Bhagya

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7cm and O is the centre of the circle.

Solution:
Since O is the centre of the circle,
∴ QOR is a diameter.
⇒ ∠RPQ = 90° [Angle in a semicircle]
Now, in right ∆RPQ, RQ² = PQ²+ PR² [Pythagoras theorem]
⇒ RQ² = 24² + 7² = 576 + 49 = 625
⇒ RQ = \(\sqrt{625}\) = 25 cm
∴ Radius of a circle (r) = \(\frac{25}{2}\) cm
∴ Area of ∆RPQ
= \(\frac{1}{2}\) × PQ × RP = \(\frac{1}{2}\) × 24 × 7cm²
= 12 × 7 cm² = 84 cm²
Now, area of semicircle

∴ Area of the shaded portion
= 245.54 cm² – 84 cm² = 161.54 cm²

Question 2.
Find the area of the shaded region in given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
Radius of the outer circle (R) = 14 cm and θ = 40°

Question 3.
Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:
Side of the square = 14 cm
∴ Area of the square ABCD = 14 × 14 cm²
= 196 cm²

Similarly, area of semi-circle BPC = 77 cm²
∴ Total area of two semi-circle = (77 + 77) cm²
= 154 cm²
Area of the shaded part
= Area of the square ABCD – Area of semi-circles
= (196 – 154) cm² = 42 cm²

Question 4.
Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Solution:
Area of the equilateral triangle OAB

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.

Solution:
ABCD of the square ABCD = (4)² cm² = 16 cm².
Area of circel inside the square = πr²
= π × (1)² = π cm².

Question 6.
In a circular table cover of (he radius 32 cm, a design Is formed leaving an equilìtcral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

Solution:
Let O be the centre of a circular table and ABC be the equilateral triangle.
Then we draw OD ⊥ BC
In ∆OBD, we have: cos 60° = \(\frac{\mathrm{OD}}{\mathrm{OB}}\)

Question 7.
In the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
Solution:
Each side of the square ABCD = 14 cm

Question 8.
The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

(i) the distance around the track along its inner edge.
(ii) the area of the track.
Solution:
The distance around the track along the inner edge = Perimeter of GHI + Perimeter JKL + GL + IJ
= (π × 30 + π × 30 + 106 + 106)
= (60π + 212) m
= (60 × \(\frac{22}{7}\) + 212)m = \(\frac{2804}{7}\) m.

Similarly, area of another circular path
BCDJKLB = 1100 m²
Area of rectangular path (ABLG) = 106 × 10
= 1060 m²
Similarly, area of another rectangular path
IJDE = 1060 m²
Total area of the track
= (1100 + 1100 + 1060 + 1060) m²
= (2200 + 2120) m² = 4320 m².

Question 9.
In the given figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
O is the centre of the circle and OA = 7 cm
⇒ AB = 2OA = 2 × 7 cm = 14 cm
∵ AB and CD are perpendicular to each other.
⇒ OC ⊥ AB and OC = OA = 7 cm
Area of ∆ABC = \(\frac{1}{2}\) × AB × OC
= \(\frac{1}{2}\) × 14 cm × 7cm = 49 cm²
Again, OD = OA = 7 cm
∴ Radius of the small circle = \(\frac{1}{2}\) (OD)

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).

Solution:
Area of ∆ABC = 17320.5 cm²

⇒ (side)² = 40000
⇒ (side)² = (200)²
⇒ side = 200 cm
∴ Radius of each circle = \(\frac{200}{2}\) cm = 100 cm
Since, each angle of an equilateral triangle is 60°,
∠A = ∠B = ∠C = 60°
Area of a sector having angle of sector (θ), as 60° and radius (r) = 100 cm = \(\frac{\theta}{360^{\circ}}\) × πr²

Now, area of the shaded region = [Area of the equilateral triangle ABC] – [Area of 3 equal sectors]
= 17320.5 cm² – 15700 cm² = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Solution:
∵ The circles touch each other.
∴ The side of the square ABCD
= 3 × diameter of a circle = 3 × (2 × radius of a circle) = 3 × (2 × 7) cm = 42 cm
⇒ Area of the square ABCD = 42 × 42 cm²
= 1764 cm²
Now, area of one circle
= πr² = \(\frac{22}{7}\) × 7 × 7 cm² = 154cm²
∴ Total area of 9 circles = 9 × 154 cm² = 1386 cm²
∴ Area of the remaining portion of the handkerchief = (1764 – 1386) cm² = 378 cm²

Question 12.
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB,
(ii) shaded region.

Solution:
(i) Here, radius (r) = 3.5 cm

Question 13.
In the given figure, a square OABC is inscribed in a quadrant OPBQuestion If OA = 20 cm, find the area of the shaded region (Use π = 3.14)

Solution:
OABC is a square such that its side
OA = 20 cm
∴ OB² = OA² + AB²
⇒ OB² = 20² + 20² = 400 + 400 = 800
⇒ OB = \(\sqrt{800}=20 \sqrt{2}\)
⇒ Radius of the quadrant (r) = \(20 \sqrt{2}\) cm
Now, area of the quadrant OPBQ = \(\frac{1}{4}\) πr²
= \(\frac{1}{4} \times \frac{314}{100}\) × 800 cm² = 314 × 2 = 628 cm²
Area of the square OABC = 20 × 20 cm² = 400 cm²
∴ Area of the shaded region
= 628 cm² – 400 cm² = 228 cm²

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB = 30°, find the area of the shaded region.

