Class 10 Maths HOTS Questions Chapterwise with Solutions
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Class 10 Maths HOTS (Higher Order Thinking Skills) Questions Chapterwise with Solutions are specially designed to enhance critical thinking and problem-solving abilities among students. Covering every chapter like Real Numbers, Polynomials, Quadratic Equations, Triangles, Circles, and more, these HOTS questions help students tackle complex problems effectively. With detailed chapterwise solutions, students can strengthen their conceptual understanding and score higher marks in board exams. Solving HOTS questions regularly ensures better analytical skills, improves accuracy, and prepares students for competitive exams too.
When it comes to mastering Class 10 Maths, practicing HOTS (Higher Order Thinking Skills) questions chapterwise with solutions is crucial for achieving excellence. These questions are not just standard textbook problems; they are thoughtfully designed to challenge students' conceptual clarity, logical reasoning, and application skills. Every chapter from Chapter 1: Real Numbers to Chapter 15: Probability includes a set of HOTS questions that push the boundaries of regular understanding and encourage deeper analysis. By solving these questions along with step-by-step solutions, students can build a stronger foundation for both board exams and entrance tests like NTSE and Olympiads.
Class 10 Maths HOTS questions are crafted to cover all important chapters such as Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions, Circles, Constructions, Surface Areas and Volumes, Statistics, and Probability. Each question tests different skills — from mathematical modeling, critical analysis, logical deduction, to real-life application of mathematical concepts. Having access to chapterwise solutions helps students to not only attempt these questions confidently but also to understand different methods of approaching a problem.
One major advantage of practicing HOTS questions is that they expose students to unseen problems, which are often part of the CBSE board exam pattern. Regular practice of chapterwise HOTS builds exam temperament, boosts confidence, and enhances time management skills. The detailed solutions act as a guiding tool where students can check their approach, rectify mistakes, and learn alternative problem-solving strategies. With comprehensive explanations, students also develop better answer writing skills, which is important for scoring full marks in the board exams.
Moreover, the Class 10 Maths HOTS Questions with Solutions Chapterwise make revision easier and organized. Students can pick any chapter, solve the HOTS questions, and quickly revise the core concepts associated with it. This method of structured preparation is highly recommended by toppers and teachers alike. It ensures complete coverage of the CBSE syllabus while pushing students towards excellence.
If you are aiming for a top score in your CBSE Class 10 Maths exam, integrating chapterwise HOTS questions and solutions into your daily study routine can be a game changer. It transforms the learning experience from rote memorization to analytical application. Always ensure you practice enough HOTS questions from each chapter to gain confidence in tackling even the trickiest problems in the exam hall.
Boost your High Order Thinking Skills (HOTS) for Class 10 Maths with this comprehensive guide. This article covers chapterwise HOTS questions along with detailed solutions to help you tackle advanced problems in your CBSE board exams.
Table of Contents
- Chapter 1: Real Numbers
- Chapter 2: Polynomials
- Chapter 3: Pair of Linear Equations in Two Variables
- Chapter 4: Quadratic Equations
- Chapter 5: Arithmetic Progressions
- Chapter 6: Triangles
- Chapter 7: Coordinate Geometry
- Chapter 8: Introduction to Trigonometry
- Chapter 9: Some Applications of Trigonometry
- Chapter 10: Circles
- Chapter 11: Constructions
- Chapter 12: Areas Related to Circles
- Chapter 13: Surface Areas and Volumes
- Chapter 14: Statistics
- Chapter 15: Probability
Chapter 1: Real Numbers
Here are 5 HOTS questions from Chapter 1: Real Numbers with solutions.
Q1: Prove that the square root of 3 is irrational.
Solution:
Assume √3 is rational. Then, √3 = a/b, where a and b are coprime integers. Squaring both sides gives 3 = a²/b² or a² = 3b². This shows that a² is divisible by 3, so a must be divisible by 3. Let a = 3k. Substituting into the equation, we get (3k)² = 3b² → 9k² = 3b² → b² = 3k², implying b is also divisible by 3. This contradicts the assumption that a and b are coprime. Hence, √3 is irrational.
Q2: If a and b are irrational numbers such that their sum is rational, explain with an example.
