low, the segments $overline{rs}$ and $overline{rt}$ are tangent to the circle centered at $o$. given that $rt = 6$ and $or = 6.8$, find $os$.

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Accepted Answer

# Explanation:
## Step1: Use the property of tangent to a circle
A tangent to a circle is perpendicular to the radius at the point of tangency. So, $\angle{OTS} = \angle{OSR}=90^{\circ}$. Also, by the tangent - segment congruence theorem, $RT = RS = 6$ (tangents from the same external point $R$ to a circle are congruent).
## Step2: Apply the Pythagorean theorem
In right - triangle $RTO$ (or $RSO$), by the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $c = OR = 6.8$, $a=RT = 6$ (or $RS = 6$), and $b = OT$ (or $OS$). Let $x = OS$. Then $x^{2}+6^{2}=6.8^{2}$.
$$
\begin{align*}
x^{2}&=6.8^{2}-6^{2}\
x^{2}&=(6.8 + 6)(6.8 - 6)\quad(\text{using }a^{2}-b^{2}=(a + b)(a - b))\
x^{2}&=(12.8)(0.8)\
x^{2}&=10.24\
x&=\sqrt{10.24}\
x& = 3.2
\end{align*}
$$

# Answer:
$3.2$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the property of tangent to a circle

Step2: Apply the Pythagorean theorem

Answer: