Question

A point is 5 units away from the vertical plane and horizontal plane 4 units away from profile plane in 3rd quadrant then the projections are drawn on paper the distance between the side view and top view of point is _________________

(a) 13.45

(b) 12.72

(c) 19

(d) 12.04

The question was asked by my college professor while I was bunking the class.

My doubt is from Projection of Points in Third Quadrant in division Projection of Points of Engineering Drawing

Answer 1

Correct choice is (a) 13.45

The explanation is: Since here distance from side view and top view is asked for that we need  the distance between the front view and side view (4+5); front view and top view (5+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+9^2 ) = 13.45 units.