What is the area of a figure with the measurements of 3,6, and 5?

Answer:
I think it is 181m.
Step-by-step explanation:
[tex]by \: imagination \: you \: separate \: all \: the \: shapes[/tex]
you also name it .
I name the triangle area A
I name the rectangle area B
it is a triangle because it has different measurements on its sides.
I name the semi- circle area C.
[tex]area \: of \: a \: triangle \: = \\ \frac{1}{2} \times b \times h \\ \frac{1}{2} \times 5 \times 3 \\ \frac{1}{ 2} \times 15 \\ 7.5[/tex]
[tex]aea \: of \: a \: rectangle \: = l \times b \\ 5 \times 6 = \\ 30[/tex]
[tex]area \: of \: a \: semi \: circle \: = \frac{1}{2 } \times \pi \times {r}^{2} \\ \frac{1}{2} \times 3.14 \times 3 \times 3 \\ \frac{1}{2} \times 3.14 \times 9 \\ \frac{1}{2} \times \frac{314}{10} \times 9 \\ [/tex]
[tex]you \: then \: cancle \: 134 \: by \: 2[/tex]
[tex] = \frac{154}{10} \times 9[/tex]
[tex] = \frac{1431}{10} \\ 143.1[/tex]
[tex]you \: now \: add \: all \: the \: ares \: to \: get \: your \: full \: area \\ 7.5 + 30 + 143.1 \\ = 180.6 \\ to \: the \: nearest \: hundred \: will \: be \: 181m[/tex]