Class 10 Math Workbook – Real Numbers

HowQsir Maths Workbook

Class X | NCERT Solutions

Chapter 1: Real Numbers

Exercise 1.1 – The Fundamental Theorem of Arithmetic

Mathematics,Class 10, Ex-1.2

Class 10 Math Workbook – Real Numbers

Howqsir Maths Workbook

Class 10 | Chapter 1: Real Numbers

EXERCISE 1.2

Q1.

Prove that √5 is irrational.

Logic Used: If p is a prime number and p divides a², then p divides a. (Theorem 1.3)

Step 1: Let us assume, to the contrary, that √5 is rational.

Step 2: Therefore, we can find integers a and b (where b ≠ 0) such that:

√5 = a/b

Suppose a and b have a common factor other than 1. Then we divide by the common factor to get a and b as co-prime integers.

Step 3: Rearranging the equation, we get:

b√5 = a

Squaring both sides, we get:

5b² = a² …(1)

Step 4: From (1), we can say that 5 divides a².

By the fundamental theorem (If 5 divides a², then 5 must divide a), it follows that 5 divides a.

So, we can write a = 5c for some integer c.

Step 5: Substituting a = 5c in equation (1), we get:

5b² = (5c)²

5b² = 25c²

b² = 5c²

This means 5 divides b², and so 5 divides b.

Conclusion: Therefore, a and b have at least 5 as a common factor. But this contradicts the fact that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that √5 is rational. So, we conclude that √5 is irrational.

Q2.

Prove that 3 + 2√5 is irrational.

Logic: If (a – 3b)/2b is a ratio of integers, it is rational.

Step 1: Let 3 + 2√5 be rational.

Then 3 + 2√5 = a/b, where a, b are co-prime integers.

Step 2: Isolate the irrational part:

2√5 = a/b – 3

2√5 = (a – 3b) / b

Step 3: Final Rearrangement:

√5 = (a – 3b) / 2b

Analysis: Since a and b are integers, (a – 3b) / 2b is a rational number.

This means √5 is rational. But this contradicts the fact that √5 is irrational.

Hence, 3 + 2√5 is irrational.

Q3.

Prove that the following are irrationals:

(i) 1 / √2

Assume 1/√2 = a/b (where a, b are co-prime).

Reciprocating both sides: √2 = b/a.

Since b/a is a rational number, √2 is rational, which is a contradiction.

Hence, 1/√2 is irrational.

(ii) 7√5

Assume 7√5 = a/b (where a, b are co-prime).

Rearranging: √5 = a / (7b).

Since a/7b is rational, √5 is rational, which is a contradiction.

Hence, 7√5 is irrational.

(iii) 6 + √2

Assume 6 + √2 = a/b (where a, b are co-prime).

Rearranging: √2 = a/b – 6 = (a – 6b) / b.

Since (a – 6b)/b is rational, √2 is rational, which is a contradiction.

Hence, 6 + √2 is irrational.

Mathematics,Class 10, Ex-12.1

Class 10 Maths: Surface Areas and Volumes

Question 1

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

STEP 1: FIND SIDE OF CUBE

Volume = a³ = 64 cm³
Therefore, a = 4 cm.

STEP 2: DIMENSIONS OF CUBOID

Length (l) = 4 + 4 = 8 cm
Breadth (b) = 4 cm, Height (h) = 4 cm.

STEP 3: CALCULATION

Surface Area = 2(lb + bh + hl)
= 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32)
= 2(80) = 160 cm².

Question 2

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

STEP 1: RADII

Diameter = 14 cm, so Radius (r) = 7 cm.

STEP 2: HEIGHTS

Total Height = 13 cm
Cylinder height (h) = 13 – 7 = 6 cm.

STEP 3: INNER AREA

Area = 2πrh + 2πr² = 2πr(h+r)
= 2 × 22/7 × 7 × (6 + 7) = 572 cm².

Question 3

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

STEP 1: FIND CONE HEIGHT

Radius (r) = 3.5 cm
Height of cone (h) = 15.5 – 3.5 = 12 cm.

STEP 2: SLANT HEIGHT

l = √(h² + r²) = √(12² + 3.5²) = 12.5 cm.

STEP 3: TOTAL AREA

Area = πrl + 2πr²
= 22/7 × 3.5 × 12.5 + 2 × 22/7 × (3.5)² = 214.5 cm².

