Hello my dear students! Welcome to today's mathematics class. I am so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 11 of your NCERT textbook, which is about Surface Areas and Volumes. Now, this is a very interesting chapter because it deals with three-dimensional shapes that we see all around us in our daily life. We have already learned about cubes, cuboids, and cylinders in earlier classes. Now, we are going to add two more important three-dimensional shapes to our knowledge: cones and spheres. So, let's begin our journey!
In the first section of this chapter, we will learn about the Surface Area of a Right Circular Cone. Now, what is a cone? You must have seen an ice-cream cone, haven't you? That pointed ice-cream container is shaped like a cone. Also, think about a birthday cap - that is also a cone shape. So, let's understand how a cone is formed.
Let me tell you about an interesting activity. Imagine you have a right-angled triangle, like the one you see in your geometry box. Now, take a long thick string and paste it along one of the perpendicular sides of this triangle. Hold the string with your hands on either side of the triangle and rotate the triangle about the string several times. What do you think will happen? The triangle will start spinning around the string, and as it spins, it will create a new shape. Can you guess what shape it creates? Yes, it creates a cone! This is exactly how a cone is generated. When you rotate a right-angled triangle around one of its perpendicular sides, the shape that is formed is called a cone.
Now, let's understand the different parts of a cone. In this cone that we have formed, there is a point at the top - that is called the vertex of the cone. The distance from the vertex to the center of the circular base is called the height of the cone. The distance from the vertex to any point on the circular edge of the base is called the slant height of the cone. And the radius, as you might already know, is the distance from the center of the circular base to its edge. We usually denote the height by h, the radius by r, and the slant height by l.
Now, it is very important to understand what we mean by a "right circular cone." A cone is called a right circular cone when the line joining its vertex to the center of its base is at a right angle to the base, and the base is circular. If either of these conditions is not satisfied, then it is not a right circular cone. For example, if the vertex is not directly above the center of the base, or if the base is not a perfect circle, then it is not a right circular cone. In this chapter, we will be studying only about right circular cones, so whenever we say "cone," we mean "right circular cone."
Now, let's learn how to find the surface area of a cone. This is where it gets really interesting! Let me tell you about another activity. Take a paper cone - make sure it is neatly made without any overlapped paper. Now, cut it open along its side, starting from the vertex and going all the way down to the base. When you open it out flat, what shape do you think you will get? You will get a part of a circle! It looks like a slice of a round cake. The line along which you cut the cone is actually the slant height of the cone.
Now, if you look at this opened-out shape carefully, you will see that it is actually made up of many small triangles. Imagine if we cut this shape into hundreds of little pieces along the lines drawn from the center point (which was the vertex of the cone). Each small piece would look almost like a tiny triangle, and the height of each such triangle would be the slant height l of the cone.
Now, what is the area of one such triangle? The area of a triangle is half times base times height. So, the area of each small triangle would be half times the base of that triangle times l.
If we add up the areas of all these tiny triangles, we will get the total area of the curved surface of the cone. Now, what is the sum of all the bases of these triangles? The bases of all these triangles, when put together, form the curved boundary of the opened-out shape. And what does this curved boundary represent? It represents the circumference of the base of the cone! And the circumference of the base of the cone is equal to 2πr, where r is the radius of the base.
So, the total curved surface area of the cone would be half times l times the sum of all the bases, which is half times l times 2πr. This simplifies to πrl. So, the curved surface area of a cone is πrl, where r is the base radius and l is the slant height.
Now, there is an important relationship between the slant height, the radius, and the height of the cone. By applying the Pythagoras Theorem to the right-angled triangle formed by the height, radius, and slant height, we get l squared equals r squared plus h squared. So, l equals the square root of r squared plus h squared. This is very useful because sometimes we know the height and radius but not the slant height, and we can easily find it using this formula.
Now, what if we want to find the total surface area of the cone? The total surface area includes both the curved surface area and the area of the circular base. The area of the circular base is πr squared. So, the total surface area of a cone is πrl plus πr squared, which can also be written as πr times (l plus r).
Now, let me give you a quick recap of what we have learned so far about cones. The curved surface area of a cone is πrl. The total surface area of a cone is πrl plus πr squared, which equals πr(l + r). And remember, l squared equals r squared plus h squared.
Now, let's look at some examples to understand these formulas better.
Example 1: Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Solution: The curved surface area is πrl. So, we substitute the values: 22/7 × 7 × 10, which equals 220 cm squared. That's it!
Example 2: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. Use π = 3.14.
