Engineering Drawing Questions and Answers – Projection of Points in Fourth Quadrant

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This set of Engineering Drawing Multiple Choice Questions & Answers (MCQs) focuses on “Projection of Points in Fourth Quadrant”.

1. A point is in 4th quadrant 10 units away from the horizontal plane and 20 units away from the vertical plane. Orthographic projection is drawn. What is the distance from point of front view to reference line, top view point to reference line?
a) 20, 10
b) 10, 20
c) 0, 20
d) 10, 0
View Answer

Answer: b
Explanation: Given object is point placed in 4th quadrant the top view gives the distance from the vertical plane (20) and front view gives the distance from horizontal plane (10) both are placed overlapped in orthographic projection since the object is placed in 4th quadrant.
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2. A point is in 4th quadrant 15 cm away from the vertical plane and 10 cm away from the horizontal plane. Orthographic projection is drawn. What is the distance from point of front view to reference line, top view point to reference line?
a) 15, 10
b) 10, 15
c) 0, 15
d) 10, 0
View Answer

Answer: b
Explanation: Given object is point the top view gives the distance from vertical plane (15) and front view gives the distance from horizontal plane (10) both are placed overlapped in orthographic projection since the planes need to rotate to draw projection as the object is placed in 4th quadrant.

3. A point is in 4th quadrant, 5 m away from the vertical plane, 1 m away from the horizontal plane and 8 units away from the profile plane. Orthographic projection is drawn. What is the distance from point of front view to point of top view?
a) 6
b) 4
c) 10
d) 2
View Answer

Answer: b
Explanation: As the point is in 4th quadrant while drawing the projections the planes should rotate along the hinges such that the plane with top view overlaps the front view. So the distance between them is difference of distances from respective planes that is 5 (5-1) here.

4. A point is in 4th quadrant, 15 dm away from the vertical plane, 10 dm away from the horizontal plane and 8 units away from the profile plane. Orthographic projection is drawn. What is the distance from point of front view to point of side view?
a) 25
b) 23
c) 18
d) 5
View Answer

Answer: b
Explanation: Side view is obtained by turning the profile plane along the hinge with vertical parallel to vertical plane. Side view and front view have the same distance from a reference line. Sum of distances from the point to vertical plane and profile plane gives the following that is 15+8 = 23 dm.

5. A point in 4th quadrant is 30 mm away from both the horizontal plane and vertical plane and orthographic projections are drawn. The distance between the points formed by front view and top view is ______________
a) 0
b) 30
c) 15
d) 15+ distance from profile
View Answer

Answer: a
Explanation: Given the point is in the 4th quadrant. While drawing orthographic projections the front view and top view overlaps and also the distance of point is same from planes of projections so the distance between them is zero.

6. A point in 4th quadrant is 13 inches away from the horizontal plane and 10 inches away from both the vertical plane and profile plane. Orthographic projections are drawn find the distance from side view and front view.
a) 10
b) 13
c) 20
d) 26
View Answer

Answer: c
Explanation: Given the point is in 4th quadrant. The front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. The distance from side view to vertical reference is 10 and distance from front view to profile plane is 10. Sum is 10+10= 20 inches.
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7. A point in 4th quadrant is 10 units away from both the horizontal plane and profile plane and 15 units away from the vertical plane. Orthographic projections are drawn find the distance from side view and front view.
a) 25
b) 15
c) 30
d) 40
View Answer

Answer: a
Explanation: Given the point is in the 4th quadrant. The front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. The distance from side view to vertical reference is 15 and distance from front view to profile plane is 10. Sum is 15+10 =25 units.

8. A point in 4th quadrant is 18 units away from the horizontal plane and vertical plane and 17 units away from both the profile plane. Orthographic projections are drawn find the distance from side view and front view.
a) 1
b) 24
c) 35
d) 36
View Answer

Answer: c
Explanation: Given the point is in 4th quadrant. The front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. The distance from side view to vertical reference is 12 and distance from front view to profile plane is 13. Sum is 18 + 17 =35 units.

9. A point in 4th quadrant is 8 inches away from the horizontal plane and 20 inches away from both the vertical plane and profile plane. Orthographic projections are drawn find the distance from side view and top view.
a) 41.76
b) 20
c) 43.08
d) 16
View Answer

Answer: a
Explanation: Given the point is in the 4th 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 (20+20); front view and top view (20-8) and these lines which form perpendicular to each other gives needed distance, answer is √(402+122 ) = 41.76 units.

10. A point in 4th quadrant is 5 m away from both the horizontal plane and profile plane 3 m away from the vertical plane. Orthographic projections are drawn find the distance from side view and top view.
a) 8
b) 8.2
c) 10.19
d) 12.8
View Answer

Answer: b
Explanation: Given the point is in 4th 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 (5+3); front view and top view (5-3) and these lines which form perpendicular to each other gives needed distance, answer is √(8^2+2^2 ) = 8.2 m.

11. A point in 4th quadrant is 13 inches away from the horizontal plane and vertical plane 10 inches away from both the profile plane. Orthographic projections are drawn find the distance from side view and top view.
a) 26
b) 25.6
c) 17.69
d) 13
View Answer

Answer: a
Explanation: Given the point is in 4th 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 (13+10); front view and top view (13-13) and these lines which form perpendicular to each other gives needed distance, answer is √(262+02 ) = 26 inches.

12. A point in 4th quadrant is 15 cm away from the vertical plane and 10 cm away from the horizontal plane, orthographic projections are drawn. What is the distance from side view of point to line of vertical reference?
a) 10
b) 15
c) 25
d) Can’t found
View Answer

Answer: b
Explanation: Given the point is in 4th quadrant. The distance from the side view of point to line of vertical reference will be the distance from the point to the vertical plane in plane of projection that is as given 15 cm.
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13. A point is in 4th quadrant which is 15 inches away from horizontal and 30 inches away from profile plane. Orthographic projections are drawn. What is the distance from the top view to xy reference line?
a) 5
b) 3
c) 8
d) Can’t found
View Answer

Answer: d
Explanation: Given the point is in 4th quadrant. The xy reference line is between the vertical plane and horizontal plane but distance from vertical point is not given in question so we can’t found some given information.

14. A point is in 4th quadrant which is 17 dm away from horizontal and 12 dm away from profile plane. Orthographic projections are drawn. What is the distance from the front view to xy reference line?
a) 17
b) 12
c) 5
d) 29
View Answer

Answer: a
Explanation: Given the point is in 4th quadrant. The distance from front view is given by distance between point and horizontal plane here it is given 17 dm. And distance from vertical reference will be 12 dm.

15. A point is in 4th quadrant which is 18 mm away from vertical and 20 mm away from profile plane. Orthographic projections are drawn. What is the distance from the side view to vertical reference line?
a) 18
b) 2
c) 20
d) Can’t found
View Answer

Answer: a
Explanation: Given the point is in 4th quadrant. The distance from side view is given by distance between point and vertical plane here it is given 18 mm. And distance from front view will be 20 mm.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn