Summative Assessment Ii

Let the roots of the given quadratic equation 3x² – (3k – 2)x – (k – 6)=0 be α and β.

Now,

sum of roots = α + β = (3k – 2)/3 and,

product of roots = αβ = – (k – 6)/3

[∵ If α and β are the roots of quadratic equation ax² + bx + c=0 then α + β = – b/a and αβ = c/a]

According to question –

sum of roots = product of roots

∴ α + β = αβ

⇒ (3k – 2)/3 = – (k – 6)/3

⇒ 3k – 2 = – k + 6

⇒ 4k = 8

∴ k = 2

Hence, The value of k is 2.

All 2-digit numbers divisible by 6 are as follows: –

6, 12, ………., 96

The above series of numbers forms an arithmetic progression with

first term(a) = 6 and,

common difference(d) = (n + 1)th term – nth term = 12 – 6 = 6

last term or nth term(a_n) = 96

Let the number of terms in above series be n.

∵ a_n = a + (n – 1) × d

⇒ 96 = 6 + (n – 1) × 6

⇒ 90 = 6n – 6

⇒ 6n = 96

∴ n = 16

Thus, total no. of all 2-digit numbers divisible by 6.

Let P be the event of getting a composite number while throwing a dice.

Total no. of outcomes when n number of die are thrown = 6ⁿ

∴ no. of total outcomes = n(S) = 6

Sample Space = {1, 2, 3, 4, 5, 6}

favourable elementary events = getting a composite number

= {4, 6}

∴ no. of favourable elementary events = n(P) = 2

Thus, the probability of getting a composite number = n(P)/n(S)

= 2/6

= 1/3

A Tangent is a line that intersects a circle at exactly one point.

The tangents drawn at the end points of a chord of a circle can be parallel only if that chord is the diameter of the circle. This will be clear from the fig.1 and fig.2 shown below.

Thus, statement (a) is incorrect.

A secant is a segment that intersects a circle twice.

From a point P in the exterior of a circle, infinite no. of secants can be drown through P to the circle. This can be shown in the fig.3 drawn below.

Thus, statement (b) is incorrect.

A Tangent is a line that intersects a circle at exactly one point.

From a point P in the plane of a circle, two tangents can be drawn to the circle only if point P is exterior to the circle. This can be shown in the fig.4 drawn below.

Thus, statement (c) is incorrect.

In the above fig.5, we take a point Q on the tangent XY to the circle with centre O. Obviously, this point Q should lie outside to the circle otherwise XY will become secant. And, P is the point of contact. Clearly,

OQ > OP

Also, this is also true for all the points lying on the tangent XY except point P. And,

we know that perpendicular distance is always the shortest distance.

OP is shortest of all the distances b/w points O and any other points on XY i.e.

OP ⊥ XY

Hence, The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

Thus, statement (d) is correct.

In Δ PAB,

∠ APB = 60° and PA = 8 cm [given]

∴ PB = PA = 8 cm

[∵ tangents drawn from an exterior point to the circle are equal in length]

⇒ ∠ PAB = ∠ PBA = θ [LET]

Now, In Δ PAB

∠ APB + ∠ PAB + ∠ PBA = 180° [∵ Sum of all the angles of a Δ is 180°]

⇒ 60° + θ + θ = 180°

⇒ 60° + 2θ = 180°

⇒ 2θ = 120°

∴ θ = 60°

Thus, Δ PAB is an equilateral Triangle.

∴ Length of chord AB = 8 cm.

Let the Distance of the object from the tower be x meters.

∴ BC = x m

Given –

height of tower = AB = 60 m

Angle of depression = ∠ DAC = 30°

∴ ∠ BCA = ∠ DAC = 30°

[∵ When two‖lines are intersected by a third line then theAlternate interior angles will be equal.]

Now, In Δ ABC

tan 30° = AB/BC = 60/x [∵ tan θ = perpendicular/base]

⇒ 1/√3 = 60/x

∴ x = 60√3 meters

Let the point P (2, – 5) divide the line segment joining A (– 3,5) and B(4, – 9) in the ratio m:n.

Let (x,y) ≡ (2, – 5)

(x₁,y₁) ≡ (– 3,5)

and (x₂,y₂) ≡ (4, – 9)

Using Section Formula,

⇒ 2 × (m + n) = 4m – 3n

⇒ 2m + 2n = 4m – 3n

⇒ 5n = 2m

∴ m:n = 5:2

Since the ratio is positive, Point P divides the line segment AB internally in the ratio 5:2.

Let the radius of the sphere so formed be r cm.

Given –

Radius of 1st sphere(r₁) = 6 cm

Radius of 2nd sphere(r₂) = 8 cm

Radius of 3rd sphere(r₃) = 10 cm

After Melting all these spheres, the volume will remain unchaged.

∴ Vol. of 1st sphere + Vol. of 2nd sphere + Vol. of 3rd sphere

= Vol. of new sphere so formed

⇒ (4/3)π(r₁)³ + (4/3)π(r₂)³ + (4/3)π(r₃)³ = (4/3)π(r)³

Taking out (4/3)π from both sides, we get –

⇒ (r₁)³ + (r₂)³ + (r₃)³ = (r)³

⇒ (6)³ + (8)³ + (10)³ = (r)³

⇒ 216 + 512 + 1000 = (r)³

⇒ (r)³ = 1728

∴ r = 12 cm

Thus, the radius of new sphere is 12 cm.

The given quadratic equation is x² – 2px + 1 = 0.

And, Discriminant D of the quadratic equation ax² + bx + c = 0 is given by –

D = b² – 4ac

Comparing the equation ax² + bx + c = 0 with given quadratic equation is x² – 2px + 1 = 0, we get –

a = 1, b = – 2p and, c = 1

∴ D = (– 2p)² – 4(1)(1) = 4p² – 4 = 4(p² – 1)

For no real roots,

D < 0

⇒ 4(p² – 1) < 0

⇒ p² – 1 < 0

⇒ (p + 1)(p – 1) < 0

∴ p ∈ (– 1,1)

Thus, p can take any values between – 1 and 1 for no real roots of given quadratic equation.

The above series of numbers forms an arithmetic progression with

first term(a) = 4 and,

common difference(d) = (n + 1)th term – nth term = 9 – 4 = 5

last term or nth term(a_n) = 254

Let the total no. of terms in above A.P be n.

∴ a_n = a + (n – 1) × d

⇒ 254 = 4 + (n – 1) × 5

⇒ 250 = 5n – 5

⇒ 5n = 255

∴ n = 51

∴ 10th term from the end of AP = 51 – 10 + 1 = 42th term from the beginning

∴ 42th term = a₄₂ = a + (42 – 1)d

= 4 + 41 × 5

= 209

Hence, 10th term from the end of AP is 209.

Let the nth term of the AP be the first negative term.

In the given AP –

first term(a) = 24 and,

common difference(d) = (n + 1)th term – nth term = 21 – 24 = – 3

According to question –

∴ a_n < 0

⇒ a + (n – 1) × d < 0

⇒ 24 + (n – 1) × (– 3) < 0

⇒ – 3n + 27 < 0

⇒ 3n > 27

∴ n > 9

Thus, the first negative term of given AP is 10th term.

In the given figure,

AQ and AR are two tangents drawn from an exterior point A at contact points Q and R on the circle.

∴ AQ = AR

⇒ AQ = AC + CR…..(1)

Similarly,

BQ and BP are two tangents drawn from an exterior point B at contact points Q and P on the circle.

∴ BQ = BP…..(2)

And,

CR and CP are two tangents drawn from an exterior point C at contact points R and P on the circle.

∴ CR = CP…..(3)

Now, Equation (1) can be written as –

AQ = (AC + CR + AC + CR)/2

⇒ AQ = (AC + CP + AC + CR)/2 [using(3)]

⇒ AQ = (AC + CP + AR)/2

⇒ AQ = (AC + CP + AQ)/2

⇒ AQ = (AC + CP + AB + BQ)/2

⇒ AQ = (AC + CP + AB + BP)/2 [using(2)]

⇒ AQ = (AB + BC + AC)/2 [∵BP + CP=BC]

Thus, AQ = (1/2) × perimeter of Δ ABC

Let the third vertex C ≡ (x₃,y₃)

In a Δ ABC,

Vertex A ≡ (x₁,y₁) ≡ (6,4)

Vertex B ≡ (x₂,y₂) ≡ (– 2,2)

Centroid(G) ≡ (x,y) ≡ (3,4)

Centroid of a Δ ABC is given by –

x = (x₁ + x₂ + x₃)/3

⇒ 3 = (6 – 2 + x₃)/3

⇒ 9 = 4 + x₃

∴ x₃ = 5

And,

y = (y₁ + y₂ + y₃)/3

⇒ 4 = (4 + 2 + y₃)/3

⇒ 12 = 6 + y₃

∴ y₃ = 6

Thus, the coordinates of third vertex C is (5,6).

Total no. of Oranges = 150

Probability of rotten oranges =0.06

∴ Probability of good oranges = 1 – 0.06 = 0.94

⇒ (no. of good oranges)/(no. of total oranges) = 0.94

⇒ no. of good oranges = 0.94 × 150 = 141

Thus, the number of good orange in the box = 141

fig.10

Given –

Total Height of cone = 31 cm

Radius of hemisphere(r) = Base Radius of Cone

= Height of hemisphere

= 7 cm

∴ Height of cone(h) = 31 – 7 = 24 cm

Slant Height of Cone(l) = √(h² + r²) = √(24² + 7²) = 25 cm

Now,

Total Surface Area of the Toy

= Curved Surface Area of Cone + Curved Surface Area of Hemisphere

= πrl + 2πr²

= π(rl + 2r²)

= π(7 × 25 + 2 × (7)²)

= π(175 + 98)

= π(273)

= 3.14 × 273

= 857.22 cm²

The given quadratic equation is –

a²b²x² –(4b^{4 –} 3a⁴) x – 12a²b² = 0

Discriminant D of the quadratic equation ax² + bx + c = 0 is given by –

D = b² – 4ac

Comparing the equation ax² + bx + c = 0 with given quadratic equation is a²b²x² –(4b⁴ –3a⁴) x – 12a²b² = 0,we get –

a = a²b² , b = –(4b^{4 –} 3a⁴) and, c = – 12a²b²

∴ The roots of the given quadratic equation is given by –

Thus, the roots of the given quadratic equation are (4b²/a²) and (– 3a²/b²).

Let the first term and common difference of given AP be a and d respectively.

According to question –

8th term of AP = a₈ = 31 [Given]

⇒ a + (8 – 1)d = 31

⇒ a + 7d = 31…..(1)

15th term of AP = a₁₅ = 16 + a₁₁

⇒ a + (15 – 1)d = 16 + a + (11 – 1)d

⇒ 14d = 16 + 10d

⇒ 4d = 16

∴ d = 4

Substituting the value of d in equation(1), we get –

a = 31 – 7 × 4 = 31 – 28 = 3

Thus, the required AP is 3,7,11,15,…….

All the two-digit odd positive numbers are –

11,13,15,17,……….,99

The above series of numbers forms an arithmetic progression with

first term(a) = 11 and,

common difference(d) = (n + 1)th term – nth term = 13 – 11 = 2

last term or nth term(a_n) = 99

Let the total no. of terms in above A.P be n.

∴ a_n = a + (n – 1) × d

⇒ 99 = 11 + (n – 1) × 2

⇒ 88 = 2n – 2

⇒ 2n = 90

∴ n = 45

Sum of all the 45 terms of the AP is given by –

S₄₅ =(45/2)(11 + 99)

[∵S_n = (n/2)(a + l) =(n/2)[(2a + (n – 1)d]

=(45/2) × 110

=45 × 55

=2475

Thus, the sum of all two-digit odd positive numbers = 2475.

fig.11

Let ∠ APB = θ

In Δ APB,

PA = PB

[∵ Tangents drawn from an exterior point to the circle are equal in length]

⇒ Δ APB is an isoceles triangle.

∴ ∠ PAB = ∠ PBA = α [LET]

Now,

∠ APB + ∠ PAB + ∠ PBA = 180° [∵ sum of all the angles of Δ=180°]

⇒ θ + α + α = 180°

⇒ 2α = 180° – θ

∴ α = ∠ PAB = 90° – (θ/2)

Also, OA⊥AP

[∵ radius of a circle is always⊥to the tangent at the point of contact.]

∴ ∠ PAB + ∠ OAB = 90°

⇒ 90° – (θ/2) + ∠ OAB = 90°

⇒ ∠ OAB = (θ/2) = (1/2)∠ APB

∴ ∠ APB = 2 ∠ OAB

Hence, Proved.

