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A point in 2^nd quadrant is 12 units away from the horizontal plane and vertical plane 13 units away from both the profile plane. Orthographic projections are drawn find the distance from the side view and top view.

(a) 25.6

(b) 25

(c) 17.69

(d) 13

I have been asked this question by my college director while I was bunking the class.

Question is taken from Projection of Points in Second Quadrant in section Projection of Points of Engineering Drawing

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Right answer is (b) 25

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 (12+13); front view and top view (12-12) and these lines which form perpendicular to each other gives needed distance, answer is  √(25^2+0^2 ) = 25units.

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