will recommend to keep in touch with picture as name of sides are based on it!.
It may not get messy so i will divide whole answer into small parts.
Now Let's move to solution :)
[tex] \\ \\ [/tex]
[tex] \gray{ \bf \dag \frak{Part 1}}[/tex]
In this part we will just focus on Semicircle with diameter 22 ft.
Given :-
[tex] \\ [/tex]
To find:-
[tex] \\ [/tex]
So first of all let's convert diameter into radius.
we know:-
[tex] \boxed{ \rm{}radius = \frac{Diameter}{2} }[/tex]
So:-
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{Diameter}{2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{22}{2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{2 \times 11}{2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{\cancel2 \times 11}{\cancel2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{11}{1} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = 11[/tex]
[tex] \\ \\ [/tex]
Now let's find area of semicircle!
We know:-
[tex] \boxed{ \rm Area~of~semicircle = \frac{\pi {r}^{2} }{2} }[/tex]
[tex] \\ [/tex]
so:-
[tex] \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _1= \dfrac{\pi {r}^{2} }{2} [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _1= \dfrac{22 \times {11}^{2} }{7 \times 2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _1= \dfrac{\cancel{22} \times {11}^{\cancel2} }{7 \times \cancel2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _1= \dfrac{11\times {11}^{\cancel2} }{7} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _1= \dfrac{1331}{7} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\bf Area~of~semicircle _1= 190.14 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \gray{ \bf \dag \frak{Part 2}}[/tex]
So in this part we will find area of other semicircle.
Now as u can see BC isn't given , but AC and AB are given.
So let's find BC:-
- AC = AB + BC
- BC = AC - AB
- BC = 30 - 22
- BC = 8 ft
[tex] \\ [/tex]
Now we got the diameter, so let's change it into radius.
we know:-
[tex] \boxed{ \rm{}radius = \frac{Diameter}{2} }[/tex]
[tex] \\ [/tex]
so :-
[tex] \hookrightarrow\sf{}radius = \dfrac{Diameter}{2} [/tex]
[tex] \\[/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{8}{2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{2 \times 4}{2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = \dfrac{\cancel2 \times 4}{\cancel2} [/tex]
[tex] \\ [/tex]
[tex] \hookrightarrow\sf{}radius = 4[/tex]
[tex] \\ [/tex]
Now let's find area of semicircle!
We know:-
[tex] \boxed{ \rm Area~of~semicircle = \frac{\pi {r}^{2} }{2} }[/tex]
[tex] \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _2= \dfrac{\pi {r}^{2} }{2} [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _2= \dfrac{22 \times {4}^{2} }{7 \times 2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _2= \dfrac{\cancel{22} \times {4}^{\cancel2} }{7 \times \cancel2} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _2= \dfrac{11\times {4}^{2} }{7} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _2= \dfrac{11\times 16}{7} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\sf Area~of~semicircle _2= \dfrac{176}{7} \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \dashrightarrow\bf Area~of~semicircle _2= 25.14 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \gray{ \bf \dag \frak{Part 3}}[/tex]
Now Finally in this part we'll add both the areas of these semicircles.
Let :-
- Area of semicircle with diameter AB be A₁
- Area of semicircle with diameter BC be A₂
[tex] \\ \\ [/tex]
[tex] \sf: \implies Area~of~figure=A_1+A_2 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \sf: \implies Area~of~figure=190.14+25.14 \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \bf: \implies Area~of~figure=215.28 {ft}^{2} \\ [/tex]
[tex] \\ \\ [/tex]