In the given figure,

DS and DR are the two tangents drawn from an external point D at the point of contacts S and R respectively. And,

OS ⊥ DS and OR ⊥ DR

[∵ radius of a circle is always⊥to the tangent at the point of contact.]

⇒ OSDR is a square [∵ AD ⊥ DC (Given)]

∴ DR = 10 cm

Similarly,

BP and BQ are the two tangents drawn from an external point B at the point of contacts A and Q respectively.

∴ BP = BQ = 27 cm

[∵ Tangents drawn from an exterior point to the circle are equal in length]

⇒ QC = BC – BQ = 38 – 27 = 11 cm

Also, CR and CQ are the two tangents drawn from an external point C at the point of contacts R and Q respectively.

∴ CR = CQ = 11 cm

[∵ Tangents drawn from an exterior point to the circle are equal in length]

∴ DC = x = DR + CR = 10 + 11 = 21 cm.

Thus, the value of x is 21 cm.

Steps of Construction:

1. Draw a circle with centre O with radius OL and a point P outside it. Join PO and bisect it. Let M be the midpoint of PO.

fig.13

2. Taking M as centre and MO as radius, we will draw a circle.

Let it intersect the given circle at the points Q and R.

fig.14

3. Join PQ and PR.

Then PQ and PR are the required two tangents.

fig.15

4. Join OQ. Then ∠ PQO is an angle in the semicircle and,

∴ ∠ PQO = 90°

fig.16

Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle.

Similarly,

PR is also a tangent to the circle.

After measuring the lenghts of tangents using scale, we find that both the tangents are equal in length which concludes that all the measurements and steps done correctly.

Length of Each Tangent = 8 cm

Let the coordinates of the fourth vertex D be (x₄,y₄).

We know that –

Diagonals of a∥gm bisect each other.

∴ Mid – point of diagonal AC ≡ Mid – point of diagonal BD

⇒ 6 = 3 + x₄ and 8 = 6 + y₄

⇒ x₄ = 3 and y₄ = 2

Thus, the coordinates of the fourth vertex D is (3,2).

Let the third vertex A ≡ (x₁,y₁)

In a Δ ABC,

Vertex B ≡ (x₂,y₂) ≡ (– 3,1)

Vertex C ≡ (x₃,y₃) ≡ (0, – 2)

Centroid(G) ≡ (x,y) ≡ (0,0)

Centroid of a Δ ABC is given by –

x = (x₁ + x₂ + x₃)/3

⇒ 0 = (x₁ – 3 + 0)/3

⇒ 0 = x₁ – 3

∴ x₁ = 3

And,

y = (y₁ + y₂ + y₃)/3

⇒ 0 = (y₁ + 1 – 2)/3

⇒ 0 = y₁ – 1

∴ y₁ = 1

Thus, the coordinates of third vertex A is (3,1).

Sample Space = Cards marked with 2-digit numbers

= {10,11,12,……..,99)

No. of Sample Space = n(S) = 90

(a) Let P be the event of getting a card marked with 2-digit numbers which is divisible by 10.

∴ favourable elementary events = {10,20,30,…………,90}

no. of favourable elementary events = n(P) = 9

Thus, Probability of getting a card marked with number divisible by 10 = n(P)/n(S) = 9/90 = 1/10

(b) Let P be the event of getting a card marked with 2-digit square numbers.

∴ favourable elementary events = {16,25,36,…….,81}

no. of favourable elementary events = n(P) = 6

Thus, Probability of getting a card marked with number divisible by 10 = n(P)/n(S) = 6/90 = 1/15

(c) Let P be the event of getting a card marked with 2-digit prime numbers less than 25.

∴ favourable elementary events = {11,13,17,19,23}

no. of favourable elementary events = n(P) = 5

Thus, Probability of getting a card marked with number divisible by 10 = n(P)/n(S) = 5/90 = 1/18

fig.17

Let the radius of the circular park be r meters.

Given –

Circumference of circular park = 352 m

⇒ 2 × π × r = 352

⇒ 2 × (22/7) × r = 352

⇒ r = (7/44) × 352

∴ r = 7 × 8 = 56 m

⇒ outer radius = 56 + 7 = 63 m

∴ Area of the road = π(63² – 56²)

= (22/7)(63 + 56)(63 – 56)

= (22/7)(119)(7)

= 22 × 119

= 2618 m²

In the given figure, all the six desings covering equal area of the circle, therefore each design will subtend equal angles at the centre which is equal to (360°/6) i.e. 60°.

Also, the six triangles will be equal in area which is obtained by joining vertices of hexagon to the centre.

The triangle obtained will be equilateral because adjacent sides will be equal to the radius i.e. base angles will be equal and angle b/w them is 60° which concludes that other two angles will also be equal to 60° each.

∴ Area of six equilateral Δ = 6 × (√3/4) × (radius)²

= (3√3/2) × (28)²

= 1.5 × 1.73 × 784

= 2034.48 cm²

Area of Circle = π × (radius)² = (22/7) × (28)² = (22/7) × 784

= 22 × 112

= 2464 cm²

Area of the designs = Area of Circle – Area of six equilateral Δ

= (2464 – 2034.48) cm²

= 429.52 cm²

∴ Cost of making designs =Rs. (0.50 × 429.52) = Rs. 214.76

fig.19

Let the radius of the circle be r cm.

In the fig.19,

AR and AQ are making a pair of tangents drawn from vertex A of Δ ABC on the circle.

∴ AR = AQ = x [LET]

BR and BP are making a pair of tangents drawn from vertex B of Δ ABC on the circle.

∴ BR = BP = y [LET]

CP and CQ are making a pair of tangents drawn from vertex C of Δ ABC on the circle.

∴ CP = CQ = z [LET]

Given –

Δ ABC is an equilateral triangle.

∴ AB = BC = AC = 12 cm

⇒ AR + BR = BP + CP = AQ + CQ = 12

⇒ x + y = y + z = x + z = 12…..(1)

Now,

(x + y + y + z + x + z) = (12 + 12 + 12)

⇒ 2 × (x + y + z) = 36

⇒ x + y + z = 18…..(2)

Subtracting equation(1) from equation(2), we get –

x = y = z = 6 cm

Also, the line joining the centre the circle to the vertices of Δ which circumscribes the circle bisects the angles of a Δ.

