RD Sharma Class 8 Maths Chapter 20 Mensuration I (Area of a Trapezium and a Polygon)

Mensuration is an important topic in mathematics that helps students understand how to measure the size, area, and volume of different shapes and objects. In RD Sharma Class 8 Chapter 20, students explore Mensuration I, focusing on how to calculate the area of a trapezium and a polygon. This chapter introduces key formulas and methods that are essential for solving real-life problems involving irregular and regular figures.

The RD Sharma Class 8 Maths Chapter 20 simplifies these complex concepts with step-by-step explanations and practical examples. For students looking to strengthen their understanding, the RD Sharma Class 8 Solutions Chapter 20 provide detailed answers to each question in the textbook. These solutions are aligned with the CBSE Syllabus for Class 8 Maths and support students in preparing for both school exams and competitive assessments.

Whether you're studying from the NCERT Solutions for Class 8 Maths Mensuration I (Area of a Trapezium and a Polygon) or reviewing through the RD Sharma Class 8 Maths Solutions Chapter 20, this chapter is a must-learn for building a solid foundation in geometry. The content is designed to match the latest CBSE curriculum and is helpful for students following the RD Sharma Class 8 Chapter 20 2025-26 edition.

With the right guidance and clear understanding of the formulas, students can master the art of measuring and calculating areas in different figures. The RD Sharma Chapter 20 Class 8 not only helps in concept clarity but also improves problem-solving speed and accuracy—skills that are vital for academic success.

RD Sharma Class 8 Maths Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) PDF

Understanding areas of shapes is an important part of geometry. The RD Sharma Class 8 Maths Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) PDF helps students learn how to find the area of different polygons and trapeziums using simple formulas. This chapter is explained clearly in the RD Sharma Class 8 Mensuration I (Area of a Trapezium and a Polygon) book, making it easier for Class 8 students to follow.

Students can download the Class 8 Maths Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) PDF RD Sharma for quick reference and revision. The Class 8 Maths Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) solutions provide step-by-step answers, helping students improve their understanding. Whether you're revising for school exams or learning the basics, the RD Sharma Chapter 20 Class 8 Mensuration I (Area of a Trapezium and a Polygon) is a helpful resource aligned with CBSE guidelines.

RD Sharma Class 8 Solutions Mensuration I (Area of a Trapezium and a Polygon) Questions with Solutions

Q1. Find the area of a trapezium whose parallel sides are 12 cm and 8 cm, and the height is 6 cm.

Solution: Area = ½ × (12 + 8) × 6 = ½ × 20 × 6 = 60 cm²

Q2. A trapezium has parallel sides of 15 cm and 25 cm and height 10 cm. Find its area.

Solution: Area = ½ × (15 + 25) × 10 = ½ × 40 × 10 = 200 cm²

Q3. Find the area of a regular hexagon with side 6 cm.

Solution: Area = (3√3 / 2) × a² = (3√3 / 2) × 36 = 54√3 ≈ 93.53 cm²

Q4. Find the area of a square with side 9 cm.

Solution: Area = side² = 9² = 81 cm²

Q5. A parallelogram has base 12 cm and height 5 cm. Find the area.

Solution: Area = base × height = 12 × 5 = 60 cm²

Q6. A rhombus has diagonals 10 cm and 8 cm. Find its area.

Solution: Area = ½ × d₁ × d₂ = ½ × 10 × 8 = 40 cm²

Q7. Find the area of a triangle with base 10 cm and height 7 cm.

Solution: Area = ½ × base × height = ½ × 10 × 7 = 35 cm²

Q8. A trapezium has one parallel side 14 cm, another side 6 cm, and the height is 5 cm. Find its area.

Solution: Area = ½ × (14 + 6) × 5 = ½ × 20 × 5 = 50 cm²

Q9. What is the perimeter of a regular pentagon with each side of length 7 cm?

Solution: Perimeter = 5 × side = 5 × 7 = 35 cm

Q10. Find the area of a regular pentagon with side 10 cm and apothem 6.88 cm.

Solution: Area = ½ × Perimeter × Apothem = ½ × (5 × 10) × 6.88 = ½ × 50 × 6.88 = 172 cm²

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