Solution: First, we need to find the slant height l. Using l squared equals h squared plus r squared, we get l equals the square root of 16 squared plus 12 squared, which equals the square root of 256 plus 144, which equals the square root of 400, which is 20 cm. Now, curved surface area equals πrl, which is 3.14 × 12 × 20, which equals 753.6 cm squared. For total surface area, we add πr squared, which is 3.14 × 12 × 12, which equals 452.16. So, total surface area is 753.6 plus 452.16, which equals 1205.76 cm squared.
Example 3: A corn cob, shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 cm² of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob.
Solution: Since the grains are found only on the curved surface, we need to find the curved surface area. First, find the slant height: l equals the square root of r squared plus h squared, which equals the square root of (2.1)² + 20², which equals the square root of 4.41 + 400, which equals the square root of 404.41, which is approximately 20.11 cm. Now, curved surface area equals πrl, which is 22/7 × 2.1 × 20.11, which equals approximately 132.73 cm². Since each cm² has 4 grains, the total number of grains is 132.73 × 4, which is approximately 531 grains.
So, students, this is how we calculate the surface areas of cones. Now, let's move on to the next topic: Surface Area of a Sphere.
What is a sphere? Is it the same as a circle? This is a very common question, and the answer is no! A circle is a two-dimensional figure - it is flat, like the shape of a coin placed on a table. But a sphere is a three-dimensional figure - it has depth, like a ball. You can hold a sphere in your hand, but you cannot hold a circle.
Now, how do we form a sphere? Let me tell you about another interesting activity. Take a circular disc, like a round piece of cardboard. Now, paste a string along a diameter of this disc. Now, rotate this disc around the string, just like we rotated the triangle to form a cone. What shape do you think will be formed? Yes, it forms a sphere! The center of the circle becomes the center of the sphere. So, a sphere is a three-dimensional figure that is made up of all points in space that lie at a constant distance called the radius from a fixed point called the center.
Now, let's learn how to find the surface area of a sphere. This is a very interesting activity. Take a rubber ball and drive a nail into it. Now, wind a string around the ball, starting from the nail and going around and around until you have completely covered the ball with string. Make sure you cover every part of the ball. Now, mark the starting and finishing points on the string and carefully unwind it from the ball. Now, on a sheet of paper, draw four circles with the same radius as the ball. Now, start filling these circles with the string you wound around the ball. What do you think will happen? The string that completely covered the surface of the sphere will exactly fill the regions of four circles, all of the same radius as the sphere!
This is amazing, isn't it? This experiment tells us that the surface area of a sphere is equal to four times the area of a circle with the same radius. And since the area of a circle is πr², the surface area of a sphere is 4πr², where r is the radius of the sphere.
Now, what is a hemisphere? If you take a sphere and slice it exactly through the middle with a plane that passes through its center, you get two equal parts. Each part is called a hemisphere. The word "hemi" means "half," so hemisphere literally means "half a sphere."
Now, how many faces does a hemisphere have? It has two faces: a curved face and a flat face (the base). The curved surface area of a hemisphere is half of the surface area of the sphere, which is half of 4πr², which equals 2πr². And the total surface area of a hemisphere includes both the curved surface and the flat base, so it is 2πr² plus πr², which equals 3πr².
Let me recap what we have learned about spheres. The surface area of a sphere is 4πr². The curved surface area of a hemisphere is 2πr². The total surface area of a hemisphere is 3πr².
Now, let's look at some examples.
Example 4: Find the surface area of a sphere of radius 7 cm.
Solution: Surface area equals 4πr², which is 4 × 22/7 × 7 × 7, which equals 616 cm².
Example 5: Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Solution: (i) Curved surface area equals 2πr², which is 2 × 22/7 × 21 × 21, which equals 2772 cm². (ii) Total surface area equals 3πr², which is 3 × 22/7 × 21 × 21, which equals 4158 cm².
Example 6: The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Solution: Diameter is 7 m, so radius is 3.5 m. The riding space is the surface area of the sphere, which is 4πr², which is 4 × 22/7 × 3.5 × 3.5, which equals 154 m².
Example 7: A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹5 per 100 cm².
Solution: First, find the radius from the circumference. Circumference is 2πr, so 17.6 = 2 × 22/7 × r, which gives r = 2.8 m. The curved surface area of the hemisphere is 2πr², which is 2 × 22/7 × 2.8 × 2.8, which equals 49.28 m². Now, cost of painting 100 cm² is ₹5, so cost of 1 m² (which is 10,000 cm²) is ₹500. So, total cost is 500 × 49.28, which equals ₹24,640.
Now, students, let's move on to the third section: Volume of a Right Circular Cone.
We have already learned about volumes of cubes, cuboids, and cylinders in earlier classes. Now, we will learn about the volume of a cone.