Question 4

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

STEP 1: MAX DIAMETER

Greatest diameter = side of cube = 7 cm.

STEP 2: AREA LOGIC

Total Area = (Area of 6 faces) + (CSA of Hemisphere) – (Area of circular base of hemisphere).

STEP 3: RESULT

Area = 6(7²) + 2π(3.5)² – π(3.5)²
= 294 + 38.5 = 332.5 cm².

Question 5

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

STEP 1: SURFACE AREA FORMULA

Remaining Area = 6l² + 2π(l/2)² – π(l/2)².

STEP 2: SIMPLIFICATION

Area = 6l² + πl²/4.

STEP 3: FINAL ANSWER

Area = l²/4 (24 + π) square units.

Question 6

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

STEP 1: DIMENSIONS

Radius (r) = 2.5 mm
Height of cylinder (h) = 14 – 5 = 9 mm.

STEP 2: SURFACE COMPONENTS

Total Area = CSA of cylinder + 2 × CSA of hemisphere.

STEP 3: RESULT

Area = 2πr(h + 2r) = 2 × 22/7 × 2.5 × 14 = 220 mm².

Question 7

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)

STEP 1: FIND SURFACE AREA

CSA = 2πrh + πrl = πr(2h + l)
= 22/7 × 2 × (2 × 2.1 + 2.8) = 44 m².

STEP 2: COST CALCULATION

Rate = ₹ 500/m²
Cost = Area × Rate = 44 × 500.

STEP 3: TOTAL

Total Cost = ₹ 22,000.

Question 8

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

STEP 1: FIND SLANT HEIGHT

r = 0.7 cm, h = 2.4 cm
l = √(0.7² + 2.4²) = 2.5 cm.

STEP 2: COMBINED AREA

Area = CSA Cylinder + CSA Cone + Base Area
= 2πrh + πrl + πr².

STEP 3: CALCULATION

Area = 17.6 cm²
To the nearest cm², Area = 18 cm².

Question 9

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, if the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

STEP 1: AREA COMPONENTS

Total Area = CSA of cylinder + 2 × CSA of hemisphere.

STEP 2: SUBSTITUTION

Area = 2πrh + 2(2πr²) = 2πr(h + 2r)
= 2 × 22/7 × 3.5 × (10 + 7).

STEP 3: RESULT

Area = 22 × 17 = 374 cm².

Class 10 Maths – Circles Workbook

Mathematics – Class X (Important Question)

Q. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Step 1: Given Information

Let the circle touch sides BC, AB, and AC at points D, E, and F respectively.
Radius of circle (r) = 4 cm.
BD = 8 cm and DC = 6 cm.

Step 2: Theorem of Tangents

Theorem: Lengths of tangents from an external point to a circle are equal.

1. CD = CF = 6 cm (Tangents from C)

2. BD = BE = 8 cm (Tangents from B)

3. Let AE = AF = x cm (Tangents from A)

Step 3: Calculating Sides & s

Side a (BC) = 6 + 8 = 14 cm

Side b (AC) = 6 + x cm

Side c (AB) = 8 + x cm

Semi-perimeter (s) = (14 + 6 + x + 8 + x) / 2

s = (28 + 2x) / 2 = (14 + x) cm

Step 4: Area using Heron’s Formula

Area = √[s(s-a)(s-b)(s-c)]

s – a = (14+x) – 14 = x
s – b = (14+x) – (6+x) = 8
s – c = (14+x) – (8+x) = 6

Area of ΔABC = √[(14+x) · x · 8 · 6]
Area = √[48x(14+x)] … (i)

Step 5: Area using Sub-triangles

Area(ΔABC) = Area(ΔOBC) + Area(ΔOCA) + Area(ΔOAB)

Area = [1/2 · BC · r] + [1/2 · AC · r] + [1/2 · AB · r]
Area = 1/2 · 4 · (BC + AC + AB)
Area = 2 · (14 + 6+x + 8+x)
Area = 2 · (28 + 2x) = 4(14+x) … (ii)

Step 6: Final Calculation

Equating (i) and (ii):

√[48x(14+x)] = 4(14+x)
48x(14+x) = 16(14+x)² (Squaring sides)
3x = 14 + x
2x = 14
x = 7 cm

AB = 8 + 7 = 15 cm
AC = 6 + 7 = 13 cm

Mathematics,Class 10, Ex-12.2

Class X Math Workbook – Exercise 12.2

Class X Mathematics

Surface Areas and Volumes

Exercise 12.2 Unless stated otherwise, take \(\pi = \frac{22}{7}\)

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to \(1\text{ cm}\) and the height of the cone is equal to its radius. Find the volume of the solid in terms of \(\pi\).

