RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.4 Solutions

RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.4 Solutions: Wherever we look, usually we see solids. So far, in all our studies, we have been dealing with figures that can be easily drawn on our notebooks or blackboards. These are called plane figures. We have understood what rectangles, squares, and circles are, what we mean by their perimeters and areas, and how we can find them. We have learned these in earlier classes. It would be interesting to see what happens if we cut out many of these plane figures of the same shape and size from a cardboard sheet and stack them up in a vertical pile. By this process, we shall obtain some solid figures (briefly called solids) such as a cuboid, a cylinder, etc. Know more about this chapter here.

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Download RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.4 Solutions

EXERCISE – 15D

Important Definition for RS Aggarwal Chapter 15 Class 9 Maths Ex 15d Solutions

Name of the Solid Figure	Formulas
Cuboid	LSA: 2h(l + b) TSA: 2(lb + bh + hl) Volume: l × b × h l = length, b = breadth, h = height
Cube	LSA: 4a2 TSA: 6a2 Volume: a3 a = sides of a cube
Right Circular Cylinder	LSA: 2(π × r × h) TSA: 2πr (r + h) Volume: π × r2 × h r = radius, h = height
Right Pyramid	LSA: ½ × p × l TSA: LSA + Area of the base Volume: ⅓ × Area of the base × h p = perimeter of the base, l = slant height, h = height
Prism	LSA: p × h TSA: LSA × 2B Volume: B × h p = perimeter of the base, B = area of base, h = height
Right Circular Cone	LSA: πrl TSA: π × r × (r + l) Volume: ⅓ × (πr2h) r = radius, l = slant height, h = height
Hemisphere	LSA: 2 × π × r2 TSA: 3 × π × r2 Volume: ⅔ × (πr3) r = radius
Sphere	LSA: 4 × π × r2 TSA: 4 × π × r2 Volume: 4/3 × (πr3) r = radius

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