∴ ∠ OBP = 30°

In Δ BOP,

tan ∠ OBP = OP/BP = r/6

⇒ tan 30° = r/6

⇒ 1/√3 = r/6

∴ r = 6/√3 = 2√3 = 3.46 cm

Area of Δ ABC = (√3/4) × (side)²

=(1.73/4) × (12)²

= 1.73 × 36

= 62.28 cm²

Area of circle = π × (radius)²

= 3.14 × (3.46)²

= 37.59 cm²

Thus, Area of the triangle which is not included in the circle

= Area of Δ ABC – Area of circle

= (62.28 – 37.59) cm²

= 24.69 cm²

Let the radius of the sphere be r cm.

Given –

Height of cone(h) = 40 cm

Radius of cone(r) = 30 cm

∴ Slant height of cone(l) = √(h² + r²) = √(40² + 30²) = 50 cm

According to question –

Surface Area of Sphere = Total Surface Area of Circular Cone

⇒ 4 × π × r² = π × r × (r + l)

⇒ 4r = r + l

⇒ 3r = l

∴ r = (l/3) = (50/3) cm

Thus, the radius of the Sphere = (50/3) cm

Let the two-digit number be xy(i.e. 10x + y).

After reversing the digits of the number xy, the new number becomes yx (i.e. 10y + x).

According to question –

xy = 35…..(1)

And,

(10x + y) + 18 = (10y + x)

⇒ 9x – 9y = – 18

⇒ x – y = – 2…..(2)

From equation(2), we get –

x = y – 2…..(3)

Substitute the value of x in equation(1), we get –

y(y – 2) = 35

⇒ y² – 2y – 35 = 0

⇒ y² – 7y + 5y – 35 = 0

⇒ y(y – 7) + 5(y – 7) = 0

⇒ (y – 7)(y + 5) = 0

∴ y = 7 [∵ y = – 5 is invalid because digit of a number can't be – ve.]

Substituting the value of y in equation (3), we get –

x = 5

Thus, the required number is 57.

Let the tap of the smaller diameter and larger diameter fills the tank alone in x and (x – 10) hours respectively.

In 1 hour, the tap of the smaller diameter can fill 1/x part of the tank.

In 1 hour, the tap of the larger diameter can fill 1/(x – 10) part of the tank.

Two water taps together can fill a tank in hours = 75/ 8 hours.

But in 1 hour the taps fill 8/75 part of the tank.

⇒

⇒

⇒ 4x² – 40x = 75x – 375

⇒ 4x² – 115x + 375 = 0

⇒ 4x² – 100x – 15x + 375 = 0

⇒ 4x(x – 25) – 15( x – 25) = 0

⇒ (4x -15)( x – 25) = 0

⇒ x = 25, 15/4

Taking x = 15 / 4

⇒ x – 10 = -25 /4 (But, time cannot be negative)

Now, taking x = 25

⇒ x – 10 = 15

Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill

the tank in 25 hours.

fig.20

In the fig.20, PA and PB are the two tangents drawn from an external point P at the point of contacts A and B on the circle with centre O respectively.

∴ OA ⊥ PA and OB ⊥ PB

[∵ radius of a circle is always⊥to the tangent at the point of contact.]

∴ ∠ OAP = ∠ OBP = 90°

we know that –

Sum of all the angles of a quadrilateral = 360°

In quadrilateral OAPB,

∠ OAP + ∠ OBP + ∠ APB + ∠ AOB = 360°

⇒ 180° + ∠ APB + ∠ AOB = 360°

∴ ∠ APB + ∠ AOB = 180°

Hence, the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.

fig.21

Given –

Angle of Elevation = ∠ EAC = 60°

Angle of Depression = ∠ EAD = ∠ BDA = 45°

Height of Building = AB = ED = 7 m

In Δ ABD,

tan 45° = AB/BD

⇒ 1 = 7/BD

⇒ BD = 7 m

∴ AE = BD = 7 m [from fig.21]

And, In Δ ACE

tan ∠ CAE = CE/AE

⇒ tan 60° = CE/7

⇒ √3 = CE/7

⇒ CE = 7√3 m

Thus, Height of Tower = CE + ED = 7√3 + 7

= 7(1.73 + 1)

= 7 × 2.73

= 19.11 m

Given –

Monthly Salary = Rs. 35000

∴ Annual Salary = Rs. (12 × 35000) = Rs. 420000

Annual Increment = Rs. 1500

Let us consider this situation as an AP with

first term = a = Rs. 420000

and, Common Difference = d = Rs. 1500

∴ Salary in 10th year is given by –

a₁₀ = a + (10 – 1)d = 420000 + 9 × 1500 = Rs. 433500

Thus, Monthly Salary in 10th year = Rs. (433500/12)

= Rs. 36125

Area of quadrant CAB = (π/4) × (radius)²

= (22/28) × (7)²

= 37.5 cm²

Area of Δ EAB = (1/2) × base × height

= (1/2) × 7 × 2

= 7 cm²

Thus, Area of shaded Region

= Area of quadrant CAB – Area of Δ EAB

= (37.5 – 7) cm²

= 30.5 cm²

fig.23

Given –

MC = 27 cm, NE = 33 cm and CE = 10 cm

Let AM = h cm, AN = H cm and AC = l cm

∴ AE = AC + CE = (l + 10) cm

In the above fig.19,

Δ AMC and Δ ANE are similar triangles because their corresponding angles are equal.

…..(1)

On cross multiplying last two fractional parts of equation(1), we get –

33l = 27l + 270

⇒ 6l = 270

∴ l = 45 cm

∴ AE = 45 + 10 = 55 cm

In Δ ANE,

AN² + NE² = AE² [by using pythagoras theorem]

⇒ H² + (33)² = (55)²

⇒ H² + 1089 = 3025

⇒ H² = 1936

∴ H = 44 cm

From first and last fractional parts of equation(1), we get –

h = (27/33) × 44 = 36 cm

∴ Height of frustum = H – h = 44 – 36 = 8 cm

Now,

Capacity of Frustum = Vol. of Cone ADE – Vol. of cone ABC

= (1/3)π × (NE)² × (AN) – (1/3)π × (MC)² × (AM)

= (1/3)π × [(33)² × (44) – (27)² × (36)]

= (22/21) × [47916 – 26244]

= (22/21) × 21672

= 22 × 1032

= 22704 cm³

Total Surface Area of Frustum

= Area of Curved Part(Trapezium)

+ Area of Upper Circular Part

+ Area of lower Circular Part

= [(1/2) × (sum of parallel sides) × (height of frustum)]

+ [π × (MC)²] + [π × (NE)²]

= [(1/2) × 2π(27 + 33) × 8] + [(22/7) × (27)²] + [(22/7) × (33)²]

= 480(22/7) + (22/7) × [(27)² + (33)²]

= 480(22/7) + (22/7) × 1818

= (22/7) × 2298

= 22 × 328.28

= 7222.16 cm²

Thus, Capacity of Frustum = 22704 cm³

and, Total Surface Area of Frustum = 7222.16 cm²

In the given fig.,

CA and CE are the two tangents drawn from an external point C at the point of contacts A and E respectively.

