surface area

Geometry Level 3

The dimensions of a large water tank is shown above. The upper and middle parts are in the shape of a frustum of a right circular cone while the lower part is in the shape of a right circular cylinder. Find the surface area of the whole water tank.

Note:

$m$ means meter.
The water tank is closed on top.

4\pi \sqrt{13}+7\pi \sqrt{11}+\dfrac{132\pi}{2}+45\pi

4\pi \sqrt{10}+7\pi \sqrt{13}+\dfrac{45\pi}{2}+45\pi

4\pi \sqrt{10}+7\pi \sqrt{13}+\dfrac{132\pi}{2}

11\pi (\sqrt{10}+\sqrt{13})+\dfrac{45\pi}{2}

1 solution

A Former Brilliant Member
Oct 10, 2017

Consider the diagram above. For the frustum of a right circular cone, we can use the formula $s=\dfrac{(c+C)L}{2}$ where $c$ and $C$ are the circumferences of the two bases and $L$ is the slant height which can be computed by pythagorean theorem. For the right circular cylinder, we can use the formula $s=ch$ where $c$ is the circumference of the base and $h$ is the height. Threfore,

$S=S_1+S_2+S_3+S_4+S_5$

$S=\dfrac{(3\pi +5\pi)\sqrt{10}}{2}+\dfrac{(5\pi +9\pi)\sqrt{13}}{2}+9\pi(5)+\pi (1.5^2)+\pi (4.5^2)=$ $\boxed{4\pi \sqrt{10}+7\pi \sqrt{13}+45\pi+\dfrac{45\pi}{2}}$

surface area

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