Step 1: Identify Given Data

Radius of the hemisphere part \((r) = 1\text{ cm}\)
Radius of the cone part \((r) = 1\text{ cm}\)
Height of the cone part \((h) = \text{radius} = 1\text{ cm}\)

Step 2: Logic & Formula

The solid is composed of two parts: a cone and a hemisphere. To find the total volume, we add the volumes of these two shapes.

\text{Total Volume } (V) = \text{Volume of Cone} + \text{Volume of Hemisphere}

V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3

Step 3: Substitution

Substitute \(r = 1\) and \(h = 1\) into the formula:

V = \left[ \frac{1}{3}\pi(1)^2(1) \right] + \left[ \frac{2}{3}\pi(1)^3 \right]

V = \frac{1}{3}\pi(1) + \frac{2}{3}\pi(1)

Step 4: Result

V = \frac{\pi}{3} + \frac{2\pi}{3}

V = \frac{\pi + 2\pi}{3} = \frac{3\pi}{3} = \pi

Conclusion: The volume of the solid is \(\pi\text{ cm}^3\).

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is \(3\text{ cm}\) and its length is \(12\text{ cm}\). If each cone has a height of \(2\text{ cm}\), find the volume of air contained in the model.

Step 1: Extract Dimensions

Diameter of model \(= 3\text{ cm} \implies\) Radius \((r) = \frac{3}{2} = 1.5\text{ cm}\)
Height of each cone \((h_1) = 2\text{ cm}\)
Total length of model \(= 12\text{ cm}\)
Height of cylinder \((h_2) = \text{Total length} – 2 \times (\text{height of cone})\)
\(h_2 = 12 – (2 + 2) = 8\text{ cm}\)

Step 2: Logical Combination

Volume of air = Volume of cylindrical portion + Volume of 2 conical ends.

V = (\pi r^2 h_2) + 2 \times \left( \frac{1}{3}\pi r^2 h_1 \right)

Factor out common terms to simplify: \(\pi r^2\)

V = \pi r^2 \left( h_2 + \frac{2}{3}h_1 \right)

Step 3: Solve

V = \frac{22}{7} \times (1.5)^2 \times \left( 8 + \frac{2}{3}(2) \right)

V = \frac{22}{7} \times 2.25 \times \left( 8 + \frac{4}{3} \right)

V = \frac{22}{7} \times \frac{9}{4} \times \frac{28}{3}

V = \frac{22 \times 9 \times 28}{7 \times 4 \times 3} = 22 \times 3 = 66\text{ cm}^3

Answer: The volume of air contained in the model is \(66\text{ cm}^3\).

3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length \(5\text{ cm}\) and diameter \(2.8\text{ cm}\).

Step 1: Single Unit Dimensions

Radius of hemisphere part \((r) = \frac{2.8}{2} = 1.4\text{ cm}\)
Radius of cylinder part \((r) = 1.4\text{ cm}\)
Length of cylindrical part \((h) = \text{Total length} – 2r\)
\(h = 5 – (1.4 + 1.4) = 2.2\text{ cm}\)

Step 2: Formula for 1 Gulab Jamun

One unit = 1 Cylinder + 2 Hemispheres (which equals 1 Sphere).

V_1 = \pi r^2 h + \frac{4}{3}\pi r^3

V_1 = \pi r^2 \left( h + \frac{4}{3}r \right)

V_1 = \frac{22}{7} \times 1.4 \times 1.4 \times \left( 2.2 + \frac{4}{3} \times 1.4 \right)

V_1 = 6.16 \times \left( \frac{6.6 + 5.6}{3} \right) = 6.16 \times \frac{12.2}{3} \approx 25.05\text{ cm}^3

Step 3: Total Syrup Calculation

Volume of 45 gulab jamuns \(= 45 \times 25.05 = 1127.25\text{ cm}^3\)

Quantity of syrup \(= 30\% \text{ of the total volume}\)

\text{Syrup} = \frac{30}{100} \times 1127.25

\text{Syrup} \approx 338.175\text{ cm}^3

Conclusion: There is approximately \(338\text{ cm}^3\) of syrup.

