Question

A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the Dome is equal to 2/3 of the total height of the building. Find the surface area of the building, if it contains 2816/21 cubic meter of air.

Answer 1

hope it helps...........

Answer 2

Answer:

$119.528 m^2$

Step-by-step explanation:

Let the radius of the cylinder be r

Radius of cylinder = radius of hemisphere = r

Let the height of the cylinder = h

Total height of the structure = h + r

We are given that The base diameter of the Dome is equal to 2/3 of the total height of the building.

Let d be the diameter of dome

So, $d = \frac{2}{3}(h+r)$

Diameter = $2 \times radius = 2r$

$2r= \frac{2}{3}(h+r)$

$r= \frac{1}{3}(h+r)$

$3r= h+r$

$2r= h$

Total volume of the air  = volume of cylinder + volume of dome

[tex]\frac{2816}{21} =\pi r^2 h + \frac{2}{3} \pi r^3[/tex]

$\frac{2816}{21} =\frac{22}{7} \times 2r^3+ \frac{2}{3} \times \frac{22}{7} \times r^3$

$\frac{2816}{21} =r^3(\frac{22}{7} \times 2+ \frac{2}{3} \times \frac{22}{7})$

$\frac{\frac{2816}{21}}{\frac{22}{7} \times 2+ \frac{2}{3} \times \frac{22}{7}} =r^3$

$16 =r^3$

$\sqrt[3]{16} =r$

$2.519=r$

$h = 2r = 2 \times 2.519 =5.03$

Surface area of building = Surface area of cylinder +Surface area of hemisphere

Surface area of building =$2\pi rh + 2\pi r^2$

Surface area of building =$2\times \frac{22}{7}\times 2.519 \times 5.03 + 2\times \frac{22}{7}\times 2.519^2$

Surface area of building =$119.528 m^2$

Hence the surface area of the building is $119.528 m^2$