∴ CA = CE

[∵ Tangents drawn from an exterior point to the circle are equal in length]

Similarly, DE and DB are the two tangents drawn from an external point D at the point of contacts E and B respectively.

∴ DE = DB

[∵ Tangents drawn from an exterior point to the circle are equal in length]

Perimeter of Δ PCD = PC + CD + PD

= PC + CE + DE + PD

= PC + CA + BD + PD

= PA + PB [∵ PA = PC + CA and PB = PD + BD]

= 14 + 14 [∵ PA=PB]

= 28 cm

Steps of Construction :

1. Draw a line Segment BC = 5.4 cm and draw an angle of 60°

at point B and mark a length of AB = 4.5 cm on the line passing through B. Then join AC.

fig.25

2. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

fig.26

3. Locate 4 points [the greater of 4 and 3 in (3/4)] B₁, B₂, B₃, B₄ on BX so that

BB₁ = B₁B₂ = B₂B₃ = B₃B₄

fig.27

4. Join B₃ [the 3rd point, 3 being smaller of 3 and 4 in (3/4)] to C and draw a line through B₄ parallel to B₃C, intersecting the extended line segment BC at C′.

fig.28

5. Draw a line through C′ parallel to CA intersecting the extended line segment BA at A′.

fig.29

Then A′BC′ is the required triangle.

Let the no. of blue balls in the bag be x.

Let B and R be the event of drawing a blue and red ball respectively.

∴ total no. of balls in the bag = x + 5

According to question –

Probability of drawing blue ball

= 3 × Probability of drawing blue ball

⇒ [no. of blue balls/total no. of balls]

= 3 × [no. of red balls/total no. of balls]

⇒ (x/x + 5) =3 × (5/x + 5)

∴ x = 15

Thus, the no. of blue balls in the bag is 15.

Let the point on the y-axis which divides the line segment joining the points A(– 2, – 3) and B(3, 7) be C(0,y).

Let the ratio in which y-axis divides AB line segment be m:n.

Let (x,y) ≡ (0,y)

(x₁,y₁) ≡ (– 2, – 3)

and (x₂,y₂) ≡ (3,7)

Using Section Formula,

⇒ 3m = 2n

∴ m:n = 2:3

Now,

⇒ y = (5/5) = 1

Thus, the line segment joining the points (– 2, – 3) and (3, 7) divided by the y-axis in the ratio 2:3 internally and the coordinates of the point of division is (0,1).

Any quadratic equation in the form ax² + bx + c = 0 has equal roots if and only if Discriminant, D = 0

Where, D = b² – 4ac

In the given equation,

a = 2

b = k

c = 3

Now, above equation will have equal roots if

D = 0

i.e.

(k)² – 4(2)(3) = 0

⇒ k² = 24

⇒ k= ±2√6

In the given AP,

First term, a = 7

Common difference, a₂ – a₁ = (11 – 7) = 4

Let the no of terms be n

Nth term, a_n = 139

We know that, For any AP

a_n = a + (n – 1)d

where,

a_n = nth term

d = common difference

n = no of terms

using the above formula for given AP, we have

139 = 7 + (n – 1)(4)

⇒ 4(n – 1) = 132

⇒ n – 1 = 33

⇒ n = 34

Hence, there are 34 terms in given AP.

We know that,

Probability

Now,

No of total outcomes i.e. total no of cards = 52

No of favourable outcomes i.e. no of black suits of 10 = 2

Probability (Getting a 10 of black suit)

Given, A circle with center O and radius, OT = 7 cm and PT = 24 cm

Now, we know that

Tangent at a point on the circle is perpendicular to the radius through the point of contact.

i.e.

OT ⏊ OP

By Pythagoras Theorem in ΔOTP [ i.e. Hypotenuse² = Base² + Height²]

(OP)² = (OT)² + (PT)²

⇒ (OP)² = (7)² + (24)²

⇒ (OP)² = 49 + 576 = 625

⇒ OP = 25 cm

We know that any point on y axis is in the form (0, x) where x is any real number, let y axis intersect the line segment AB at point P with coordinates (0, c)

And we have

Coordinates of A = (– 3, 2)

Coordinates of B = (6, 1)

Let P divides AB in K:1

Now, By using section formula i.e.

The coordinates of Point P which divides line AB in a ration m : n is

Where, (x₁, y₁) and (x₂, y₂) are the coordinates of points A and B respectively.

So,

Coordinates of P

So,

P divides AB in 2:1

Coordinates of Given Point (say P) = (6, – 6)

Coordinates of Origin (say O) = (0, 0)

By using distance formula i.e

Distance

Where, (x₁, y₁) and (x₂, y₂) are the coordinates of points A and B respectively.

So, we have

⇒ OP =√(36 + 36)

⇒ OP =6√2 units

Consider, the situation in the form of a triangle ABC where A is the kite and AC shows the height of kite i.e.

AC = 75 cm

And

AB be the string with angle of inclination i.e.

∠CAB = θ = 60°

We have to find length of string i.e. AB

Clearly, ABC is a right – angled triangle

So, we have

On cross multiplying we get,

On rationalizing we get,

For solid metal cone,

Height, h = 24 cm

Base radius, b = 12 cm

We know,

Where r is base radius and h is the height of cone.

Putting the values,

For a solid spherical ball,

Diameter = 6 cm

Radius, r = 3 cm

We know,

Where, r is radius of sphere.

Putting the values, we have

On solving, we get

No of balls = 32

As the equation is in the form Ax² + Bx + C = 0 with non-zero A.