Solution:
Let a = √2 and b = -√2. Both √2 and -√2 are irrational numbers, but their sum is: √2 + (-√2) = 0, which is rational.
Q3: Prove that between any two rational numbers, there exists an irrational number.
Solution:
Let p and q be two rational numbers such that p < q. Consider the number (p + q)/2, which is rational. Now, consider the number (p + q)/2 + √2. Since √2 is irrational, (p + q)/2 + √2 is irrational. Therefore, there exists an irrational number between any two rational numbers.
Q4: Show that the product of a rational and an irrational number is always irrational.
Solution:
Let r be a rational number and i be an irrational number. Suppose their product ri is rational. Then, we can express i as i = r⁻¹ * (ri). Since r⁻¹ is rational and ri is rational, this would imply that i is rational. This contradicts the assumption that i is irrational. Therefore, the product of a rational and an irrational number is always irrational.
Q5: Is the sum of two irrational numbers always irrational? Justify your answer.
Solution:
No, the sum of two irrational numbers is not always irrational. For example, let a = √2 and b = -√2. Both are irrational numbers, but their sum is a + b = √2 + (-√2) = 0, which is rational.
Chapter 2: Polynomials
Here are 5 HOTS questions from Chapter 2: Polynomials with solutions.
Q1: Find the value of the constant k for which the polynomial 2x² + kx + 5 has equal roots.
Solution:
For equal roots, the discriminant must be 0. The discriminant of a quadratic equation ax² + bx + c is Δ = b² - 4ac. For 2x² + kx + 5, a = 2, b = k, and c = 5. The discriminant is Δ = k² - 4(2)(5) = k² - 40. For equal roots, Δ = 0, so k² - 40 = 0 → k² = 40 → k = ±√40 = ±2√10.
Q2: Show that the sum of the roots of the quadratic polynomial 3x² + 7x + 5 is -7/3.
Solution:
The sum of the roots of a quadratic polynomial ax² + bx + c is given by -b/a. For 3x² + 7x + 5, a = 3, b = 7, and c = 5. The sum of the roots is -7/3, which is the required result.
Q3: Find the roots of the polynomial x² - 5x + 6 and verify their sum and product.
Solution:
The roots of x² - 5x + 6 = 0 can be found using factorization: x² - 5x + 6 = (x - 2)(x - 3) = 0. Therefore, the roots are x = 2 and x = 3. The sum of the roots is 2 + 3 = 5, and the product is 2 * 3 = 6, which is correct.
Q4: Find the quadratic polynomial whose roots are 3 and -4.
Solution:
The quadratic polynomial with roots α and β is given by x² - (sum of the roots)x + (product of the roots). The sum of the roots is 3 + (-4) = -1, and the product is 3 * (-4) = -12. Therefore, the polynomial is x² + x - 12.
Q5: Prove that the product of the roots of the quadratic equation x² - 5x + 6 = 0 is equal to the constant term 6.
Solution:
The product of the roots of the quadratic equation ax² + bx + c = 0 is given by c/a. For x² - 5x + 6 = 0, a = 1, b = -5, and c = 6. The product of the roots is 6/1 = 6, which is equal to the constant term.
Chapter 3: Pair of Linear Equations in Two Variables
Here are 5 HOTS questions from Chapter 3: Pair of Linear Equations in Two Variables with solutions.
Q1: Solve the system of equations 3x + 4y = 8 and 5x - 2y = 6 using substitution method.
Solution:
Solve 3x + 4y = 8 for x: x = (8 - 4y)/3. Substitute this into 5x - 2y = 6 to get 5[(8 - 4y)/3] - 2y = 6. Simplifying: (40 - 20y)/3 - 2y = 6, multiply through by 3 to eliminate the fraction: 40 - 20y - 6y = 18. Simplifying further: 40 - 26y = 18, so -26y = -22, hence y = 22/26 = 11/13. Substituting y = 11/13 into x = (8 - 4y)/3, we get x = (8 - 4(11/13))/3 = (104/13 - 44/13)/3 = 60/13 × 1/3 = 20/13. So, x = 20/13 and y = 11/13.
Q2: Solve the system of equations 2x + 3y = 12 and 3x + 4y = 13 using elimination method.
Solution:
Multiply the first equation by 3 and the second by 2: 6x + 9y = 36 and 6x + 8y = 26. Subtracting the second from the first: (6x + 9y) - (6x + 8y) = 36 - 26, which gives y = 10. Substitute y = 10 into 2x + 3y = 12: 2x + 30 = 12, so 2x = -18, hence x = -9. Therefore, the solution is x = -9 and y = 10.
Q3: Find the values of x and y for the equations 5x + 2y = 12 and 3x - 4y = 5.
Solution:
Multiply the first equation by 2 and the second by 5 to eliminate y: 10x + 4y = 24 and 15x - 20y = 25. Add the two equations: (10x + 4y) + (15x - 20y) = 24 + 25, which simplifies to 25x - 16y = 49.
Q4: Solve for x and y in the following system of equations: x + y = 6 and 2x - 3y = 3.
Solution:
Solve the first equation for y: y = 6 - x. Substitute this into the second equation: 2x - 3(6 - x) = 3, which simplifies to 2x - 18 + 3x = 3. Combining like terms gives 5x = 21, so x = 21/5. Substituting x = 21/5 into y = 6 - x gives y = 6 - 21/5 = 30/5 - 21/5 = 9/5. Therefore, the solution is x = 21/5 and y = 9/5.
Q5: Solve the system of equations 4x + 5y = 20 and 2x + 3y = 12 using the substitution method.
Solution:
From the first equation, solve for x: x = (20 - 5y)/4. Substitute this into the second equation: 2[(20 - 5y)/4] + 3y = 12. Simplifying: (40 - 10y)/4 + 3y = 12. Multiply through by 4 to eliminate the fraction: 40 - 10y + 12y = 48. Simplifying further gives 40 + 2y = 48, so 2y = 8, hence y = 4. Substitute y = 4 into x = (20 - 5y)/4, which gives x = (20 - 20)/4 = 0. Therefore, the solution is x = 0 and y = 4.
Chapter 4: Quadratic Equations
Here are 5 HOTS questions from Chapter 4: Quadratic Equations with solutions.
Q1: Solve the quadratic equation x² - 7x + 10 = 0 using factorization method.
Solution:
The quadratic equation is x² - 7x + 10 = 0. Factorizing, we get (x - 2)(x - 5) = 0. Therefore, x = 2 or x = 5.
Q2: Find the nature of the roots of the quadratic equation 2x² + 3x - 2 = 0.
Solution:
The discriminant of the equation is Δ = b² - 4ac. For 2x² + 3x - 2 = 0, a = 2, b = 3, and c = -2. Δ = (3)² - 4(2)(-2) = 9 + 16 = 25. Since the discriminant is positive, the equation has real and distinct roots.
Q3: Solve the quadratic equation 3x² - 4x - 5 = 0 using the quadratic formula.
Solution:
The quadratic formula is x = (-b ± √(b² - 4ac))/2a. For 3x² - 4x - 5 = 0, a = 3, b = -4, and c = -5. The discriminant is Δ = (-4)² - 4(3)(-5) = 16 + 60 = 76. Therefore, x = [4 ± √76]/6. Simplifying further: x = (4 ± 2√19)/6. The solutions are x = (2 ± √19)/3.
Q4: If the roots of the quadratic equation ax² + bx + c = 0 are real and equal, prove that b² = 4ac.
Solution:
For real and equal roots, the discriminant must be zero. The discriminant is Δ = b² - 4ac. For equal roots, Δ = 0, so b² - 4ac = 0, which gives b² = 4ac.
Q5: Solve the equation x² + 6x + 8 = 0 by completing the square.
Solution:
The given equation is x² + 6x + 8 = 0. To complete the square, take half of the coefficient of x, which is 6/2 = 3, and square it to get 9. Add and subtract 9 to the equation: x² + 6x + 9 - 9 + 8 = 0. This simplifies to (x + 3)² - 1 = 0. Therefore, (x + 3)² = 1, and x + 3 = ±1. Solving gives x = -2 or x = -4.
Chapter 5: Arithmetic Progressions
Here are 5 HOTS questions from Chapter 5: Arithmetic Progressions with solutions.
Q1: Find the 10th term of the arithmetic progression 3, 7, 11, 15, ...
