Chapter 9.2, Problem 10WE

a.

To determine

To explain: TS is the geometric mean between some described lengths in the figure.

Answer to Problem 10WE

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

Explanation of Solution

Given information:

PT is tangent to the circle at O at T and

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

b.

To determine

To explain: TO is the geometric mean between.

Answer to Problem 10WE

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

Explanation of Solution

Given information:

PT is tangent to the circle at O at T and

TO is the geometric mean between OP and OS

c.

To determine

To Calculate: TS and TP. 

Answer to Problem 10WE

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

Explanation of Solution

Given information:

PT is tangent to the circle at O at T and

  $SP=6$ and $OS=24$ .

Calculation:

  $\begin{array}{l}\text{In }\Delta STP,\text{ by Pythagorean theorem,}\\ S{P}^2+S{T}^2=P{T}^2\text{. }\mathrm{...}(1)\\ \text{In }\Delta OTP,\text{ by Pythagorean theorem,}\\ {\text{OT}}^{\text{2}}+P{T}^2=O{P}^2.\\ \Rightarrow O{P}^2-{\text{OT}}^{\text{2}}=P{T}^2\mathrm{...}(2)\\ \text{So, by eq - (1) and (2), we can write:}\\ S{P}^2+S{T}^2=O{P}^2-{\text{OT}}^{\text{2}},\\ \Rightarrow {24}^2+S{T}^2={30}^2-O{T}^2,\\ \Rightarrow S{T}^2+O{T}^2={30}^2-{24}^2,\\ \Rightarrow OT=\frac{5}{6}ST.\\ \text{Thus, we have:}\\ S{T}^2+\frac{25}{36}S{T}^2=324,\\ \Rightarrow \frac{11}{36}S{T}^2=324,\\ \Rightarrow S{T}^2=1060,\\ ST=32.5.\\ \text{Now, by Plugging the value of }ST\text{ in eq - (1), we obtain:}\\ {24}^2+{32.5}^2=P{T}^2,\\ \Rightarrow PT=40.4.\end{array}$