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RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 Area Of A Trapezium And A Polygon

Here we have provided RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 Area Of A Trapezium And A Polygon for the ease of the students. From this article you can also download this chapter's PDF.
authorImageAnanya Gupta8 Aug, 2024
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RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2

RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 : For Class 8 Maths Chapter 18 Exercise 18.2, the Physics Wallah team has developed comprehensive and simple solutions. This exercise helps students practice calculating the area of different trapeziums and polygons, building on what was learned in Exercise 18.1.

It is important to first understand the basics of the shapes and their areas before starting Exercise 18.2. Reading through the chapter’s theory will make solving these problems easier. The solutions provided are designed to guide students step-by-step, making it simpler to apply the concepts and solve the exercises accurately.

RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 Area Of A Trapezium And A Polygon Overview

In RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2, students practice calculating the area of trapeziums and polygons. The exercise includes a variety of problems that help students apply their knowledge of geometric shapes.

For trapeziums , the solutions guide students through finding the area by using specific dimensions given in the problems.

For polygons , the exercise focuses on solving problems related to finding areas of different types of polygons based on the information provided.

The solutions prepared by the Physics Wallah team, are detailed and easy to follow, helping students grasp the concepts clearly and efficiently.

RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 PDF

RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 provide clear and detailed explanations for calculating the area of trapeziums and polygons. This exercise helps students grasp the concepts of geometry through well-structured solutions and examples. For easy reference and further study, the complete exercise PDF is available for download from the link provided below.

RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 PDF

RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 (Ex 18B)

Below we have provided RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 Area Of A Trapezium And A Polygon-

(Question 1) In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ⊥ AC and DM ⊥ AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.

Solution:

Area of quad. ABCD = ar (∆ADC) + ar (∆ABC)

(Question 2) In the given figure, ABCD is a quadrilateral – shaped field in which diagonal BD is 36 m, AL ⊥ BD and CM ⊥ BD such that AL = 19 m and CM = 11m. Find the area of the field.

Solution:

Area of the field ABCD = ar (∆ ABD) + ar (∆ BCD)

(Question 3) Find the area of pentagon ABCDE in which BL ⊥ AC, DM ⊥ AC and EN ⊥ AC and FP ⊥ AD such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.

Solution:

Area of the given pentagon ABCDE = ar (∆AEN) + ar (trap. EDMN) + ar (∆DCM) + ar (∆ACE)

(Question 4) Find the area of hexagon ABCDEF in which BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL = 8 cm and CM = 6 cm.

Solution:

Area of hexagon ABCDEF, = ar (∆AFP) + ar (trap. EFNP) + ar (∆DEN) + ar (∆ ABL) + ar (trap. BCLM) + ar (∆DCM)

(Question 5) Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5cm.

Solution:

Area of pentagon ABCDE= Area of Δ A B C + Area of Δ A C D + Area of Δ A D E
Here , B L ⊥ A C , C M ⊥ A D , E N ⊥ A D
So, BL,AM and EN perpendiculars to Δ A B C , Δ A C D Δ A D E respectively.
Area of triangle with perpendicular = 1 2 × b a s e × h e i g h t
∴ A r e a o f Δ A B C = 1 / 2 × A C × B L
A r e a o f Δ A C D = 1 / 2 × A D × C M
A r e a o f Δ A D E = 1 / 2 × A D × E N
Now area of pentagon A B C D E = 1 / 2 A C × B L + 1 / 2 × A D × C M + 1 / 2 × A D × E N
= ( 1 / 2 × 10 × 3 ) + ( 1 / 2 × 12 × 7 ) + ( 1 / 2 × 12 × 5 )
= 15 + 42 + 30
= 87

(Question 6) Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.

