Question

A point in 2^nd quadrant is 15 units away from the horizontal plane and 10 units away from both the vertical plane and profile plane. Orthographic projections are drawn find the distance from side view and top view.

(a) 25

(b) 20.6

(c) 25.49

(d) 15.8

The question was posed to me during an interview.

I'd like to ask this question from Projection of Points in Second Quadrant in chapter Projection of Points of Engineering Drawing

Answer 1

The correct option is (b) 20.6

Easy explanation: 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 (10+10); front view and top view (10-15) and these lines which form perpendicular to each other gives needed distance, answer is  √(20^2+5^2 ) = 20.6 units.