4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are \(15\text{ cm}\) by \(10\text{ cm}\) by \(3.5\text{ cm}\). The radius of each of the depressions is \(0.5\text{ cm}\) and the depth is \(1.4\text{ cm}\). Find the volume of wood in the entire stand.

Step 1: Calculate Raw Volume

First, find the volume of the solid wooden cuboid before holes were drilled.

V_{cuboid} = l \times b \times h = 15 \times 10 \times 3.5

V_{cuboid} = 525\text{ cm}^3

Step 2: Volume of Depressions

The holes are conical. Radius \((r) = 0.5\text{ cm}\) and Depth \((d) = 1.4\text{ cm}\).

V_{hole} = \frac{1}{3} \pi r^2 d

V_{4holes} = 4 \times \left[ \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 \right]

V_{4holes} = 4 \times \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4

V_{4holes} \approx 1.47\text{ cm}^3

Step 3: Subtraction Logic

The wood was removed to make the holes, so we subtract the volume of the 4 cones from the cuboid.

V_{wood} = 525 – 1.47 = 523.53\text{ cm}^3

Answer: The volume of wood in the entire stand is \(523.53\text{ cm}^3\).

5. A vessel is in the form of an inverted cone. Its height is \(8\text{ cm}\) and the radius of its top, which is open, is \(5\text{ cm}\). It is filled with water up to the brim. When lead shots, each of which is a sphere of radius \(0.5\text{ cm}\) are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Step 1: Water in Cone

The vessel is a cone with \(R=5, H=8\).

V_{cone} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \times \pi \times 5^2 \times 8 = \frac{200\pi}{3}\text{ cm}^3

Step 2: Physics Logic (Archimedes)

One-fourth of water flows out. The volume of this water equals the total volume of all lead shots dropped.

V_{displaced} = \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}\text{ cm}^3

Step 3: Volume of 1 Lead Shot

Each shot is a sphere with \(r=0.5\text{ cm}\).

V_{shot} = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times (0.5)^3

V_{shot} = \frac{4}{3} \times \pi \times \frac{1}{8} = \frac{\pi}{6}\text{ cm}^3

Step 4: Algebraic Equation

Let \(n\) be the number of shots. Then \(n \times V_{shot} = V_{displaced}\).

n \times \frac{\pi}{6} = \frac{50\pi}{3}

n = \frac{50\pi \times 6}{3 \times \pi} = \frac{300}{3} = 100

Answer: 100 lead shots were dropped.

6. A solid iron pole consists of a cylinder of height \(220\text{ cm}\) and base diameter \(24\text{ cm}\), which is surmounted by another cylinder of height \(60\text{ cm}\) and radius \(8\text{ cm}\). Find the mass of the pole, given that \(1\text{ cm}^3\) of iron has approximately \(8\text{ g}\) mass. (Use \(\pi = 3.14\))

Step 1: Vol. of Base Cylinder

Height \((h_1) = 220\text{ cm}\)
Radius \((r_1) = 24/2 = 12\text{ cm}\)

V_1 = \pi \times (12)^2 \times 220 = 144 \times 220 \times 3.14

V_1 = 31680 \times 3.14\text{ cm}^3

Step 2: Vol. of Top Cylinder

Height \((h_2) = 60\text{ cm}\)
Radius \((r_2) = 8\text{ cm}\)

V_2 = \pi \times 8^2 \times 60 = 64 \times 60 \times 3.14

V_2 = 3840 \times 3.14\text{ cm}^3

Step 3: Mass Calculation

\text{Total Volume} = (31680 + 3840) \times 3.14 = 35520 \times 3.14 = 111532.8\text{ cm}^3

\text{Mass} = \text{Volume} \times \text{Density} = 111532.8 \times 8\text{ g}

\text{Mass} = 892262.4\text{ g} \approx 892.26\text{ kg}

Conclusion: The mass of the pole is approximately \(892.26\text{ kg}\).

7. A solid consisting of a right circular cone of height \(120\text{ cm}\) and radius \(60\text{ cm}\) standing on a hemisphere of radius \(60\text{ cm}\) is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is \(60\text{ cm}\) and its height is \(180\text{ cm}\).

Step 1: Vol. of Container

The cylinder holds all the water initially. \(R=60, H=180\).

V_{cylinder} = \pi R^2 H = \pi \times 60^2 \times 180

V_{cylinder} = 648000\pi\text{ cm}^3

Step 2: Vol. of the Solid

Solid = Cone \((h=120, r=60)\) + Hemisphere \((r=60)\).

V_{solid} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3

V_{solid} = \frac{1}{3}\pi (60)^2 (120) + \frac{2}{3}\pi (60)^3

V_{solid} = 144000\pi + 144000\pi = 288000\pi\text{ cm}^3

Step 3: Result in \(m^3\)

V_{left} = V_{cylinder} – V_{solid} = 648000\pi – 288000\pi

V_{left} = 360000\pi = 360000 \times \frac{22}{7} \approx 1131428.57\text{ cm}^3

1\text{ m}^3 = 1000000\text{ cm}^3 \implies V \approx 1.131\text{ m}^3

Conclusion: Volume of water left is \(1.131\text{ m}^3\).

8. A spherical glass vessel has a cylindrical neck \(8\text{ cm}\) long, \(2\text{ cm}\) in diameter; the diameter of the spherical part is \(8.5\text{ cm}\). By measuring the amount of water it holds, a child finds its volume to be \(345\text{ cm}^3\). Check whether she is correct, taking the above as the inside measurements, and \(\pi = 3.14\).

Step 1: Vol. of Neck

Radius \(r = 1\text{ cm}\), Height \(h = 8\text{ cm}\).

V_1 = \pi r^2 h = 3.14 \times 1^2 \times 8 = 25.12\text{ cm}^3

Step 2: Vol. of Sphere

Radius \(R = 8.5/2 = 4.25\text{ cm}\).

V_2 = \frac{4}{3} \pi R^3 = \frac{4}{3} \times 3.14 \times (4.25)^3

V_2 = \frac{4}{3} \times 3.14 \times 76.765625 \approx 321.39\text{ cm}^3

Step 3: Comparison

V_{total} = V_1 + V_2 = 25.12 + 321.39 = 346.51\text{ cm}^3

The child measured \(345\text{ cm}^3\).

Conclusion: She is incorrect. The correct volume is \(346.51\text{ cm}^3\).

Interactive Surface Areas and Volumes Study Pack

Mathematics,Class 10, Ex-1.2

Prove that √5 is irrational.

Prove that 3 + 2√5 is irrational.

Prove that the following are irrationals:

(i) 1 / √2

(ii) 7√5

(iii) 6 + √2

Mathematics,Class 10, Ex-12.1

STEP 1: FIND SIDE OF CUBE

STEP 2: DIMENSIONS OF CUBOID

STEP 3: CALCULATION

STEP 1: RADII

STEP 2: HEIGHTS

STEP 3: INNER AREA

STEP 1: FIND CONE HEIGHT

STEP 2: SLANT HEIGHT

STEP 3: TOTAL AREA

STEP 1: MAX DIAMETER

STEP 2: AREA LOGIC

STEP 3: RESULT

STEP 1: SURFACE AREA FORMULA

STEP 2: SIMPLIFICATION

STEP 3: FINAL ANSWER

STEP 1: DIMENSIONS

STEP 2: SURFACE COMPONENTS

STEP 3: RESULT

STEP 1: FIND SURFACE AREA

STEP 2: COST CALCULATION

STEP 3: TOTAL

STEP 1: FIND SLANT HEIGHT

STEP 2: COMBINED AREA

STEP 3: CALCULATION

STEP 1: AREA COMPONENTS

STEP 2: SUBSTITUTION

STEP 3: RESULT

Mathematics – Class X (Important Question)

Q. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Step 1: Given Information

Step 2: Theorem of Tangents

Step 3: Calculating Sides & s

Step 4: Area using Heron’s Formula

Step 5: Area using Sub-triangles

Step 6: Final Calculation

Mathematics,Class 10, Ex-12.2

Class X Mathematics

Surface Areas and Volumes

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to \(1\text{ cm}\) and the height of the cone is equal to its radius. Find the volume of the solid in terms of \(\pi\).

3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length \(5\text{ cm}\) and diameter \(2.8\text{ cm}\).

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