For the base of the swimming pool:
You have the next figures as the sides:
The side number 1 and side 2 are retangles:
[tex]\begin{gathered} A_{s1}=L\cdot W \\ A_{s2}=L\cdot W \end{gathered}[/tex]
For the sides 3 and 4 you can draw the side as the sum of a recrtangle and a triangle:
Then, the area of this sides is the sum of the area of the rectangle and the area of the triangle:
[tex]\begin{gathered} A_R=L\cdot W \\ A_T=\frac{B\cdot h}{2} \\ \\ A_{s3}=(L\cdot W)+\frac{B\cdot h}{2} \end{gathered}[/tex]
You have the next figure as the base:
The area of a rectangle is:
[tex]A_b=L\cdot W[/tex]
L is length and w is width
To find the lenght you use the trangle formed in the side 3 and 4:
Pythagoras theorem:
[tex]\begin{gathered} x=\sqrt[]{0.7^2+25^2} \\ x=\sqrt[]{0.49+625} \\ x=\sqrt[]{625.49} \\ x=25.01m \end{gathered}[/tex]Then, you have the next areas:Side 1:[tex]A_{s1}=12m\cdot1.8m=21.6m^2[/tex]Side 2:[tex]A_{S2}=12m\cdot1.1m=13.2m^2[/tex]Side 3:[tex]\begin{gathered} A_{S3}=(25m\cdot1.1m)+(\frac{25m\cdot\text{0}.7m}{2}) \\ \\ A_{s3}=27.5m^2+8.75m^2=36.25m^2 \end{gathered}[/tex]Side 4:[tex]\begin{gathered} A_{S4}=(25m\cdot1.1m)+(\frac{25m\cdot\text{0}.7m}{2}) \\ \\ A_{s4}=27.5m^2+8.75m^2=36.25m^2 \end{gathered}[/tex]Base:[tex]A_B=12m\cdot25.01m=300.12m^2[/tex]The total area of the swimming pool is the sum of the areas above, approx. 407 sqare meters
