Chapter 9, Problem 27RP

To determine

Tocalculate:The value of x .

Answer to Problem 27RP

The value of xis 7.5.

Explanation of Solution

Given information:

  $\begin{array}{l}AB=8\frac{1}{2},\\ AC=x,\\ BD=3,\\ CE=12,\\ DE=13.\end{array}$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Sides DC , CB and x can be calculated by applying Pythagoras Theorem.

In right angled $\Delta DCE$ , we get

  $\begin{array}{l}{\left(DE\right)}^2={\left(DC\right)}^2+{\left(CE\right)}^2\\ {\left(13\right)}^2={\left(DC\right)}^2+{\left(12\right)}^2\\ {\left(DC\right)}^2=169-144\\ {\left(DC\right)}^2=25\\ {\left(DC\right)}^2=\sqrt{25}\\ DC=5\end{array}$

In right angled $\Delta CBD$ , we get

  $\begin{array}{l}{\left(CD\right)}^2={\left(CB\right)}^2+{\left(BD\right)}^2\\ {\left(5\right)}^2={\left(CB\right)}^2+{\left(3\right)}^2\\ {\left(CB\right)}^2=25-9\\ {\left(CB\right)}^2=16\\ {\left(CB\right)}^2=\sqrt{16}\\ CB=4\end{array}$

In right angled $\Delta ACB$ , we get

  $\begin{array}{l}{\left(AB\right)}^2={\left(AC\right)}^2+{\left(CB\right)}^2\\ {\left(8.5\right)}^2={\left(x\right)}^2+{\left(4\right)}^2\\ {\left(x\right)}^2=72.25-16\\ x^2=56.25\\ x=7.5\end{array}$