Permalink Submitted by fitzmerl duron on February 3, 2018 - 11:06am.

To answer the question above, kailangan muna natin ng figure....It looks like this:

Assuming that your reservoir is a closed tank, your reservoir looks ugly because it will be lined/covered with tiles...hehehehe...

Your reservoir is shaped in a frustrum of a cone and it is lined with tiles, we need to get the surface area of the frustrum of a cone.The surface area of the frustrum of a cone is...

where $r_1$ is radius of upper base of the frustrum of a cone, $r_2$ is radius of lower base of the frustrum of a cone, and $h$ is the height of the frustrum of a cone.

With that in mind, the surface area of the frustrum of a cone would be...

## To answer the question above,

To answer the question above, kailangan muna natin ng figure....It looks like this:

Assuming that your reservoir is a closed tank, your reservoir looks ugly because it will be lined/covered with tiles...hehehehe...

Your reservoir is shaped in a frustrum of a cone and it is lined with tiles, we need to get the surface area of the frustrum of a cone.The surface area of the frustrum of a cone is...

$$Surface \space area = \pi (r_1^2 + r_2^2 + r_1r_2\sqrt{(r_1^2 - r_2^2)^2 + h^2})$$

where $r_1$ is radius of upper base of the frustrum of a cone, $r_2$ is radius of lower base of the frustrum of a cone, and $h$ is the height of the frustrum of a cone.

With that in mind, the surface area of the frustrum of a cone would be...

$$Surface \space area = \pi (r_1^2 + r_2^2 + r_1r_2\sqrt{(r_1^2 - r_2^2)^2 + h^2})$$ $$Surface \space area = \pi ((17.5 \space ft)^2 + (34 \space ft)^2 + (17.5 \space ft)(34 \space ft)\sqrt{((17.5 \space ft)^2 - (34 \space ft)^2)^2 + (18 \space feet)^2})$$ $$Surface \space area = 50237.51 ft^2$$

The cost of lining the reservoir with tiles then is $$50237.51 ft^2 \left( \right) = \color{green}{\$ 50237.51}$$

and that is one expensive reservoir....

Alternate solutions are highly encouraged...

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