Surface Areas And Volumes

Hello students, welcome to today's mathematics class. I'm so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 12, which is about Surface Areas and Volumes. Now, I know some of you might be thinking, "Oh no, not again! We already studied this in Class 9." But trust me, students, what we are going to learn today is going to take your understanding to a completely new level. We are going to learn how to find the surface areas and volumes of objects that are made up of combinations of different basic solids. This is going to be really useful in real life, and I promise by the end of this lesson, you will be able to solve problems that you might have never thought possible.

So, let's begin, shall we?

In Class 9, you learned about some basic solids like the cuboid, the cone, the cylinder, and the sphere. You learned how to find their surface areas and their volumes. Now, think about the world around you. Do you think everything you see is just a simple cuboid or a simple cylinder? Of course not! Most objects around us are made up of combinations of these basic shapes. Let me give you some examples from everyday life in India.

Have you ever seen those big trucks carrying water or oil? You know, the ones with those big cylindrical containers on the back? Well, that container is not just a simple cylinder. If you look closely at its ends, you will see that they are curved, like hemispheres. So, the entire container is actually a combination of a cylinder with two hemispheres at either end. This is exactly the kind of object we are going to learn to handle today.

Another example is a test tube. You must have used test tubes in your science laboratory. What is the shape of a test tube? It has a cylindrical body, but at the bottom, it is rounded, like a hemisphere. So, it is a combination of a cylinder and a hemisphere.

Now, think about some beautiful buildings or monuments you might have seen. Many of them have combinations of different shapes. For instance, some temples have dome-shaped tops, which are like hemispheres, sitting on top of cylindrical or rectangular structures.

So, students, you see, in our daily life, we come across many such objects that are made by combining two or more basic solids. Now, the question is: how do we find the surface area or the volume of such objects? We cannot simply use the formulas we learned for basic solids because these are combinations. That is exactly what we are going to learn in this chapter.

Let me tell you what we will cover in this lesson. First, we will learn how to find the surface area of a combination of solids. Then, we will learn how to find the volume of such combinations. And along the way, we will work through several examples to make sure you understand each and every concept thoroughly.

So, let's start with Section 12.2: Surface Area of a Combination of Solids.

Now, students, whenever we come across a new problem, what do we do? We try to break it down into smaller problems that we already know how to solve. That's exactly what we are going to do here.

Consider that truck container I mentioned earlier. It is made up of a cylinder with two hemispheres stuck at either end. If we want to find the total surface area of this solid, we need to think about which parts are visible and which parts are hidden.

When we put the pieces together, what do we see? We see the curved surface of the cylinder, and we see the curved surfaces of the two hemispheres. The flat circular faces where the hemispheres are attached to the cylinder are not visible from the outside, are they? So, they are not part of the external surface area.

Therefore, the total surface area of this new solid is simply the sum of the curved surface area of the cylinder and the curved surface areas of the two hemispheres. We can write this as:

Total Surface Area equals Curved Surface Area of one hemisphere plus Curved Surface Area of the cylinder plus Curved Surface Area of the other hemisphere.

Here, TSA stands for Total Surface Area, and CSA stands for Curved Surface Area. Make sure you remember these abbreviations, students, as we will be using them frequently.

Now, let's consider another situation. Suppose we are making a toy by putting together a hemisphere and a cone. How would we do that? First, we take a cone and a hemisphere and bring their flat faces together. But for this to work smoothly, the base radius of the cone must be equal to the radius of the hemisphere. Otherwise, there would be a gap, and the toy wouldn't be smooth.

So, imagine we have joined a cone and a hemisphere in this way. Now, if we want to find how much paint would be required to color the surface of this toy, what would we need to know? We would need to know the total surface area of the toy. And what does this surface consist of? It consists of the curved surface area of the hemisphere and the curved surface area of the cone. The flat circular face where they are joined is hidden inside, so we don't count it.

So, we can say: Total surface area of the toy equals Curved Surface Area of hemisphere plus Curved Surface Area of cone.

