RS Aggarwal Chapter 15 Class 9 Maths Exercises 15.2 (ex 15b) Solutions

RS Aggarwal Chapter 15 Class 9 Maths Exercise 15.2 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.

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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 cardboard sheets and stack them up in a vertical pile. Through 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 15.2 Solutions

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EXERCISE – 15B

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Important Definition for RS Aggarwal Chapter 15 Class 9 Maths Ex 15b 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: 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

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