Question

A point is 20 cm away from the vertical plane and 8 units away from profile plane and 17 cm away from horizontal plane in 3rd quadrant then the projections are drawn on paper the shortest distance from top view and side view of point is _______________

(a) 37

(b) 44.65

(c) 46.40

(d) 37.53

The question was posed to me in an online quiz.

Enquiry is from Projection of Points in Third Quadrant topic in chapter Projection of Points of Engineering Drawing

Answer 1

The correct choice is (c) 46.40

For explanation I would say: Since here distance from side view and top view is asked for that we need  the distance between the front view and side view (8+20); front view and top view (17+20)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(28^2+37^2 ) =46.40 units.