Now, students, I want you to understand an important point here. When we are finding the surface area of a combination of solids, we only count the external curved surfaces. We do not count the surfaces that are hidden inside, where the solids are joined together. This is a very important concept, and you must remember it.

Now, let's work through some examples to understand this better.

Example 1: Rasheed got a playing top, which we call a lattu in India, as his birthday present. It had no color on it, and he wanted to color it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 centimeters in height, and the diameter of the top is 3.5 centimeters. We need to find the area he has to color. Take pi equal to 22/7.

Now, this is exactly like the object we just discussed: a cone with a hemisphere on top. So, we can use the same formula: Total Surface Area equals Curved Surface Area of hemisphere plus Curved Surface Area of cone.

First, let's find the radius. The diameter is 3.5 centimeters, so the radius is half of that, which is 3.5 divided by 2, equals 1.75 centimeters.

Now, the curved surface area of a hemisphere. You might remember that the surface area of a sphere is 4πr². So, half of that, for a hemisphere, is 2πr². We can write this as half of 4πr², which is 2πr².

So, curved surface area of hemisphere equals 2πr². Let's calculate this: 2 times 22/7 times 1.75 times 1.75. Now, 1.75 is the same as 7/4. So, this becomes 2 times 22/7 times 7/4 times 7/4. The 7 in the numerator and denominator cancels, and we get 2 times 22 times 7 divided by 4 times 4. Let me calculate this step by step.

2 times 22 is 44. 44 divided by 7 is not straightforward. Let me do it differently. 1.75 times 1.75 equals 3.0625. So, 2 times 22/7 times 3.0625. 22/7 is approximately 3.1429. So, 2 times 3.1429 times 3.0625 equals approximately 19.24 square centimeters. But let's keep it in fractional form for accuracy.

Actually, let's do it properly: 2 × 22/7 × (3.5/2)². Now, (3.5/2)² = (3.5²)/(2²) = 12.25/4 = 3.0625. So, 2 × 22/7 × 3.0625 = (44/7) × 3.0625. 3.0625 is 49/16. So, this becomes 44/7 × 49/16 = 44 × 7 / 16 = 308/16 = 19.25 square centimeters. Yes, that's correct.

Now, we need the curved surface area of the cone. For this, we need the slant height of the cone. But first, let's find the height of the cone.

The total height of the top is 5 centimeters. This includes the hemisphere. Now, what is the height of the hemisphere? The height of a hemisphere is equal to its radius, which is 1.75 centimeters. So, the height of the cone is total height minus height of hemisphere, which is 5 minus 1.75, equals 3.25 centimeters.

Now, for the slant height of the cone, we use the Pythagorean theorem. The slant height l equals square root of (r² + h²), where r is the radius and h is the height of the cone.

So, l equals square root of (1.75² + 3.25²). Let's calculate: 1.75 squared is 3.0625, and 3.25 squared is 10.5625. Adding them gives 13.625. The square root of 13.625 is approximately 3.69, but let's take it as 3.7 centimeters as given in the book.

Now, the curved surface area of the cone equals πrl, where r is the radius and l is the slant height. So, this is 22/7 × 1.75 × 3.7. 1.75 is 7/4, so this becomes 22/7 × 7/4 × 3.7 = 22/4 × 3.7 = 5.5 × 3.7 = 20.35 square centimeters.

Now, the total surface area of the top is the sum of these two: 19.25 plus 20.35 equals 39.6 square centimeters approximately.

So, Rasheed needs to color an area of about 39.6 square centimeters.

Now, students, I want you to note something important here. The total surface area of the top is not the sum of the total surface areas of the cone and the hemisphere separately. If we had added the total surface areas, we would have included the base areas of both the cone and the hemisphere, which are not exposed when they are joined together. So, we only add the curved surface areas. This is a crucial point to remember.

Now, let's look at Example 2.

Example 2: A decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 centimeters, and the hemisphere is fixed on the top. The hemisphere has a diameter of 4.2 centimeters. We need to find the total surface area of the block. Take pi equal to 22/7.

