5.5.3 Exercises

Calculate the volume of a prism of height $h=8\hspace{0.17em}\mathrm{cm}$ with a triangle as its base. Two sides of this triangle are of length $5\hspace{0.17em}\mathrm{cm}$, and one side is of length $6\hspace{0.17em}\mathrm{cm}$.
Answer:
$\hspace{0.17em}\mathrm{cm}{}^3$

The surface area of a cylinder of height $h=6\hspace{0.17em}\mathrm{cm}$ is to be covered with a coloured sheet. The surface area shall be $O=200\hspace{0.17em}\mathrm{cm}{}^2$. Calculate the diameter $d$ of the disk and the volume of the cylinder. Use the approximate value $3.1415$ for $\pi$ and round off your result to the nearest millimetre.
Answers:

1. $d$$\hspace{0.5em}=\hspace{0.5em}$
   $\hspace{0.17em}\mathrm{cm}$
   
2. $V$$\hspace{0.5em}=\hspace{0.5em}$
   $\hspace{0.17em}\mathrm{cm}{}^3$
   

Consider a piece of wood with the shape of a rectangular cuboid with the volume $V$. The height of the cuboid is $h=120\hspace{0.17em}\mathrm{cm}$, and the base face is a square with sides of length $s=40\hspace{0.17em}\mathrm{cm}$. From the piece of wood, a cylindrical hole of height $g$ with a diameter $d=20\hspace{0.17em}\mathrm{cm}$ is drilled "centrically" (i.e. the intersection point of the diagonals of the quadratic base face is the centre of the base disk of the cylinder). Use the approximate value $3.1415$ for $\pi$ and round off your result to integers. Calculate

1. the volume $V_Z$ of the drilled hole:
   $V_Z$$\hspace{0.5em}=\hspace{0.5em}$
   $\hspace{0.17em}\mathrm{cm}{}^3$ Solution
   The volume of the cylinder is
   
   ${\displaystyle V_Z=\pi ·{\left(\frac{d}{2}\right)}^2·h=3.1415·{\left(\frac{20\hspace{0.17em}\mathrm{cm}}{2}\right)}^2·120\hspace{0.17em}\mathrm{cm}=3.1415·12000\hspace{0.17em}\mathrm{cm}{}^3=37698\hspace{0.17em}\mathrm{cm}{}^3}$
   
   
   
2. the percentage of the volume $V_1$ of the new piece of wood remaining after drilling of the volume $V_0$:
   Answer:
   $\%$ Solution
   The volume $V$ of the piece of wood is
   
   ${\displaystyle V=s^2·h={\left(40\hspace{0.17em}\mathrm{cm}\right)}^2·120\hspace{0.17em}\mathrm{cm}=1600·120\hspace{0.17em}\mathrm{cm}{}^3=16·12000\hspace{0.17em}\mathrm{cm}{}^3}$
   The percentage $p_Z$ of the drilled cylinder is
   
   ${\displaystyle p_Z=\frac{V_Z}{V}=\frac{\pi ·12000\hspace{0.17em}\mathrm{cm}{}^3}{16·12000\hspace{0.17em}\mathrm{cm}{}^3}\approx \frac{3.1415}{16}\approx 19\%}$
   and thus, $p=(100-19)\%=81\%$ is the percentage original wooden rectangular cuboid which forms the new piece of wood.