# Pythagoras in 3D

### Pythagoras in 3D

I can do question a. I got 25m. But I can't do questions b nor c at all.

Please see pictures for the questions and the diagram.
Attachments
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IMG_0278.JPG (606.91 KiB) Viewed 379 times
IMG_0276.JPG (62.59 KiB) Viewed 379 times
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### Re: Pythagoras in 3D

Good night!

a)
Let's fix the origin of the coordinates in the base of SW mast.
So, calling A the SW mast top and B the NE mast top:
A(0,0,23) and
B(16,12,8), right?

The distance between them:
$$d^2=16^2+12^2+(23-8)^2$$
$$d^2=256+144+225$$
$$d=\sqrt{625}=25$$
So, 25 is right

b)
From SW to NE, the 'high' of the middle point is:
$$\frac{23+8}{2}=\frac{31}{2}=15,5$$

From SE to NW, the 'high'of the middle point is:
$$\frac{28+5}{2}=\frac{33}{2}=16,5$$

So, the sparrow is 16,5-15,5=1,0 below the robin.

c)
Try to see that the figure is two trapezes.
They intercept one to each other 6 m distant from the East edge.
And the distance between each cable, is 3.

A figure to try to help
Captura de Tela 2019-10-22 às 21.48.08.png (256.74 KiB) Viewed 280 times

Baltuilhe

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