Chapter 9.8, Problem 23PSD

To determine

To show:The ratio $\frac{b}{a}$ is equal to $\frac{\sqrt{2}}{1}$ .

Explanation of Solution

Given information:

  $AB=a,BC=b,CE=x$ and $DE=y$ .

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,

  $a^2+b^2\text{= }{c}^2$

Proof:

  

Applying Pythagoras Theorem, we get

In right angle triangle AFB , we get

  ${\left(x\right)}^2+{(y)}^2={\left(a\right)}^2----Equation1$

In right angle triangle CEB , we get

  ${\left(x\right)}^2+{(4y)}^2={\left(b\right)}^2----Equation2$

In right angle triangle CDB , we get

  ${\left(a\right)}^2+{(b)}^2=9{(y)}^2----Equation3$

From Equation 1 and Equation 2, we get

  ${\left(a\right)}^2+{(b)}^2=2x^2+5y^2----Equation4$

From Equation 3 and Equation 4, we get

  $\begin{array}{l}9y^2=2x^2+5y^2\\ 4y^2=2x^2\\ 2y^2=x^2----Equation5\end{array}$

From Equation 1 and Equation 5, we get

  $\begin{array}{l}{\left(2y\right)}^2+{(}^y={\left(a\right)}^2\\ {\left(3y\right)}^2={\left(a\right)}^2----Equation6\end{array}$

From Equation 2 and Equation 5, we get

  $\begin{array}{l}2y^2+4y^2=b^2\\ 6y^2=b^2\end{array}$

  $\begin{array}{l}\frac{b^2}{a^2}=\frac{6y^2}{3y^2}=\frac{2}{1}\\ \frac{b}{a}=\frac{\sqrt{2}}{1}\end{array}$

The ratio $\frac{b}{a}$ is equal to $\frac{\sqrt{2}}{1}$ .

Thus, two ratios are equal.