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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EXERCISE – 15D
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: 4a2TSA: 6a2Volume: 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 × lTSA: LSA + Area of the baseVolume: ⅓ × Area of the base × h p = perimeter of the base,l = slant height, h = height |
Prism | LSA: p × hTSA: LSA × 2BVolume: B × h p = 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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