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
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?

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Answer

6250πm36250\pi m^3

Question I: What volume of water can be stored in the reservoir?

Step 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.

Step 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 \text{ m^3}

Question II: What is the surface area of the wall of the cone?

Step 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.

Step 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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Question I: What volume of water can be stored in the reservoir? Step 1: Formula for the volume of a cone.

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

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

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Question I: What volume of water can be stored in the reservoir? Step 1: Formula for the volume of a cone. V = (1)/(3) r^2 h Top diameter = 50 m, so radius r = 25 m. Depth h = 30 m. Step 2: Substitute the values. V &= (1)/(3) (25)^2 (30) \\ &= (1)/(3) (625) (30) \\ &= (1)/(3) (18750) \\ &= 6250 Volume = 6250 m³. 6250 m^3 Question II: What is the surface area of the wall of the cone? Step 1: Slant height l. l &= sqrt(r^2) + h^2 \\ &= sqrt(25^2) + 30^2 \\ &= sqrt(625 + 900) \\ &= sqrt(1525) \\ &= sqrt(25 × 61) \\ &= 5sqrt(61) Slant height = 5sqrt(61) m. Step 2: Lateral surface area of the cone. A = r l A &= (25) (5sqrt(61)) \\ &= 125 sqrt(61) Surface area = 125 sqrt(61) m². 125(61) m^2