Now, let's think about this. We have a cube of side 5 centimeters. On top of it, there is a hemisphere. The hemisphere has a diameter of 4.2 centimeters, so its radius is 2.1 centimeters.

Now, what is the total surface area of this block? We start with the total surface area of the cube. The cube has six faces, each of area side squared. So, total surface area of cube is 6 times 5 squared, which is 6 times 25, equals 150 square centimeters.

But wait, is the entire cube exposed? No, because the hemisphere is attached to the top face of the cube. So, the part of the top face that is covered by the hemisphere is not exposed. That part is a circle of radius 2.1 centimeters. Its area is πr², which is 22/7 times 2.1 times 2.1. 2.1 is 21/10, so this is 22/7 × 441/100 = 22 × 63 / 100 = 1386/100 = 13.86 square centimeters.

So, from the cube's surface area, we need to subtract this circle area because it's not exposed. But then, we also need to add the curved surface area of the hemisphere, because that is now exposed.

The curved surface area of the hemisphere is 2πr², which is 2 times 22/7 times 2.1 times 2.1. We already calculated πr² as 13.86, so 2πr² is 2 times 13.86, equals 27.72 square centimeters.

So, the total surface area of the block equals surface area of cube minus area of circle plus curved surface area of hemisphere. That is 150 minus 13.86 plus 27.72, which equals 163.86 square centimeters.

So, students, you see, when a hemisphere is placed on top of a cube, we subtract the area of the base of the hemisphere from the cube's surface area, and then add the curved surface area of the hemisphere.

Now, let's look at Example 3.

Example 3: A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 26 centimeters, while the height of the conical part is 6 centimeters. The base of the conical portion has a diameter of 5 centimeters, while the base diameter of the cylindrical portion is 3 centimeters. The conical portion is to be painted orange, and the cylindrical portion is to be painted yellow. We need to find the area of the rocket painted with each of these colors. Take pi equal to 3.14.

Now, this is a more interesting problem because the two parts have different radii. Let's understand this.

The cone has a diameter of 5 centimeters, so its radius is 2.5 centimeters. The cylinder has a diameter of 3 centimeters, so its radius is 1.5 centimeters.

The height of the cone is given as 6 centimeters. The total height is 26 centimeters, so the height of the cylinder is 26 minus 6, equals 20 centimeters.

Now, the slant height of the cone, l, equals square root of (r² + h²), where r is 2.5 and h is 6. So, l equals square root of (2.5² + 6²) = square root of (6.25 + 36) = square root of 42.25 = 6.5 centimeters.

Now, here's the tricky part. The base of the cone is larger than the base of the cylinder. The cone has a radius of 2.5, so its base area is π times 2.5 squared, which is π times 6.25. The cylinder has a radius of 1.5, so its base area is π times 1.5 squared, which is π times 2.25.

When the cone is mounted on the cylinder, the cylinder sits inside the cone, but the cone's base extends beyond the cylinder. So, there is a ring-shaped area of the cone's base that is exposed. This area needs to be painted orange along with the curved surface of the cone.

So, the area to be painted orange equals the curved surface area of the cone plus the area of the cone's base minus the area of the cylinder's base. That is πrl + πr² minus π times (r')², where r is the cone's radius and r' is the cylinder's radius.

Let's calculate this. πrl is 3.14 times 2.5 times 6.5. 2.5 times 6.5 is 16.25, so this is 3.14 times 16.25, which is 51.025 square centimeters.

πr² is 3.14 times 2.5 squared, which is 3.14 times 6.25, equals 19.625 square centimeters.

π(r')² is 3.14 times 1.5 squared, which is 3.14 times 2.25, equals 7.065 square centimeters.

So, the orange area is 51.025 plus 19.625 minus 7.065, which equals 63.585 square centimeters.

Now, for the yellow part, which is the cylinder. The cylinder has a curved surface area, which is 2πr'h', where r' is 1.5 and h' is 20. So, this is 2 times 3.14 times 1.5 times 20, which is 2 times 3.14 times 30, which is 188.4 square centimeters.

