 Hello and welcome to the session. In this session we are going to study about the surface area of two and three dimensional objects composed of presence and pyramids. Now the surface area of any solid is equal to the sum of the areas and its faces. For example, if we want to paint a house from outside we will apply the paint on the surface area of the house. A composition of solid is when two or more solids are combined together to make one solid. Like a house may be a combination of a rectangular prism and a pyramid. The staircase to a house is also a combination of solids consisting of number of rectangular prisms. Now suppose we want to find the surface area of the solid composed of a rectangular prism and a rectangular pyramid. That is, if we have to find the surface area of this solid which is composed of a rectangular prism and a rectangular pyramid where the length of the rectangular prism is 6 meters, the width is also 6 meters and the depth of height of the prism is 3 meters. Also, the slant height of the rectangular pyramid is 4 meters. Now to find the surface area of the solid we have to find the area of the prism, the area of the pyramid except the face F G D E. This face is not on the surface. Now we know that surface area of the solid is the sum of the areas of all its faces Now here the surface area of the solid which is given to us is equal to, now first of all let us start with this prism. Now we will add the area of all its faces except the face F G D E. So let us start with the area of the face A B C H plus area of the face G B C D plus area of the face E D C H, area of the face E H A F plus area of the face A B G. The area of all the faces of this pyramid except the face F E D G. Rest all of the faces of this pyramid are the triangles. So we will add the area of the triangle P F G then area of the triangle P G D plus area of the triangle P E D This area of the triangle P E F. Now area of the face A H C B is equal to 6 meters into 6 meters H C B is a square with side 6 meters. So its area will be equal to 6 meters into 6 meters which is equal to 36 square meter. Now area of the face G B C D is equal to area of the face where the dimensions are 6 by 3 meters. So area of the face G B C D is equal to area of the face E H A F which is equal to 6 meters into 3 meters which is equal to 18 square meter. Also area of the face E D C H is equal to area of the face G B A their dimensions are 6 meters by 3 meters. So the area of both these faces will be equal to 6 meters into 3 meters which is equal to 18 square meter. Now in case of this pyramid the area of all 4 triangle faces are equal as they have same base that is 6 meters that is 4 meters. So area of the triangle P F G is equal to area of the triangle P G D is equal to area of the triangle P E D is equal to area of the triangle P E F which is equal to half into base into height which is equal to half into now base is 6 meters and height is 4 meters. Now we know that 2 into 3 is 6 so this will be equal to 12 square meter. So area of these 4 triangle faces is equal to 12 square meter and area of the face E D C H area of the face F G B A area of the face G B C D area of the face E H A F are equal to area of the face A H C B is 36 square meter Now by putting all these values in this formula we can find the surface area of the given solid. So the surface area of the solid is equal to half of the face A H C D which is equal to 36 square meter plus area of the face G B C D which is equal to 18 square meter plus area of the face C H which is equal to 18 square meter plus area of the face which is again 18 square meter plus area of the face F A B G which is also 18 square meter area of the triangle P H G which is 12 square meter plus area of the triangle P G D which is again 12 square meter plus area of the triangle P E D which is again 12 square meter plus area of the triangle P E F which is again 12 square meter Now adding up all these values, we will get the surface area of the solid is equal to 156 square meter. In this way we have obtained the surface area of the given solid. So in this session we have learnt about surface area of two and three dimensional solids composed of presence and pyramids. And this completes our session. Hope you all have enjoyed the session.