Chapter 12.2, Problem 17WE

(a)

To determine

To find: The value of $VX$ and $VY$ .

Answer to Problem 17WE

The value of $VX$ and $VY$ is equal to $15\text{cm}$ and $13\text{cm}$ , respectively.

Explanation of Solution

Given information:

  $V-ABCD$ is a pyramid with rectangular base $18\text{cm}$ long and $10\text{cm}$ wide. $O$ is the center of the rectangle. The height, $VO$ , of the pyramid is $12\text{cm}$ .

  

Calculation:

Since, $O$ is the center of the rectangle. So, the length of the side $OX$ is half the length of the square base. So, the length of the side $OX$ is equal to $9\text{cm}$ .

For the length of the side $VX$ use the Pythagorean theorem in $\Delta VOX$ .

  $\begin{array}{c}{\left(VX\right)}^2={\left(VO\right)}^2+{\left(OX\right)}^2\\ {\left(VX\right)}^2={\left(12\right)}^2+{\left(9\right)}^2\\ VX=\sqrt{144+81}\\ VX=15\end{array}$

So, the length of the side $VX$ is equal to $15\text{cm}$ .

Similarly, $O$ is the center of the rectangle. So, the length of the side $OY$ is half the width of the square base. So, the length of the side $OY$ is equal to $5\text{cm}$ .

For the length of the side $VY$ use the Pythagorean theorem in $\Delta VOY$ .

  $\begin{array}{c}{\left(VY\right)}^2={\left(VO\right)}^2+{\left(OY\right)}^2\\ {\left(VY\right)}^2={\left(12\right)}^2+{\left(5\right)}^2\\ VY=\sqrt{144+25}\\ VY=13\end{array}$

So, the length of the side $VY$ is equal to $13\text{cm}$ .

Therefore, the value of $VX$ and $VY$ is equal to $15\text{cm}$ and $13\text{cm}$ , respectively.

(b)

To determine

To find: The lateral area of the pyramid.

Answer to Problem 17WE

The lateral area of the pyramid is equal to $384{\text{cm}}^2$ .

Explanation of Solution

Given information:

  $V-ABCD$ is a pyramid with rectangular base $18\text{cm}$ long and $10\text{cm}$ wide. $O$ is the center of the rectangle. The height, $VO$ , of the pyramid is $12\text{cm}$ .

  

Calculation:

The lateral area of the pyramid is the sum of the area of all lateral face of the pyramid.

The area of the face $VAB$ is the half of the product of length of $VY$ and $AB$ .

  $\begin{array}{c}\text{A}=\frac{1}{2}\left(VY\right)\left(AB\right)\\ =\frac{1}{2}\left(13\right)\left(18\right)\\ =117\end{array}$

So, the area of the face $VAB$ is equal to $117{\text{cm}}^2$ .

The area of the face $VAB$ is equal to the area of the face $VDC$ as the slant height of both the sides is equal and the length of edge along the base is same for both the faces.

So, the area of the face $VDC$ is equal to $117{\text{cm}}^2$ .

The area of the face $VBC$ is the half of the product of length of $VX$ and $BC$ .

  $\begin{array}{c}\text{A}=\frac{1}{2}\left(VX\right)\left(BC\right)\\ =\frac{1}{2}\left(15\right)\left(10\right)\\ =75\end{array}$

So, the area of the face $VBC$ is equal to $75{\text{cm}}^2$ .

The area of the face $VBC$ is equal to the area of the face $VDA$ as the slant height of both the sides is equal and also the length of edge along the base is same for both the faces.

So, the area of the face $VDA$ is equal to $75{\text{cm}}^2$ .

So, the lateral area of the pyramid is equal to $117+75+117+75=384{\text{cm}}^2$ .

The formula $\frac{1}{2}pl$ is not used for lateral area as the lateral height of pyramid is not same for all the faces.

Therefore, the lateral area of the pyramid is equal to $384{\text{cm}}^2$ .