Chapter 12.5, Problem 9PSB

To determine

To calculate: The volume of a pyramid with a diagonal of 10 and lateral edge of 13.

Answer to Problem 9PSB

The volume of a pyramid is $200$ .

Explanation of Solution

Given information:

Diagonal of pyramid = 10,

Lateral edge = 13.

Formula used:

Volume of pyramid $=\frac{1}{3}B\times h$

B = Base area of pyramid and h = height of pyramid

Area of square: $A=s^2$

s = side of square

Calculation:

  

Diagonal AC = 10.

△ACB is ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$

In △ABC

  $\begin{array}{l}\sin {45}^{\circ }=\frac{AB}{AC}\\ \frac{1}{\sqrt{2}}=\frac{AB}{10}\\ AB=\frac{10}{\sqrt{2}}=5\sqrt{2}\end{array}$

  $B={\left(5\sqrt{2}\right)}^2=50$

Side PF can be calculated by applying Pythagoras Theorem.

In right angled triangle PFA , we get

  $\begin{array}{l}{\left(PA\right)}^2={\left(PF\right)}^2+{\left(FA\right)}^2\\ FA=\frac{5\sqrt{2}}{\sqrt{2}}=5\\ {\left(PF\right)}^2={\left(13\right)}^2-{\left(5\right)}^2\\ {\left(PF\right)}^2=169-25\\ {\left(PF\right)}^2=144\\ PF=\sqrt{144}\\ PF=12\\ h=PF=12\end{array}$

Volume of pyramid:

  $\begin{array}{l}{V}_{Pyramid}=\frac{1}{3}\times B\times h\\ V_{Pyramid}=\frac{1}{3}\times \left(50\right)\times 12\\ V_{Pyramid}=200\end{array}$