Question

A point is 20 units away from the vertical plane and 12 units away from profile plane and 9 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the side view and top view of point is ________________

(a) 29 units

(b) 21 units

(c) 35.8 units

(d) 17.9 units

This question was posed to me in an online quiz.

This intriguing question comes from Projection of Points in First Quadrant in portion Projection of Points of Engineering Drawing

Answer 1

The correct choice is (c) 35.8 units

To elaborate: Since here distance from side view and top view is asked for that we need  the distance between the front view and side view (12+9); front view and top view (9+20)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(21^2+29^2 ) = 35.80 units.