space straight p open parentheses negative 3 close parentheses equals 0
rightwards double arrow straight a open parentheses 2 over 3 close parentheses squared plus 7 open parentheses 2 over 3 close parentheses plus straight b equals 0
rightwards double arrow fraction numerator 4 straight a over denominator 9 end fraction plus 14 over 3 plus straight b equals 0
rightwards double arrow 4 straight a plus 9 straight b plus 42 equals 0 space... open parentheses straight i close parentheses
Also comma space straight a open parentheses negative 3 close parentheses squared plus 7 open parentheses negative 3 close parentheses plus straight b equals 0
rightwards double arrow 9 straight a plus straight b minus 21 equals 0 space... open parentheses ii close parentheses end style

Solving (i) and (ii) simultaneously, we get

a = 3
Substituting the value of a in (ii), we get

b = -6

 

Q 23. Change the following distribution to a ‘more than type’ distribution.

Hence draw the ‘more than type’ ogive for this distribution.

Class interval: 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Frequency: 10 8 12 24 6 25 15

Solution:

Intervals f cf(more than type)  
20-30 10 100 (30, 100)
30-40 8 90 (40, 90)
40-50 12 82 (50, 82)
50-60 24 70 (60, 70)
60-70 6 46 (70, 46)
70-80 25 40 (80, 40)
80-90 15 15 (90, 15)
  100    



Q 24. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower. (Given begin mathsize 12px style square root of 3 end style= 1.732)

Solution:

From the figure, DE is the tower.
When sun’s altitude is 60˚ then ∠DPE = 60˚ and length of shadow is PE.
Shadow is 40 m more when angle changes from 60˚ to 30˚
FP = 40 m.
As tower is a vertical ground, ∠DEF is 90˚.
In right angled triangle DEF, 

begin mathsize 12px style tan space straight P equals DE over PE
therefore space tan space 60 degree equals DE over PE
therefore space square root of 3 equals DE over PE
therefore space PE equals fraction numerator DE over denominator square root of 3 end fraction space... open parentheses straight i close parentheses
In space right space angled space triangle space DEF comma
tan space straight F equals DE over FE
therefore space tan space 30 degree space equals space DE over FE
therefore space fraction numerator 1 over denominator square root of 3 end fraction equals DE over FE
therefore space FE space equals space square root of 3 DE
therefore space FP space plus space PE space equals square root of 3 DE
therefore space 40 space plus space PE space equals square root of 3 DE
therefore space PE space equals space square root of 3 DE minus 40 space space... open parentheses ii close parentheses
fraction numerator DE over denominator square root of 3 end fraction equals square root of 3 DE minus 40 space space space from space open parentheses straight i close parentheses space and space open parentheses ii close parentheses
therefore DE space equals space 3 DE space minus space 40 square root of 3
therefore space 2 DE space equals space 40 square root of 3
therefore space DE space equals space 20 square root of 3
Hence comma space the space height space of space the space tower equals DE equals 20 square root of 3 straight m.
end style

Q 25. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

OR

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Solution:

Given : ΔPQR in which XY || QR, XY intersects PQ and PR at X and Y respectively.

To prove : begin mathsize 12px style PX over XQ equals PY over YR end style

Construction : Join RX and QY and draw YN perpendicular to PQ and XM perpendicular to PR.

Proof:

begin mathsize 12px style Since comma space ar open parentheses increment PXY close parentheses equals 1 half cross times PX cross times YN space... open parentheses straight i close parentheses
ar open parentheses increment PXY close parentheses equals 1 half cross times P straight Y cross times XM space... open parentheses ii close parentheses
Similarly comma space ar open parentheses increment QXY close parentheses equals 1 half cross times QX cross times NY space... open parentheses iii close parentheses
ar open parentheses increment RXY close parentheses equals 1 half cross times YR cross times XM space... open parentheses iv close parentheses
Dividing space left parenthesis straight i right parenthesis space by space left parenthesis iii right parenthesis space we space get comma
therefore fraction numerator ar open parentheses PXY close parentheses over denominator ar open parentheses QXY close parentheses end fraction equals fraction numerator 1 half cross times PX cross times YN over denominator 1 half cross times QX cross times YN end fraction equals PX over QX space... left parenthesis straight v right parenthesis
Again space dividing space left parenthesis ii right parenthesis space by space left parenthesis iv right parenthesis
therefore fraction numerator ar open parentheses PXY close parentheses over denominator ar open parentheses RXY close parentheses end fraction equals fraction numerator 1 half cross times PY cross times XM over denominator 1 half cross times YR cross times XM end fraction equals PY over YR space... left parenthesis vi right parenthesis
end style

