Let $\overline{PQ}$, $\overline{RS}$, and $\overline{TU}$ be parallel chords of a circle. The distance between chords $\overline{PQ}$ and $\overline{RS}$ is 4, and the distance between chords $\overline{RS}$ and $\overline{TU}$ is also 4. If $PQ = 78$ and $TU = 50$, then find $RS$.
Bruh you think you tricking anyone? You had one good sentence and all the other 3 have wrong capitalization, and the fact that you sound like some goody-two-shoes 9 year old. If you know it's AoPS, and you aren't here for exposing cheaters, then you must be cheating too.
yeah like srsly
if you dont cheat then how do u know about this place
richard wouldnt waste his time on finding cheaters
also ur making it look like richard himself said those words and he worldnt be so rude
although tbh searching it up is against the honor code...
"You should not search for potential solutions to problems online or elsewhere. (Especially searching for the problem statement!) Using the internet in this way to search for answers is against our Course Policies! Instead, if you need some assistance with a particular problem, we welcome you to ask questions on your course message board. Your instructor and other AoPS staff members will be happy to help."
We aint idiots dont try to trick us. Ur not Richard.
let the distance from TU to the centre be x
then RS is x+4 units away from the centre, and
PQ is x+8 units away from the centre
draw the common perpendicular to all the chords, and draw radii to the ends of all the chords
You now have right-angled triangles
from chord TU: r^2 = x^2 + 39^2
from chord PQ: r^2 = (x+8)^2 + 25^2
sp (x+8)^2 + 625 = x^2 + 1521
x^2 + 16x + 64 + 625 = x^2 + 1521
16x = 832
x = 52
r^2 = 52^2 + 39^2 = 4225
r = 65
now to the right-triangle from chord RS
let half the chord by y
(x+4)^2 + y^2 = r^2
56^2 + y^2 = 65^2
y^2 = 1089
y = 33
so RS = 2(33) = 66
you guys are all wrong
Remember that if you get stuck on a homework problem, you can always ask on the message board! Click on the pencil icon V in the upper-right corner of the problem, and this will open a box where you can ask your question, which will be posted on the message board. You can also click on the speech bubble icon t, which will bring up any discussions on that problem. Seriously, a student showed me this and I cant believe it! Just sop and pay attention in class.
um why is everyone fighting 😀
To solve this problem, we can use the property that the perpendicular distance between two parallel chords is the same at any point along the circle.
Let's first draw a diagram to visualize the given information:
```
T ________ U
|
|
4 |
|
R ________ S 4
|
|
|
P ________ Q
```
From the diagram, we can see that the distance between $\overline{PQ}$ and $\overline{RS}$ is 4, and the distance between $\overline{RS}$ and $\overline{TU}$ is also 4.
Now, let's focus on the chord $\overline{RS}$. We can draw a perpendicular line from $\overline{RS}$ to $\overline{TU}$, and let's call the intersection point as $V$.
```
V
|
|
-----------
R ________ S -----------
| |
| |
4 | | 4
| |
| |
R ________ S ----------- + ----------- T ________ U
| |
| |
| |
| |
| |
| |
| |
P ________ Q
```
From the diagram, we can observe that we have formed two right triangles: $\triangle VRT$ and $\triangle VUS$. The perpendicular distance between $\overline{PQ}$ and $\overline{RS}$ is equal to $VT$, and the perpendicular distance between $\overline{RS}$ and $\overline{TU}$ is equal to $VS$.
Since $VT=4$ and $PT=78$, we can use the Pythagorean theorem to find $VR$:
$VR^2 = PT^2 - VT^2 = 78^2 - 4^2 = 6084 - 16 = 6068$
Taking the square root of both sides:
$VR = \sqrt{6068} \approx 78.0078$
Similarly, since $VS=4$ and $UT=50$, we can use the Pythagorean theorem to find $VS$:
$VS^2 = UT^2 - US^2 = 50^2 - 4^2 = 2500 - 16 = 2484$
Taking the square root of both sides:
$VS = \sqrt{2484} \approx 49.8369$
Finally, we need to find $RS$. Since $RVS$ is a right triangle, we can use the Pythagorean theorem:
$RS^2 = VR^2 + VS^2 = 6068 + 2484 = 8552$
Taking the square root of both sides:
$RS = \sqrt{8552} \approx 92.5303$
So, the length of chord $\overline{RS}$ is approximately 92.5303.