A tetrahedron of 5 cm long edges is resting on the HP on one of its faces, with an edge of that face parallel to the VP. Draw its projections and measure the distance of its apex from the ground.
Answers
Answer:
Explanation:
consider this question :
A tetrahedron 45 mm side has one of its edge in the HP and inclined at 450 to the VP. The triangular plane contained by the edge of a base in the HP is perpendicular to the HP. Draw its projections.
ANSWER -:
Given Data
Side of tetrahedron = 45 mm
Resting on one of its edge in HP
One of its edge is inclined 45° with VP
Stage 1
Draw XY line.
As tetrahedron is resting on HP on one of its edge, draw line 2 3 perpendicular to VP.
Mark point 1 using compass and radius 45 mm (side of tetrahedron).
Join corner points with apex O
Name all the points 1 2 3 and apex O.
Take the projections of all corner points and apex into FV.
Draw the axis and complete the FV.
Name the FV on base 1’ 2’ 3’ and apex point as O’.
Stage 2
Now the triangular plane O’2’3 contained by edge 2’3’ is perpendicular to HP. So make the line O’ 2’ perpendicular to XY line and accordingly draw the Fv in 2nd stage.
Take the projections of all 4 points 1’2’3’ and O’ in TV.
Take the horizontal projections of all corner points towards right side.
Mark intersection points as 1 2 3 and a O.
Complete the TV joining all the points.
Stage 3 14. Now in third stage the edge of base 2 3 is inclined 45° to VP.
Draw a line inclined at angle of 45° to XY line.
Mark point O, 2 and 3 on that line using compass.
Complete the TV in inclined position.
Take the projection all points along with apex into FV.
Take the projection all points along with apex towards right side from the FV of 2nd stage.
Mark intersection points 1’ 2’ 3’ and O’.
Join all the points. As edge O’3’ is not visible in FV, it will be a hidden line.
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