Since the area of triangles with same base and between same parallel lines are equal, so
∴ ar(ΔQXY) = ar(ΔRXY)  …..(vii)
As ΔQXY and ΔRXY are on same base XY and between same parallel lines XY and QR.
Therefore, from (v), (vi) and (vii) we get

begin mathsize 12px style therefore PX over XQ equals PY over YR end style

OR

Given : A right angle triangle ABC at B.
To prove : AC2 = AB2 + BC2
Construction : Draw BD perpendicular to AC.
Proof :
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
∴ ΔADB ~ ΔABC

begin mathsize 12px style therefore space AD over AB equals AB over AC space space space space space space space because space straight c. space straight p. space straight c. space straight t.
therefore space AD cross times AC equals AB squared space... left parenthesis straight i right parenthesis
increment BDC space tilde increment ABC
therefore space CD over BC equals BC over AC space space space space space space space because space straight c. space straight p. space straight c. space straight t.
therefore space CD cross times AC equals BC squared space... left parenthesis ii right parenthesis end style

Adding (i) and (ii)
∴ AD × AC + CD × AC = AB2 + BC2
∴ AC(AD + CD) = AB2 + BC2
∴ AC × AC = AB2 + BC2
∴ AB2 + BC2 = AC2

Q 26. If m times the mth term of an Arithmetic Progression is equal to n times its nth term and m ≠ n, show that the (m + n)th term of the A.P. is zero.

OR

The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.

Solution:

Let the mth term of an Arithmetic Progression be tm and nth term be tn.
Also, a and d be the first term and common difference respectively.
According to the question,
mtm = ntn
∴ m[a + (m – 1)d] = n[a + (n – 1)d]
∴ am + m2d – md = an + n2d – nd = 0
∴ a(m – n) + (m2 – n2)d – (m – n)d = 0
∴ a(m – n) + (m + n)(m – n)d – (m – n)d = 0
∴ (m – n)[a + (m + n)d – d] = 0
It is given that m ≠ n.
∴ a + (m + n – 1)d = 0
∴ tm + n = 0

OR

Let the three numbers of an arithmetic progression be a – d, a, a + d.
According to the question,
a – d + a + a + d = 18
∴ 3a = 18
∴ a = 6
Hence, three numbers are 6 – d, 6, 6 + d
According to the question,
∴  (6 – d)(6 + d) = 5d
∴ 36 – d2 = 5d
∴ d2 + 5d – 36 = 0
∴  (d + 9)(d – 4) = 0
∴ d = -9 and d = 4
If d = -9 then the three numbers are 6 + 9, 6, 6 – 9 i. e. 15, 6, -3.
If d = 4 then the three numbers are 6 – 4, 6, 6 + 4 i. e. 2, 6, 10.


 

Q 27. In Figure3, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. Find begin mathsize 12px style open parentheses Take space straight pi space equals space 22 over 7 close parentheses end style

 

(a) the total surface area of the block.

 

(b) the volume of the block formed. 