Let me tell you about a very interesting experiment. Take a hollow cylinder and a hollow cone that have the same base radius and the same height. Now, fill the cone with sand up to the brim and empty it into the cylinder. What happens? The cylinder is not full, is it? It only fills up a part of the cylinder. Now, fill the cone again and empty it into the cylinder. Still, the cylinder is not full. Fill the cone for the third time and empty it into the cylinder. Now, the cylinder is full to the brim!
What does this tell us? It tells us that three cones full of sand fill up one cylinder of the same base radius and height. In other words, the volume of a cone is one-third of the volume of a cylinder with the same base radius and height. So, the volume of a cone is one-third times πr²h, where r is the base radius and h is the height of the cone.
Let me recap: Volume of a cone equals one-third πr²h.
Now, let's look at some examples.
Example 8: The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone.
Solution: First, we need to find the radius. Using l² = r² + h², we get r² = l² - h² = 28² - 21² = 784 - 441 = 343. So, r = √343 = 7√7 cm. Now, volume = 1/3 πr²h = 1/3 × 22/7 × 7√7 × 7√7 × 21. Now, 7√7 × 7√7 = 7² × 7 = 49 × 7 = 343. So, volume = 1/3 × 22/7 × 343 × 21 = 22 × 7 × 7 = 7546 cm³.
Example 9: Monica has a piece of canvas whose area is 551 m². She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting amounts to approximately 1 m², find the volume of the tent that can be made with it.
Solution: Available canvas area is 551 - 1 = 550 m². This is the curved surface area of the tent, which is πrl. So, 22/7 × 7 × l = 550, which gives l = 25 m. Now, using l² = r² + h², we get h = √(25² - 7²) = √(625 - 49) = √576 = 24 m. So, volume = 1/3 πr²h = 1/3 × 22/7 × 7 × 7 × 24 = 1232 m³.
Now, students, let's move on to the final section: Volume of a Sphere.
How do we find the volume of a sphere? Let me tell you about an experiment. Take a container that is big enough to hold a sphere and fill it completely with water. Now, carefully place a sphere into the container. Some water will overflow. Now, measure the volume of the water that overflowed. This volume is equal to the volume of the sphere! This is because of a principle called Archimedes' principle. When you do this experiment with spheres of different radii, you will find that the volume of the sphere is always equal to 4/3 πr³, where r is the radius of the sphere.
So, the volume of a sphere is 4/3 πr³.
Now, what about the volume of a hemisphere? Since a hemisphere is half of a sphere, its volume is half of 4/3 πr³, which is 2/3 πr³.
Let me recap: Volume of a sphere equals 4/3 πr³. Volume of a hemisphere equals 2/3 πr³.
Now, let's look at some examples.
Example 10: Find the volume of a sphere of radius 11.2 cm.
Solution: Volume = 4/3 πr³ = 4/3 × 22/7 × 11.2 × 11.2 × 11.2 = 5887.32 cm³.
Example 11: A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-putt.
Solution: First, find the volume: 4/3 πr³ = 4/3 × 22/7 × 4.9 × 4.9 × 4.9 = 493 cm³ (approximately). Now, mass = volume × density = 493 × 7.8 = 3845.44 g = 3.85 kg (approximately).
Example 12: A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Solution: Volume = 2/3 πr³ = 2/3 × 22/7 × 3.5 × 3.5 × 3.5 = 89.8 cm³.
Now, students, we have covered all the topics in this chapter. Let me give you a complete summary of everything we have learned today.
In this chapter, we studied about surface areas and volumes of cones and spheres.
First, we learned about the surface area of a right circular cone. The curved surface area of a cone is πrl, where r is the base radius and l is the slant height. The total surface area of a cone is πrl + πr², which can also be written as πr(l + r). We also learned that l² = r² + h², where h is the height of the cone.
Then, we learned about the surface area of a sphere. The surface area of a sphere of radius r is 4πr². We also learned about hemispheres. A hemisphere is half of a sphere. The curved surface area of a hemisphere is 2πr², and the total surface area of a hemisphere is 3πr².
After that, we learned about the volume of a right circular cone. The volume of a cone is one-third times πr²h, where r is the base radius and h is the height of the cone.
Finally, we learned about the volume of a sphere. The volume of a sphere is 4/3 πr³, and the volume of a hemisphere is 2/3 πr³.
These are all the formulas and concepts that you need to know from this chapter. Make sure you remember them all and practice as many problems as you can. Mathematics is a subject that requires a lot of practice, and I am confident that you will do very well!
That brings us to the end of today's lesson. Thank you for listening so attentively. Keep studying hard, and I will see you in the next class. Goodbye, students!