Chapter 9.4, Problem 28PSC

To determine

To calculate: The length of ahypotenuseAB .

Answer to Problem 28PSC

The length of a hypotenuse ABis 10.

Explanation of Solution

Given information:

Side $AD=2\sqrt{13},$

Side $EB=\sqrt{73}$

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:

  

Apply Pythagoras Theorem in triangle ACD and BCE .

In right angled triangle ACD , we get

  $\begin{array}{l}{\left(2a\right)}^2+{\left(b\right)}^2={\left(2\sqrt{13}\right)}^2\\ 4a^2+b^2=52----Equation1\end{array}$

In right angled triangle BCE , we get

  $\begin{array}{l}{\left(a\right)}^2+{\left(2b\right)}^2={\left(\sqrt{73}\right)}^2\\ a^2+4b^2=73----Equation2\end{array}$

Multiplying Equation 1 by -4 on both sides, we get

  $-16a^2-4b^2=-208----Equation3$

Adding Equation 2 and Equation 3, we get

  $\begin{array}{l}-15a^2=-135\\ a^2=9\\ \therefore a=3\end{array}$

  $\begin{array}{l}{\left(2a\right)}^2+{\left(b\right)}^2={\left(2\sqrt{13}\right)}^2\\ 36+b^2=52\\ b^2=16\\ \therefore b=4\end{array}$

  $\begin{array}{l}{\left(AB\right)}^2={\left(2a\right)}^2+{\left(2b\right)}^2\\ A{B}^2=36+64\\ AB=\sqrt{100}\\ \therefore AB=10\end{array}$