Please help. Test is soon. Circumference, Tangent Radius

[Guest]

Hi, Need some help on these problems:

I tried this:

 

[pka]

The figure BPFK is a square!
Therefore, (24+x)<SUP>2</SUP>=24<SUP>2</SUP>+24<SUP>2</SUP>.
Solve for x

 

[stapel]

In case the images become unreachable:

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• 
  16) Consider a right-angled "corner"; draw this as the top-right corner of a square (that is, the top and right sides of the square), with the vertex labelled as "P". "Wedge" a circle into this right angle. The circle has diameter d = 48 centimeters.
  
  a) Consider the shortest line segment drawn from P to the circle, being the "distance" of the circle from the point P. Let this "distance" segment terminate on the circle at the point "A". Find the length of segment PA.
  
  b) Consider the point on the top "side" of the angle where the circle touches the angle. Call this point "B". Find the length of PB.
  
  17) Two pulleys are connected by a belt. The radii of the two pulleys are r = 3 cm and r = 15 cm. The distance between the centers of the two pulleys is 24 cm. Find the total length of the belt going around these pulleys, assuming no slack.

 

[Guest]

The figure BPFK is a square!
Therefore, (24+x)<SUP>2</SUP>=24<SUP>2</SUP>+24<SUP>2</SUP>.
Solve for x

Thanks, you can also use the triple: x, x, and x square root of 2 right?

Thanks stapel too, hope other people can understand to help me! Thanks!

 

[Guest]

Can anyone help me? My test is tomorrow.

 

[pka]

We did not help you?
What do you not understand?

 

[soroban]

Hello, AirForceOne!

The Pulley Problem is rather intricate . . .

            A
           /
          /   
       O /
    -  *
    :  |   *                   F
    :  |       * 24           /
    :  |12         *         /
    :  |               *    /
   15 Q+ - - - - - - - - - *P
    :  |                   |
    :  |3                  |3
    :  |                   |
    -  * - - - - - - - - - *
      J                   G

O is the center of 15-cm circle; P is the center of 3-cm circle.
The line joining the centers: \(\displaystyle \,OP\,=\,24\)
\(\displaystyle JG\) is the length of the pulley tangent to both circles.
\(\displaystyle \;\;\)(There is an identical length of pulley at the top of assembly.)

We have right triangle \(\displaystyle OQP\) and we see that is a 30-60-90 triangle.
\(\displaystyle \;\;\sin(\angle OPQ) = \frac{12}{24}\,=\,\frac{1}{2}
\;\;\Rightarrow\;\;\angle OPQ\,=\,30^o\,=\,\frac{\pi}{6},\;\;\angle POQ\,=\,\frac{\pi}{3}\)

Note that: \(\displaystyle \,QP\,=\,JG\,=\,12\sqrt{3}\) cm.


Length of arc: \(\displaystyle \,L\,=\,r\theta\)
\(\displaystyle \;\;\)where \(\displaystyle r\) is the radius and \(\displaystyle \theta\) is the central angle in radians.

Hence: \(\displaystyle \,\angle AOJ\,=\,\frac{4\pi}{3}\)

And the "rest of the angle": \(\displaystyle \,\angle(AOJ)'\; =\;2\pi\,-\,\frac{4\pi}{3}\:=\:\frac{5\pi}{3}\)

The length of the pulley from \(\displaystyle A\) counterclockwise to \(\displaystyle J\)
\(\displaystyle \;\;\) is: \(\displaystyle \,L_1\:=\:15\left(\frac{5\pi}{3}\right)\:=\:25\pi\) cm


In the smaller circle: \(\displaystyle \,\angle OPQ\,=\,\frac{\pi}{6},\:\) hence: \(\displaystyle \angle OPG\:=\:\frac{\pi}{6}\,+\,\frac{\pi}{2}\:=\:\frac{2\pi}{3}\)

Hence: \(\displaystyle \angle FPG\,=\,\frac{4\pi}{3}\)

And the "rest of the angle": \(\displaystyle \,\angle(FPG)'\:=\:2\pi\,-\,\frac{4\pi}{3}\:=\:\frac{2\pi}{3}\)

The length of the pulley from \(\displaystyle F\) clockwise to \(\displaystyle G\)
\(\displaystyle \;\;\)is: \(\displaystyle \,L_2\:=\:3\left(\frac{2\pi}{3}\right)\:=\:2\pi\) cm.


The total length of the pulley is: \(\displaystyle \,L_1\,+\,L_2\,+\,2\cdot JG\;=\;25\pi\,+\,2\pi\,+\,2(12\sqrt{3})\)

Answer: \(\displaystyle \:27\pi + 24\sqrt{3}\;\approx\;126.4\) cm.