Question

A point in 4th quadrant is 8 inches away from the horizontal plane and 20 inches away from both the vertical plane and profile plane. Orthographic projections are drawn find the distance from side view and top view.

(a) 41.76

(b) 20

(c) 43.08

(d) 16

I had been asked this question in quiz.

I would like to ask this question from Projection of Points in Fourth Quadrant topic in section Projection of Points of Engineering Drawing

Answer 1

Right option is (a) 41.76

To explain: Given the point is in the 4th quadrant. Since here distance from side view and top view is asked for that we need  the distance between the front view and side view (20+20); front view and top view (20-8) and these lines which form perpendicular to each other gives needed distance, answer is  √(40^2+12^2 ) = 41.76 units.