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}* 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=(\theta)/(360)*\pi r^2 \\ =(120)/(360)*(22)/(7)*2^2 \\ =(1)/(3)*(22)/(7)*4 \\ =(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=(1)/(2)* b* h \\ =(1)/(2)*1*\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 =}(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}*6 \\ =\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}`