Question

A greenhouse is built against a garden wall, as shown in the image attached. The greenhouse is a prism of length 6 m.The curve AD is an arc of a circle with centre B and radius 2 m. The sector angle ABD is 120° and BC has length 1 m.Calculate the volume of the greenhouse in m³. Simplify your answer.

Answer 1

Given:

The length of the greenhouse is 6m.

The sectoral angle of ABD is 120 degrees.

The radius of the sector is r=2m.

The length of the line segment BC is b=1m

To find the volume of the greenhouse:

The formula of the volume of the greenhouse is given below,

$\text{Volume=Area of the cross section}\times length$

Let us first find the area of the cross-section.

$\text{Area of the cross section=}Area\text{ of the sector}+\text{Area of the triangle}$

Using the formula of area of the sector and area of the triangle,

$\begin{gathered} \text{Area of the }\sec tor=\frac{\theta}{360}\times\pi r^2 \\ =\frac{120}{360}\times\frac{22}{7}\times2^2 \\ =\frac{1}{3}\times\frac{22}{7}\times4 \\ =\frac{88}{21}m^2\ldots\ldots\ldots\ldots.(1) \end{gathered}$

From the triangle BCD, using Pythagoras theorem to find the height CD,

$\begin{gathered} BD^2=BC^2+CD^2 \\ 2^2=1^2+CD^2 \\ 4-1=CD^2 \\ CD^2=3 \\ CD=\sqrt[]{3}m \end{gathered}$

So, the area of the triangle is,

$\begin{gathered} A=\frac{1}{2}\times b\times h \\ =\frac{1}{2}\times1\times\sqrt[]{3} \\ =\frac{\sqrt[]{3}}{2}m^2\ldots\ldots\ldots\ldots\text{.}(2) \end{gathered}$

Adding (1) and (2) we get,

$\begin{gathered} \text{Area of the cross section =}\frac{88}{21}+\frac{\sqrt[]{3}}{2} \\ =\frac{176+21\sqrt[]{3}}{42}m^2 \end{gathered}$

Using this value in the volume formula we get,

$\begin{gathered} V=\frac{(176+21\sqrt[]{3})}{42}\times6 \\ =\frac{176+21\sqrt[]{3}}{7}m^3 \end{gathered}$

Hence, the volume of green house is,

$\begin{gathered} \frac{176+21\sqrt[]{3}}{7}m^3 \\ (or) \\ 30.34m^3 \end{gathered}$