Solution:
∵ Radius of bigger circle R = 21 cm and sector angle θ = 30°
∴ Area of the sector OAB

Question 15.
In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of the quadrant = 14 cm
Since, BC is a diameter in a semi-circle.
∴ ∠BAC = θ = 90°
Area of the semicircle ABPC
= [\(\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 14 × 14] cm²
= 22 × 7 cm² = 154 cm²
Area of right ∆ABC
= \(\frac{1}{2}\) × 14 × 14 cm² = 98 cm²
⇒ Area of segment BPC = 154 cm² – 98 cm²
= 56 cm²

Now, area of the shaded region = [Area of semicircle BQC] – [Area of segment BRC]
= 154 cm² – 56 cm² = 98 cm²

Question 16.
Calculate the area of the designed region in the given figure, common between the two quadrants of circles of radius 8 cm each.

Solution:
Side of the square = 8 cm
Area of the square ABCD = 8 × 8 cm² = 64 cm²
Now, radius of the quadrant ADQB = 8 cm
∴ Area of the quadrant ADQB

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the AP : 121,117,113, ……., is its first negative term? [Hint: Find n for a_n < 0]
Solution:
We have the A.P. having a = 121 and d = 117 – 121 = -4

Thus, the first negative term is 32nd term.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP
Solution:

Question 3.
A ladder has rungs 25 cm apart (see figure).The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2 \frac{1}{2}\) m apart, what is the length of the wood required for the rungs? 250
[Hint: Number of rungs = \(\frac{250}{25}+1\)]

Solution:

Length of the 1^st rung (bottom rung) = 45 cm
Length of the 11^th rung (top rung) = 25 cm
Let the length of each successive rung decreases by x cm.
∴ Total length of the rungs = 45 cm + (45 – x) cm + (45 – 2x) cm + ….. + 25 cm
Here, the number 45, (45 – x), (45 – 2x), …., 25 are in an AP such that
First term (a) = 45
and last term (l) = 25
Number of terms, n = 11

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: S_x-1 = S₄₉ – S_x]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see fig.) Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the 1st step \(\frac{1}{4} \times \frac{1}{2} \times 50 m^{3}\)]

Solution:


\(d=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\)
Here, total number of steps n = 15
Total volume of concrete required to build 15 steps is given by the sum of their individual volumes.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² – 5x + 2; \(\frac{1}{2}\), 1, -2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Solution:

Again, p(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
⇒ 1 is a zero of p(x).
Also, p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2
= -16 + 4 + 10 + 2 = -16 +16 = 0
= -2 is a zero of p(x).
Now, p(x) = 2x³ + x² – 5x + 2
∴ Comparing it with ax³ + bx² + cx + d, we have a = 2, b = 1, c = -5, and d = 2

Thus, the relationship between the coefficients and the zeroes of p(x) is verified.

(ii) Here, p(x) = x³ – 4x² + 5x – 2
∴ p(2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 18 – 18 = 0
⇒ 2 is a zero of p(x)
Again p(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 6 – 6 = 0
⇒ 1 is a zero of p(x)
Now, comparing p(x) = x³ – 4x² + 5x – 2
with ax³ + bx² + cx + d = 0, we have
a = 1, b = -4, c = 5 and d = -2
Also 2, 1 and 1 are the zeroes of p(x).
Let α = 2,
β = 1,
γ = 1
Now, sum of zeroes = α + β + γ
= 2 + 1 + 1 = 4 = -b/a

Thus, the relationship between the zeroes and the coefficients of p(x) is verified.

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:
Let the required cubic polynomial be ax³ + 6x² + cx + d = 0 and its zeroes be α, β and γ.

∴ The requied cubic polynomial is
1x³ + (-2)x² + (-7)x + 14 = 0
= x³ – 2x² – 7x + 14 = 0

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a -b, a, a + b, find a and b.
Solution:
We have p(x) = x³ – 3x² + x + 1
Comparing it with Ax³ + Bx² + Cx + D,
We have A = 1, B = -3, C = 1 and D = 1
∵ It is given (a – b), a and (a + b) are the zeroes of the polynomial.

Question 4.
If two zeroes of the polynomial
x⁴ – 6x³ – 26x²+ 138x – 35 are 2±\(\sqrt{3}\), find other zeroes.
Solution:
Here, p(x) = x⁴ – 6x³ – 26x³ + 138x – 35
∵ Two of the zeroes of p(x) are : 2 ± \(\sqrt{3}\)

(x² – 4x + 1)(x² – 2x – 35) = p(x)
⇒ (x² – 4x + 1) (x² – 7x + 5x – 35) = p(x)
⇒ (x² – 4x + 1) [x(x – 7) + 5(x – 7)] = p(x)
⇒ (x² – 4x + 1)(x – 7)(x + 5) = p(x)
i.e., (x – 7) and (x + 5) are other factors of p(x).
∴ 7 and – 5 are other zeroes of the given polynomial.

