A)
For this tringle we can turn the figure like this:
now we have two right triangles and we can calulate the base of the first triangle with the sin law
[tex]\begin{gathered} \frac{\sin (90)}{3}=\frac{sin(a)}{2.5} \\ \sin (a)=\frac{2.5\sin (90)}{3} \\ \sin (a)=0.8 \\ a=\sin ^{-1}(0.8)=53º \end{gathered}[/tex]
the angle b is going to be:
[tex]\begin{gathered} 180=90+53+b \\ b=180-90-53 \\ b=37 \end{gathered}[/tex]
Now the base is going to be:
[tex]\begin{gathered} \frac{\sin(90)}{3}=\frac{\sin(37)}{\text{base}} \\ \text{base}=\frac{3\sin (37)}{\sin (90)}=1.8 \end{gathered}[/tex]
and the base of the secon triangle is going to be:
[tex]\text{base}2=7.2-1.8=5.4[/tex]
And the area of the triangles is going to be:
[tex]A1=\frac{base\times2.5}{2}=\frac{1.8\times2.5}{2}=2.25[/tex][tex]A2=\frac{base2\times2.5}{2}=\frac{5.4\times2.5}{2}=6.75[/tex]
so in total the area is going to be:
[tex]A1+A2=2.25+6.75=9[/tex]
B)
the procedure is similar, first we turn the tiangle like this:
the angle a is going to be:
[tex]\begin{gathered} \frac{\text{sin(a)}}{4.8\text{ }}=\frac{\sin (90)}{6} \\ \sin (a)=\frac{4.8\sin (90)}{6}=0.8 \\ a=\sin ^{-1}(0.8) \\ a=53º \end{gathered}[/tex]
the angle b is going to be:
[tex]\begin{gathered} 180=90+53+b \\ b=180-90-53 \\ b=37º \end{gathered}[/tex]
now the base is going to be:
[tex]\begin{gathered} \frac{\sin (37)}{base}=\frac{sen(90)}{4.8} \\ \text{base}=\frac{4.8\sin (37)}{\sin (90)} \\ \text{base}=2.8 \end{gathered}[/tex]
and the base of the other triangle will be:
[tex]\text{base}2=5-2.8=2.2[/tex]
And the area of the triangles will be:
[tex]\begin{gathered} A1=\frac{base\times4.8}{2}=\frac{2.8\times4.8}{2}=6.72 \\ A2=\frac{base2\times4.8}{2}=\frac{2.2\times4.8}{2}=5.28 \end{gathered}[/tex]
And the total area will be:
[tex]A1+A2=6.72+5.28=12[/tex]
