RS Aggarwal Chapter 15 Class 9 Maths Exercises 15.1 (ex 15a) Solutions

RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 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.

Download RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.1 Solutions



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

Name of the Solid Figure



LSA: 2h(l + b)
TSA: 2(lb + bh + hl)
Volume: l × b × h

l = length,
b = breadth,
h = height


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


LSA: p × h
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


LSA: 2 × π × r2
TSA: 3 × π × r2
Volume: ⅔ × (πr3)

r = radius


LSA: 4 × π × r2
TSA: 4 × π × r2
Volume: 4/3 × (πr3)

r = radius

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