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Q. |
## A point is 16 units away from the vertical plane and horizontal plane 4 units away from profile plane in 1st quadrant then the projections are drawn on paper the distance between the side view and top view of point is |

A. | 37.73 units |

B. | 32.98 units |

C. | 16 |

D. | 8 |

Answer» D. 8 | |

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+16); front view and top view (16+16)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √202+322 ) = 37.73 units. |

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