ambiguous

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)
View attachment 39755

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)
View attachment 39755

Let us assume that \(\displaystyle \overline{QR} \ \cong \overline{ST}\).

 

[logistic_guy]

Let us assume that \(\displaystyle \overline{QR} \ \cong \overline{ST}\).

Then,

\(\displaystyle 5x - 9 = 2x\)

\(\displaystyle 5x - 9 - 2x = 0\)

\(\displaystyle 3x - 9 = 0\)

\(\displaystyle 3x = 9\)

\(\displaystyle x = \frac{9}{3} = 3\)

 

[logistic_guy]

Suppose \(\displaystyle \overline{QR} = 16\)

 

[logistic_guy]

Suppose \(\displaystyle \overline{QR} = 16\)

If our aim is to find the radius of the circle, then

\(\displaystyle r^2 = \left(\frac{\overline{QR}}{2}\right)^2 + (2x)^2\)


\(\displaystyle r^2 = \left(\frac{16}{2}\right)^2 + (2 \times 3)^2\)


\(\displaystyle r^2 = 8^2 + 6^2\)


\(\displaystyle r^2 = 64 + 36\)


\(\displaystyle r^2 = 100\)


\(\displaystyle r = \sqrt{100} = \textcolor{blue}{10}\)