 Hello everyone, this is Shila Ratnabanswade from Walchand Institute of Technology, Solapur. Today we are going to see the topic sections of solids and in that we will see pyramids. At the end of this session the students will be able to draw the sectional view of a pyramid. At this point of the video I suggest you to pause it and think over the geometry of a pyramid. As we all know pyramid is a type of polyhedron or type of solid as polyhedron which has triangular faces, it has vertical edges, it has edges at the base, the base is of the given polygon either it be a pentagonal pyramid, a hexagonal, a square pyramid or a triangular pyramid and at the top it has an apex point. So let us move further, consider an example, a pentagonal pyramid, edge of base 40mm long and height 75mm is lying on HP on the triangular face with its axis parallel to VP. It is cut by the section plane perpendicular to HP inclined at 30 degrees to VP and bisect the axis of pyramid. Draw the sectional front view, top view and true shape when the apex is retained. Let us start the drawing. The first step we draw the XY line. Now to finalize the first stage look into the question, the pyramid is lying in HP on the triangular face with axis parallel to VP. So depending upon this condition we finalize that or we decide or we assume that the cone is resting on HP with its axis parallel to VP and perpendicular to HP. So in this case the top view of the pyramid will be a pentagon in this way. Now drawing of the pentagon in this fashion that is edge towards the right and corner towards the left is based on this condition that it is resting on triangular face. So as the pyramid rests on triangular face we have drawn this edge towards the right and corners towards the left. Now you can draw the pentagon by using any method either the circle method or by finding the exterior angle method. Naming of the pyramid 40mm is the base side. Let us project the points in the vertical upward direction to get the required front view. So this is the required front view of the pyramid where this is the axis height 75mm and this is the naming. So in the front view we have named as O1- O2-5- O3-4- whereas in the top view we have named as 1, 2, 3, 4 and 5 and apex as both in top view as well as front view. Now let us move to the second condition. The second condition says that the pyramid is resting in HP on the triangular face with the axis parallel to VP. Now in the first stage the axis is already parallel to VP. So only thing is that we need to tilt this pyramid in such a way that it is resting on its triangular face O3-4. I repeat we need to tilt the pyramid in such a way that it is resting on HP on the triangular face O3-4 which is seen in front view as O3-4-. So in the second stage we redraw this front view in such a way that O3-4- is coinciding with XY line in this way. So it is resting on the triangular face O3-4-. We have redrawn this front view from the first stage. Now to draw the top view in the second stage project the points vertically downwards from the second front view and top view of the first stage. So point 1, 1 dash from front view and 1 from the top view. So this is the required point. Similarly 3, 4 and 2, 5 along with the apex. So these are the points. By joining these points we will get the required top view. This is the base and this is the top surface of the pyramid. Now here you can see that the exterior is complete dark. The edge 5-1 and 1-2 is dark whereas O4 and O3 are shown in dotted. The reason is that the direction of observation is this that is above the front view. So O3 and O4 are not directly seen. So they are hidden behind the solid. So we have drawn this in dotted form. So this is the complete front view and top view in the second stage. Now let's move to the cutting plane. It is cut by the section plane perpendicular to HP. Now as the cutting plane is perpendicular to HP it should be seen as a line view in top view. And it is inclined at 30 degrees to VP. So if a plane or any object makes an angle with VP it is seen in the top view. So let us draw the cutting plane. So this is the cutting plane which makes an angle of 30 degrees with VP and it is perpendicular to HP. Consider the points where the cutting plane intersects with the edges of the solid that is 4, 5, 5, 1, O2, O3 and O4. We are concerned only the points where the cutting plane cuts the edges of the solid. Let us name the points. The points that we obtain on the generators, vertical generators we have named them as P, 2, P3, P4 and the points on the base we have named them as A and B. The point A is on the base edge of 5, 1 and point B is on the base edge of 4, 5. Now let us project these points in the vertical upward direction to get the sectional front view as the direction of observation is from below. Before that let us darken the remaining part of the solid. As I said previously the observer is at this position that is below top view. So this portion is removed whereas this portion is retained. That is the portion above the cutting plane is retained which we have shown in dark lines. Now let us project the points. So point B will be projected on 4, 5, point A on 5, 1 similarly P3, P2 and P4 on the respective edges. By joining this we get the required sectional view of the pyramid. So this is the complete projection of the section of the solid which is cut by a section plane inclined at 30 degrees to VP. Now as the cutting plane is inclined one so the obtained section is not the true shape. Now to obtain the true shape of the section let us reproduce this cutting plane parallel to XY this way. Before that we have to dark the remaining part of the solid in the front view in this way. Now coming to the true shape this is the cutting plane parallel to XY. Now plot these points B, A, P3, P2 and P4 on this line. You can take the distances with the help of compass and plot the same over this line. Project these distances in the vertical upward direction and take the horizontal projections from the previous front view like point B from the front view and point B from the cutting plane similarly point A and point A similarly P1 and P1, P3, P3 and P2, P2. Point 3 being on XY line. These are the points by joining this point we get the required true shape of the section. So as you can see this is the required true shape of the section. So this is the complete projection as far as the given question is concerned. Thank you.