| Q. |
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 |
| Answer» A. 13.45 | |
| 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 (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 √(102+92 | |
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