Pythagoras 3D: Solve Problems with Ease

In summary, the hawk sees two cables running across a horizontal roof. The two cables have different ends at the two corners of the roof, with one end being at the top of a mast on the south-west corner, and the other end being at the top of a mast on the north-east corner. The length of the masts and edges of the roof are given in the diagram below.
  • #1
Natasha1
493
9
Homework Statement
Pythagoras in 3D
Relevant Equations
See pictures and questions attached
I can do question a but not b nor c.

Any help would be greatly appreciated
 

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  • #2
Try drawing a picture diagram for B) and C)

The hint of what does some bird flying overhead see could imply that the points in question are on the same horizontal plane. I know question C) says they are at the same height so height doesn't factor into solving that problem.

LASTLY, Please type in the question don't take the easy way and post a poor quality picture of the question. This allows us to quote portions of the question instead of having to refer to your image and retype was found there.
 
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  • #3
Since one bird is directly above the other bird, they must be at the points where one cable crosses over the other cable.

The hawk overhead sees everything projected down into the plane of the roof. What does this problem look like when projected down into the plane? What do the corners and cables look like? What can you figure out about that cross-over point? Where is it in the picture where everything is flattened as the hawk sees it?

For part c, the far away bird is seeing everything flattened into a different plane.
 
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  • #4
For b I drew it and get a rectangle of 16m by 12m with both diagonals being 20m and the birds must be at the crossing of these two diagonals only that I don't know how to determine how far from each other since they are not on the diagonals of 20m but on the cables one being 25m long (from question a) and the other is 30.48m to 2d.p.
 
  • #5
Please see picture
 

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  • #6
jedishrfu said:
LASTLY, Please type in the question don't take the easy way and post a poor quality picture of the question.
Amen to that!
IMG_0277.jpg is very poor quality, and IMG_0276.jpg is completely unreadable. Many members won't even bother to try to decipher images this bad.
 
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  • #7
Here it is then:

The roof of a tall building is horizontal and rectangular. Vertical masts are fixed to the corners of the roof. A straight cable runs from the top of the mast on the south-west corner to the top of the mast on the north-east corner; another runs from the top of the mast on the north-west corner of the top of the mast on the south-east corner. The lengths of the masts and edges of the roof are given in the diagram below (see picture attached)

My question is this:

b) A robin lands on one of the cables. A sparrow lands on the other cable, perching directly below the robin. How far apart are the birds?

c) A blackbird lands on one of the cables and a thrush lands on the other. The birds are the same height above the roof and the same distance from its eastern edge. How far apart are the birds?
 

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  • #8
Natasha1 said:
For b I drew it and get a rectangle of 16m by 12m with both diagonals being 20m and the birds must be at the crossing of these two diagonals only that I don't know how to determine how far from each other since they are not on the diagonals of 20m but on the cables one being 25m long (from question a) and the other is 30.48m to 2d.p.

On the flat picture can you say anything about where that intersection point is? Relative to anything else?
 
  • #9
Yes, the diagonals intersection must meet half way e.g. 10 m
 
  • #10
Natasha1 said:
Yes, the diagonals intersection must meet half way e.g. 10 m

OK, now consider the 3D picture. Consider one cable. You know the height of one end. You know the height of the other end. What is the height of the point in the middle?
 
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  • #11
Wait, i think I am on to something...
 
  • #12
23-8=15m on one side and 0m on the other so that makes it 15/2 = 7.5m heigh and
28-5= 23m on one side and 0m on the other so that makes it 23/2 = 11.5m heigh so the difference is 11.5-7.5 = 4m is this correct?

If it is correct can someone please tell me the answer to C or at least all the steps to tackle this tricky question! Please :)
 
  • #13
Could someone kindly check my answer of 4m for question b and help a lot of question c?
 
  • #14
7.5m is not the height of the point. It is the height of this point above the lower end of the cable, which is at 8m. So the height is?
Similarly with the other cable.
 
  • #15
I don't understand why it is 8m...
 
  • #16
Natasha1 said:
Here it is then:

The roof of a tall building is horizontal and rectangular. Vertical masts are fixed to the corners of the roof. A straight cable runs from the top of the mast on the south-west corner to the top of the mast on the north-east corner; another runs from the top of the mast on the north-west corner of the top of the mast on the south-east corner. The lengths of the masts and edges of the roof are given in the diagram below (see picture attached)

My question is this:

b) A robin lands on one of the cables. A sparrow lands on the other cable, perching directly below the robin. How far apart are the birds?

c) A blackbird lands on one of the cables and a thrush lands on the other. The birds are the same height above the roof and the same distance from its eastern edge. How far apart are the birds?
Thanks for typing the problem statement. In the interest of completeness, part a) reads as follows.
a) Find the length of the cable that runs from the SW mast to the NE mast.​

However, the image you display is same image that @Mark44 referred to as being of poor quality. To read the values on the figure in the image, I had to display that image as full size on a computer screen, then zoom that image to 200%. Here is a snip of the figure in the image.
from img-0277-ipg.PNG
 
  • #17
Natasha1 said:
I don't understand why it is 8m...
The height (above the roof level) of the north-east mast is ...
 

What is Pythagoras 3D?

Pythagoras 3D is a computer software that uses the Pythagorean theorem to solve three-dimensional geometry problems.

How does Pythagoras 3D work?

Pythagoras 3D takes in the given measurements of a three-dimensional shape and uses the Pythagorean theorem to calculate the missing measurement or solve for a specific angle.

Who can use Pythagoras 3D?

Pythagoras 3D is designed for anyone who needs to solve three-dimensional geometry problems, such as students, teachers, engineers, and architects.

Is Pythagoras 3D accurate?

Yes, Pythagoras 3D uses the well-known and proven Pythagorean theorem to solve problems, ensuring accuracy in its calculations.

Can Pythagoras 3D solve any three-dimensional geometry problem?

No, Pythagoras 3D is designed to solve problems that involve right triangles in three-dimensional space. It may not be able to solve problems involving other shapes or non-right triangles.

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