Solution:
The nth term of an arithmetic progression is given by aₙ = a + (n-1) × d, where a is the first term and d is the common difference. Here, a = 3 and d = 7 - 3 = 4. The 10th term is a₁₀ = 3 + (10 - 1) × 4 = 3 + 36 = 39.
Q2: If the sum of the first n terms of an arithmetic progression is given by Sₙ = 3n² + n, find the 5th term of the progression.
Solution:
The nth term of an arithmetic progression is given by Tₙ = Sₙ - Sₙ₋₁. Using the given sum formula, S₅ = 3(5)² + 5 = 75 + 5 = 80 and S₄ = 3(4)² + 4 = 48 + 4 = 52. Therefore, the 5th term is T₅ = S₅ - S₄ = 80 - 52 = 28.
Q3: In an arithmetic progression, the 5th term is 10 and the 10th term is 25. Find the common difference.
Solution:
Let the nth term of the arithmetic progression be given by Tₙ = a + (n - 1) × d. For the 5th term, T₅ = a + 4d = 10, and for the 10th term, T₁₀ = a + 9d = 25. Subtracting these two equations: (a + 9d) - (a + 4d) = 25 - 10, we get 5d = 15, hence d = 3.
Q4: The 8th term of an arithmetic progression is 20, and the 15th term is 35. Find the first term.
Solution:
Let the nth term be Tₙ = a + (n - 1) × d. For the 8th term, T₈ = a + 7d = 20, and for the 15th term, T₁₅ = a + 14d = 35. Subtracting the first equation from the second: (a + 14d) - (a + 7d) = 35 - 20, we get 7d = 15, so d = 15/7. Now, substitute d = 15/7 into a + 7d = 20 to find a: a + 7(15/7) = 20 → a + 15 = 20 → a = 5.
Q5: If the sum of the first n terms of an arithmetic progression is given by Sₙ = 2n² + 3n, find the 4th term of the progression.
Solution:
The nth term of the arithmetic progression is given by Tₙ = Sₙ - Sₙ₋₁. Using the sum formula, S₄ = 2(4)² + 3(4) = 32 + 12 = 44 and S₃ = 2(3)² + 3(3) = 18 + 9 = 27. Therefore, the 4th term is T₄ = S₄ - S₃ = 44 - 27 = 17.
Chapter 6: Triangles
Here are 5 HOTS questions from Chapter 6: Triangles with solutions.
Q1: In a right-angled triangle ABC, the hypotenuse is 10 cm and one of the legs is 6 cm. Find the length of the other leg.
Solution:
Using the Pythagorean theorem, we know that in a right-angled triangle, a² + b² = c², where c is the hypotenuse and a, b are the legs. Here, c = 10 and one leg (a) = 6. Therefore, 6² + b² = 10², which simplifies to 36 + b² = 100. Solving for b², we get b² = 64, so b = 8 cm.
Q2: Prove that the sum of the angles of a triangle is 180°.
Solution:
Consider a triangle ABC. Draw a line parallel to the base BC through vertex A. This will create alternate interior angles, which are congruent to the angles of triangle ABC. Since the sum of the interior angles on a straight line is 180°, the sum of the three angles of triangle ABC is 180°.
Q3: In triangle ABC, if angle A = 30° and angle B = 60°, find angle C.
Solution:
The sum of the angles of a triangle is always 180°. Therefore, angle C = 180° - (30° + 60°) = 90°. Hence, angle C = 90°.
Q4: Prove that if two triangles are congruent, then their corresponding sides and angles are equal.
Solution:
By the definition of congruence, two triangles are congruent if one can be superimposed on the other. This means their corresponding sides and angles coincide exactly. Therefore, the corresponding sides and angles of congruent triangles must be equal.
Q5: In a triangle ABC, the lengths of the sides are given by AB = 6 cm, BC = 8 cm, and CA = 10 cm. Prove that triangle ABC is a right-angled triangle.
Solution:
To prove that triangle ABC is a right-angled triangle, use the Pythagorean theorem. If AB² + BC² = CA², the triangle will be right-angled. Here, AB = 6 cm, BC = 8 cm, and CA = 10 cm. Therefore, 6² + 8² = 36 + 64 = 100 and 10² = 100. Since AB² + BC² = CA², triangle ABC is a right-angled triangle.
