 Hello and welcome to the session. In this session, we will discuss a question which says that calculate how much cardboard is required to make a box of holes which is of the shape of a rectangular prism. The length of the box should be 4 cm, width should be 5 cm and height should be 10 cm. Now, before solving the solution of this question, we should know some results. And the first result is, surface area of ever coming hydrogen is the sum of the areas of its natural faces and the faces certainly in rectangular prism are all rectangular nodes. Now, let us see the definition of a net. Now, net is a three-dimensional representation of a three-dimensional figure. Now, it will work out as a key area for solving out the given question. Now, let us start with the solution of the given question. We have to calculate that how much cardboard is required to make a box of holes which is of the shape of a rectangular prism. The length of the box should be 4 cm, width should be 5 cm and height should be 10 cm. So, given length of the box is equal to 4 cm, width of the box is equal to 5 cm and height of the box is equal to 10 cm. Now, this is the box whose length is 4 cm, width is 5 cm and height is 10 cm. Now, we have to calculate how much cardboard will be required to make this box. For this, we will find the area of this box, the rectangular prism. So, we will find the area of this rectangular prism and for finding out the area, we should first draw the net of this prism. Now, here we have drawn the net and for drawing the net, we have cut the box and open it such that this A, B, C, D is joined with four rectangles. That is the rectangles B, H, E, C, D, C, E, F, D, F, G, A, E, G, H, B. Now, we have a rectangle F, E, H, G attached to the H, E, H of the rectangle, E, H, B, C. Now, from the clear idea, we know that surface area of any polyhedron is the sum of the area of its lazy faces and the braces. Now, this is the rectangular prism and here area of this prism will be equal to F, D, C, D of the face H, B, A, S area of the face. B, C, E is equal to A, D because both have same dimensions is equal to area of the face G, H, B, A. Here also both have the same dimensions to area of the face H, B, C, E is equal to area of the face F, G, A, D of the face H, B, C, E plus area of the face F, D, A, G will become of the face H, B, C, E of the face F, D, C, E. Our 10 centimeter face F, G, A, D is equal to length into width that is 10 centimeters into 5 centimeters H, B, A and 4 centimeters respectively. D, C, E is equal to area of the face H, B, A is equal to 10 centimeters into 4 centimeters which is equal to 40 square centimeters of the face is A, B, C, D. A, B, C, D is equal to area of the face G, H, E, F is equal to just into 4 centimeters which is equal to 20 square centimeters. Now putting all these values, number 1 is equal to 2 into 50 the whole, 30 the whole plus 2 into 20 the whole which is equal to 140. Now the area of the oint depth is equal to 220 square centimeters so the area of cardboard requires the depth of oint. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed this session.