Chapter 9.4, Problem 16PSB

To determine

To Calculate: The length of CD .

Answer to Problem 16PSB

The length of CD is $\frac{24}{5}$

Explanation of Solution

Given information: Measures of the given lengths are as follows:

AC = $6$

BC = $8$

Formula used:

Here, we are using Pythagoras theorem.

  ${\left(Hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

Calculation:

Here, the formula is

  ${\left(Hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

In $\Delta ABC$

  $A{B}^2=A{C}^2+B{C}^2$

  $A{B}^2=64+36$

  $AB=10$

Similarly,

In $\Delta CDA$

  $36=C{D}^2+x^2$

  $C{D}^2=36-x^2$

And, In $\Delta CDB$

  $64=C{D}^2+{(10-x)}^2$

  $64=C{D}^2+100+x^2-20x$

So that, $64=36-x^2+100+x^2-20x$

  $20x=72$

  $x=\frac{18}{5}$

Hence, $C{D}^2=36-\frac{324}{25}$

  $C{D}^2-\frac{900-324}{25}=\frac{576}{25}$

  $CD=\frac{24}{5}$

Conclusion: Hence,the length of CD is $\frac{24}{5}$