In which,

A = a - b

B = b - c

C = c - a

And we know that if the roots of a equation are equal then we have

Discriminant, D = 0

Where, D = b² - 4ac

⇒ b² - 4ac = 0

⇒ (b - c)² - 4(a - b)(c - a) = 0

⇒ (b - c)² + 4(a - b)(a - c) = 0

⇒ b² + c² - 2bc + 4(a² - ac - ab + bc) = 0

⇒ b² + c² - 2bc + 4a² - 4ac - 4ab + 4bc = 0

⇒ 4a²+ b² + c² - 4ac + 2bc - 4ac = 0

⇒ (-2a)²+ b² + c² + 2(-2a)c + 2bc + 2(-2a)c = 0

⇒ (-2a + b + c)² = 0

[using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2za ]

⇒ -2a + b + c = 0

⇒ b + c = 2a

Hence Proved.

First term, a = 4

Common difference, a₂ – a₁ = 14 – (4) = 10

Let the no of terms be n

We know, that nth term of an AP is

a_n =a + (n – 1)d

where a is first term and d is common difference.

254 = 4 + (n – 1)10

⇒ 250 = (n – 1)10

⇒ n – 1 = 25

⇒ n = 26

10^th term from last will be 17^th term from starting

And a₁₀ = a + 16d

= 4 + 16(10)

= 164

Given AP = 3, 15, 27, 39, …

First term, a = 3

Common difference, a₂ - a₁ = 15 - 3 = 12

And we know

Nth term of an AP, a_n = a + (n - 1)d

Where a is first term and d is common difference.

Now, let the m^th term be 132 more than 54^th term

In that case,

a_m = a₅₄ + 132

⇒ a + (m - 1)d = a + 53d + 132

⇒ (m - 1)12 = 53(12) + 132

⇒ 12m - 12 = 636 + 132

⇒ 12m = 768 + 12

⇒ 12m = 780

⇒ m = 65

hence, 65^th term will be 132 more than its 54^th term.

Let AB be the diameter of a circle with center O.

CD and EF are two tangents at ends A and B respectively.

To Prove : CD || EF

Proof :

OA ⏊ CD and OB ⏊ EF [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAD = ∠OBE = 90°

∠OAD + ∠OBE = 90° + 90° = 180°

Considering AB as a transversal

⇒ CD || EF

[Two sides are parallel, if any pair of the interior angles on the same sides of transversal is supplementary]

Given: From an external point P, two tangents, PA and PB are drawn to a circle with center O. At a point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. And PA = 14 cm

To Find: Perimeter of ΔPCD

As we know that, Tangents drawn from an external point to a circle are equal.

So we have

AC = CE …[1] [Tangents from point C]

ED = DB …[2] [Tangents from point D]

Now Perimeter of Triangle PCD

= PC + CD + DP

= PC + CE + ED + DP

= PC + AC + DB + DP [From 1 and 2]

= PA + PB

Now,

PA = PB = 14 cm as tangents drawn from an external point to a circle are equal

So we have

Perimeter = PA + PB = 14 + 14 = 28 cm

Area of circular base = 616 cm²

We know that,

Area of circle = πr²

Where r is the radius of circle

Let the radius of circular base be r

We have,

πr² = 616 cm²

⇒ r² = 196

⇒ r = 14 cm

Now, Height = 48 cm [Given]

And we know,

Slant height, l =√(r² + h²)

Where r is radius and h is the height of the cone

l=√(14² + 48²)=√(196 + 2304)=√2500=50 cm

Now,

Total surface area of a cone = πr(l + r)

Where r is radius and l is slant height.

So, Putting values we have

Total surface of cone = π(14)(50 + 14)

Given,

Outer radius of circle, R = 21 cm

Area of enclosed region = 770 cm²

Let the radius of inner circle be r.

Area of enclosed region = Area of outer circle – Area of inner circle

⇒ 770 = πR² – πr²

⇒

⇒ 35(7) = (21)² – r²

⇒ r² = 441 – 245

⇒ r² = 196

⇒ r = 14 cm

12abx² – (9a² – 8b²)x – 6ab = 0

12abx² – 9a²x + 8b²x– 6ab = 0

3ax(4bx – 3a) + 2b(4bx – 3a) = 0

(3ax + 2b)(4bx – 3a) = 0

So, we have

3ax + 2b = 0 or 4bx – 3a = 0

Let the a be first term and d be common difference

As we know

a_n = a + (n – 1)d

Given,

⇒ a₈ = 31

⇒ a + 7d = 31

⇒ a = 31 – 7d …[1]

Also, As 15^th term is 16 more than 11^th term

⇒ a₁₅ = a₁₁ + 16

⇒ a + 14d = a + 10d + 16

⇒ 4d = 16

⇒ d = 4

Using this value in equation [1]

a = 31 – 7(4) = 3

So, AP is

a, a + d, a + 2d, …

3, 3 + 4, 3 + 2(4), …

3, 7, 11, …

Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.

To Proof : ABCD is a rhombus.

As ABCD is a parallelogram

AB = CD and BC = AD [opposite sides of a parallelogram are equal] …[1]

Now, As tangents drawn from an external point are equal.

We have

AP = AS [tangents from point A]

BP = BQ [tangents from point B]

CR = CQ [tangents from point C]

DR = DS [tangents from point D]

Add the above equations

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ AB + CD = AS + DS + BQ + CQ

⇒ AB + CD = AD + BC

⇒ AB + AB = BC + BC [From 1]

⇒ AB = BC …[2]

From [1] and [2]

AB = BC = CD = AD

And we know,

A parallelogram with all sides equal is a rhombus

So, ABCD is a rhombus.

Hence Proved !

Given: ΔABC that is drawn to circumscribe a circle with radius r = 4 cm and BD = 6 cm DC = 8 cm

To Find: AB and AC

Now,

As we know tangents drawn from an external point to a circle are equal.

Then,

FB = BD = 6 cm [Tangents from same external point B]

DC = EC = 8 cm [Tangents from same external point C]

AF = EA = x (let) [Tangents from same external point A]

Using the above data we get

AB = AF + FB = x + 6 cm

AC = AE + EC = x + 8 cm

BC = BD + DC = 6 + 8 = 14 cm

Now we have heron's formula for area of triangles if its three sides a, b and c are given

ar = √(s(s – a)(s – b)(s – c))

Where,

So for ΔABC

a = AB = x + 6

b = AC = x + 8

c = BC = 14 cm

And

ar(ΔABC) = √((x + 14)(x + 14 – (x + 6))(x + 14 – (x + 8))(x + 14 – 14))

= √((x + 14)(8)(6)(x)) [1]

ar(ABC) = ar(AOB) + ar(BOC) + ar(AOC)

at, tangent at a point on the circle is perpendicular to the radius through point of contact,

So, we have

OF ⏊ AB, OE ⏊ AC and OD ⏊ BC

Therefore, AOB, BOC and AOC are right – angled triangles.

