91Ó°ÊÓ

We are given a diagram,

Area A₂ consists of two equal sectors with radius 90ft and central angle is given by $\mathrm{δ}$

Now we find $\mathrm{δ}$

We get

role="math" localid="1646790759664" $$\begin{gathered} \mathrm{δ}+40.5^o={90}^o \\ \mathrm{δ}=49.5^o \end{gathered}$$

Now we find the area A₂by doubling the area of one of the sectors

role="math" localid="1646790934563" $$\begin{gathered} A_2=2\left(\frac{\mathrm{δ}}{360}\right)r^2\mathrm{π} \\ {A}_2=2\frac{49.5^o}{{360}^o}{\left({90}^o\right)}^2\mathrm{π} \\ {A}_2=2\left(\frac{11}{80}\right)8100\mathrm{π} \\ {A}_2=6997.9f{t}^2 \end{gathered}$$

we have,

Area A₃is divided into two equal triangles dashed by a line

We find area of one triangle and then double it

First find the measure of angle $\mathrm{Î\pm }$ which completes a straight line with 45^o angle

We have,

$$\begin{gathered} \mathrm{Î\pm }+{45}^o={180}^o \\ \mathrm{Î\pm }={135}^o \end{gathered}$$

Now we find the values of $\mathrm{β},\mathrm{γ}$

$$\begin{gathered} \frac{\sin \mathrm{γ}}{10}=\frac{\sin 135}{90} \\ \sin \mathrm{γ}=\frac{\sin 135}{90}\left(10\right) \\ \sin \mathrm{γ}=0.0785 \\ \mathrm{γ}=4.5^o \end{gathered}$$

Now

$$\begin{gathered} \mathrm{Î\pm }+\mathrm{β}+\mathrm{γ}={180}^o \\ \mathrm{β}={180}^o-{135}^o-4.5^o \\ \mathrm{β}=40.5^o \end{gathered}$$

And finally the area can be given as

$$\begin{gathered} A_3=2\left(\frac{1}{2}\right)\left(10\right)\left(90\right)(\sin 40.5) \\ {A}_3=584.5f{t}^2 \end{gathered}$$

Area A₁Consists of $\frac{3}{4}th$of the circle

Hence

$$\begin{gathered} A_1=\frac{3}{4}\left(\mathrm{Ï€}{r}^2\right) \\ {A}_1=\frac{3}{4}\left(\mathrm{Ï€}{100}^2\right) \\ {A}_1==23561.945f{t}^2 \end{gathered}$$

Now find the total area

$$\begin{gathered} A=A_1+A_2+A_3 \\ A=23561.945+6997.9+584.5 \\ A=31144.345f{t}^2 \end{gathered}$$