 Hello everyone, I am Rahul Patil, assistant professor, department of mechanical engineering Walshchand Institute of Technology, Solapur. Today we will discuss about projection of square plane. At the end of this session, student will be able to draw the projection of square plane. For understanding the projection of a square plane, consider the problem. A square plane A B C D with sides equal to 30 mm has its surface inclined at an angle of 30 degree to the HP. One of the sides say A B is in HP and makes an angle of 45 degree with the VP. Draw its projection. Now before starting the problem, read the problem carefully and note down the given data. Here side for the square is 30 mm, second surface inclined at an angle of 30 degree to the HP. Now inclination with the HP is given. Now inclination with the HP is drawn in the VP. Now second condition is the side A B is in HP and makes an angle of 45 degree to the VP. Now second condition with the VP is given. Inclination with the VP is drawn in the HP. This is the basic criteria for drawing the problem. Now start the drawing problem. First draw X Y line name as X and Y then mark VP and HP. Then first for first stretch the true shape of the object is drawn that is square and name it as A B C D. Now pause the video and try to answer the question what is the front view for the given object or given square. It is simple, it is a line view. Now how line view is appear, just go through the procedure. Here draw the projection from point A B and point C D so that we can complete the front view. Name it as A B and C D because A and B lie on a single line that why projection from point A and B appear at a single point. Similar way projection from point C and D will appear at a single point. Now in front view you will get two points A B and C D join this point you will get the line view. This is the first stage for solving the problem. Now in second stage before starting the second stage first find out which condition is fulfilled that is in first stage true shape is drawn HP that is why in second stage condition with the HP is drawn that is inclination with the HP is given. If inclination with the HP is given then it is appear in the VP that is why on XY line only take a first point as A B from this point draw a inclined line with same distance as a C D this A to A B to C D distance must be same with inclination as per given problem that is 30 degree. Now for completion of the top view in second stage draw the projection from point A B and C D similar way draw the projections from A D and B C so that you will get projections from both front view and top view so that you will get top view for the second stage. Now intersection from the point A and here again point A you will get first point as a point A similar way intersection from projection from point B and point B you will get second point as a point B similar way you will get point C and point D. Now join this A B, B C, C D and A D here second stage for the problem is completed. Now again for third stage condition with the VP is prepared because in second stage condition with the HP is carried out for the third stage condition with the VP is considered. Now for third stage again inclination is given and for the third stage as per the problem side A B is on the inclined line which is given in the problem so that draw a line with the inclination of 45 degree and draw the second stage top view of a second stage that is this A B C D object on this 45 degree line such a way that A B must be lie on this 45 degree mark this will appear like this mark it as A B C and D now similar way draw the projection so that you will complete the front view for the stage third now draw the projection from point C and D here A and B lie on x-axis so that similar way for obtaining point A B C D in the front view draw the projections from point A intersection between projection point A and here again from point A you will get first point again from point B and name it as a B again from point C and point D now join A B this is A B again BC CD and AD this is the third stage for the problem side A B is inclined 45 degree to the VP because inclination with the VP is drawn into the HP and inclination with the HP is drawn in the VP now total solution is like this where square A B CD is given with first inclination with the HP is 30 degree and inclination with the VP is 45 degree this is the total solution for the given problem these are the references thank you