Answer
The surface area of the figure is\[84\text{ c}{{\text{m}}^{\text{2}}}\].
Work Step by Step
Let the base and perpendicular of the right angle triangle be \[a\]and \[b\]respectively. Let the unknown side be\[c\]. Now, in order to find the surface area of the figure.Firstly, compute the length of the unknown side using Pythagoras theorem as follows:
\[\begin{align}
& \text{Hypotenus}{{\text{e}}^{2}}=\text{Perpendicula}{{\text{r}}^{2}}+\text{Bas}{{\text{e}}^{2}} \\
& {{c}^{2}}={{a}^{2}}+{{b}^{2}} \\
& =\left( {{4}^{2}}\text{ }+{{3}^{2}}\text{ } \right)\text{c}{{\text{m}}^{2}} \\
& =\left( 16\text{ }+9 \right)\text{ c}{{\text{m}}^{2}}
\end{align}\]
Simplifying further
\[\begin{align}
& c=\sqrt{16\text{ }+9\text{ }}\text{cm} \\
& =\sqrt{25\text{ }}\text{cm} \\
& =5\text{ cm}
\end{align}\]
Now, there are three rectangular face shaving different dimensions and two triangular faces having same dimensions in the given figure. In order to compute the total surface area of the figure, it is required to compute the area of each of the rectangular and triangular face.
One of the rectangular face is having dimensions that is length and width as \[5\text{ cm}\]and \[6\text{ }\operatorname{cm}\]respectively.
Compute the area of the rectangular face using the equation as shown below:
\[\begin{align}
& \text{Area of the rectangular face }({{A}_{1}})=l\times b \\
& =5\text{ cm}\times 6\text{ cm} \\
& =30\text{ c}{{\text{m}}^{2}}
\end{align}\]
Now, other rectangular face is having dimensions that is length and width as \[6\text{ cm}\] and \[4\text{ }\operatorname{cm}\]respectively.
Compute the area of the rectangular face using the equation as shown below:
\[\begin{align}
& \text{Area of the rectangular face}\left( {{A}_{2}} \right)=l\times w \\
& =6\text{ cm}\times 4\text{ cm} \\
& =24\text{ c}{{\text{m}}^{2}}
\end{align}\]
Now, other rectangular face is having dimensions that is length and width as \[6\text{ cm}\] and \[3\text{ }\operatorname{cm}\]respectively.
Compute the area of the rectangular face using the equation as shown below:
\[\begin{align}
& \text{Area of the rectangular face }({{A}_{3}})=l\times w \\
& =6\text{ cm}\times 3\text{ cm} \\
& =18\text{ c}{{\text{m}}^{2}}
\end{align}\]
Now, there are two triangular faces each of which is having dimension that is base and height as \[3\text{ cm}\] and \[4\text{ }\operatorname{cm}\]respectively.
\[\begin{align}
& \text{Area of Triangular face }({{A}_{4}})=\frac{1}{2}\times b\times h \\
& =\frac{1}{2}\times 3\text{ cm}\times 4\text{ cm} \\
& =6\text{ c}{{\text{m}}^{2}}
\end{align}\]
Now, compute the total surface area of the figure using the equation as shown below:
\[\begin{align}
& \text{Total surface area of the figure}={{A}_{^{1}}}+{{A}_{^{2}}}+{{A}_{^{3}}}+2\times \left( {{A}_{^{4}}} \right) \\
& =\left( 30+24+18+\left( 2\times 6 \right) \right)\text{c}{{\text{m}}^{2}} \\
& =84\text{ c}{{\text{m}}^{2}}
\end{align}\]
