Chapter 10.6, Problem 15PSB

To determine

Find the length of the given segment using given data.

Answer to Problem 15PSB

The length of $\overline{AD}=10\sqrt{3}$ .

Explanation of Solution

Given:

The given statement is that the quadrilateral $ABCD$ inscribed in a circle. The given measurements are $AB=12,BC=16,CD=10$ and $\angle ABC$ is a right angle. The given segment is $\overline{AD}$ .

Calculation:

Draw a diagram using the given statement.

  

In diagram, $O$ is the center of the circle.

  $AC$ is the diameter of the circle.

The triangle $\Delta ABC$ is a right angle triangle.

Use Pythagorean formula.

  $\text{hypotenuse}=\sqrt{{\text{perpendicular}}^2+{\text{base}}^2}$ .

  $\begin{array}{l}AC=\sqrt{A{B}^2+B{C}^2}\\ =\sqrt{{12}^2+{16}^2}\\ =\sqrt{144+256}\\ =\sqrt{400}\\ =20\end{array}$

Use theory, the angle inscribed in a semi-circle is a right angle.

  $\angle ADC={90}^{°}$ .

So, the triangle $\Delta ADC$ is a right angle triangle.

Use Pythagorean formula.

  ${\text{hypotenuse}}^2={\text{perpendicular}}^2+{\text{base}}^2$ .

  $\begin{array}{l}A{C}^2=A{D}^2+C{D}^2\\ \Rightarrow A{D}^2=A{C}^2-C{D}^2\\ \Rightarrow AD=\sqrt{A{C}^2-C{D}^2}\\ =\sqrt{{20}^2-{10}^2}\\ =\sqrt{400-100}\\ =\sqrt{300}\\ =10\sqrt{3}\end{array}$

Hence the length of $\overline{AD}=10\sqrt{3}$ .