Question

Really confused here.

Answer 1

#18 is question. I believe it has to do with "If two chords intersect, then the measure of any one of the vertical angles formed is equal to half the sum of the measures of the two arcs intercepted by the two vertical angles".

Answer 2

for question no \(18.\) first start with finding \(\angle EFG\) AND \(\angle EGF\)

they are part of iscoceles triangle as \(EF \cong EG\)

Answer 3

\(\large \color{black}{\angle EFG+\angle EGF+\angle FEG=180\hspace{.33em}\\~\\
\angle EFG+\angle EGF+\angle FEG=180\hspace{.33em}\\~\\
x+x+\angle BEC=180\hspace{.33em}\\~\\
( \angle EFG=\angle EGF )\text{as EF=EG}\therefore \triangle EFG\text{is iscoceles triangle } \hspace{.33em}\\~\\
x+x+71=180\hspace{.33em}\\~\\
2x+71=180\hspace{.33em}\\~\\
2x=109\hspace{.33em}\\~\\
x=\dfrac{109}{2}
}\)
see if that makes sense

Answer 4

Yes that makes sense.

Answer 5

So do I use the 54.5 to figure out Arc AE and Arc ED for next step?

Answer 6

Or does <CGF = 125.5 because its a line and 180 -54.5 = 125.5. And do same for other angles

Answer 7

ok so now by applying formula u will get
\(\large \color{black}{
\dfrac{m(arc AB)+m(arc ED)}{2}=\angle EFG\hspace{.33em}\\~\\
\dfrac{18+m(arc ED)}{2}=\dfrac{109}{2}\hspace{.33em}\\~\\
}\)

Answer 8

by similar method u can find
\(\large \color{black}{
m(arc\ AE)\quad and\quad m(arc\ CD) \hspace{.33em}\\~\\
}\)
and with information that

\(\large \color{black}{
m(arc\ AE)=3m(arc\ CD) \hspace{.33em}\\~\\
}\)

Answer 9

Ok. thanks, let me see if i can work this and find answers. thank you

Answer 10

Little confused still about how you got 109/2 for first answer.

Answer 11

\(\large \color{black}{\angle EFG+\angle EGF+\angle FEG=180\hspace{.33em}\\~\\
\angle EFG+\angle EGF+\angle FEG=180\hspace{.33em}\\~\\
\color{red}{ x+x+\angle BEC=180}\hspace{.33em}\\~\\
( \angle EFG=\angle EGF )\text{as EF=EG}\therefore \triangle EFG\text{is iscoceles triangle } \hspace{.33em}\\~\\
x+x+71=180\hspace{.33em}\\~\\
2x+71=180\hspace{.33em}\\~\\
2x=109\hspace{.33em}\\~\\
x=\dfrac{109}{2}
}\)

did u get the red part ?

Answer 12

oh sorry, I meant the arc 109/2

Answer 13

u mean this part ?

\(\large \color{black}{
\dfrac{m(arc AB)+m(arc ED)}{2}=\angle EFG\hspace{.33em}\\~\\
\dfrac{18+m(arc ED)}{2}=\dfrac{109}{2}\hspace{.33em}\\~\\
}\)

Answer 14

we don't know arcED yet, so don't understand why that formula = 109/2

Answer 15

do u understand this formula that if we have this figure then

\(\large \color{black}{
\dfrac{1}{2}\cdot m(arcA)+m(arcB)=\angle x \hspace{.33em}\\~\\
}\)