 

 

OR

A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use 𝜋 = 3.14)

Solution:

Let the side of the cube be a.
∴ a = 6 cm
Let radius of the hemisphere be r.
∴ r = 2.1 cm
(a)
The total surface area of the block
= Total surface area of a cube – base area of hemisphere + Curved surface area of hemisphere
= 6a2 – πr2 + 2πr2
= 6 × 62 +πr

begin mathsize 12px style equals 216 plus 22 over 7 cross times 2.1 squared end style

= 216 + 13.86

= 229.86 cm2

(b)
Volume of the clock
= Volume of a cube + volume of hemisphere

begin mathsize 12px style equals a cubed plus 2 over 3 πr cubed
equals 6 cubed plus 2 over 3 cross times 22 over 7 cross times open parentheses 2.1 close parentheses cubed end style

= 216+19.404
= 235.404 cm3

OR

Volume of the bucket = 12308.8 cm3, r1 = 20 cm and r2 = 12 cm, h = ?

begin mathsize 12px style therefore space V o l u m e space o f space b u c k e t equals πh over 3 open parentheses r subscript 1 superscript 2 plus r subscript 2 superscript 2 plus r subscript 1 r subscript 2 close parentheses
therefore space πh over 3 open parentheses r subscript 1 superscript 2 plus r subscript 2 superscript 2 plus r subscript 1 r subscript 2 close parentheses equals 12308.8
therefore space fraction numerator 22 cross times h over denominator 7 cross times 3 end fraction open parentheses 20 squared plus 12 squared plus 20 cross times 12 close parentheses equals 12308.8
therefore space fraction numerator 22 h over denominator 21 end fraction open parentheses 400 plus 144 plus 240 close parentheses equals 12308.8
therefore space fraction numerator 22 h over denominator 21 end fraction cross times 784 equals 12308.8
therefore space h equals fraction numerator 12308.8 cross times 21 over denominator 22 cross times 784 end fraction

therefore space h equals 15 space c m
S u r f a c e space a r e a space o f space t h e space m e t a l space s h e e t space u s e d space equals πr subscript 2 superscript 2 plus straight pi open parentheses straight r subscript 1 plus straight r subscript 2 close parentheses space straight l
straight l equals square root of straight h squared plus open parentheses straight r subscript 1 plus straight r subscript 2 close parentheses squared end root equals square root of 15 squared plus 8 squared end root equals 17 space cm
therefore space Surface space area space of space the space metal space sheet space used space equals πr subscript 2 superscript 2 plus straight pi open parentheses straight r subscript 1 plus straight r subscript 2 close parentheses space straight l
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 3.14 cross times 12 squared plus 3.14 cross times 32 cross times 17
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2160.32 space cm squared end style

Hence, height of the bucket = 15 cm, area of the metal sheet used = 2160.32 cm2.

 

Q 28. Construct a triangle, the lengths of whose sides are 5 cm, 6 cm and 7 cm. Now construct another triangle whose sides begin mathsize 12px style 5 over 7 end style are  times the corresponding sides of the first triangle.

Solution:

Steps of construction :

  1. Draw a line segment AB of 5 cm. Taking A and B as centres, draw two arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm and 7 cm respectively.
  2. Draw a ray AX making acute angle with the line AB on opposite side of vertex C.
  3. Locate 7 points A1, A2, A3, A4, A5, A6, A7 (as 7 is greater between 5 and 7) on line AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.
  4. Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B’.
  5. Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’. ΔAB’C’ is the required triangle.

 

Q 29. Prove that :

begin mathsize 12px style fraction numerator tan cubed straight theta over denominator 1 plus tan squared straight theta end fraction plus fraction numerator cot cubed straight theta over denominator 1 plus cot squared straight theta end fraction equals secθ space space cosecθ minus 2 sinθ space cosθ. end style

