Chapter SR21, Problem 6P

Solution

Find the value of x.

  $x=8\sqrt{2}in.$

Given:

The given triangle.

  

Concept Used:

• The Pythagoras theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Calculation:

The given right angle triangle is -

  

Hypotenuse = xin.

Base = 8in.

  $Perpendicular$ = 8in.

According to Pythagoras Theorem −

  $\begin{array}{c}$ {(Hypotenuse)}^2={(Base)}^2+{(Perpendicular)}^2$$ {(x)}^2={(8)}^2+{(8)}^2$$ {(x)}^2=64+64$$ {(x)}^2=128$x=\sqrt{128}\\ x=\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2}\\ x=2\cdot 2\cdot 2\sqrt{2}\\ x=8\sqrt{2}\end{array}$

  $x=8\sqrt{2}in.$