Chapter 10, Problem 3RP

a)

To determine

To find: The length of BD

Answer to Problem 3RP

BD = $16$

Explanation of Solution

Given Information :

The given length as AE = $6$ , EC = $8$ and DE = $4$

  

Formula Used:

The products of lengths of the line segments on each chords are equal.

Calculation :

  $\begin{array}{l}AE\times EC=DE\times EB\\ \Rightarrow 6\times 8=4\times EB\\ \Rightarrow EB=8\end{array}$

Now $BD=BE+ED$

  $BE=12+4=16$

b)

To determine

To find : The length of PT

Answer to Problem 3RP

PT = $8$

Explanation of Solution

Given Information :

We know that , PQ = $4$ and QR = $12$

  

Formula Used:

The product of the lengths of the secant segment and its external part equals to the square of the length of the tangent segment.

Calculation :

  $\begin{array}{l}\text{PR}\times \text{PQ}={(\text{PT})}^2\\ (4+12)\times 4={(\text{PT})}^2\\ $ \Rightarrow {(\text{PT})}^2=16\times 4$\Rightarrow \text{PT}=8\end{array}$

c)

To determine

To find : the length of WX

Answer to Problem 3RP

WX = 4

Explanation of Solution

Given Information :

The given lengths SX = $5$ , SY = $3$ and YZ = $12$ .

  

Formula Used:

(Whole Secant) × (External part) = (Whole secant) × (External part)

Calculation :

(WS) × (XS) = (SY) × (SZ)

  $\begin{array}{l}\Rightarrow (5+\text{WX})\times 5=3\times 15\\ \Rightarrow (5+WX)=9\\ \Rightarrow WX=9-5\\ \Rightarrow WX=4\end{array}$