Question

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Answer 1

@stupidinmath Are you given in the problem statement that segment TO is congruent to segment SO ?

I ask because in a circle, chords that are equally distant from the center of the circle are congruent.

Answer 2

Nope. but I guess it is?
Yeah..

Answer 3

Was this given or did you assume it? > MN=62=PQ

I cannot tell the difference between what was given in the problem and what you might have assumed.

Answer 4

this was the only ones stated
"In Circle O, MN+PQ, MN= 7x + 13, and PQ = 10x - 8. Find PS."

Answer 5

When I did this:

7x + 3 = 10x - 8

I did not get x = 7.

I got x = 11/3.

So, I can't help with this problem until I know what the given information is.

Answer 6

Does "+" mean "and" in this --> MN+PQ, ?

There is a piece of information missing from the given problem.

Answer 7

oh, sorry, I mean.. MN=PQ.. typo

Answer 8

PS = (1/2)*PQ because if a diameter (radius) is perpendicular to a chord, it bisects the chord.

But, we cannot get the measure of PS until we know the measure of PQ.

Answer 9

Oh, well, what about the 62? Is it a typo?

Answer 10

No, that's my answer..

Answer 11

So, 62/2 = PS?

Answer 12

See, I don't know how you got the 62.

Answer 13

Let me check my work.

Answer 14

MN = PQ
7x+13 = 10x -8
13+8 = 10x - 7x
21 = 3x
7 = x

PQ = 10x -8
PQ = 10 (7) -8
PQ= 62..

Answer 15

13+8 = 21
10 - 7 = 3x

so.. 21 = 3x
x=7..

why x=6?

Answer 16

3rd Attempt :)

Let's take it from the beginning.

MN = PQ

Given 7x + 13 = 10x - 8

x = 7

PQ = 10 * (7) - 8

PQ = 62

PS = (1/2)* 62 A diameter that is perpendicular to a chord bisects the chord.

PS = 31

Answer 17

yeah, that's my answer too.

Answer 18

Back to your original question:

how do I find PS? Use this theorem:

A diameter that is perpendicular to a chord bisects the chord.

Answer 19

oh, thank you :)

Answer 20

Sorry to be so bad on my computations this morning. Thanks for your patience.

Answer 21

can it also be "A radius that is perpendicular to a chord bisects the chord"?
NAh, its alrght :)

Answer 22

Yes. In this case, you can use radius and diameter interchangeably.

Answer 23

hmm, alright, got it :))

Answer 24

That is good. If you can remember why something is true (the theorems), you can see how to set up the problems.