Answer:
[tex]x\approx10.8[/tex]
Step-by-step explanation:
[tex]\text{1. In triangles ABC and ABD,}\\\text{i. }\angle\text{B}=\angle\text{B}\quad (A)\ \ \ [\text{Common angle in both triangles.}]\\\text{ii. }\angle\text{ADB}=\angle\text{A}\quad(A)\ \ \ [\text{Both angles are equal to 90}^\circ.]\\\text{iii. }\triangle\text{ABC}\sim\triangle\text{ABD}\ \ \ [\text{By A.A. axiom.}][/tex]
[tex]\text{2. In triangles ACD and ABC,}\\\text{i. }\angle\text{C}=\angle\text{C}\ \ \ [\text{Common angle in both triangles.}]\\\text{ii. }\angle\text{ADC}=\angle\text{A}\ \ \ [\text{Both angles are }90^\circ.]\\\text{iii. }\triangle\text{ACD}\sim\triangle\text{ABC}\ \ \ [\text{By A.A. axiom}][/tex]
[tex]\text{3. }\triangle\text{ABD}\sim\triangle\text{ACD}\ \ \ [\triangle\text{ABC}\sim\triangle\text{ACD}\text{ and }\triangle\text{ABC}\sim\triangle\text{ABD}][/tex]
[tex]\text{4. }\dfrac{\text{AD}}{\text{CD}}=\dfrac{\text{BD}}{\text{AD}}\ \ \ [\text{Corresponding sides of similar triangles are proportional.}]\\\\\text{or, }\dfrac{x}{13}=\dfrac{9}{x}\\\\\text{or, }x^2=13(9)=117\\\\\text{or, }x=3\sqrt{13}\\\\\therefore\ x\approx10.8[/tex]
If there was no compulsion to use the concept of similar triangles, this problem can be solved in another method as follows:
[tex]\text{A. Using Pythagoras theorem in: }\\[/tex]
[tex]\text{i. }\triangle\text{ABD,}\\\text{AB}^2=\text{AD}^2+\text{BD}^2\\\text{or, }y^2=x^2+9^2.......(1)[/tex]
[tex]\text{ii. }\triangle\text{ACD,}\\\text{AC}^2=\text{AD}^2+\text{CD}^2\\\text{or, }z^2=x^2+13^2......(2)[/tex]
[tex]\text{iii. }\triangle\text{ABC,}\\\text{CB}^2=\text{AB}^2+\text{AC}^2\\\text{or, }(9+13)^2=y^2+z^2\\\text{or, }22^2=y^2+z^2\\\text{Using equations(1) and (2),}\\22^2=x^2+9^2+x^2+13^2\\\text{or, }22^2-9^2-13^2=2x^2\\\text{or, }2x^2=234\\\text{or, }x^2=117\\\therefore\ x\approx10.8[/tex]