And area of right angled triangle =1/2 × Base × Height

Using the formula,

Using [1] we have,

Squaring both side

⇒ 48x(x + 14) = (2x + 6 + 28 + 2x + 16)²

⇒ 48x² + 672x = (56 + 4x)²

⇒ 48x² + 672x = (4(14 + x))²

⇒ 48x² + 672x = 16(196 + x² + 28x)

⇒ 3x² + 42x = 196 + x² + 28x

⇒ 2x² + 14x – 196 = 0

⇒ x² + 7x – 98= 0

⇒ x² + 14x – 7x – 98 = 0

⇒ x(x + 14) – 7(x + 14) = 0

⇒ (x – 7)(x + 14) = 0

⇒ x = 7 or x = – 14 cm

Negative value of x is not possible, as length can't be negative

Therefore,

x = 7 cm

⇒ AB = x + 6 = 7 + 6 = 13 cm

⇒ AC = x + 8 = 7 + 8 = 15 cm

Steps of Construction:

1. Take a point O and draw a circle of radius 6 cm [ i.e. diameter 12 cm]

2. Mark a point P at a distance of 10 cm from O in any direction. Join OP

3. Draw right bisector of OP, intersecting OP at O'

4. Taking O' as center and O'O=O'P as radius, draw a circle to intersect the previous circle at T and T'.

5. Join PT and PT', which are required tangents.

6. Measured PT and PT' by a ruler and we get PT = PT' = 8 cm

For the points A, B and C to be vertices of an equilateral triangle,

AB = BC = CA and we have distance formula,

For two point P(x₁, y₁) and Q(x₂, y₂)

PQ= √((x₂ – x₁)² + (y₂ – y₁)²)

Using the above formula, and coordinates we have

AB=√((– a – a)² + (– a – a)²)

⇒ AB= √(4a² + 4a² )=√8 a

As AB = BC = AC

ABC is an equilateral triangle.

As the diagonal of rhombus divides it into two parts, it is sufficient to calculate the area of one part and double it.

Consider, the Diagonal AC,

Then,

ar(ABCD) = 2× ar(ΔABC)

Now,

A = (3,0); B = (4, 5); C = (– 1, 4)

As we know area of triangle formed by three points (x₁, y₁), (x₂, y₂) and (x₃, y₃)

=6 square units

⇒ ar(ABCD) = 2× ar(ΔABC) = 2(6) = 12 square units

Total no of numbers = 60 – 13 + 1 = 48

[As total no's from a to b are (b – a + 1)]

(a) No Divisible by 5 in the given sequence = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60}

So, we have

No of favourable outcomes = 10

No of total outcomes = 48

And,

Probability of an event

Therefore,

P(Getting a card having no divisible by 5)

(b) Perfect squares in the given sequence = {16, 25, 36, 49}

So, we have

No of favourable outcomes = 4

No of total outcomes = 48

And,

Probability of an event

Therefore,

P(Getting a card having a perfect square)

Let us consider this situation by a diagram as shown, in which AB is a building and C depicts the window and A be the top.

Now Given,

Height of window from the ground, BC = 10 m

Angle of depression of point P from window, ∠XCP = 30°

⇒ ∠XCP = ∠CPB = θ₁ = 30° [Alternate Angles]

Angle of elevation of top of the building from point P, ∠APB = 60°

⇒ ∠ APB = θ₂ = 60°

Now, In Δ BCP

Cross – Multiplying we get,

BP=10√3 meters

Now, In ΔABP

⇒ AB = 10√3 × √3 = 30 meters

So, Height of building is 30 meters.

Let AB be a tree, and P be the point of break,

And As tree falls, we can consider the situation as a right angled triangle at B

Given,

Angle of broken tree with ground, θ = 30°

Distance of top of broken tree from root, AB = 30 m

In ΔAPB

So, tree bents at a height ofmeters from the ground.

Also, In Δ APB

On cross – multiplying

AP = 10√3 × 2 = 20√3 meters

Original height of tree = AP + BP

= 20√3 + 10√3

= 30√3 meters

Given,

Radius of circle made by wire, r = 42 cm

Circumference of circle of radius r = 2πr

Circumference of circle made by wire

As, the same wire is bent to make a square the perimeter of square will be equal to circumference of circle.

Let the side of square be a.

Perimeter of square of side 'a' = 4a

We have,

4a = 264

a = 66 cm

Now,

Ratio of areas

Area of circle of radius r = πr²

Area of square of radius a = a²

Putting value, we get

Ratio of areas

Required ratio is 14 : 11

We know volume of sphere of radius R is

And

Volume of cone of radius r and height h is

So, Given,

Radius of sphere, R = 10.5 cm

Radius of cone, r = 3.5 cm

Height of cone, h = 3 cm

No of cones can be made by melting sphere

Using formulas, and putting values

No of cones

Hence, 126 cones can be made.

Let semicircle I, II and III are semicircles with diameters AB, AC and BC respectively

Area of shaded region =

Area of semicircle I + Area of semicircle II + Area of triangle ABC – Area of semicircle III

As, ∠BAC is in semicircle,

∠BAC = 90° [Angle in a semicircle is right angle]

And ABC is a right – angled triangle at A

By Pythagoras Theorem

(Hypotenuse)² = (Base)² + (Perpendicular)²

(BC)² = (AB)² + (AC)²

⇒ (BC)² = 3² + 4² = 9 + 16 = 25

⇒ BC = 5 cm

Now, For semicircle I

Diameter = AB = 3 cm

Radius

Area of semicircle of radius r

Area of semicircle I

For semicircle II

Diameter = AC = 4 cm

Radius

Area of semicircle of radius r

Area of semicircle II

For semicircle III

Diameter = BC = 5 cm

Radius,

Area of semicircle of radius r

Area of semicircle I

Area of a right – angled triangle

Area of ΔABC

Required area (From eqn [1])

Let the no of children is x and amount given to each child is y

As the total amount is 250 ₹

We have,

xy = 250

…[1]

Also, given if no of children is increased by 25, the amount to each get less by 50 paise i.e. 0.5 ₹

So, we have

(x + 25)(y – 0.5) = 250

[By 1]

⇒ 500x – x² + 12500 – 25x = 500x

⇒ x² + 25x – 12500 = 0

⇒ x² + 125x – 100x – 12500 = 0

⇒ x(x + 125) – 100(x + 125) = 0

⇒ (x – 100)(x + 125) = 0

so,

⇒ x – 100 = 0 or x + 125 = 0

⇒ x = 100 or – 125

However, no. of students can't be negative

Hence, x = 100

So, there were 100 students.

