Chapter 9.6, Problem 3EX

To determine

To measure: ${\left(BE\right)}^2$ and $ED,CE$ and to compute ${\left(BE\right)}^2$ and $ED\cdot CE$

Answer to Problem 3EX

  $\begin{array}{l}{\left(BE\right)}^2=31.36\\ ED\cdot CE=28.2\end{array}$

${\left(BE\right)}^2$ is greater than $ED\cdot CE$ because $2BE$ is greater than $CE$ .

Explanation of Solution

Given information:

Given figure,

Calculation:

Consider a circle.

Draw a circle. Choose three points $B,C$ and $D$ on circle. Draw a tangent $\overline{BE}$ and draw $\overline{CD}$ and extend it which meet $E$ , like below:

Here the measurements are:

  $\begin{array}{l}BE=5.6\text{cm}\\ ED=4\text{cm}\\ CE=7.2\text{cm}\end{array}$

And,

  $\begin{array}{c}{\left(BE\right)}^2=\left(5.6\right)\left(5.6\right)\\ =31.36\\ ED\cdot CE=\left(4\right)\left(7.2\right)\\ =28.2\end{array}$

Here, ${\left(BE\right)}^2$ is greater than $ED\cdot CE$ because $2BE$ is greater than $CE$ .