Chapter 12.2, Problem 10PSC

a.

To determine

To find: The measure of a diagonal of the base of a regular square pyramid.

Answer to Problem 10PSC

The measure of a diagonal of the base is $4\sqrt{2}.$

Explanation of Solution

Given Information:

Lateral edge of a regular square pyramid $\left(l\right)=3$

Height of the pyramid $\left(h\right)=1$

Formula used:

Using Pythagoras theorem,

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

Calculation:

Lateral edge of a regular square pyramid $\left(l\right)=3$

Height of the pyramid $\left(h\right)=1$

Now, the lateral edge, height and half the diagonal of a square form a right angled triangle.

Let the diagonal be $d$

Using Pythagoras theorem,

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

  $\begin{array}{l}\therefore l^2={\left(\frac{d}{2}\right)}^2+h^2\\ \therefore {\left(\frac{d}{2}\right)}^2=l^2-h^2\end{array}$

  $\therefore {\left(\frac{d}{2}\right)}^2=4^2-1^2$

  $\begin{array}{l}\therefore \frac{d}{2}=\sqrt{8}\\ \therefore \frac{d}{2}=2\sqrt{2}\\ \therefore d=4\sqrt{2}\end{array}$

Hence, the measure of a diagonal of the base is $4\sqrt{2}.$

b.

To determine

To calculate: The slant height of the pyramid.

Answer to Problem 10PSC

The slant height of the pyramid is $\sqrt{5}.$

Explanation of Solution

Given Information:

From part a, diagonal of the base is $4\sqrt{2}$

Height of the pyramid $\left(h\right)=1$

Formula used:

Side of a square $\left(s\right)=\frac{\text{Diagonal}}{\sqrt{2}}$

Using Pythagoras theorem,

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

Calculation:

Height of the pyramid $\left(h\right)=1$

From part a, diagonal of the base is $4\sqrt{2}$

As we know that, side of a square $\left(s\right)=\frac{\text{Diagonal}}{\sqrt{2}}$

  $\therefore s=\frac{4\sqrt{2}}{\sqrt{2}}=4$

Now, the slant height, height and half the side of a square form a right angled triangle.

Let the slant height be $h\text{'}$

Using Pythagoras theorem,

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

  $\begin{array}{l}\therefore h{\text{'}}^2={\left(\frac{s}{2}\right)}^2+h^2\\ \therefore h{\text{'}}^2=2^2+1^2\end{array}$

  $\begin{array}{l}\therefore h\text{'}=\sqrt{2^2+1^2}\\ \therefore h\text{'}=\sqrt{5}\end{array}$

Hence, the slant height of the pyramid is $\sqrt{5}.$

c.

To determine

To calculate: The area of the square base of a pyramid.

Answer to Problem 10PSC

The area of the base is 16.

Explanation of Solution

Given Information:

From part a, diagonal of the base is $4\sqrt{2}$

Formula used:

Side of a square $\left(s\right)=\frac{\text{Diagonal}}{\sqrt{2}}$

Area of the square $\left(A\right)=sid{e}^2=s^2$

Calculation:

From part a, diagonal of the base is $4\sqrt{2}$

As we know that, side of a square $\left(s\right)=\frac{\text{Diagonal}}{\sqrt{2}}$

  $\therefore s=\frac{4\sqrt{2}}{\sqrt{2}}=4$

Now, area of the base = Area of the square

We know that, area of the square $\left(A\right)=sid{e}^2=s^2$

  $\begin{array}{l}\therefore A=4^2\\ \therefore A=16\end{array}$

Hence, area of the base is 16.

d.

To determine

To calculate: The lateral area of a square base pyramid.

Answer to Problem 10PSC

The area of the base is 16.

Explanation of Solution

Given Information:

From part b, slant height of the pyramid is $\sqrt{5}.$

From part a, diagonal of the square is $4\sqrt{2}.$

Formula used:

Area of a triangle $=\frac{1}{2}\times b{h}_1$

  $"b"$ is the base of a triangle

  $"h_1"$ is the height of a triangle

Calculation:

From part a, diagonal of the base is $4\sqrt{2}$

As we know that, side of a square $\left(s\right)=\frac{\text{Diagonal}}{\sqrt{2}}$

  $\therefore s=\frac{4\sqrt{2}}{\sqrt{2}}=4$

Now, base of the triangle = side of the square

Therefore, base of the triangle $\left(b\right)=4$

Height of the triangle = Slant height of the pyramid

Therefore, height of the triangle $\left(h_1\right)=\sqrt{5}$

We know that, Area of a triangle $=\frac{1}{2}\times b{h}_1$

  $\begin{array}{l}\therefore A=\frac{1}{2}\times 4\times \sqrt{5}\\ \therefore A=2\sqrt{5}\end{array}$

Lateral area of a pyramid $\left(L\right)=4\times$ Area of a triangle

  $\therefore L=4A$

  $\begin{array}{l}\therefore L=4\times 2\sqrt{5}\\ \therefore L=8\sqrt{5}\end{array}$

Hence, lateral area of the pyramid is $8\sqrt{5}.$