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The diagram below shows a rope connecting the top of a pole to the ground. The rope is 26 yd long and touches the ground 22 yd from the pole. How tall is the pole? Round approximations to the nearest tenth.

 May 7, 2014

Best Answer 

 #1
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+11

If you draw a model of the pole and the rope touching the ground, a right triangle is formed. The rope would be the hypotenuse, which is 26yards long. The distance between the pole and rope forms one side, 22 yards long. The length of the pole would form the third side. A^2+B^2=C^2

A^2+22^2=26^2     

A^2+484=676             Subtract 486 from both sides.

A^2=192                     Take the square root of 192.

A= 13.86

The length of the pole would be approximately 13.86 yards.

 May 7, 2014
 #1
avatar
+11
Best Answer

If you draw a model of the pole and the rope touching the ground, a right triangle is formed. The rope would be the hypotenuse, which is 26yards long. The distance between the pole and rope forms one side, 22 yards long. The length of the pole would form the third side. A^2+B^2=C^2

A^2+22^2=26^2     

A^2+484=676             Subtract 486 from both sides.

A^2=192                     Take the square root of 192.

A= 13.86

The length of the pole would be approximately 13.86 yards.

Guest May 7, 2014
 #3
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+3

Sorry! :) 

No hard feelings.

 May 7, 2014
 #4
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+8

This problem is a Pythagorean Theorem problem, I assume?

 

  • The rope is the hypotenuse (h = 26 yards)
  • Length of the ground from the pole is side 'a' (a = 22 yards)

 

Thus, the set-up for this problem is:

 

$${{\mathtt{22}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{26}}}^{{\mathtt{2}}}$$

 

Taking these values, the next step is to square 22 and 26.

 

22^2 = 484

26^2 = 676

 

The equation after simplifying the numbers is:

 

$${\mathtt{484}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {\mathtt{676}}$$

 

Now, it is necessary to isolate the 'b^2'.

 

$$\left({\mathtt{484}}{\mathtt{\,-\,}}{\mathtt{484}}\right){\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = \left({\mathtt{676}}{\mathtt{\,-\,}}{\mathtt{484}}\right)$$

$${{\mathtt{b}}}^{{\mathtt{2}}} = {\mathtt{192}}$$

 

Now, in order to completely isolate the 'b^2', we must take the square root of it, and in turn, 192.

 

$${\sqrt{{{\mathtt{b}}}^{{\mathtt{2}}}}} = {\sqrt{{\mathtt{192}}}}$$

$${\mathtt{b}} = {\sqrt{{\mathtt{192}}}}$$

 

The square root of 192 is 13.8564064605510183, which, when rounded to the nearest tenth, is about 13.9

 

@Anon: You rounded to the nearest hundredth, but you summed it up pretty well. I do have a habit of extracting it to the barebones 

 May 7, 2014

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