Question

Given: $\overline{PT}$is tangent to $\odot O$at $T;$$\overline{TS}\perp \overline{PO}$

Complete the following statements.

$\left(a\right).$ $TS$ is the geometric mean between $?$ and $?$

$\left(b\right).$ $TO$ is the geometric mean between $?$ and $?$

$\left(c\right).$ If $OS=6$ and $SP=24,TS=?$ and $TP=?$

Short answer

a. $TS$is the geometric mean between the lengths $OP$and $SP$.

b. $TO$is the geometric mean between the lengths $OP$and $OS$.

c.The required value is,

$\begin{array}{l}ST=32.5\text{\hspace{0.17em}.}\\ PT=40.4\text{\hspace{0.17em}.}\end{array}$

Step-by-step solution

Part a. Step 1. Given information.

$PT$is tangent to the circle at $O$at $T$and $TS\perp PO$.

Therefore,

$TS$ is the geometric mean between the lengths $OP$ and $SP$.

Part b. Step 1. Use Pythagoras theorem. 

$PT$is tangent to the circle at $O$ at $T$and $TS\perp PO$.

Therefore,

$TO$ is the geometric mean between the lengths $OP$and $OS$.

Part c. Step 1. Given information.

$PT$ is tangent to the circle at $O$ at $T$and $TS\perp PO$.

$SP=6$ and $OS=24$.

Step 2. Use Pythagoras theorem. 

$\begin{array}{l}\text{In}\Delta STP,\text{byPythagoreantheorem,}\\ S{P}^2+S{T}^2=P{T}^2\text{.}\mathrm{...}\left(1\right)\\ \text{In}\Delta OTP,\text{byPythagoreantheorem,}\\ {\text{OT}}^{\text{2}}+P{T}^2=O{P}^2.\\ \Rightarrow O{P}^2-{\text{\hspace{0.17em}\hspace{0.17em}OT}}^{\text{2}}=P{T}^2\mathrm{...}\left(2\right)\\ \text{So,byeq-(1)and(2),wecanwrite:}\\ S{P}^2+S{T}^2=O{P}^2-{\text{\hspace{0.17em}\hspace{0.17em}OT}}^{\text{2}},\\ \Rightarrow {24}^2+S{T}^2={30}^2-O{T}^2,\\ \Rightarrow S{T}^2+O{T}^2={30}^2-{24}^2,\\ \Rightarrow OT=\frac{5}{6}ST.\\ \text{Thus,wehave:}\\ S{T}^2+\frac{25}{36}S{T}^2=324,\\ \Rightarrow \frac{11}{36}S{T}^2=324,\\ \Rightarrow S{T}^2=1060,\\ ST=32.5.\\ \text{Now,byPluggingthevalueof}ST\text{ineq-(1),weobtain:}\\ {24}^2+{32.5}^2=P{T}^2,\\ \Rightarrow PT=40.4.\end{array}$