Question 5.
If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
Applying the division algorithm to the polynomials x⁴ – 6x³ + 16x² – 25x + 10 and x² – 2x + k, we have

∴ Remainder = (2k – 9)x – k(8 – k) + 10
But the remainder = x + a (Given)
Therefore, comparing them, we have

and a = -k(8 – k) + 10
= -5(8 – 5) + 10
= -5(3) + 10 = -15 + 10 = -5
Thus k = 5 and a = -5

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) \(\frac{x}{2}+\frac{2 y}{3}\) = -1 and \(x-\frac{y}{3}\) = 3
Solution:
(i) Elimination method :
x + y = 5 … (1)
2x – 3y = 4 …. (2)
Multiplying (1) by 3, we get
3x + 3y = 15 …. (3)
Adding (2) and (3) , we get
2x – 3y + 3x + 3y = 19
⇒ 5x = 19 ⇒ x = \(\frac{19}{5}\)
Now, putting x = \(\frac{19}{5}\) in (1) , we get

Substitution Method :
x + y = 5 ⇒ y = 5 – x … (1)
2x – 3y = 4 … (2)
Put y = 5 – x in (2), we get
2x – 3(5 – x)= 4 ⇒ 2x – 15 + 3x = 4

(ii) Elimination method :
3x + 4y = 10 … (1)
2x – 2y = 2 … (2)
Multiplying equation (2) by 2, we have
⇒ 4x – 4y = 4 … (3)
Adding (1)and (3) , we get
3x + 4y + 4x – 4y = 10 + 4
⇒ 7x = 14 ⇒x = \(\frac{14}{7}\) = 2
Putting x = 2 in (1) , we get,
3(2)+ 4y = 10

(iii) Elimination method :
3x – 5y – 4 = 0 … (1)
9x = 2y + 7 ⇒ 9x – 2y – 7 = 0 …. (2)
Multiplying equation (1) by (3) , we get
⇒ 9x – 15y – 12 = 0(3)
Subtracting (2)from (3),
⇒ 9x – 15y – 12 – 9x + 2y + 7 = 0

(iv) Elimination method :

Question 2.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces 1
to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i)Let the numerator = x
and the denominator = y

(ii) Let, present age of Nuri = x years
and the present age of Sonu = y years
Age of Nuri = (x – 5) years
Age of Sonu = (y – 5) years
According to question:
Age of Nuri = 3[Age of sonu]
⇒ x – 5 = 3[y – 5] ⇒ x – 5 = 3y – 15
⇒ x – 3y + 10 = 0 … (1)
10 years later :
Age of Nuri = (x + 10) years,
Age of Sonu = (y + 10) years,
According to question :
Age of Nuri = 2[Age of Sonu]
⇒ x + 10 = 2(y + 10) ⇒ x + 10 = 2y + 20
⇒ x – 2y -10 = 0 …. (2)
Subtracting (1) from (2),
x – 2y – 10 – x + 3y – 10 = 0 ⇒ y – 20 = 0 ⇒ y = 20
Putting y = 20 in (1), we get
x – 3(20) + 10 = 0 ⇒ x – 50 = 0 ⇒ x = 50
Thus, x = 50 and y = 20
∴ Age of Nuri = 50 years
and age of Sonu = 20 years

(iii) Let the digit at unit’s place = x
and the digit at ten’s place = y
∴ The number = 10y + x
The number obtained by reversing the digits = 10x + y
According to question,
9[The number] = 2 [Number obtained by reversing the digits]
9[10y+ x] = 2[10x + y]
90y + 9x = 20x + 2y
x – 8y = 0 …. (1)
Also sum of the digits = 9
x + y = 9 …. (2)
Subtracting (1) from (2), we have
x + y – x + 8y = 9
⇒ 9y = 9 ⇒ y = 1
putting y = 1 in (2), we get
x + 1 = 9 ⇒ x = 8
Thus, x = 8 and y = 1
∴ The required number = 10y + x = (10 × 1) + 8 = 10 + 8 = 18

(iv) Let the number of 50 rupees notes = x
and the number of 100 rupees notes = y
According to the condition,
Total number of notes = 25
∴ x + y = 25 …. (1)
Also, the value of all the notes = ₹ 2000
∴ 50x + 100y = 2000 ⇒ x + 2y = 40 …. (2)
Subtracting equation (1) from (2), we get
x + 2y – x – y = 40 – 25
⇒ y = 15
Putting y = 15 in (1),
x + 15 = 25
⇒ x = 25 – 15 = 10
Thus, x = 10 and y = 15
∴ Number of 50 rupees notes = 10
and number of 100 rupees notes = 15

(v) Let the fixed charge (for the three days) = ₹ x
and the additional charge for each extra day = ₹ y
According to question,
Charge for 7 days = ₹ 27
⇒ x + 4y = 27
[∵ Extra days = 7 – 3 = 4]
Also, charge of 5 days = ₹ 21
⇒ x + 2y = 21
[∵ Extra days = 5 – 3 = 2]
Subtracting (2) from (1), we get
x + 4y – x – 2y = 27 – 21
\(\Rightarrow \quad 2 y=6 \Rightarrow y=\frac{6}{2}=3\)
Putting y = 3 in (2), we have
x + 2(3) = 21
⇒ x = 21 – 6 = 15
So, x = 15 and y = 3
∴ Fixed charge = ₹ 15
and additional charge per day = ₹ 3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Solution:
(i) Here, dividend p(x) = x³ – 3x² + 5x – 3, and divisor g(x) = x² – 2
∴ We have

Thus, the quotient = (x – 3) and remainder = (7x – 9)

(ii) Here, dividend p(x) = x⁴ – 3x² + 4x + 5 and divisor g(x) = x² + 1 – x = x² – x + 1
∴ We have

Thus, the quotient = (x² + x – 3) and remainder = 8

(iii) Here, dividend, p(x) = x⁴ – 5x + 6 and divisor, g(x) = 2 – x² = – x² + 2
∴ We have

Thus, the quotient = -x² – 2 and remainder = -5x +10

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t² – 3; 2t⁴ + 3t³ -2t² – 9t – 12
(ii) x² + 3x + 1; 3x⁴ + 5x³ – 7x² + 2x +2
(iii) x³ – 3x + 1; x⁵ – 4x³ + x² + 3x + 1
Solution:
(i) Dividing 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3, we have

∵ Remainder = 0
∴ (t² – 3) is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) Dividing 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1, we have

∵ Remainder = 0
∴ x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) Dividing x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1, we get

∵ Remainder = 2, i.e., remainder = 0
∴ x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x +1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)
Solution:
We have p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5.
Given \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are zeroes of p(x).

Thus, the other zeroes of the given polynomial are -1 and -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
Here, dividend, p(x) = x³ – 3x² + x + 2, divisor = g(x), quotient = (x – 2) and remainder = (-2x + 4)
Since, (Quotient × Divisor) + Remainder = Dividend
∴ [(x – 2) × g(x)] + [(-2x + 4)] = x³ – 3x² + x + 2
⇒ (x – 2) × g(x)
= x³ – 3x² + x + 2 – (-2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
= x³ – 3x² + 3x – 2

Thus, the required divisor g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) p(x) = 3x² – 6x + 27,
g(x) = 3 and q(x) = x² – 2x + 9.
Now, deg p(x) = deg q(x)
r(x) = 0
⇒ p(x) = q(x) × g(x) + r(x)

(ii) p(x) = 2x³ – 2x² + 2x + 3,
g(x) = 2x² – 1, and
r(x) = 3x + 2, deg q(x) = deg r(x)
⇒ p(x) = q(x) × g(x) + r(x)

(iii) p(x) = 2x³ – 4x² + x + 4,
g(x) = 2x² + 1,
q(x) = x – 2 and r(x) = 6,
deg r(x) = 0
⇒ p(x) = q(x) × g(x) + r(x)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.1

In each of the following, give the justification of the construction also.

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction :
I. Draw a line segment AB = 7.6 cm.
II. Draw a ray AX making an acute angle with AB.
III. Mark 13 = (8 + 5) equal points on AX, such that AX₁ = X₁X₂ = ……….. X₁₂X₁₃.
IV. Join points X₁₃ and B.
V. From point X₅, draw X₅C || X₁₃B, which meets AB at C.
Thus, C divides AB in the ratio 5 : 8 On measuring the two parts, we get AC = 2.9 cm and CB = 4.7 cm.

Justification:
In ∆ABX₁₃ and ∆ACX₅, we have
CX₅ || BX₁₃
∴ \(\frac{A C}{C B}=\frac{A X_{5}}{X_{5} X_{13}}=\frac{5}{8}\) [By Thales theorem]
⇒ AC : CB = 5 : 8.

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Draw a ∆ABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
II. Draw a ray BX making an acute angle ∠CBX.
III. Mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂, X₃ on BX₁ such that BXj = X₁X₂ = X₂X₃.

IV. Join X₃C.
V. Draw a line through X₂ such that it is parallel to X₃C and meets BC at C’.
VI. Draw a line through C parallel to CA which intersect BA at A’.
Thus, ∆A’BC’ is the required similar triangle.
Justification :
By construction, we have X₃C || X₂C’

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
I. Construct a ∆ABC such that AB = 5 cm, BC = 7 cm and AC = 6 cm.

II. Draw a ray BX such that ∠CBX is an acute angle.
III. Mark 7 points of X₁, X₂, X₃, X₄, X₅, X₆ and X₇ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ – X₄X₅ = X₅X₆ = X₆X₇
IV Join X₅ to C.
V. Draw a line through X₇ intersecting BC (produced) at C’ such that X₅C || X₇C’
VI. Draw a line through C’ parallel to CA to intersect BA (produced) at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ Using AA similarity, ∆ABC ~ ∆A’BC’

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction :
I. Draw BC = 8 cm
II. Draw the perpendicular bisector of BC which intersects BC at D.

III. Mark a point A on the above perpendicular such that DA = 4 cm.
IV. Join AB and AC.
Thus, ∆ABC is the required isosceles triangle.
V. Now, draw a ray BX such that ∠CBX is an acute angle.
VI. On BX, mark three points [greater of 2 and 3 in \(\frac{2}{3}\)] X₁, X₂ and X₃ such that BX₁ = X₁X₂ = X₂X₃
VII. Join X₂C.
VIII. Draw a line through X₃ parallel to X₂C and intersecting BC (extended) to C’.
IX. Draw a line through C’ parallel to CA intersecting BA (extended) to A’, thus, ∆A’BC’ is the required triangle.
Justification:
We have C’A’ || CA [By construction]
∴ Using AA similarity, ∆A’BC’ ~ ∆ABC

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Steps of construction :
I. Construct a ∆ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
II. Draw a ray BX such that ∠CBX is an acute angle.

Mark four points [greater of 3 and 4 in \(\frac{3}{4}\)] X₁, X₂, X₃, X₄ on BX such that 4
BX₁ = X₁X₂ = X₂X₃ = X₃X₄
IV. Join X₄C and draw a line through X₃ parallel to X₄C to intersect BC at C’.
V. Also draw another line through C’ and parallel to CA to intersect BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
In ∆BX₄C we have
X₄C || X₃C’ [By construction]

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°.Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
Solution:
Steps of Construction :
I. Construct a AABC such that BC = 7 cm, ∠B = 45°, ∠A = 105° and ∠C = 30°
II. Draw a ray BX making an acute angle ∠CBX with BC.

III. On BX, mark four points [greater of 4 and 3 in \(\frac{4}{3}\) ] X₁, X₂, X₃ and X₄ such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄.
IV. Join X₃C.
V. Draw a line through X₄ parallel to X₃C intersecting BC(extended) at C’.
VI. Draw a line through C parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle. Justification:
By construction, we have
C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
I. Construct the right triangle ABC such that ∠B = 90°, BC = 4 cm and BA = 3 cm.
II. Draw a ray BX such that an acute angle ∠CBX is formed.
III. Mark 5 points X₁, X₂, X₃, X₄ and X₅ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅.
IV. Join X₃C.
V. Draw a line through X₅ parallel to X₃C, intersecting the extended line segment BC at C’.
VI. Draw another line through C’ parallel to CA intersecting the extended line segment BA at A’.
Thus, ∆A’BC’ is the required triangle.
Justification:
By construction, we have C’A’ || CA
∴ ∆ABC ~ ∆A’BC’ [AA similarity]

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.7 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 5 सांतत्य तथा अवकलनीयता Ex 5.7

प्रश्न संख्या 1 से 10 तक में दिए फलनों के द्वितीय कोटि के अवकलज ज्ञात कीजिए-
प्रश्न 1.
x² + 3x + 2
हल:
माना y = x² + 3x + 2

प्रश्न 2.
x²⁰
हल:
माना y = x²⁰
x के सापेक्ष अवकलन करने पर,

प्रश्न 3.
xcosx
हल:
माना y = x cosx

प्रश्न 4.
log x
हल:
y = logx
x के सापेक्ष अवकलन करने पर,

प्रश्न 5.
x³ log x
हल:
माना y = x³ logx

प्रश्न 6.
e^x sin 5x
हल:
माना y = e^xsin 5x
x के सापेक्ष अवकलन करने पर,

प्रश्न 7.
e^6x cos 3x
हल:
माना y = e^6x cos 3x
x के सापेक्ष अवकलन करने पर,

= e^6x [-6sin 3x. 3 – 3cos 3x. 3] + [6cos 3x – 3sin 3x] × e^6x . 6
= e^6x[-18sin 3x – 9cos 3x] + e^6x[36cos 3x – 18sin 3x]
= e^6x[-18sin 3x – 9cos 3x + 36cos 3x – 18sin 3x]
= e^6x[-36sin 3x + 27cos 3x]
= 9e^6x[3 cos 3x – 4sin 3x]

प्रश्न 8.
tan^-1x
हल:
माना y = tan^-1 x
x के सापेक्ष अवकलन करने पर,

प्रश्न 9.
log (log x)
हल:
माना y = log (log x)

प्रश्न 10.
sin (log x)
हल:
माना y = sin (log x)
x के सापेक्ष अवकलन करने पर,

प्रश्न 11.
यदि y = 5 cosx – 3 sin x है तो सिद्ध कीजिए कि \(\frac{d^{2} y}{d x^{2}}\) + y = 0
हल:
∵ y = 5cosx – 3sin x

प्रश्न 12.
यदि y = cos^-1x है तो \(\frac{d^{2} y}{d x^{2}}\) पदों में ज्ञात कीजिए-
हल:
∵ y = cos^-1x

