Question

A point q is situated in first quadrant.It is 40mm above h.P. And 30mm in front of v.P.Draw its projections and find its shortest distance from the intersection pf h.P.,v.P. And auxiliary plane.

Answer 1

Answer:

d = 50 mm is the required distance.

Step-by-step explanation:

Given that

⇒ Point is in the 1st quadrant

⇒ It is 40 mm above h.P.  ⇒ P = 40 mm

⇒ It is 30 mm in front of v.P.  ⇒ B = 30 mm

⇒ We need to find its shortest distance from the intersection pf h.P.,v.P.

Required distance is

d² = B² + P²

d² = 30² + 40²

d² = 900 + 1600

d² = 2500

d = 50 mm is the required distance.

Kindly see the Attachment.

Answer 2

Answer:

d=50cm

Step-by-step explanation:

It is required distance..

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