 Hello and welcome to the session. In this session we will discuss about the surface area of various basic solids. Let's recall the formula for the surface area of cuboid which is equal to 2 multiplied by LB plus BH plus HL. Where we have L is the length of the cuboid, B is the breadth of the cuboid and H is the height of the cuboid. Next we have the curved surface area or you can say CSA of cone is equal to pi RL and then the total surface area or TSA of cone is given by pi R into L plus R. Where R is the radius of base of the cone, L is the slant height of the cone. Next we have curved surface area of cylinder is given by 2 pi RH then the total surface area of cylinder is given by 2 pi R into R plus H where we have R is the radius of base of cylinder and H is the height of cylinder. Then the surface area of sphere is given by 4 pi R square where this R is the radius of sphere. Then curved surface area of hemisphere is given by 2 pi R square and total surface area of hemisphere is given by 3 pi R square where R is the radius of hemisphere. In our day to day life we come across a number of solids which are made up of combinations of two or more of the basic solids like sphere, hemisphere, cuboid, cone, cylinder, etc. So now we shall discuss the surface area of a combination of solids. Consider this object as you can see this is a combination of a cone and a hemisphere. So to find the surface area of this whole object what we do is we find the curved surface area of the cone and the curved surface area of the hemisphere and the sum of these two curved surface area would give us the surface area of this object. Let the height H of this cone be given as 4 centimeters. Now radii of the base of the cone and the hemisphere is given as 3 centimeters. Let's find out the slant height which is the length of this portion. Slant height L is given by square root of X square plus R square that is equal to 5 centimeters. So slant height L is 5 centimeters. Now let's find out the curved surface area of this cone which is given by pi RL. This comes out to be equal to 47.14 centimeters square. Now the curved surface area of the hemisphere is given by 2 pi R square which comes out to be equal to 56.57 centimeters square. Thus the surface area, so now the total surface area of the given object is equal to the curved surface area of this cone plus the curved surface area of this hemisphere and this is equal to 103.71 centimeters square. Thus we conclude that when we are given an object which is a combination of two or more basic solids then we can find its total surface area by finding the surface areas or curved surface areas of its constituents and adding them. Now we shall discuss about the volumes of basic solids. Volume of cuboid as you know is given by L into B into H where this L is the length of cuboid, B is the breadth of cuboid and H is the height of cuboid. Then we have volume of cone is given as 1 upon 3 pi R square H where this R is the radius of base of the cone and H is the height of the cone. Next is the volume of cylinder is given as pi R square H where R is the radius of base of cylinder and H is the height of cylinder. Then we have the volume of sphere which is given by 4 by 3 pi R cube. This R is the radius of sphere and also volume of hemisphere is equal to 2 upon 3 pi R cube. This R is the radius of hemisphere. Now the volume of the solid which is formed by joining two basic solids will actually be the sum of the volume of the constituents. So now let's see how do we find the volume of a combination of solids. The same object which is the combination of a cone and a hemisphere, height of the cone is given as 4 cm and radii of the base of cone and hemisphere is 3 cm. Volume of this cone is given by 1 by 3 pi R square H which comes out to be equal to 37.71 cm cube. Now the volume of this hemisphere is given by 2 by 3 pi R cube which comes out to be equal to 56.57 cm cube. Now the volume of this whole object which is a combination of a cone and a hemisphere is equal to volume of the cone plus the volume of the hemisphere which comes out to be equal to 94.28 cm cube. So whenever we are given an object which is a combination of two or more solids we can easily find its volume by finding the individual volumes of the basic solids and adding them. Now we discuss about the conversion of solid from one shape to another. Whenever we come across objects which are converted from one shape to another then the volume of both these objects would remain the same. Consider a sphere of radius R1 equal to 4.2 cm. Suppose this sphere is melted and recast into a shape of cylinder of radius R2 equal to 6 cm. Then we can find the height of the cylinder because we know that when one solid is melted and recast into a shape of another solid then the volume of both these solids would be same. So in this case volume of sphere would be equal to volume of the cylinder that is 4 by 3 pi R1 cube is equal to pi R2 square H. Now from here we can find the value of this H that is the height of the cylinder. This would be given by 4 by 3 R1 cube upon R2 square. It comes out to be equal to 2.744 cm that is we have got the height of the cylinder H as this. So we see when one solid is converted to another solid the volume remains the same and this would help us determine the unknown dimension. This completes the session. Hope you understood how to find the surface area of combination of solids and the volume of combination of solids.