Chapter 7: Coordinate Geometry
Here are 5 HOTS questions from Chapter 7: Coordinate Geometry with solutions.
Q1: Find the distance between the points (3, 4) and (7, 1).
Solution:
The distance formula is given by: d = √[(x₂ - x₁)² + (y₂ - y₁)²]. Substituting the values (x₁, y₁) = (3, 4) and (x₂, y₂) = (7, 1), we get: d = √[(7 - 3)² + (1 - 4)²] = √[16 + 9] = √25 = 5 units.
Q2: Find the coordinates of the midpoint of the line segment joining the points (4, -1) and (-2, 3).
Solution:
The midpoint formula is given by: M = [(x₁ + x₂)/2, (y₁ + y₂)/2]. Substituting the coordinates (x₁, y₁) = (4, -1) and (x₂, y₂) = (-2, 3), we get: M = [(4 + (-2))/2, (-1 + 3)/2] = [2/2, 2/2] = (1, 1).
Q3: Find the area of the triangle with vertices A(2, 3), B(4, 5), and C(6, 7).
Solution:
The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by the formula: Area = 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|. Substituting the coordinates, we get: Area = 1/2 |2(5 - 7) + 4(7 - 3) + 6(3 - 5)| = 1/2 |-4 + 16 - 12| = 1/2 * 8 = 4 square units.
Q4: Find the equation of the line passing through the point (2, 3) and having a slope of 4.
Solution:
The equation of a line in point-slope form is given by: y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line. Substituting m = 4 and (x₁, y₁) = (2, 3), we get: y - 3 = 4(x - 2). Simplifying, we get the equation of the line as y = 4x - 5.
Q5: If the coordinates of the midpoint of a line segment are (5, 7), and one endpoint is (3, 5), find the other endpoint.
Solution:
Let the coordinates of the other endpoint be (x, y). The midpoint formula gives: [(x + 3)/2, (y + 5)/2] = (5, 7). This gives the equations: (x + 3)/2 = 5 and (y + 5)/2 = 7. Solving for x and y, we get x + 3 = 10, hence x = 7, and y + 5 = 14, hence y = 9. Therefore, the other endpoint is (7, 9).
Chapter 8: Introduction to Trigonometry
Here are 5 HOTS questions from Chapter 8: Introduction to Trigonometry with solutions.
Q1: In a right-angled triangle, if the length of the opposite side is 5 and the length of the hypotenuse is 13, find the sine of the angle.
Solution:
Sine of an angle in a right-angled triangle is given by: sin(θ) = opposite/hypotenuse. Here, the opposite side is 5 and the hypotenuse is 13, so sin(θ) = 5/13.
Q2: Find the cosine of an angle in a right-angled triangle if the adjacent side is 12 and the hypotenuse is 13.
Solution:
Cosine of an angle in a right-angled triangle is given by: cos(θ) = adjacent/hypotenuse. Here, the adjacent side is 12 and the hypotenuse is 13, so cos(θ) = 12/13.
Q3: If tan(θ) = 3/4, find the value of sin(θ) and cos(θ).
Solution:
tan(θ) = opposite/adjacent = 3/4. To find sin(θ) and cos(θ), use the identity tan²(θ) + 1 = sec²(θ). First, calculate sec(θ) = √(1 + tan²(θ)) = √(1 + 9/16) = √(25/16) = 5/4. Therefore, cos(θ) = 1/sec(θ) = 4/5, and sin(θ) = tan(θ) × cos(θ) = 3/4 × 4/5 = 3/5.
Q4: If sin(θ) = 5/13, find cos(θ) and tan(θ).
Solution:
Using the Pythagorean identity sin²(θ) + cos²(θ) = 1, we get cos²(θ) = 1 - (5/13)² = 1 - 25/169 = 144/169. Therefore, cos(θ) = 12/13. Now, tan(θ) = sin(θ)/cos(θ) = 5/13 ÷ 12/13 = 5/12.
Q5: In a right-angled triangle, if tan(θ) = 3/4, find sin(θ) and cos(θ).
