Question

A point is 5 units away from the vertical plane and profile plane and 10 units away from the 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) 15

(b) 10

(c) 32.5

(d) 18.02 units

The question was asked in an interview for job.

This key question is from Projection of Points in First Quadrant in portion Projection of Points of Engineering Drawing

Answer 1

The correct answer is (d) 18.02 units

Easy 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 (5+5); front view and top view (10+5)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(10^2+15^2 ) = 18.02 units.