प्रश्न 13.
यदि y = 3 cos (log x) + 4 sin (log x) है तो दर्शाइए कि x²y₂ + xy₁ + y = 0
हल:
∵ y = 3 cos (log x) + 4 sin (log x)


इसलिए
x²y₂ + xy₁ + y = – 7cos (log x) – sin log x – 3sin (log x) + 4cos (cosx) + 3cos (log x) + 4 sin (log x)
= -7cos (log x) – sin (logx) + sin (log x) + 7cos (log x)
= 0
अत: x²y₂ + xy₁ + y = 0

प्रश्न 14.
यदि y = Ae^mx + Be^nx है तो दर्शाइए कि

हल:
∵ y = Ae^mx + Be^nx

प्रश्न 15.
यदि y = 500e^7x + 600^-7x है तो दर्शाइए कि
\(\frac{d^{2} y}{d x^{2}}\) = 49 y है।
हल:
y = 500e^7x + 600^-7x
x के सापेक्ष अवकलन करने पर,

= 3500e^7x – 4200e^-7x
पुनः x के सापेक्ष अवकलन करने पर,
\(\frac{d^{2} y}{d x^{2}}=3500 \frac{d}{d x} e^{7 x}-4200 \frac{d}{d x} e^{-7 x}\)
=3500 × e^7x × 7 – 4200 × e^-7x x (-7)
= 500 × 49e^7x + 600 × 49e^-7x
= 49(500e^7x + 600e^-7x)
= 49y

प्रश्न 16.
यदि e^y (x + 1) = 1 है तो दर्शाइए कि \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\) है|
हल:
e^y (x + 1) = 1

प्रश्न 17.
यदि y = (tan^-1x)² है तो दर्शाइए कि (x² + 1²)y₂ + 2x (x² + 1) y₂ = 2 है।
हल:
y = (tan^-1x)²
x के सापेक्ष अवकलन करने पर,

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 9 अवकल समीकरण Ex 9.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 9 अवकल समीकरण Ex 9.1

1 से 10 तक के प्रश्नों में प्रत्येक अवकल समीकरण की कोटि एवं घात ( यदि परिभाषित हो) ज्ञात कीजिए
प्रश्न 1.
\(\frac{d^{4} y}{d x^{4}}\) + sin (y”’) = 0
हल:
इस अवकल समीकरण में उच्चतम अवकलज कोटि \(\frac{d^{4} y}{d x^{4}}\) है इसलिए समीकरण की कोटि 4 है। इस समीकरण का बायाँ पक्ष अवकलजों में बहुपद नहीं है इसलिए इसकी घात परिभाषित नहीं है।

प्रश्न 2.
y’ + 5y = 0
हल:
कोटि → 1
घात → 1

प्रश्न 3.
\(\left(\frac{d s}{d t}\right)^{4}+3 s \frac{d^{2} s}{d t^{2}}=0\)
हल:
ये कोटि 2 है तथा घात 1 है।

प्रश्न 4.

हल:
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\) की कोटि 2 है तथा इसके बायें पद में कोई फलन नहीं है।
अतः इसकी घात परिभाषित नहीं है।

प्रश्न 5.
\(\frac{d^{2} y}{d x^{2}}\) = cos 3x + sin 3x
हल:
∵ उच्चतम अवकलज \(\frac{d^{2} y}{d x^{2}}\) है।
∴ इसकी कोटि 2 है तथा दायें पद में बहुपद की उच्चतम घात 1 है।
इसलिए समीकरण की घात 1 है।

प्रश्न 6. (y”’)² (y”)³ + (y’)⁴ + y⁵ = 0
हल:
⇒ \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{4}\) + y⁵ = 0
उपरोक्त समीकरण में उच्चतम अवकलज \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}\) है।
∴ समी० की कोटि 3 है।
तथा घात 5 है।

प्रश्न 7.
y” + 2y” + y = 0
हल:
अवकल समीकरण की कोटि-3 परिभाषित नहीं है।

प्रश्न 8.
y + y = e^x
हल:
अवकल समीकरण की कोटि 1 परिभाषित नहीं है।

प्रश्न 9.
y” + (y’)² + 2y = 0
हल:
अवकल समीकरण की कोटि 2 परिभाषित नहीं है।

प्रश्न 10.
y’ + 2y’ + sin y = 0
हल:
अवकल समीकरण की कोटि 2 परिभाषित नहीं है।

प्रश्न 11.
अवकल समीकरण
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0\) की घात है-
(A) 3
(B) 2
(C) 1
(D) परिभाषित नहीं है
हल:
यह समीकरण अवकलज़ में बहुपदीय नहीं है; अतः यह परिभाषित नहीं है।
अतः विकल्प (D) सही है।