Solution:

In the right triangle Δ A B C , We have, from the Pythagoras theorem, A B 2 = B C 2 + A C 2 ⇒ ( 5 ) 2 = ( 4 ) 2 + A C 2 ⇒ 25 = 16 + A C 2 A C 2 = 25 − 16 = 9 = ( 3 ) 2 ∴ A C = 3 c m ∴ A D = A C + C D = 3 + 4 = 7 c m and also ∴ B G = B C + C F + F G = 4 + 8 + 4 = 16 c m A H = D E = 8 c m Now area of rect. CDEF = C D × D F = 8 × 4 = 32 c m 2 and area of trapezium ABGH = 1 2 ( A H + B G ) × A C = 1 2 × ( 8 + 16 ) × 3 c m 2 = 1 2 × 24 × 3 = 36 c m 2 ∴ Total area of the figure = area of trapezium ABGH + area of rect. CDEF = 32 + 36 = 68 c m 2

(Question 7) Find the area of given figure ABCDEFGH as per dimensions given in it.

Solution:
In the right triangle Δ A B C , We have, from the Pythagoras theorem, A B 2 = B C 2 + A C 2 ⇒ ( 5 ) 2 = ( 4 ) 2 + A C 2 ⇒ 25 = 16 + A C 2 A C 2 = 25 − 16 = 9 = ( 3 ) 2 ∴ A C = 3 c m ∴ A D = A C + C D = 3 + 4 = 7 c m and also ∴ B G = B C + C F + F G = 4 + 8 + 4 = 16 c m A H = D E = 8 c m Now area of rect. CDEF = C D × D F = 8 × 4 = 32 c m 2 and area of trapezium ABGH = 1 2 ( A H + B G ) × A C = 1 2 × ( 8 + 16 ) × 3 c m 2 = 1 2 × 24 × 3 = 36 c m 2 ∴ Total area of the figure = area of trapezium ABGH + area of rect. CDEF = 32 + 36 = 68 c m 2

(Question 8) Find the area of regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.

Solution:
AD = 23 cm, LM = 13 cm ∴ A L = M D = 23 − 13 2 = 10 2 = 5 cm In Δ ALB A B 2 = A L 2 + L B 2 ( 13 ) 2 = ( 5 ) 2 + L B 2 169 = 25 + L B 2 ⇒ L B 2 = 169 − 25 = 144 = ( 12 ) 2 ∴ L B = 12 c m ∴ F B = E C = 2 × 12 = 24 c m AF =BC = LM = 13 cm Area of ABCDEF = area of rect. ECBF + area Δ A F B + a r e a Δ D C E = 24 × 13 + 1 2 B F × A L + 1 2 E C × D M = 312 + 1 2 × 24 × 5 + 1 2 E C × D M = 312 + 1 2 × 24 × 5 + 1 2 × 24 × 5 = 312 + 60 + 60 = 432 c m 2

Benefits of RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2

  • Clear Explanations : Each solution is presented with step-by-step explanations making complex concepts more accessible and easier to understand.
  • Enhanced Understanding : By working through the detailed solutions students can gain a deeper understanding of how to calculate the area of trapeziums and polygons, which strengthens their grasp of geometry.
  • Error Correction : Solutions provide insight into common mistakes and how to avoid them, helping students to improve their accuracy and confidence in solving similar problems.
  • Exam Preparation : The detailed solutions can help in exam preparation by familiarizing students with the types of questions that may appear and how to approach them effectively.

RS Aggarwal Solutions for Class 8 Maths Chapter 18 Exercise 18.2 FAQs

What topics are covered in Exercise 18.2 of Chapter 18?

Exercise 18.2 focuses on problems related to the calculation of the area of trapeziums and polygons. It includes a variety of questions that require applying geometric principles to determine the area of these shapes.

How can I use these solutions to improve my understanding?

The solutions provide detailed steps and explanations for each problem, which can help clarify the process of calculating areas. Studying these solutions can enhance your understanding of geometric formulas and improve problem-solving skills.

What should I do if I have difficulty understanding a problem in Exercise 18.2?

If you encounter difficulties, review the detailed solutions provided for similar problems. Additionally, practicing more problems and seeking help from teachers or tutors can further assist in overcoming any challenges.

How can I effectively use the solutions for Exercise 18.2 to prepare for exams?

To effectively use the solutions for exam preparation, start by thoroughly studying the step-by-step explanations for each problem in Exercise 18.2.
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