RS Aggarwal Class 8 Maths Chapter 20 Ex 20.3 Solutions 2022 | Download Free PDF

RS Aggarwal Class 8 Maths Chapter 20 Ex 20.3 Solutions

RS Aggarwal Class 8 Maths Chapter 20 Ex 20.3 Solutions: This exercise is based on word problems of different solid figures such as a cube, cylinder, & cuboid. The questions on this exercise can be tricky so the students need a lot of practice to solve all the problems accurately. These solutions offer students an answer to all the questions in an easy step-wise manner.

RS Aggarwal Class 8 Maths Chapter 20 Ex 20.3 Solutions assists the students to manage their timings of solving Maths paper in the Class 8th final exam. These solutions are reliable & are designed by our top Mathematics experts that assist the students to solve all their doubts related to different solid figures such as a cube, cylinder, & cuboid. These solutions are prepared as per the guidelines of the CBSE.

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RS Aggarwal Class 8 Maths Chapter 20 Ex 20.3 Solutions

Important Definition for RS Aggarwal Class 8 Maths Chapter 20 Ex 20.3 Solutions

  • Solid figures

They have 3-dimensions. They have curved surfaces and flat or lateral sides. Their areas are in square units & the total surface area is equal to the sum of the curved surface area & areas of flat sides. They hold space equal to their volumes.

  • Cylinder

It has a round shape with flat faces on top and bottom. Assume a cylinder of radius r and height h. It can be considered to have a surface of rectangular height ‘h’ and width 2πr which is the circumference of the circular top and bottom end faces when it is opened and spread.

Therefore, the Total Surface Area of a cylinder = Area of curved surface + Areas of 2 circular faces

                 = 2πrh + 2πr2 = 2πr(h + r) square Units

Curved Surface Area of a Cylinder = 2πrh square units

Volume of a cylinder = πr2h cubic units.    

  • Cube

It is a solid shaped object which has 6 flat sides of a square shape, each side meeting the adjacent sides at the edge and vertices. All sides are at 90⁰ to the adjacent sides & have equal length & make a square face.

Surface Area of a Cube = areas of 6 square sides of length ‘a’

                 = a2 + a2 + a2 + a2 + a2 + a2 = 6 a2 square units.

Volume of a cube = area of 1 side × length (height) of the cube

                = a2 × a = a3 cubic units.

  • Cuboid

The area = 2ab for cuboid shape, opposite sides area of flat rectangular faces of equal length and breadth or height with length ‘a’ and width ‘b’. All sides form 90⁰ with adjacent sides.

Suppose a cuboid of 2 faces with length ‘a’ and width ‘b’ the area = 2ab

2 faces with height ‘h’ & length ‘b’, the area = 2bh

2 faces with height ‘h’ & length ‘b’, the area = 2ah

Surface Area of a Cuboid = 2(ab + ah + bh) square Units

Volume of a cuboid = a × b × h square units with length ‘a’ and width ‘b’ the area = 2ab

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