Question

A point is 16 units away from the vertical plane and horizontal plane 4 units away from profile plane in 1st quadrant then the projections are drawn on paper the distance between the side view and top view of point is ______________

(a) 37.73 units

(b) 32.98 units

(c) 16

(d) 8

This question was posed to me in examination.

The query is from Projection of Points in First Quadrant topic in section Projection of Points of Engineering Drawing

Answer 1

Correct answer is (d) 8

The best explanation: Since here distance from side view and top view is asked for that we need  the distance between the front view and side view (4+16); front view and top view (16+16)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √20^2+32^2 ) = 37.73 units.