Chapter 9.7, Problem 21WE

(a)

To determine

To Explain: why ${\left(r+h\right)}^2=r^2+d^2$ in a circle shown.

Explanation of Solution

Given:

Circle is shown below.

  

  ${\left(r+h\right)}^2=r^2+d^2$

Calculation:

Figure is redrawn as.

  

Since, line from center of a circle to the tangent is perpendicular and it makes a right angle.

Triangle OBC shown is a right triangle therefore using Pythagoras theorem.

  $\begin{array}{l}O{C}^2=O{B}^2+B{C}^2\\ {\left(r+h\right)}^2=r^2+d^2\end{array}$

(b)

To determine

To Simplify: the equation ${\left(r+h\right)}^2=r^2+d^2$ to show that $d^2=h\left(2r+h\right)$ .

Answer to Problem 21WE

No

Explanation of Solution

Given:

  ${\left(r+h\right)}^2=r^2+d^2$

Calculation:

  $\begin{array}{l}{\left(r+h\right)}^2=r^2+d^2\\ r^2+h^2+2rh=r^2+d^2\\ h^2+2rh=d^2\\ d^2=h^2+2rh\\ d^2=h\left(h+2r\right)\\ d^2=h\left(2+hr\right)\end{array}$

Therefore, it has been proved.

(c)

To determine

To Find: which theorem justifies the proof $d^2=h\left(2r+h\right)$ in figure shown.

Answer to Problem 21WE

No

Explanation of Solution

Given Circle is shown below.

  

  ${\left(r+h\right)}^2=r^2+d^2$

Calculation:

Theorem 9-13 state that When a secant segment and a tangent segment are drawn to a circle from an external point, the product of the secant segment and its external segment is equal to the square of the tangent segment.