Chapter 7, Problem 1PS

To determine

To show: if the 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.

Explanation of Solution

Given:

The givencircle is $x^2+y^2=100$

And the triangle formed by the lines

  $\begin{array}{l}y=0\\ y=\frac{1}{2}x+5\end{array}$

And

  $y=-2x+20$

Calculation:

Let us consider the following equation of the circle:

  $x^2+y^2=100$

The equation of lines forming the triangles are:

  $\begin{array}{c}y=0\\ y=\frac{1}{2}x+5\\ y=-2x+20\end{array}$

Let us draw the graph of the given equation and the graph is as shown below:

  

From the figure let us find out the length of $AB,AC$ and $BC$ .

Length of $AB=20$ units

Length of $BC$

  $\begin{array}{c}BC=\sqrt{{\left(10-6\right)}^2+{\left(0-8\right)}^2}\\ =\sqrt{4^2+8^2}\\ =\sqrt{16+64}\\ =\sqrt{80}\text{units}\end{array}$

Length of $AC$

  $\begin{array}{c}AC=\sqrt{{\left(-10-6\right)}^2+{\left(0-8\right)}^2}\\ =\sqrt{{16}^2+8^2}\\ =\sqrt{256+64}\\ =\sqrt{320}\text{units}\end{array}$

Hence,

  $\begin{array}{c}B{C}^2+A{C}^2=320+80\\ =400\\ ={20}^2\\ =A{B}^2\end{array}$

Therefore, triangle $ABC$ inscribed in the circle given is a right triangle at $C$ .