Chapter 9.8, Problem 18PSC

a.

To determine

To calculate: The length of a diagonal in terms of x.

Answer to Problem 18PSC

The length of a diagonal is $x\sqrt{3}$ .

Explanation of Solution

Given information:

Side $AB=x,$

Diagonal $AC=x\sqrt{2}.$

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:

  

Side DCcan be calculated by applying Pythagoras Theorem.

In right angled triangle DAC , we get

  $\begin{array}{l}{\left(AD\right)}^2+{\left(CA\right)}^2={\left(DC\right)}^2\\ AD=AB\\ {\left(x\right)}^2+{\left(x\sqrt{2}\right)}^2={\left(DC\right)}^2\\ 3{\left(x\right)}^2={\left(DC\right)}^2\\ \therefore DC=x\sqrt{3}\end{array}$

b.

To determine

To find: The length of a diagonal in terms of x.

Answer to Problem 18PSC

The length of a diagonal is $\frac{x\sqrt{6}}{2}$ .

Explanation of Solution

Given information:

Diagonal $AC=x$

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:

  

Diagonal of a square is $s\sqrt{2}$ .

If $s\sqrt{2}$ = x

  $s=\frac{x}{\sqrt{2}}$

Side $=\frac{x\sqrt{2}}{2}$

  $DC=x\sqrt{3}$

x = side

  $\begin{array}{l}DC=(\frac{x\sqrt{2}}{2})\sqrt{3}\\ =\frac{x\sqrt{6}}{2}\end{array}$