+0

# geometry

+1
334
8

As shown in the diagram, one end of a 11 foot long electric cable is fixed at point A on the wall, and the other end of the cable (denoted by C) is a sensor. There is a pin attached to the cable at a point B that is 4 units away from C, and the pin can be fixed to any point on the wall reachable by B. Due to gravity, the free part of the cable (i.e. BC) is always perpendicular to the ground (line OD). A few other measurements are provided in the diagram. What is the minimum distance between C and D?

Jan 28, 2022

#8
+117821
+2

Here is the really simple way to do it.

No algebra AT ALL!  You dont even have to work out the equation of any of the circles!

The shortest difference between two points is a straight line.

But in this case a part of the problem is that a 4 foot vertical drop must be included.

I have included it at the start but it can be included at the end if you want. I called the end point of it A'

Now use pythagoras to find distance A'D     5,12,13 is a pythagorean triad so  A'D=13

The rope is 7 foot

13 - 7 = 6 feet left!

Easy peazy!

Jan 29, 2022
edited by Melody  Jan 29, 2022

#2
+13883
+4

What is the minimum distance between C and D?

Hello Guest!

The sensor describes an arc of a circle:

$$f(x)=\sqrt{7^2-x^2}+9-4$$

The point of closest distance from the sensor to D is on the line (0,5) --> D:

$$g(x)= -\frac{5}{12}x+5\\ f(x)=g(x)$$

$$\sqrt{7^2-x^2}+5=-\frac{5}{12}x+5\\ 49-x^2=\frac{25x^2}{144}\\ \frac{169x^2}{144}=49\\ x=\frac{7\cdot 12}{13}\\ x=6.\overline{461538}\\ y=-\frac{5}{12}\cdot x+5\\ y=2.\overline{307692}$$

What is the minimum distance between C and D?

$$\overline{CD}_{min}=\sqrt{(12-x)^2+y^2}=\sqrt{(12-6.4615)^2+2.3077^2}\\ \color{blue}\overline{CD}_{min}=6$$

!

Jan 28, 2022
#3
+117821
+3

Thanks asinus,

I got the same answser as you but my answer invlolved some calculus.

To find the minimum distance between the points   $$(12,0)\;\; and \;\;(x, \sqrt{49-x^2}+5)$$

I also graphed it to as a second way to solve it.

Here is my gaphical solution.  If you right click on the point C you can drag it and see what difference it makes to the distance CD.

https://www.geogebra.org/classic/vbr4yrcr

The pic without the interaction is below.

Jan 28, 2022
edited by Melody  Jan 28, 2022
#5
+13883
+1

Hi Melody!

Graphic: I would have extended the DC line to the center of the lower circle. So she is the image of my thoughts on the solution.

!

asinus  Jan 29, 2022
edited by asinus  Jan 29, 2022
#6
+117821
+1

Thanks asinus,

I had not actually understood what you had said before but now this comment totally explaines it!

Yes that would be a good idea!

Melody  Jan 29, 2022
#4
+117821
0

Jan 28, 2022
#7
+117821
+3

What asinus has said (sort of)  is that a part of the vertical drop is the 4feet.

Which on my number plane is the point (0,5)

THEN

the the shortest path to D is a straight line.  The rope will cover the first 7 feet.  then you have to work out what is left.

That is much easier than how I did it.

EXCELLENT thinking asinus!

Jan 29, 2022
#8
+117821
+2

Here is the really simple way to do it.

No algebra AT ALL!  You dont even have to work out the equation of any of the circles!

The shortest difference between two points is a straight line.

But in this case a part of the problem is that a 4 foot vertical drop must be included.

I have included it at the start but it can be included at the end if you want. I called the end point of it A'

Now use pythagoras to find distance A'D     5,12,13 is a pythagorean triad so  A'D=13

The rope is 7 foot

13 - 7 = 6 feet left!

Easy peazy!

Melody Jan 29, 2022
edited by Melody  Jan 29, 2022