Solution:
Given tan(θ) = 3/4, the opposite side is 3 and the adjacent side is 4. Using the Pythagorean theorem, the hypotenuse is √(3² + 4²) = √(9 + 16) = √25 = 5. Therefore, sin(θ) = 3/5 and cos(θ) = 4/5.
Chapter 9: Some Applications of Trigonometry
Here are 5 HOTS questions from Chapter 9: Some Applications of Trigonometry with solutions.
Q1: A 10 m long ladder is placed against a wall such that the angle of elevation is 60°. Find the height at which the ladder touches the wall.
Solution:
Using trigonometry, we know that sin(θ) = opposite/hypotenuse. Here, the angle of elevation θ = 60° and the hypotenuse (the length of the ladder) is 10 m. Therefore, sin(60°) = height/10. Since sin(60°) = √3/2, we get height = (√3/2) × 10 = 5√3 ≈ 8.66 m.
Q2: A person is standing 30 m away from a tower. The angle of elevation to the top of the tower is 45°. Find the height of the tower.
Solution:
Using tan(θ) = opposite/adjacent, we get tan(45°) = height/30. Since tan(45°) = 1, we get height = 30 m. Therefore, the height of the tower is 30 m.
Q3: From a point 30 m away from the base of a building, the angle of elevation to the top of the building is 60°. Find the height of the building.
Solution:
Using tan(60°) = height/30, and tan(60°) = √3, we get height = 30 × √3 ≈ 51.96 m. Therefore, the height of the building is approximately 51.96 m.
Q4: A kite is flying at a height of 20 m above the ground. The string of the kite makes an angle of 45° with the horizontal. Find the length of the string.
Solution:
Using sin(45°) = opposite/hypotenuse, we have sin(45°) = 20/hypotenuse. Since sin(45°) = √2/2, we get 20 = (√2/2) × hypotenuse, so hypotenuse = 20 × 2/√2 = 20√2 ≈ 28.28 m. Therefore, the length of the string is approximately 28.28 m.
Q5: From the top of a 60 m high tower, the angle of depression to a person standing on the ground is 30°. Find the distance of the person from the base of the tower.
Solution:
Using tan(30°) = height/distance, we get tan(30°) = 60/distance. Since tan(30°) = 1/√3, we have 1/√3 = 60/distance, so distance = 60√3 ≈ 103.92 m. Therefore, the distance of the person from the base of the tower is approximately 103.92 m.
Chapter 10: Circles
Here are 5 HOTS questions from Chapter 10: Circles with solutions.
Q1: A tangent PQ to a circle with center O and radius r meets a line through O at an angle of 90°. If PQ = 5 cm and OP = 12 cm, find the radius of the circle.
Solution:
Using the Pythagorean theorem in the right-angled triangle OPQ, where OP is the hypotenuse and PQ is one of the sides, we have OP² = PQ² + radius². Substituting the values, we get 12² = 5² + radius², which simplifies to 144 = 25 + radius², so radius² = 119. Therefore, radius = √119 ≈ 10.91 cm.
Q2: The length of a chord AB of a circle is 10 cm. The distance from the center O of the circle to the chord AB is 6 cm. Find the radius of the circle.
Solution:
In the right-angled triangle OMA, where M is the midpoint of the chord AB, OM = 6 cm and AM = AB/2 = 5 cm. Using the Pythagorean theorem, we get OM² + AM² = radius². Substituting the values, we get 6² + 5² = radius², which simplifies to 36 + 25 = radius², so radius² = 61. Therefore, radius = √61 ≈ 7.81 cm.
Q3: In a circle with center O, two chords AB and CD intersect at a point E. If AE = 4 cm, BE = 3 cm, CE = 5 cm, and DE = x cm, find the value of x.
Solution:
According to the intersecting chord theorem, AE × BE = CE × DE. Substituting the known values, we get 4 × 3 = 5 × x, which simplifies to 12 = 5x, so x = 12/5 = 2.4 cm. Therefore, DE = 2.4 cm.
Q4: The radius of a circle is 7 cm. Find the length of the chord at a distance of 4 cm from the center of the circle.
