Chapter 12, Problem 5RP

a.

To determine

To find: The height of a box of volume 100, length 15 and width $1\frac{1}{3}.$

Answer to Problem 5RP

The height of the box formed is 5.

Explanation of Solution

Given Information:

Length of a cuboid $\left(l\right)=15$

Width of a cuboid $\left(b\right)=1\frac{1}{3}$

Volume of a cuboid $\left(V\right)=100$

Formula used:

Volume of a cuboid $\left(V\right)=lbh$

  $l,b\text{and}h$ are the length, breadth and height of a cuboid respectively

Calculation:

We can find the height of the box using the formula for volume of a cuboid.

Now we know that, Volume of a cuboid $\left(V\right)=lbh$

As $V=100,l=15,b=1\frac{1}{3}=\frac{4}{3}$

  $\begin{array}{l}\therefore 100=15\times \frac{4}{3}\times h\\ \therefore h=\frac{100\times 3}{15\times 4}\\ \therefore h=\frac{100}{20}\\ \therefore h=5\end{array}$

Hence, the height of the box formed is 5.

b.

To determine

To find: The height of a cube with volume 216.

Answer to Problem 5RP

The height of the cube is 6.

Explanation of Solution

Given Information:

Volume of a cube $\left(V\right)=216$

Formula used:

Volume of a cube $\left(V\right)=a^3$

  $a$ is the edge of a cube

Calculation:

In a cube, all the edges are equal.

So, we can find the height of a cube by finding the value of an edge.

Now we know that, Volume of a cube $\left(V\right)=a^3$

As $V=216$

  $\begin{array}{l}\therefore 216=a^3\\ \therefore a=6\end{array}$

Therefore, measure of each edge of a cube is 6.

Hence, the height of the cube is 6.