Chapter 9.8, Problem 22PSC

To determine

To calculate: Thealtitude ( PY ) and slant height ( PR ).

Answer to Problem 22PSC

The altitude ( PY ) is $3\sqrt{2}cm$ and slant height ( PR ) is $3\sqrt{3}cm$ .

Explanation of Solution

Given information:

A pyramid is formed by assembling four equilateral triangles and a square having a side 6 cm long.

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:

  

Since $\Delta PDA$ isa equilateral with sides 6 cm, $DR=3$ cm.

If $DR=3cm,PR=3\sqrt{3}cm$ (slant height).

Side PY can be calculated by applying Pythagoras Theorem.

In right angle triangle PYR ,

  $PR=3\sqrt{3}cm$ and $YR=3cm$ .

  $\begin{array}{l}{\left(PY\right)}^2+3^2={\left(3\sqrt{3}\right)}^2\\ {\left(PY\right)}^2+9=27\\ \therefore PY=\sqrt{18}\end{array}$

  $\therefore PY=3\sqrt{2}cm$ (altitude)