Solution:
Let the length of the chord be 2x. Using the Pythagorean theorem in the right-angled triangle formed by the radius, the perpendicular distance from the center to the chord, and half the length of the chord, we have radius² = (distance from center)² + x². Substituting the values, we get 7² = 4² + x², which simplifies to 49 = 16 + x², so x² = 33. Therefore, x = √33 ≈ 5.74 cm, and the length of the chord is 2x ≈ 2 × 5.74 = 11.48 cm.
Q5: A circle with center O has a radius of 6 cm. A chord AB of the circle subtends an angle of 60° at the center. Find the length of the chord AB.
Solution:
Using the formula for the length of a chord subtending an angle θ at the center of a circle, the length of the chord is given by: AB = 2r × sin(θ/2), where r is the radius and θ is the angle at the center. Substituting the values, we get AB = 2 × 6 × sin(60°/2) = 12 × sin(30°) = 12 × 1/2 = 6 cm. Therefore, the length of the chord AB is 6 cm.
Chapter 11: Areas of Parallelograms and Triangles
Here are 5 HOTS questions from Chapter 11: Areas of Parallelograms and Triangles with solutions.
Q1: The area of a triangle is 60 cm², and its base is 12 cm. Find the height of the triangle.
Solution:
The area of a triangle is given by: Area = 1/2 × base × height. Substituting the known values, we get 60 = 1/2 × 12 × height, which simplifies to 60 = 6 × height. Therefore, height = 60/6 = 10 cm.
Q2: If the area of a parallelogram is 120 cm² and its base is 15 cm, find the height.
Solution:
The area of a parallelogram is given by: Area = base × height. Substituting the known values, we get 120 = 15 × height, so height = 120/15 = 8 cm.
Q3: A parallelogram has base 14 cm and height 9 cm. Find its area.
Solution:
The area of a parallelogram is given by: Area = base × height. Substituting the known values, we get Area = 14 × 9 = 126 cm².
Q4: The area of a triangle is 45 cm². The base of the triangle is 9 cm. Find its height.
Solution:
Using the formula for the area of a triangle: Area = 1/2 × base × height, we get 45 = 1/2 × 9 × height. Solving for height, we get height = 45 × 2 / 9 = 10 cm.
Q5: A triangle has base 8 cm and height 5 cm. Find its area.
Solution:
Using the formula for the area of a triangle: Area = 1/2 × base × height, we get Area = 1/2 × 8 × 5 = 20 cm².
Chapter 12: Surface Areas and Volumes
Here are 5 HOTS questions from Chapter 12: Surface Areas and Volumes with solutions.
Q1: A cone has a radius of 7 cm and a slant height of 25 cm. Find the surface area of the cone.
Solution:
The surface area of a cone is given by: Surface Area = πr(r + l), where r is the radius and l is the slant height.
Substituting the given values, we get Surface Area = π × 7 × (7 + 25) = π × 7 × 32 = 224π ≈ 703.72 cm².
Q2: A cylinder has a radius of 3 cm and a height of 10 cm. Find its total surface area.
Solution:
The total surface area of a cylinder is given by: Total Surface Area = 2πr(h + r), where r is the radius and h is the height.
Substituting the given values, we get Total Surface Area = 2π × 3 × (10 + 3) = 2π × 3 × 13 = 78π ≈ 245.04 cm².
Q3: Find the volume of a hemisphere with radius 7 cm.
Solution:
The volume of a hemisphere is given by: Volume = (2/3)πr³.
Substituting the given value of radius, we get Volume = (2/3)π × 7³ = (2/3)π × 343 ≈ 718.4 cm³.
Q4: A rectangular box has dimensions 5 cm × 6 cm × 7 cm. Find its volume and surface area.
Solution:
Volume of the box = length × width × height = 5 × 6 × 7 = 210 cm³.
Surface Area of the box = 2lw + 2lh + 2wh = 2(5×6) + 2(5×7) + 2(6×7) = 60 + 70 + 84 = 214 cm².
Q5: Find the volume of a cone with radius 4 cm and height 9 cm.
Solution:
The volume of a cone is given by: Volume = (1/3)πr²h.
Substituting the given values, we get Volume = (1/3)π × 4² × 9 = (1/3)π × 16 × 9 = 48π ≈ 150.8 cm³.
Chapter 13: Statistics
Here are 5 HOTS questions from Chapter 13: Statistics with solutions.
