Q10. - Geometry

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 $TP=?$

Step-by-Step Solution

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,

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

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.

$\begin{array}{l}\text{In }\Delta STP,\text{ by Pythagorean theorem,}\\ S{P}^2+S{T}^2=P{T}^2\text{. }\mathrm{...}\left(1\right)\\ \text{In }\Delta OTP,\text{ by Pythagorean theorem,}\\ {\text{OT}}^{\text{2}}+P{T}^2=O{P}^2.\\ \Rightarrow O{P}^2-{\text{\hspace{0.17em}\hspace{0.17em}OT}}^{\text{2}}=P{T}^2\mathrm{...}\left(2\right)\\ \text{So, by eq - (1) and (2), we can write:}\\ S{P}^2+S{T}^2=O{P}^2-{\text{\hspace{0.17em}\hspace{0.17em}OT}}^{\text{2}},\\ \Rightarrow {24}^2+S{T}^2={30}^2-O{T}^2,\\ \Rightarrow S{T}^2+O{T}^2={30}^2-{24}^2,\\ \Rightarrow OT=\frac{5}{6}ST.\\ \text{Thus, we have:}\\ S{T}^2+\frac{25}{36}S{T}^2=324,\\ \Rightarrow \frac{11}{36}S{T}^2=324,\\ \Rightarrow S{T}^2=1060,\\ ST=32.5.\\ \text{Now, by Plugging the value of }ST\text{ in eq - (1), we obtain:}\\ {24}^2+{32.5}^2=P{T}^2,\\ \Rightarrow PT=40.4.\end{array}$