Chapter 9.4, Problem 32PSD

To determine

To calculate:The length of a sideCE .

Answer to Problem 32PSD

The length of a sideCEis 2.68 m.

Explanation of Solution

Given information:

Side $AC=3,$

Side $CB=4,$

Side $EB=2.$

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:

  

By theorem, AC is the mean proportion between AB and AD .

  $\begin{array}{l}AB={\left(3\right)}^2+{\left(4\right)}^2\\ AB=\sqrt{25}\\ \therefore AB=5\end{array}$

  $\begin{array}{l}{\left(AC\right)}^2=(AB)(AD)\\ {\left(3\right)}^2=5\ast AD\\ \therefore AD=\frac{9}{5}=1.8\\ DE=5-2-\frac{9}{5}=\frac{6}{5}=1.2\end{array}$

Apply Pythagoras Theorem in triangle ACD and CDE .

In right angle triangle ACD , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(CD\right)}^2+{\left(AD\right)}^2\\ $ 3^2={(CD)}^2+{\left(1.8\right)}^2$$ 9={(CD)}^2+3.24$$ \therefore {(CD)}^2=5.76$\therefore CD=\sqrt{5.76}\\ \therefore CD=2.4\end{array}$

In right angletriangle CDE , we get

  $\begin{array}{l}{\left(CE\right)}^2={\left(CD\right)}^2+{\left(DE\right)}^2\\ $ {\left(CE\right)}^2={(2.4)}^2+{\left(1.2\right)}^2${\left(CE\right)}^2=5.76+1.44\\ $ \therefore {(CE)}^2=7.2$\therefore CE=\sqrt{7.2}\\ \therefore CE=2.68\end{array}$