But wait, we also need to paint the top of the cylinder, because it is exposed. The cylinder's top is a circle of radius 1.5, so its area is π times 1.5 squared, which is 7.065 square centimeters.

So, the yellow area is 188.4 plus 7.065, equals 195.465 square centimeters.

So, students, you see, in this problem, we had to be careful about which areas are exposed and which are not. The key is to always think about what is visible from the outside.

Now, let's look at Example 4.

Example 4: Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 meters, and its radius is 30 centimeters. We need to find the total surface area of the bird-bath. Take pi equal to 22/7.

Now, this is interesting. The bird-bath has a hemispherical depression. That means instead of having a hemisphere sticking out, it has a hemisphere scooped out. So, it's like a cylinder with a hollow hemisphere at one end.

But wait, the problem says "hemispherical depression at one end." So, it's a cylinder with a hemisphere removed from one end. But then, what is the other end? Is it open or closed? Let me re-read the problem. It says "a cylinder with a hemispherical depression at one end." So, it seems like the cylinder is open at both ends, but one end has a hemisphere scooped out, meaning it curves inward.

Actually, looking at the figure, it would be a cylinder with a hemisphere cut out from one end. So, the exposed surface consists of the curved surface of the cylinder and the curved surface of the hemisphere, but the hemisphere is curved inward.

Wait, let me think about this more carefully. If we have a cylinder and we scoop out a hemisphere from one end, what surfaces are exposed? The outer curved surface of the cylinder is exposed. And the inner curved surface of the hemisphere is also exposed. But the flat circular face where the hemisphere was cut out is not there anymore because it's been scooped out.

So, the total surface area would be the curved surface area of the cylinder plus the curved surface area of the hemisphere. But since the hemisphere is a depression, its curved surface is facing inward, but it is still exposed, so we count it.

So, total surface area equals curved surface area of cylinder plus curved surface area of hemisphere. The curved surface area of cylinder is 2πrh, and curved surface area of hemisphere is 2πr².

So, this equals 2πr (h + r).

Now, let's plug in the values. The radius is 30 centimeters. The height is 1.45 meters, which is 145 centimeters. So, h is 145 cm, and r is 30 cm.

So, total surface area equals 2 times 22/7 times 30 times (145 + 30). 145 + 30 is 175. So, this is 2 times 22/7 times 30 times 175.

Let's calculate: 2 times 22 is 44. 44/7 times 30 is 1320/7. 1320/7 times 175. Now, 175 is 25 times 7. So, this becomes 1320/7 times 25 times 7, which is 1320 times 25, which is 33,000 square centimeters.

Now, we need to convert this to square meters. 10,000 square centimeters equals 1 square meter. So, 33,000 square centimeters equals 3.3 square meters.

So, the total surface area of the bird-bath is 3.3 square meters.

Now, students, I hope you are getting a good understanding of how to find the surface area of combinations of solids. The key points to remember are:

First, identify all the basic solids that make up the given object. Second, determine which surfaces are exposed and which are hidden. Third, add the curved surface areas of all the exposed parts. If any flat surfaces are exposed, add those too. Fourth, subtract the areas of any surfaces that are hidden or covered by other solids.

Now, let's move on to Section 12.3: Volume of a Combination of Solids.

In the previous section, we learned how to find the surface area of solids made up of combinations. Now, we are going to learn how to find their volumes.

Now, here's an important difference between surface area and volume. When we calculate surface area, we have to be careful about not counting hidden surfaces. But when we calculate volume, this is not an issue. The volume of a solid formed by joining two basic solids is simply the sum of the volumes of the individual solids. This is because volume is about how much space is inside, and when we join solids, the spaces inside them add up. There is no overlapping of volumes, unlike with surfaces.

So, to find the volume of a combination of solids, we simply find the volumes of each part and add them together. This is much simpler than finding surface areas, isn't it?

Now, let's work through some examples to understand this better.

