Question

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold Pens. The dimensions of the cuboid are 15cm by 10cm by 3.5 cm . The radius of each of the depression is 0.5cm and the depth is 1⋅4cm .Find the volume of the wood in the entire stand. $$$$

Answer 1

Hint: We find the volume of the stand as $V=V_1-4V_2$ where $V_1$ is the volume of the cuboid without depressions and $V_2$ is the volume of the conical depressions. We find the volume of cuboid using the formula $V_1=l\times b\times h$ where $l,b,h$ are the length, breadth and height of the cuboid. We find $V_2={\displaystyle \frac{1}{3}}\pi r^{2}h$ where $r$ the radius of depression and $h$ is the depth. $$$$

Complete step by step answer:
We know that a cuboid is a three-dimensional object with six rectangular faces joined by 8 vertices. It has three different types of sides called length, breadth and height denoted $l,$ $b$ and $h$ respectively. So the volume of the cuboid is given by;

$$V_{\text{cuboid}}=l\times b\times h$$

We are given the question that the wooden pen stand is in the shape of a cuboid with The dimensions of the cuboid are 15cm by 10cm by 3.5 cm. So let us assign $l=15,b=10,h=3.5$ . $$$$

So the volume of the cuboid is
$$V_1=15\times 10\times 3.5=525\text{ c}{\text{m}}^3$$
We know from the right circular cone that the line segment joining the apex to the center is the height of the cone denoted as $h$ . the volume of a cone with radius at the base $r$ and height $h$ is given by
$$V_2={\displaystyle \frac{1}{3}}\pi r^{2}h$$
We see that there are 4 conical depressions on the wooden pen stand. We are given that the radius of each of the depressions is 0.5cm and the depth is 1⋅4cm. Here the radius of the depression is the radius of the base that is $r=0.5$ cm and the height of the cone is the depth of the depression that is $h=1.4$ cm. $$$$
 So volume of one conical depression is ;
$$\begin{array}{rl}& V_2={\displaystyle \frac{1}{3}}\pi r^{2}h\\ & \Rightarrow V_2={\displaystyle \frac{1}{3}}\times 3.14\times {\left(0.5\right)}^2\times 1.4\\ & \Rightarrow V_2=0.3663\text{ c}{\text{m}}^3\end{array}$$
So the volume of the wooden pen stand volume of the cuboid minus the volumes of 4 conical depressions that is
$$V=V_1-4V_2=525-1.4665=523.535\text{ c}{\text{m}}^3$$
Note:
 We have taken the value of $\pi$ here as 3.14 , we can also take $\pi ={\displaystyle \frac{22}{7}}$ . If we want to approximate up to two decimals we have to take $\pi =3.141$ . We can also find the surface area of the top surface as $l\times b-4\times \pi r^2$ . If we are not given the height of the cone but slant height $l$ is given , we can find the volume as $V={\displaystyle \frac{1}{3}}\pi r^2\sqrt{r^2+l^2}$ .We should take care of the units during calculation.