Solution:

begin mathsize 12px style LHS equals space fraction numerator tan cubed straight theta over denominator 1 plus tan squared straight theta end fraction plus fraction numerator cot cubed straight theta over denominator 1 plus cot squared straight theta end fraction
space space space space space space space space equals space fraction numerator begin display style fraction numerator sin cubed straight theta over denominator cos cubed straight theta end fraction end style over denominator sec squared straight theta end fraction plus fraction numerator begin display style fraction numerator cos cubed straight theta over denominator sin cubed straight theta end fraction end style over denominator cosec squared straight theta end fraction
space space space space space space space space equals space fraction numerator begin display style fraction numerator sin cubed straight theta over denominator cos cubed straight theta end fraction end style over denominator begin display style fraction numerator 1 over denominator cos squared straight theta end fraction end style end fraction plus fraction numerator begin display style fraction numerator cos cubed straight theta over denominator sin cubed straight theta end fraction end style over denominator begin display style fraction numerator 1 over denominator sin squared straight theta end fraction end style end fraction
space space space space space space space space equals space fraction numerator sin cubed straight theta over denominator cos cubed straight theta end fraction cross times cos squared straight theta plus fraction numerator cos cubed straight theta over denominator sin cubed straight theta end fraction cross times sin squared straight theta
space space space space space space space space equals space fraction numerator sin cubed straight theta over denominator cosθ end fraction plus fraction numerator cos cubed straight theta over denominator sinθ end fraction
space space space space space space space space equals space fraction numerator sin to the power of 4 straight theta plus cos to the power of 4 straight theta over denominator sinθ space cosθ end fraction
space space space space space space space space equals space fraction numerator sin to the power of 4 straight theta plus cos to the power of 4 straight theta plus 2 sin squared straight theta plus cos squared straight theta minus 2 sin squared straight theta plus cos squared straight theta over denominator sinθ plus cosθ end fraction
space space space space space space space space space equals space fraction numerator open parentheses sin squared straight theta plus cos squared straight theta close parentheses minus 2 sin squared straight theta space cos squared straight theta over denominator sinθ plus cosθ end fraction
space space space space space space space space space equals space fraction numerator 1 minus 2 space sin squared straight theta plus cos squared straight theta over denominator sinθ plus cosθ end fraction
space space space space space space space space space equals space fraction numerator 1 over denominator sinθ plus cosθ end fraction minus fraction numerator 2 space sin squared straight theta plus cos squared straight theta over denominator sinθ plus cosθ end fraction
space space space space space space space space space equals space sinθ space cosecθ minus 2 space sinθ plus cosθ
space space space space Hence comma space fraction numerator tan cubed straight theta over denominator 1 plus tan squared straight theta end fraction fraction numerator cot cubed straight theta over denominator 1 plus cot cubed straight theta end fraction equals secθ space cosecθ space minus space 2 space sinθ space cosθ end style


Q 30. A motorboat whose speed in still water is 9 km/h, goes 15 km downstream and comes to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.

Solution:

Let T be the time for downstream and t be the time for upstream, travel.
T + t = 3 hrs 45 min = 15/4 hrs   ….(i)
For downstream, distance/velocity = time

begin mathsize 12px style therefore space fraction numerator 15 over denominator straight V plus straight v end fraction equals straight T space where space straight V space is space the space speed space of space boat space and space straight v space is space the space speed space of space stream.
therefore space fraction numerator 15 over denominator 9 plus straight v end fraction equals straight T space... open parentheses ii close parentheses
Similarly comma
therefore space fraction numerator 15 over denominator 9 plus straight v end fraction equals straight t space... open parentheses iii close parentheses
Additing space open parentheses ii close parentheses space and space open parentheses iii close parentheses
therefore space fraction numerator 15 over denominator 9 plus straight v end fraction plus fraction numerator 15 over denominator 9 minus straight v end fraction equals straight T plus straight t
therefore 15 open parentheses fraction numerator 9 minus straight v plus 9 plus straight v over denominator open parentheses 9 plus straight v close parentheses open parentheses 9 minus straight v close parentheses end fraction close parentheses equals 15 over 4 space space space from open parentheses straight i close parentheses
therefore fraction numerator 15 cross times 18 over denominator 81 minus straight v squared end fraction equals 15 over 4
therefore fraction numerator 18 over denominator 81 minus straight v squared end fraction equals 1 fourth
therefore 81 minus straight v squared equals 72
therefore space straight v squared equals 9
therefore straight v equals 3 space km divided by hr
Hence comma space the space speed space of space the space stream space is space 3 space km divided by hr. end style

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