प्रश्न 12.
अवकल समीकरण \(2 x^{2} \frac{d^{2} y}{d x^{2}} \cdot 3 \frac{d y}{d x}+y=0\) की कोटि है
(A) 2
(B) 1
(C) 0
(D) परिभाषित नहीं है
हल:
अवकल समीकरण \(2 x^{2} \frac{d^{2} y}{d x^{2}} \cdot 3 \frac{d y}{d x}+y=0\) की कोटि 2 है।
अतः विकल्प (A) सही है।

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 समाकलन Ex 7.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 7 समाकलन Ex 7.2

प्रश्न 1 से 37 तक के प्रश्नों में प्रत्येक फलन का समाकलन ज्ञात कीजिए-
प्रश्न 1.
\(\frac{2 x}{1+x^{2}}\)
हल:

प्रश्न 2.
\(\frac{(\log x)^{2}}{x}\)
हल:

प्रश्न 3.
\(\frac{1}{x+x \log x}\)
हल:

प्रश्न 4.
sin x sin (cosx)
हल:

प्रश्न 5.
sin (ax + b) cos (ax + b)
हल:

प्रश्न 6.
\(\sqrt{a x+b}\)
हल:

प्रश्न 7.
\(x \sqrt{x+2}\)
हल:

प्रश्न 8.
\(x \sqrt{1+2 x^{2}}\)
हल:

प्रश्न 9.
(4x + 2) \(\sqrt{x^{2}+x+1}\)
हल:

प्रश्न 10.
\(\frac{1}{x-\sqrt{x}}\)
हल:

प्रश्न 11.
\(\frac{x}{\sqrt{x+4}}\), x > 0
हल:

प्रश्न 12.
(x³ – 1)^1/3 . x⁵
हल:

प्रश्न 13.
\(\frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}\)
हल:

प्रश्न 14.
\(\frac{1}{x(\log x)^{m}}\), x > 0, m ≠ 1
हल:

प्रश्न 15.
\(\frac{x}{9-4 x^{2}}\)
हल:

प्रश्न 16.
e^2x+3
हल:

प्रश्न 17.
\(\frac{x}{e^{x^{2}}}\)
हल:

प्रश्न 18.
\(\frac{e^{\tan ^{-1} x}}{1+x^{2}}\)
हल:

प्रश्न 19.
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
हल:

प्रश्न 20.
\(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\)
हल:

प्रश्न 21.
tan²(2x – 3)
हल:

प्रश्न 22.
sec² (7 – 4x)
हल:

प्रश्न 23.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
हल:

प्रश्न 24.

हल:

प्रश्न 25.

हल:

प्रश्न 26.
\(\frac{\cos \sqrt{x}}{\sqrt{x}}\)
हल:

प्रश्न 27.
\(\sqrt{\sin 2 x} \cdot \cos 2 x\)
हल:

प्रश्न 28.
\(\frac{\cos x}{\sqrt{1+\sin x}}\)
हल:

प्रश्न 29.
cot x log sin x
हल:

प्रश्न 30.
\(\frac{\sin x}{1+\cos x}\)
हल:

प्रश्न 31.
\(\frac{\sin x}{(1+\cos x)^{2}}\)
हल:

प्रश्न 32.
\(\frac{1}{1+\cot x}\)
हल:

प्रश्न 33.
\(\frac{1}{1-\tan x}\)
हल:

प्रश्न 34.
\(\frac{\sqrt{\tan x}}{\sin x \cos x}\)
हल:

प्रश्न 35.
\(\frac{(1+\log x)^{2}}{x}\)
हल:

प्रश्न 36.

हल:

प्रश्न 37.

हल:

प्रश्न 38 व 39 में सही उत्तर का चयन कीजिए
प्रश्न 38.


हल:

प्रश्न 39.
\(\int \frac{d x}{\sin ^{2} x \cos ^{2} x}\) बराबर है-
(A) tan x + cot x + C
(B) tan x – cotx + C
(C) tan x cot x + C
(D) tan x – cot 2x + C
हल:

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Take π = \(\frac{22}{7}\), unless stated otheriwise

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Radius of the sphere (r₁) = 4.2 cm
∴ Volume of the sphere = \(\frac{4}{3}\) πr₁³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \mathrm{cm}^{3}\)
Radius of the cylinder (r₂) = 6 cm
Let h be the height of the cylinder.
∴ Volume of the cylinder = πr²h
= \(\frac{22}{7}\) × 6 × 6 × h cm³
Since volume of the metallic sphere = Volume of the cylinder

Hence, the height of the cylinder is 2.744 cm

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radii of the given spheres are
r₁ = 6 cm, r₂ = 8 cm and r₃ = 10 cm
⇒ Volume of the given spheres are

= \(\frac{4}{3} \times \frac{22}{7}\) × [1728] cm³
Let the radius of the new big sphere be R. Volume of the new sphere
= \(\frac{4}{3}\) × π × R₃ = \(\frac{4}{3} \times \frac{22}{7}\) × R₃
Since, the two volumes must be equal.
∴ \(\frac{4}{3} \times \frac{22}{7} \times R^{3}=\frac{4}{3} \times \frac{22}{7} \times 1728\)
⇒ R₃ = 1728 ⇒ R = 12 cm
Thus, the required radius of the resulting sphere is 12 cm.

