 Hello, everyone. This is Shila Ratna Banshwade from Walshan Institute of Technology, Solapur. Today we are going to see the topic sections of solids and in that we will be particularly looking into prism. At the end of this session, students will be able to draw the sectional view of prism. At this point of the video, I suggest you to pause it and think over the geometry of a prism. As we all know, solids are of two types that is polyhedrons and solids of revolution. So prism is a type of polyhedron which has bottom base as well as top base. It has vertical edges or longer edges. It has vertical faces. It does not have an apex. So it has a top base and a bottom base. Let us move further. Consider an example. A pentagonal prism edge of base 30 mm and length of axis 65 mm rests on one of its rectangular face on HP. Such that its axis is parallel to vertical plane AB. An auxiliary inclined plane perpendicular to VP and inclined at 30 degrees to HP cuts the prism by secting axis. Draw the front view and sectional top view. So let us proceed with the drawing. Now this is concerned with the pentagonal prism. Let us draw the XY line. That is the basic thing what we do before starting the drawing. Now as the question says that a pentagonal prism edge of base so and so given as 30 and 65 rest on one of its rectangular face on HP such that its axis is parallel to vertical plane. You can see the condition. It is resting on rectangular face on HP and the axis is parallel to VP. In this condition when the prism is totally horizontal and its axis is parallel to VP it is considered that or it is the fact that the axis is also parallel to HP. So in this condition the axis is parallel to both HP as well as VP and the axis is perpendicular to the profile plane. So instead of starting with front view or top view we start with the side view. So in the side view we have to draw a pentagonal structure this way. So this is the rectangular face on which it is resting on HP. This is the pentagon of 30 mm. These are the namings. So this is the side view of a pentagonal prism which is resting on triangular face 34 with axis parallel to HP as well as VP. Now as the axis is perpendicular profile plane it will be seen as a point view in the side view. Let us move further to draw the front view. We project the points from the side view in the horizontal fashion towards the front view. Now the height of prism is given as 65. So we draw a rectangular of 65 with this height in this way. Naming 1 will be 1 dash 1 dash. 2 and 5 will be 2 dash 5 dash and 2 dash 5 dash at both top and bottom. 3 and 4 will be 3 dash 4 dash at both top and bottom bases. The same naming we have used for top and bottom. Now to complete the top view let us make the construction of orthographic projections that is draw a line at 45 degrees as you can see on the screen. Project the points from the side view on the 45 degrees line and rotate it through 90 degrees in the top view. Project the height of the prism from the front view. So this is the required rectangle which is of the given size of the prism. Further the sides 3 and 4 that is 3 3 4 4 will be seen dotted in the top view as they are at the bottom and direction of observation is from the top. So they are shown as dotted. Naming you can see 3 3 is dotted 4 4 is dotted 1 being at the top 2 and 5 being at the sides. So this is the projection of a solid which is resting on rectangular face and axis parallel to both HP and VP. Further an auxiliary inclined plane perpendicular to VP and inclined at 30 degrees to HP cuts the prism. When a plane is perpendicular to VP it is seen as a line view in the front view and inclination with HP is seen in front view. So we draw a cutting plane in this fashion. So the cutting plane is perpendicular to VP and inclined at 30 degrees to HP. Now consider the points where this cutting plane cuts the edges of the prism that is the bottom edge the vertical edge and this bottom edge. So here we have two sides two here we have two edges and here we have two edges. So in all total we get six points where the cutting plane cuts the solids. We name them as PAPB, PCD, PEF. Now let us project these points in the top view. Before that we project the points in the side view to obtain the section view. Now as this part of the solid has been removed and the direction of observation is this we project the points like PEF will be on 1 2 and 1 5. So 1 2 and 1 5. Similarly PCD will be on 2 and 5. Similarly A and B will be on 3 4 and 2 5. So 2 3 1 point and 4 5 1 point. So these are the points and this is the required cut section. Further project the points PAPB, PC and PD on the 45 degrees line and project it in the top view. So these are the points that is PA will be projected through 45 degrees line and it will be projected at this that is 2 3. Similarly PB 4 5 and we obtain the points that is PC, PD, PE and PEF. By joining these points we get the section view in the top view. So this is the complete projection of the solid when it is cut by a cutting plane making angle of 30 degrees. So these are the darkening of the edges that is remaining portion of the solid. Now let us move for the true shape of the section. As the cutting plane here is inclined it does not give you the true shape. You can draw the true shape by the following method. Take and one more xy line x1 y1 parallel to the cutting plane. Project the points PA, PB, PC, PD, PE and PF perpendicular to the new x1 y1. Now we have to transfer the distances of these particular points from the previous view that is the points on PA, PB will be at a distance of from xy to this PA from xy to this PB. So you have to take the distances of points PA, PB, PC, PD, PE and PF from the top view with respect to xy. So from xy the distance of PA is plotted at this from x1 y1. Similarly distance of PB from xy will be taken and it will be plotted with reference to x1 y1 on the projector of PA. So this is the point PA. So this distance from xy is equal to this distance from xy. Similarly PB, so x1 y1 to PB is equal to the distance of xy to PB in the top view. Similarly we obtain different points PC that is xy to PC is equal to x1 to PC in the auxiliary view. The remaining points PD is xy to PD. Similarly x1 y1 PE is xy to PE, x1 y1 to PF is equal to xy to PF. By joining these points we get the required true shape of the section. So this is the projection of a prism and this is the way you obtain the true shape of a solid. Thank you.