Question

A point in 2^nd quadrant is 25 units away from both the horizontal plane and profile plane 15 units away from the vertical plane. Orthographic projections are drawn find the distance from the side view and top view.

(a) 40

(b) 50.99

(c) 33.54

(d) 41.23

I have been asked this question in an internship interview.

The query is from Projection of Points in Second Quadrant topic in division Projection of Points of Engineering Drawing

Answer 1

Correct choice is (d) 41.23

To explain: Given the point is in 2^nd 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 (25+15); front view and top view (25-15) and these lines which form perpendicular to each other gives needed distance, answer is  √(40^2+10^2 ) = 41.23units.