Question

Can anyone explain this to me? I'm pretty confused.

Answer 1

Math aint my strong suit so im gonna go

Answer 2

if not one sec

Answer 3

no direct answers never

Answer 4

how you ve get this result ?

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
no direct answers never
\(\color{#0cbb34}{\text{End of Quote}}\)
I wasn't looking for x, so it isn't a direct answer.

Answer 6

i gace her the value of x that is only the first step in finding the radius

Answer 7

this not is the radius

Answer 8

Can someone just help me?

Answer 9

moment pls

Answer 10

maybe
jhonyy look at the question i think you can figure it faster

Answer 11

than i

Answer 12

recognise that QA = QB

from pure symmetry

does that help?

Answer 13

\(\color{#0cbb34}{\text{Originally Posted by}}\) @SemiDefinite
recognise that QA = QB

from pure symmetry

does that help?
\(\color{#0cbb34}{\text{End of Quote}}\)
QA and QB being what ?

Answer 14

but qa dose not equal r

Answer 15

4x+3=
7x-6

Answer 16

x =3 as i said earlier

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nevaehhh
\(\color{#0cbb34}{\text{Originally Posted by}}\) @SemiDefinite
recognise that QA = QB

from pure symmetry

does that help?
\(\color{#0cbb34}{\text{End of Quote}}\)
QA and QB being what ?
\(\color{#0cbb34}{\text{End of Quote}}\)

the length QA & QB on the diagram

i am deliberately avoiding geometry terms.

Answer 18

the rest of this step is arithmetic

Answer 19

\(\color{#0cbb34}{\text{Originally Posted by}}\) @SemiDefinite
\(\color{#0cbb34}{\text{Originally Posted by}}\) @nevaehhh
\(\color{#0cbb34}{\text{Originally Posted by}}\) @SemiDefinite
recognise that QA = QB

from pure symmetry

does that help?
\(\color{#0cbb34}{\text{End of Quote}}\)
QA and QB being what ?
\(\color{#0cbb34}{\text{End of Quote}}\)

the length QA & QB on the diagram

i am deliberately avoiding geometry terms.
\(\color{#0cbb34}{\text{End of Quote}}\)
Oh I'm sorry, I didn't notice that the center was labeled Q .

Answer 20

lmao

Answer 21

any idea now ?

Answer 22

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
Created with RaphaëlrrrrEFQHG4x +37x -6Reply Using Drawing

any idea now ?
\(\color{#0cbb34}{\text{End of Quote}}\)
The diagram helped a little, but my teacher hasn't explained this at all, mind you.

Answer 23

nice job jhonyy

Answer 24

ohh sorry i ve forget HG and EF = 16

Answer 25

yeah - ty

Answer 26

Do we solve the triangles to find the radius ?

Answer 27

exactly

Answer 28

how you think these triangles QEF and QHG what style of triangles are ?

Answer 29

with two equal sides

Answer 30

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
how you think these triangles QEF and QHG what style of triangles are ?
\(\color{#0cbb34}{\text{End of Quote}}\)
an isosceles ?

Answer 31

exactly

Answer 32

No giving of any answers allowed. If your start to helping a user begins with "x=a", you're giving the answer. Explain, demonstrate, conceptualize.

Answer 33

Could you possibly explain how to find the two sides of the isosceles triangle ?

Answer 34

side a = side b

Answer 35

Well ig no one is going to explain how to solve the triangle lmao

Answer 36

ok moment pls

Answer 37

x=3, which is the height of the triangle

Answer 38

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nevaehhh
\(\color{#0cbb34}{\text{Originally Posted by}}\) @SemiDefinite
\(\color{#0cbb34}{\text{Originally Posted by}}\) @nevaehhh
\(\color{#0cbb34}{\text{Originally Posted by}}\) @SemiDefinite
recognise that QA = QB

from pure symmetry

does that help?
\(\color{#0cbb34}{\text{End of Quote}}\)
QA and QB being what ?
\(\color{#0cbb34}{\text{End of Quote}}\)

the length QA & QB on the diagram

i am deliberately avoiding geometry terms.
\(\color{#0cbb34}{\text{End of Quote}}\)
Oh I'm sorry, I didn't notice that the center was labeled Q .
\(\color{#0cbb34}{\text{End of Quote}}\)

QA = QB allows you to solve for x

Then: AE = AF = BH = BG = 8

that allows you to solve for QF or QG, simply using Pythagoreas, eg:

- \(QA^2 + AF^2 = QF^2 \qquad [= r^2]\)
- \(QB^2 + BG^2 = QG^2 \qquad [= r^2]\)

etc

Answer 39

yes so than you know the height and you know that the base is halfed so 16/2 and from this triangle you can calcule the r

Answer 40

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
yes so than you know the height and you know that the base is halfed so 16/2 and from this triangle you can calcule the r
\(\color{#0cbb34}{\text{End of Quote}}\)
I'm sorry, but I'm confused as to how we can find the radius of the circle from the triangle.

Answer 41

ok you know the height yes that is 3 and the base halved is 8 so r^2 = 3^2 +8^2

Answer 42

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
ok you know the height yes that is 3 and the base halved is 8 so r^2 = 3^2 +8^2
\(\color{#0cbb34}{\text{End of Quote}}\)
I already solved the triangle, i got 16 as the base, and 8.54 as the two sides.

Answer 43

sorry x = 3 so the height is 4*3+3 = 15

Answer 44

Ok, Imma redo the the triangle right quick.

Answer 45

I got 16 as the base, and 17 as the two sides of the triangle, for the top triangle.

Answer 46

r^2 = 15^2 +8^2

Answer 47

r = ?

Answer 48

r = 17

Answer 49

The radius of the circle is 17 ? I don't understand how the radius is 17, when the line doesn't completely touch half through the circle.

Answer 50

look please on my posted image

Answer 51

Thank you for helping me.

Answer 52

do you understand it now to all ?

Answer 53

Yes, I understand it now, thank you again.

Answer 54

anytime

bye bye