Chapter 7.PS, Problem 1PS

To determine

The theorem is true for given circle and triangle.

Explanation of Solution

Given:

If a triangle is inscribed in a circle such that one side of the triangle is a diameter of the circle, then the triangle is a right triangle.

Approach:

If ${\left(AB\right)}^2={\left(AC\right)}^2+{\left(BC\right)}^2$, then the triangle $ABC$ is a right triangle.

Figure$\left(1\right)$

Calculation:

The triangle given by the question is given in Figure $\left(2\right)$.

Figure $\left(2\right)$

Calculate the length of each side of the triangle.

$\begin{array}{rll}AB& =& \sqrt{{\left(-10-10\right)}^2}\\ & =& 20\\ {\left(AB\right)}^2& =& 400\end{array}$

$\begin{array}{rll}AC& =& \sqrt{{\left(-10-6\right)}^2+{\left(0-8\right)}^2}\\ & =& 8\sqrt{5}\\ {\left(AC\right)}^2& =& 320\end{array}$

$\begin{array}{rll}BC& =& \sqrt{{\left(10-6\right)}^2+{\left(0-8\right)}^2}\\ & =& \sqrt{80}\\ {\left(BC\right)}^2& =& 80\end{array}$

Since ${\left(AB\right)}^2={\left(AC\right)}^2+{\left(BC\right)}^2$, the triangle inside the circle is a right triangle.

Therefore, the theorem is true for given circle and triangle.