# RS Aggarwal Chapter 15 Class 9 Maths Exercises 15.4 (ex 15d) Solutions

RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.4 Solutions: Wherever we look, usually we see solids. So far, in all our study, 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 learnt 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 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 on this chapter here.

Table of Contents

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 × hl = length,b = breadth,h = height Cube LSA: 4a2TSA: 6a2Volume: a3a = sides of a cube Right Circular Cylinder LSA: 2(π × r × h)TSA: 2πr (r + h)Volume: π × r2 × hr = radius,h = height Right Pyramid LSA: ½ × p × lTSA: LSA + Area of the baseVolume: ⅓ × Area of the base × hp = perimeter of the base,l = slant height, h = height Prism LSA: p × hTSA: LSA × 2BVolume: B × hp = perimeter of the base,B = area of base, h = height Right Circular Cone LSA: πrlTSA: π × r × (r + l)Volume: ⅓ × (πr2h)r = radius,l = slant height,h = height Hemisphere LSA: 2 × π × r2TSA: 3 × π × r2Volume: ⅔ × (πr3)r = radius Sphere LSA: 4 × π × r2TSA: 4 × π × r2Volume: 4/3 × (πr3)r = radius

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