Question

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Answer 1

Answer:

• Question 1a. i) $x=1.8m$

• Question 1a. ii)   $Volume=27.6m^3$

• Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

         $tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

        $tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

      $tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

        $\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

        $2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

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:

         $Volume=(1/3)\pi \times radius^2\times height$

Substitute:

         $Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

• Find the volume of a cone with heigth = 1.8m and radius = 1m

        $Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

• Subtract the volume of the small cone from the volume of the big cone

        $Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

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

            $\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

           $2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

        $Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

       $Volume=38.3m^3$

iii) Add the volume of the two frustrums

• $Volume=27.6m^3+38.3m^3=65.9m^3$