Example 5: Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder. If the base of the shed is of dimension 7 meters by 15 meters, and the height of the cuboidal portion is 8 meters, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 cubic meters, and there are 20 workers, each of whom occupy about 0.08 cubic meters space on average. Then, how much air is in the shed? Take pi equal to 22/7.

Now, this is a practical problem. The shed is made of two parts: a cuboid and a half cylinder on top.

The cuboid has length 15 meters, breadth 7 meters, and height 8 meters. So, its volume is length times breadth times height, which is 15 times 7 times 8, equals 840 cubic meters.

Now, the half cylinder is on top. Its diameter is equal to the breadth of the cuboid, which is 7 meters. So, its radius is 3.5 meters. Its height is the same as the length of the cuboid, which is 15 meters.

The volume of a full cylinder is πr²h. So, the volume of a half cylinder is half of that, which is 1/2 πr²h.

So, the volume of the half cylinder is 1/2 times 22/7 times (3.5)² times 15. 3.5 is 7/2, so (3.5)² is 49/4. So, this is 1/2 times 22/7 times 49/4 times 15.

Let's simplify: 1/2 times 22/7 is 11/7. 11/7 times 49/4 is 11 times 7 / 4, which is 77/4. 77/4 times 15 is 77 times 15 / 4, which is 1155 / 4, equals 288.75 cubic meters.

So, the total volume of the shed is 840 plus 288.75, equals 1128.75 cubic meters. This is the volume of air inside the shed when there is no machinery or people.

Now, the machinery occupies 300 cubic meters. And the 20 workers occupy 20 times 0.08, which is 1.6 cubic meters.

So, the total occupied space is 300 plus 1.6, equals 301.6 cubic meters.

Therefore, the volume of air in the shed when there is machinery and workers is 1128.75 minus 301.6, equals 827.15 cubic meters.

So, students, you see how we used the volume formulas for basic solids and simply added them to find the total volume.

Now, let's look at Example 6.

Example 6: A juice seller was serving his customers using glasses as shown in the figure. The inner diameter of the cylindrical glass was 5 centimeters, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 centimeters, find the apparent capacity of the glass and its actual capacity. Use pi equal to 3.14.

Now, this is a very practical example. The glass looks like a cylinder from the outside, but inside, at the bottom, there is a hemispherical raised portion. This means the actual capacity is less than what it would be if it were a plain cylinder.

The apparent capacity is what it would hold if there were no hemisphere at the bottom. This is simply the volume of a cylinder with radius 2.5 centimeters and height 10 centimeters.

So, apparent capacity equals πr²h, which is 3.14 times 2.5 times 2.5 times 10, which is 3.14 times 6.25 times 10, which is 3.14 times 62.5, which is 196.25 cubic centimeters.

But the actual capacity is less because of the hemisphere at the bottom. The hemisphere takes up some space, so the glass can hold less liquid.

The volume of a hemisphere is 2/3 πr³. So, the volume of the hemisphere is 2/3 times 3.14 times 2.5 times 2.5 times 2.5. Let's calculate: 2.5 cubed is 15.625. So, this is 2/3 times 3.14 times 15.625.

3.14 times 15.625 is approximately 49.0625. 2/3 of that is about 32.71 cubic centimeters.

So, the actual capacity is the apparent capacity minus the volume of the hemisphere, which is 196.25 minus 32.71, equals 163.54 cubic centimeters.

So, students, you see, the actual capacity is about 163.54 cubic centimeters, while the apparent capacity is 196.25 cubic centimeters. The difference is about 32.71 cubic centimeters, which is the volume of the hemisphere.

Now, let's look at Example 7.

Example 7: A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 centimeters, and the diameter of the base is 4 centimeters. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. Take pi equal to 3.14.

Now, this is an interesting problem. We have a toy that is a combination of a hemisphere and a cone. The cone is on top of the hemisphere.

The diameter of the base is 4 centimeters, so the radius is 2 centimeters. This radius is the same for both the hemisphere and the cone because they are joined at their bases.

