Chapter 8.3, Problem 8CE

To determine

To prove: $x=5$

Explanation of Solution

Given:

The figure:

Concept Used:

Theorem 8-3: A triangle is right triangle if the square of hypotenuse is equal to the square of other sides.

Theorem 5-15: The midpoint of the hypotenuse of a any right triangle is equidistant from the three vertices of triangle.

Consider the given figure:

From the given figure it can be observed that,

  $\begin{array}{c}AB=AK+KB\\ =5+5\\ =10\end{array}$

Now consider,

  $\begin{array}{c}A{B}^2={10}^2\\ =100\end{array}$

  $\begin{array}{c}A{C}^2+B{C}^2=8^2+6^2\\ =64+36\\ =100\end{array}$

That is,

  $A{B}^2=A{C}^2+B{C}^2$

Then by theorem 8-3, the triangle $△ACB$ is a right triangle with right angle at C.

Also from the given figure it is clear that K is the midpoint of hypotenuse $\overline{AB}$ of a right triangle $△ACB$ .

Now, by theorem 5-15, the midpoint K is equidistant from the three vertices.

Thus,

  $CK=AK=5$

Hence proved that $x=5$