Answer:
- Question 1a. i) [tex]x=1.8m[/tex]
- Question 1a. ii) [tex]Volume=27.6m^3[/tex]
- Question 1b) [tex]Volume=65.9m^3[/tex]
Explanation:
Question 1 a. i) Find the value of x.
[tex]tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}[/tex]
For the smalll triangle you can write:
[tex]tan(\theta )=\dfrac{x}{1m}[/tex]
For tthe big triangle:
[tex]tan(\theta )=\dfrac{x+2.7m}{2.5m}[/tex]
Substitute:
[tex]\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}[/tex]
Solve for x:
[tex]2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m[/tex]
Question 1a ii) Find the volume of the frustrum
- Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m
Formula:
[tex]Volume=(1/3)\pi \times radius^2\times height[/tex]
Substitute:
[tex]Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3[/tex]
- Find the volume of a cone with heigth = 1.8m and radius = 1m
[tex]Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3[/tex]
- Subtract the volume of the small cone from the volume of the big cone
[tex]Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3[/tex]
Question 1b. Calculate the volume of the bin
i) Upper frustrum
This is the same frustrum from the equation of above, thus ist volume is 27.6m³.
ii) Lower frustrum
[tex]\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}[/tex]
[tex]2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m[/tex]
[tex]Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)[/tex]
[tex]Volume=38.3m^3[/tex]
iii) Add the volume of the two frustrums
- [tex]Volume=27.6m^3+38.3m^3=65.9m^3[/tex]