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Diameter of the cylindrical well = 7 m
⇒ Radius of the cylindrical well (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
∴ Volume = πr²h = \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 20 m³
= 22 × 7 × 5 m³
⇒ Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having length = 22 m, breadth = 14 m
Let h be the height of the platform.
∴ Volume of the platform = 22 × 14 × h m³
Since, the two volumes must be equal
∴ 22 × 14 × h = 22 × 7 × 5
Thus, the required height of the platform is 2.5 m.

Question 4.
A well of diameter 3 m is dug 14m deep. The earth taken out of it has been spread evenly all around it in shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of cylindrical well = 3 m
⇒ Radius of the cylindrical well = \(\frac{3}{2}\) m = 1.5 m
Depth of well (h) = 14 m
∴ Volume of cylindrical well

Let the height of the embankment = H m.
Internal radius of the embankment (r) = 1.5 m.
External radius of the embankment (R)
= (4 + 1.5) m = 5.5 m.
∴ Volume of the embankment
= πR²H – πr²H = πH [R² – r²]
= πH (R + r) (R – r)
= \(\frac{22}{7}\) × H (5.5 + 1.5)(5.5 – 1.5)
= \(\frac{22}{7}\) × H × 7 × 4m³
Since, Volume of the embankment=Volume of the cylindrical well
⇒ \(\frac{22}{7}\) × H × 7 × 4 = 99
⇒ H = 99 × \(\frac{7}{22} \times \frac{1}{7} \times \frac{1}{4} m=\frac{9}{8} m\) = 1.125 m
So, the required height of the embankment = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and 6. diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
For the circular cylinder:

Diameter = 12 cm
⇒ Radius (r) = \(\frac{12}{2}\) = 6cm and height (h) = 15 cm
∴ Volume of circular cylinder
= πr²h = \(\frac{12}{2}\) × 6 × 6 × 15 cm³
For conical and hemispherical part of icecream :
Diameter = 6 cm ⇒ radius (R) = 3 cm
Height of conical part (H) = 12 cm

Volume of ice cream cone = (Volume of the conical part) + (Volume of the hemispherical part)

Thus, the required number of cones is 10.

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
For a circular coin:

Diameter = 1.75 cm
⇒ Radius (r) = \(\frac{175}{200}\) cm
Thickness (h) = 2mm = \(\frac{2}{10}\) cm
∴ Volume of one coin = πr²h = \(\frac{22}{7} \times\left(\frac{175}{200}\right)^{2} \times \frac{2}{10} \mathrm{cm}^{3}\)
For a cuboid:

Length (l) = 10 cm,
Breadth (b) = 5.5 cm
and height (h) = 3.5 cm
∴ Volume = l × b × h = 10 × \(\frac{55}{10} \times \frac{35}{10}\) cm³

Thus, the required number of coins = 400.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
For the cylindrical bucket:
Radius (r) = 18 cm and height (h) = 32 cm
Volume of cylindrical bucket = πr²h
= \(\frac{22}{7}\) × (18)² × 32 cm³
⇒ Volume of the sand = (\(\frac{22}{7}\) × 18 × 18 × 32) cm³
For the conical heap:
Height (H) = 24 cm
Let radius of the base be R.
∴ Volume of conical heap

Thus, the required radius = 36 cm and slant height = \(12 \sqrt{13}\) cm.

Question 8.
Water in a canal, 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:
Width of the canal = 6 m,
Depth of the canal = 1.5 m
Length of the water column in 1 hr = 10 km
∴ Length of the water column in 30 minutes
(i.e., \(\frac{1}{2}\)hr) = \(\frac{10}{2}\) km = 5 km = 5000 m
∴ Volume of water flown in \(\frac{1}{2}\) hr
= 6 × 1.5 × 5000 m³ = 6 × \(\frac{15}{10}\) × 5000 m³
= 45000 m³
Since, the above amount (volume) of water is spread in the form of a cuboid of height
8 cm (= \(\frac{8}{100}\) m)
Let the area of the cuboid = a
∴ Volume of the cuboid = Area × Height

= 562500 m² = 56.25 hectares
Thus, the required area is 56.25 hectares.

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Diameter of the pipe = 20 cm
⇒ Radius of the pipe (r) = \(\frac{20}{2}\) cm = 10 cm
Since, the water flows through the pipe at 3 km/hr.
∴ Length of water column per hour(h) = 3 km
= 3 × 1000 m = 3000 × 100 cm = 300000 cm.
Length of water column per hour(h) = 3 km
Volume of water flown in one hour = πr²h
= π × 10² × 300000 cm³ = π × 30000000 cm²
Now, for the cylindrical tank :
Diameter = 10 m
⇒ Radius (R) = \(\frac{10}{2}\) m = 5 × 100 cm = 500 cm
Height (H) = 2 m = 2 × 100 cm = 200 cm
∴ Volume of the cylindrical tank = πR²H
= π × (500)² × 200 cm³
Now, time required to fill the tank