Q1: The mean of the following data is 24. Find the missing frequency:
| Class Interval | Frequency |
| 10 - 20 | f1 |
| 20 - 30 | f2 |
| 30 - 40 | 10 |
| 40 - 50 | 8 |
Solution:
The formula for the mean of grouped data is:
Mean = (Σf × x) / Σf, where f is the frequency and x is the midpoint.
Substituting the known values and using the mean = 24, solve for the missing frequencies.
Q2: Find the mode of the following data: 5, 8, 12, 8, 15, 9, 12, 8, 8.
Solution:
The mode is the most frequently occurring number in a set of data. In this case, the mode is 8, as it appears most frequently.
Q3: Calculate the mean of the following data:
| Class Interval | Frequency |
| 10 - 20 | 5 |
| 20 - 30 | 10 |
| 30 - 40 | 15 |
| 40 - 50 | 10 |
Solution:
To find the mean, we first calculate the midpoints and then use the formula Mean = (Σf × x) / Σf.
Q4: Find the range of the following data: 2, 5, 7, 10, 4, 12, 15, 3.
Solution:
The range is the difference between the highest and lowest values in a data set. Here, the highest value is 15, and the lowest value is 2. Therefore, the range = 15 - 2 = 13.
Q5: A die is rolled 100 times. The frequency distribution of the number of times each face appears is as follows:
| Face | Frequency |
| 1 | 16 |
| 2 | 14 |
| 3 | 18 |
| 4 | 19 |
| 5 | 18 |
| 6 | 15 |
Solution:
To find the mean, multiply the frequency of each face by the number on the die and then divide by the total number of trials (100).
Chapter 14: Probability
Here are 5 HOTS questions from Chapter 14: Probability with solutions.
Q1: A coin is tossed 3 times. Find the probability of getting at least one head.
Solution:
The total number of outcomes when tossing 3 coins is 2³ = 8. The favorable outcomes for getting at least one head are: HHT, HTH, THH, HTT, THT, TTH, and HHH. Therefore, the probability is 7/8.
Q2: A card is drawn from a deck of 52 cards. What is the probability of drawing a red card?
Solution:
There are 26 red cards in a deck of 52 cards (13 hearts and 13 diamonds). Therefore, the probability of drawing a red card is 26/52 = 1/2.
Q3: A bag contains 5 red, 7 green, and 8 blue balls. One ball is drawn at random. Find the probability that the ball is green.
Solution:
The total number of balls in the bag is 5 + 7 + 8 = 20. The favorable outcomes for drawing a green ball are 7. Therefore, the probability is 7/20.
Q4: A die is rolled once. What is the probability of rolling an even number?
Solution:
The total number of outcomes is 6. The favorable outcomes for an even number are 2, 4, and 6. Therefore, the probability is 3/6 = 1/2.
Q5: A bag contains 6 red balls and 4 black balls. Two balls are drawn without replacement. Find the probability that both balls are red.
Solution:
The total number of ways to draw 2 balls from 10 is C(10, 2) = 45. The number of favorable outcomes for drawing 2 red balls is C(6, 2) = 15. Therefore, the probability is 15/45 = 1/3.
Frequently Asked Questions (FAQs)
What is the difference between surface area and volume?
Answer:
Surface area refers to the total area of the outer surface of a three-dimensional object, while volume is the amount of space occupied by the object. Surface area is measured in square units, whereas volume is measured in cubic units.
What is the formula for the volume of a cone?
Answer:
The formula for the volume of a cone is:
Volume = (1/3)πr²h, where r is the radius of the base and h is the height of the cone.
How do you calculate the mean in statistics?
Answer:
To calculate the mean, sum all the values in the data set and divide by the number of values.
Formula: Mean = (Σx) / N, where Σx is the sum of all data points and N is the number of data points.
What is the probability of drawing a red card from a deck of cards?
Answer:
A standard deck of cards has 52 cards, 26 of which are red (13 hearts and 13 diamonds). The probability of drawing a red card is:
Probability = 26/52 = 1/2.
What is the formula for the surface area of a cylinder?
Answer:
The formula for the surface area of a cylinder is:
Surface Area = 2πr(h + r), where r is the radius of the base and h is the height of the cylinder.