Surface Areas and Volumes

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1. S urface Areas And Volumes Cdt . No.- 4757 By- Cdt. Gavin Chandel Class- 10 th A

2. INTRODUCTION In earlier classes, we have studied about the surface areas and volumes of different 3-D figures. But, the figures about which we studied were single figures. In this chapter, we will study about the surface areas and volumes of some joined figures such as the surface area and volume of a cuboid joined with a hemisphere. We can notice these type of figures in our daily life also. For ex.- an oil tanker, which is made up of a cylinder joined with two hemispheres at it’s both ends.

3. Surface Area of a Combination of Soli ds To solve a complex problem, we first try to break it down into simpler and more simpler parts, so that we can solve the problem easily. Similarly, to find the surface areas and volumes of the joint figures, we first break it down into different figures.

4. For example, to find the Total Surface Area of these two joined figures, we first find the T.S.A of the first figure and then of the another.

5. Volume of a Combination of Soli ds In finding the surface areas of some combinations, we found the individual surface areas of the figures and then, added them. Similarly, in order to find the volume of the combinations, we first find the volume of the first figure and then of the other. Then, we add the these two volumes to get the volume of the combination of solids. For Ex-

6. Suppose we have to find the volume of that figure, we would first find the volume of the cylinder and then of the cuboids. Then we would add these up, in order to find the total volume of the figure.

7. Conversion of Solid from One Shape to Another We all have seen candles of different shapes, like cylindrical, in the shape of some animals, statues etc. You might be thinking that how they are made? First a normal candle of a cylindrical shape is taken, then it is heated and molten in a container, and then it is allowed to cool in a container which of the shape of a bird or an animal. But do you think that if there’s a change in the dimensions of the candle, the SA and the Volume of the candle will also change. No, the surface and the volume of the candle will remain same, irrespective of its dimensions.

8. Frustum Of A Cone We have seen objects which are made up of the combination of two solids. Now, let us take an example of the right circular cone. A B C

9. A Cone is also made up of two parts, a smaller right circular cone and an another solid. First let’s do the practical: A Right Circular Cone The Two parts Separated Small Right Circular Cone Frustum of the cone

10. So, given a cone, when we slice through with a plane parallel to its base and remove the cone that is formed on one side of that plane, the part that is now left over is called the Frustum Of The Cone .

11. Some Questions Related To The Chapter The decorative block here, is made up of two solids – a cube and a hemisphere. The base of the block is a cube with edge 5cm, and the hemisphere fixed on the top has a diameter of 4.2cm. Find the total surface area of the block. The total surface area of the cube = 6 x (edge)² = 6 x 5 x 5 cm² = 150 cm² Curved surface area of the hemisphere = 2 Π r² = 2 x 22/7 x 2.1 x 2.1 cm² = 13.86 cm² .˙. Total Surface Area of the Block = (150 + 13.86) cm² = 163.86cm² 5 cm 4.2 cm

12. A juice seller was serving his customers using glasses as shown in the fig. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised which reduced the capacity of the glass. If height the glass was 10 cm, find apparent capacity of the glass and it’s capacity. ( Taking Π =3.14). The inner diameter of the glass = 5 cm and height of the glass = 10 cm .˙. The Apparent Capacity of the glass = Π r²h = 3.14 x 2.5 x 2.5 x 10 cm³ = 196.25 cm³ But, because of the hemisphere, the actual capacity of the glass is less than it’s volume. .˙. The volume of the hemisphere = 2/3 Π r²h = 2/3 x 196.25 cm³ = 32.71 cm³ Now, the actual capacity of the glass = apparent capacity – volume of the hemisphere = (196.25 – 32.71) cm³ = 163.54 cm³

13. A Cone of height 24 cm and of base diameter 12 cm is made up of modelling clay. A Child reshapes it in the form of a sphere. Find the radius of the sphere. The Diameter of the cone = 12 cm so, radius of the cone = 12/2 = 6 cm .˙.Volume of the cone = 1/3 Π r²h = 1/3 x Π x 6 x 6 x 24 cm³ Now, taking ‘r’ as the radius of the sphere, the volume will be = 4/3 Π r³ Since, the volume of clay in the form of the cone and the sphere remains the same, we have : 4/3 x Π x r³ = 1/3 x Π x 6 x 6 x 24 cm r³ = 3 x 3 x 24 = 3² x 2³ .˙. r = 3 x 2 = 6 24 cm 12 cm _ cm

14. Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have processed the sugarcane juice to the molasses, which is poured into molds in the shape of a frustum of a cone having the diameters of it’s two circular faces as 30 cm and 35 cm and the vertical height of the mold is 14cm. If each cm³ of molasses has about 1.2 g, find the mass of molasses that can be poured into each mold. (Taking Π = 22/7) Since the mold is in the shape of a frustum of a cone, the volume of molasses that can be poured into it = Π /3 x h( r ₁ ² + r ₂ ² + r₁ r₂ ) where, ‘r’ is the radius of the target base and ‘r₂’ is the radius of the smaller base = 1/3 x 22/7 x 14[(35/2)² + (30/2)² + (35/2 x 30/2)] cm³ = 11641.7 cm³ Now, we know that : 1cm³ of molasses = 1.2 g .˙. The mass of molasses that can be poured into each mold =(11641 x 1.2) g = 13970.04 g = 13.97 kg = 14 kg (approx.)

15. New Formulas Learnt In This Chapter Volume of the Frustum = 1/3 Π h( r ₁² + r₂² + r₁ r₂ ) Curved Surface Area of the Frustum = Π l(r₁+ r ₂) Total Surface Area of the Frustum = Π l(r₁+ r₂ ) + Π (r ₁+ r₂ ) where, h = vertical height of the frustum l = slant height of the frustum r ₁+ r ₂ = radii of the two bases(ends) of the frustum

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