Question 1167805: The figure below shows two pulleys of radii 6cm and 4cm with centres A and B respectively. AB = 8cm. The pulleys are connected by a string PQXRSY
https://www.easyelimu.com/images/topicalrevision/maths/form_3/circles_chords_and_tangents/circle_q5.PNG
calculate
(a) Length PQ
(b) PAS reflex
(c) Length of arc PYS and QXR
(d) The total length of the string PQXRSY
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's a step-by-step calculation for each part of the problem based on the provided figure and common geometry principles for pulleys and tangents.
First, let's define the radii and distance between centers:
* Radius of pulley A ($R_A$) = 6 cm
* Radius of pulley B ($R_B$) = 4 cm
* Distance between centers AB ($d$) = 8 cm
We will use a reference angle $\theta$ for calculations related to the tangent lines. This angle is formed by the line connecting the centers (AB) and the radius to the tangent point (e.g., AP).
In the right-angled triangle formed by drawing a line from the center of the smaller pulley (B) perpendicular to the radius of the larger pulley (AP), let's call the intersection point L.
The sides of this triangle are:
* Hypotenuse = AB = 8 cm
* Side adjacent to $\angle BAL$ = $AL = R_A - R_B = 6 - 4 = 2$ cm.
Therefore, $\cos(\angle BAL) = \frac{AL}{AB} = \frac{2}{8} = \frac{1}{4}$.
Let $\theta = \angle BAL = \arccos(1/4)$ radians.
Numerically, $\theta \approx 1.3181$ radians (or $75.52^\circ$).
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**(a) Calculate Length PQ**
The length of the common external tangent (PQ) can be found using the Pythagorean theorem:
$PQ = \sqrt{d^2 - (R_A - R_B)^2}$
$PQ = \sqrt{8^2 - (6 - 4)^2}$
$PQ = \sqrt{64 - 2^2}$
$PQ = \sqrt{64 - 4}$
$PQ = \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$ cm.
**Length PQ = $2\sqrt{15}$ cm**
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**(b) PAS reflex**
The angle $\theta = \arccos(1/4)$ is the angle formed by the line of centers AB and the radius AP (or AS).
Due to the symmetry of the tangents, the angle $\angle PAB = \theta$ and $\angle SAB = \theta$.
So, the smaller angle $\angle PAS = 2\theta = 2\arccos(1/4)$ radians.
The question asks for the **reflex** angle PAS. This is the larger angle around center A.
Reflex $\angle PAS = 2\pi - 2\theta$ radians.
In degrees: $360^\circ - (2 \times 75.52^\circ) = 360^\circ - 151.04^\circ = 208.96^\circ$.
**PAS reflex = $2\pi - 2\arccos(1/4)$ radians (or approximately $208.96^\circ$)**
---
**(c) Length of arc PYS and QXR**
* **Length of arc PYS (on pulley A):**
This arc is on the larger pulley with radius $R_A = 6$ cm. The central angle for this arc is the reflex angle PAS calculated in part (b).
Arc length $L_{PYS} = R_A \times (\text{Reflex } \angle PAS \text{ in radians})$
$L_{PYS} = 6 \times (2\pi - 2\arccos(1/4))$ cm.
$L_{PYS} = 12(\pi - \arccos(1/4))$ cm.
* **Length of arc QXR (on pulley B):**
This arc is on the smaller pulley with radius $R_B = 4$ cm. From the diagram, arc QXR is the minor arc.
The central angle for arc QXR is formed by radii BQ and BR. In the context of direct tangents, the angle for the minor arc is $\pi - 2\theta$, where $\theta = \arccos(1/4)$.
Arc length $L_{QXR} = R_B \times (\pi - 2\theta)$ radians.
$L_{QXR} = 4 \times (\pi - 2\arccos(1/4))$ cm.
**Length of arc PYS = $12(\pi - \arccos(1/4))$ cm**
**Length of arc QXR = $4(\pi - 2\arccos(1/4))$ cm**
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**(d) The total length of the string PQXRSY**
The total length of the string is the sum of the two straight tangent segments (PQ and RS) and the two arc lengths (PYS and QXR).
Due to symmetry, $PQ = RS$.
Total Length = $2 \times PQ + L_{PYS} + L_{QXR}$
Total Length = $2 \times (2\sqrt{15}) + 12(\pi - \arccos(1/4)) + 4(\pi - 2\arccos(1/4))$
Total Length = $4\sqrt{15} + 12\pi - 12\arccos(1/4) + 4\pi - 8\arccos(1/4)$
Total Length = $4\sqrt{15} + 16\pi - 20\arccos(1/4)$ cm.
**The total length of the string PQXRSY = $4\sqrt{15} + 16\pi - 20\arccos(1/4)$ cm**
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**Numerical Approximations (for reference, if needed):**
* $\sqrt{15} \approx 3.873$
* $\pi \approx 3.1416$
* $\arccos(1/4) \approx 1.3181$ radians
* Length PQ $\approx 2 \times 3.873 = 7.746$ cm.
* Reflex PAS $\approx 2\pi - 2(1.3181) = 6.2832 - 2.6362 = 3.6470$ radians.
* $L_{PYS} \approx 6 \times 3.6470 = 21.882$ cm.
* $L_{QXR} \approx 4 \times (\pi - 2 \times 1.3181) = 4 \times (3.1416 - 2.6362) = 4 \times 0.5054 = 2.022$ cm.
* Total Length $\approx (2 \times 7.746) + 21.882 + 2.022 = 15.492 + 21.882 + 2.022 = 39.396$ cm.
Answer by ikleyn(52855)  (Show Source):
You can put this solution on YOUR website! .
Regarding this post and the " solution " by @CPhill, I have two notices.
(1) The plot in the attached file does not correspond / (is not adequate) to the given data.
Indeed, if the radii are 6 cm and 4 cm and the circles do not overlay,
then the distance between the centers can not be less than 6 + 4 = 10 cm,
but with the given data the relevant distance is less than 10 cm.
(2) The solution in the post by @CPhill is INCORRECT.
Indeed, in his solution, @CPhill considers the distance of 8 cm long
as the hypotenuse of the right angled triangle.
But it is incorrect. This distance of 8 cm is one of the LEGS of the right-angled triangle.
Therefore, all that follows in the solution by @CPhill is WRONG.
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Regarding the post by @CPhill . . .
Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it made an error.
Although the @CPhill' solutions are copy-paste Google AI solutions, there is one essential difference.
Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.
Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.
And the last my comment.
When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.
Without it, their reliability is ZERO and their creadability is ZERO, too.
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