The triangle contains 3 right-angle triangles
Triangle ABC, ACD, CDB
Using Pythagoras theorem
[tex]\text{hyp}^2=opp^2+adj^2[/tex]
In triangle ACD
[tex]\begin{gathered} x^2=8^2+6^2 \\ x^2\text{ = 64 + 36} \\ x^2\text{ = 100} \\ x=\sqrt[]{100\text{ }} \\ x\text{ = 10} \end{gathered}[/tex]
In triangle CDB
[tex]\begin{gathered} y^2=6^2+(b+3)^2 \\ y^2\text{ = 36 + (b+3) (b+3)} \\ y^2=36+b^2\text{ +3b +3b + 9} \\ y^2=36+b^2\text{ +6b +9} \\ y^2=45+b^2\text{ + 6b} \end{gathered}[/tex]
the total lenght AB = 8 + b+3 = 11+b
Therefore, in triangle ABC
[tex]\begin{gathered} (11+b)^2=x^2+y^2\text{ } \\ (11+b)(11+b)=100+45+b^2\text{ +6b} \\ 121+11b+11b+b^2=145+b^2+6b \\ 121+22b+b^2=145+b^2\text{ +6b} \\ \text{collect like terms} \\ 22b-6b+b^2-b^2=145\text{ -121} \end{gathered}[/tex][tex]\begin{gathered} 16b\text{ = 24} \\ \text{divide both side by 16} \\ \frac{16b}{16}=\text{ }\frac{24}{16} \\ b\text{ = 1.5} \end{gathered}[/tex]
The answer is option A
