Q20: A cone of base diameter 50 mm and axis 60 mm is resting on its base on the H.P. A section plane perpendicular to V.P. and inclined at 450 to H.P., bisects the axis of the cone. Draw the development of its lateral surface.
Solution:
Steps :
(i) Here generators o ‘2’, o ‘3’, o ‘4’, o ‘6’, o ‘7 ‘ and o ‘8’ are not of the true length. Therefore, intersection points on these generators need to be projected on the generator o‘1’ which is of true length. Point b ‘, c ‘, d ‘ ……e ‘ are thus projected as b “, c “, d “….. on generator o ‘1’.
- The angle subtended by an arc at the center is given as, rs 3600 .
- Where, s slant height of cone = 65 mm o ‘1’ , r radius of base = 25 mm,Theta = 25/ 65 X 360= 138.50
- Mark point O and draw an arc of circle of radius = o‘1’ with angle 50 . Divide the arc 1o1 into 8 equal parts.
- Mark point A on O1 such that OA o ‘ a ‘ and point B on O2 such that OB O ‘ b “. Similarly, mark point C , D , E , F , G and H on O 3, O 4, O 5, O 6, O7 and O8
- Draw smooth curve passing through points A, B, C, D….. etc.
- Mark point O and draw an arc of circle of radius = o‘1’ with angle 50 . Divide the arc 1o1 into 8 equal parts.
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