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Your city council has decided to build a cone shaped reservoir. The plan that was given to the building contractor indicated that the top diameter of the reservoir must be 50m. The depth of the reservoir is indicated at 30m. 1. What volume of water can be stored in the reservoir? II. What is the surface area of the wall of the cone?

Mathematics

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Answers

2 parts
i

6250π m36250\pi \text{ m}^3

ii

125π61 m2125\pi\sqrt{61} \text{ m}^2

Step-by-step solution

4 steps

Part i· 2 steps

What volume of water can be stored in the reservoir?

  1. Formula for the volume of a cone.

    V=13πr2hV = \frac{1}{3} \pi r^{2} h

    Top diameter = 50 m, so radius r=25r = 25 m.
    Depth h=30h = 30 m.

  2. Substitute the values.

    V=13π(25)2(30)=13π(625)(30)=13π(18750)=6250π\begin{align*} V &= \frac{1}{3} \pi (25)^{2} (30) \\ &= \frac{1}{3} \pi (625) (30) \\ &= \frac{1}{3} \pi (18750) \\ &= 6250 \pi \end{align*}

    Volume = 6250π6250 \pi m³.

    6250\pi m^3

Part ii

125π61 m2125\pi\sqrt{61} \text{ m}^2

What is the surface area of the wall of the cone?

  1. Slant height ll.

    l=r2+h2=252+302=625+900=1525=25×61=561\begin{align*} l &= \sqrt{r^{2} + h^{2}} \\ &= \sqrt{25^{2} + 30^{2}} \\ &= \sqrt{625 + 900} \\ &= \sqrt{1525} \\ &= \sqrt{25 \times 61} \\ &= 5\sqrt{61} \end{align*}

    Slant height = 5615\sqrt{61} m.

  2. Lateral surface area of the cone.

    A=πrlA = \pi r l A=π(25)(561)=125π61\begin{align*} A &= \pi (25) (5\sqrt{61}) \\ &= 125 \pi \sqrt{61} \end{align*}

    Surface area = 125π61125 \pi \sqrt{61} m².

    125\pi\sqrt{61} m^2

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