The height of the cone is given as 2 centimeters.

Now, the volume of the toy is the sum of the volume of the hemisphere and the volume of the cone.

The volume of a hemisphere is 2/3 πr³. So, this is 2/3 times 3.14 times 2³. 2 cubed is 8. So, this is 2/3 times 3.14 times 8, which is 16.746... approximately 16.75 cubic centimeters.

The volume of a cone is 1/3 πr²h. So, this is 1/3 times 3.14 times 2² times 2. 2² is 4, times 2 is 8. So, this is 1/3 times 3.14 times 8, which is 8.37 approximately.

So, the total volume is 16.75 plus 8.37, equals 25.12 cubic centimeters.

Now, we need to find a right circular cylinder that circumscribes this toy. What does "circumscribe" mean? It means the cylinder completely contains the toy, and the toy touches the cylinder at certain points.

In this case, the cylinder has the same radius as the hemisphere and the cone, which is 2 centimeters. And the height of the cylinder is the total height of the toy.

What is the total height? The hemisphere has a height of 2 centimeters (its radius), and the cone has a height of 2 centimeters. So, the total height is 2 plus 2, equals 4 centimeters.

So, the volume of the cylinder is πr²h, which is 3.14 times 2² times 4, which is 3.14 times 4 times 4, which is 3.14 times 16, equals 50.24 cubic centimeters.

Now, the difference between the volume of the cylinder and the volume of the toy is 50.24 minus 25.12, equals 25.12 cubic centimeters.

Interestingly, the difference is exactly equal to the volume of the toy itself! This is a coincidence in this particular case, but it's a nice result.

So, students, now you have learned how to find both the surface area and the volume of combinations of solids. Let me summarize what we have learned in this chapter.

First, we learned about surface areas. When we have a solid that is made up of a combination of basic solids, we find its surface area by adding the curved surface areas of all the exposed parts. We must be careful not to include the areas where the solids are joined together, as those are not exposed.

Second, we learned about volumes. Finding the volume of a combination is simpler: we just add the volumes of all the constituent solids. There is no need to worry about overlapping because volumes simply add up.

We worked through several examples to understand these concepts better. We looked at objects like playing tops, decorative blocks, toy rockets, bird-baths, sheds, glasses, and toys. In each case, we identified the basic solids involved, determined what surfaces or volumes needed to be calculated, and then applied the appropriate formulas.

Now, let me give you a quick recap of the key formulas we used:

For a cuboid, the total surface area is 6 times side squared, and the volume is length times breadth times height.

For a cylinder, the curved surface area is 2πrh, the total surface area is 2πr (r + h), and the volume is πr²h.

For a cone, the curved surface area is πrl, where l is the slant height, and the volume is 1/3 πr²h.

For a sphere, the surface area is 4πr², and the volume is 4/3 πr³.

For a hemisphere, the curved surface area is 2πr², and the volume is 2/3 πr³.

And when we have combinations, we simply add or subtract these as needed, depending on which parts are exposed.

Students, I hope this lesson has helped you understand how to find the surface areas and volumes of combinations of solids. Remember, the key is to break down the problem into smaller parts, identify the basic solids, and then apply the appropriate formulas. With practice, you will become very good at this.

Now, before I conclude, let me remind you of some important points to keep in mind while solving problems:

Always read the problem carefully and visualize the object. Draw a rough sketch if needed.

Identify all the basic solids that make up the object.

For surface area: determine which surfaces are exposed and which are hidden. Add the areas of all exposed curved surfaces and flat surfaces. Subtract the areas of any hidden surfaces.

For volume: simply add the volumes of all the constituent solids.

Make sure you use consistent units. If the radius is in centimeters, keep everything in centimeters. If the height is in meters, convert everything to meters before calculating.

And finally, always double-check your calculations and make sure your answer makes sense.

That's all for today, students. Thank you for your attention and patience. I hope you enjoyed this lesson and feel confident about solving problems on surface areas and volumes of combinations of solids. Keep practicing, and you will master this topic in no time. Good luck!