Let the shortest side be x cm [Let it be base]

Length of hypotenuse = 2x + 6 [in cm]

Length of other side = Length of hypotenuse – 2 = 2x + 6 – 2 = 2x + 4 [in cm] [Let it be perpendicular]

As we know, By Pythagoras Theorem

(hypotenuse)² = (base)² + (perpendicular)²

⇒ (2x + 6)² = x² + (2x + 4)²

⇒ 4x² + 36 + 24x = x² + 4x² + 16 + 16x

[(a + b)² = a² + b² + 2ab]

⇒ x² – 8x – 20 = 0

⇒ x² – 10x + 2x – 20 = 0

⇒ x(x – 10) + 2(x – 10) = 0

⇒ (x + 2)(x – 10) = 0

⇒ x = – 2 or x = 10 cm

However, Length can't be negative hence x = – 2 is not possible

Therefore,

x = 10 cm

we have,

Shortest Side = x = 10 cm

Hypotenuse = 2x + 6 = 2(10) + 6 = 26 cm

Third side = 2x + 4 = 2(10) + 4 = 24 cm

We know that sum of first n terms of an AP is

Where a is first term and d is common difference

So,

Sum of first 2n terms

Sum of first 3n terms

Now, Taking RHS

RHS = LHS

Hence Proved

Let the jet plane goes from point P to point C and we have given,

Initially angle of elevation from point A, ∠PAQ = θ₁ = 60°

After 15 seconds,

Angle of elevation from point A, ∠CAB = θ₂ = 30°

As the plane is flying at a constant height,

BC = PQ = 1500√3 m

Now,

In ΔABC

In ΔAPQ

So, we have

QB = AB – AQ = 4500 – 1500 = 3000 m

And

QB = PC

So, jet plane travels 3000 m in 15 seconds

And we know,

Given: A circle with center O and P be any point on a circle and XY is a tangent on circle passing through point P.

To prove : OP⏊XY

Proof :

Take a point Q on XY other than P and join OQ .

The point Q must lie outside the circle. (because if Q lies inside the circle, XY will become a secant and not a tangent to the circle).

Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP.

Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY.

So OP is perpendicular to XY.

[As Out of all the line segments, drawn from a point to points of a line not passing through the point, the smallest is the perpendicular to the line.]

Given: A quadrilateral ABCD, And a circle is circumscribed by ABCD

Also, Sides AB, BC, CD and DA touch circle at P, Q, R and S respectively.

To Prove: AB + CD = AD + BC

Proof:

In the Figure,

As tangents drawn from an external point are equal.

We have

AP = AS [tangents from point A]

BP = BQ [tangents from point B]

CR = CQ [tangents from point C]

DR = DS [tangents from point D]

Add the above equations

⇒ AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ AB + CD = AS + DS + BQ + CQ

⇒ AB + CD = AD + BC

Hence Proved.

Total area to be painted = TSA of cube + CSA of hemisphere – Base area of hemisphere

[TSA = Total surface area & CSA = Curved surface area]

Given,

Diameter of hemisphere = 4.2 cm

Radius of hemisphere, r = 2.1 cm

Side of cube, a = 5 cm

And

TSA of cube = 6a², where a is the side of cube

CSA of hemisphere = 3πr², where r is the base radius

Base area = πr² [As base is circular]

Therefore,

Total area to be painted = 6a² + 3πr² – πr² = 6a²+ 2πr²

= 6(25) + (2 × 22 × 0.3 × 2.1)

= 177.72 cm²

Given,

The diameter of lower end = 10 cm

As Radius = Diameter/2

Radius of lower end, r₂ = 5 cm

The diameter of upper end = 30 cm
Radius of upper end, r₁ = 15 cm

Height of bucket, h = 24 cm

(i) As we know

volume of frustum of a cone =

Where, h = height, r₁ and r₂ are radii of two ends (r₁ > r₂)

Capacity of bucket

=3.14 × 8 × (25 + 225 + 75)

= 3.14 × 8 × 325 = 8164 cm³

(ii) Area of metal used to make bucket = CSA of frustum + base area

We know that,

Curved surface area of frustum = πl(r₁ + r₂)

Where, r₁ and r₂ are the radii of two ends (r₁ > r₂)

And l = slant height and

l = √(h²+ (r₁ – r₂ )²)

So, we have

Slant height, l=√(24² + (15 – 5)²)

⇒ l =√(576 + 100)

⇒ l =√676

⇒ l = 26 cm

And as the base has lower end,

Base area = πr₂², where r₂ is the radius of lower end

Therefore,

Area of metal sheet used = πl(r₁ + r₂) + πr₂²

= π(26)[15 + 5] + π(5)²

= 520π + 25π

Three points A, B and C are collinear if and only if

Area(ΔABC) = 0

As we know area of triangle formed by three points (x₁, y₁), (x₂,y₂) and (x₃, y₃)

⇒ 0= – 6k + 6

⇒ 6k = 6

⇒ k = 1

So, For k = 1, A, B and C are collinear.

When two dice are thrown, the possible outcomes are

{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6), (2,1) (2,2) (2,3) (2,4) (2,5) (2,6), (3,1) (3,2) (3,3) (3,4) (3,5) (3,6), (4,1) (4,2) (4,3) (4,4) (4,5) (4,6), (5,1) (5,2) (5,3) (5,4) (5,5) (5,6), (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

The outcomes in which sum of no's is 9 are ={(3,6) (4,5) (5,4) (6,3)}

No of Total possible outcomes = 36

No of favourable outcomes = 4

And, Probability of an event

Therefore,

P(Getting sum 9)

Total area of canvas required = CSA of cylindrical part + CSA of conical part [CSA = Curved surface area]

Now,

Radius of cone = Radius of cylinder = r = 52.5 m

Height of cylindrical part, h = 3 cm = 0.03 m [As 1 m = 100 cm]

Lateral height of conical part, l = 53 m

Now, we know

CSA of cylinder = 2πrh

Where, r is base radius and h is height of cylinder and

CSA of cone = πrl

Where, r is base radius and l is slant height.

Area of canvas required = 2πrh + πrl

= πr(2h + l)

= 8754.9 m²