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Answer:
5
Step-by-step explanation:
Refer to attachment for marking of sides.
In the given figure , ∆ABC , ∆ABD and ∆ADC are right angled triangles . Therefore here we can use the Pythagoras theorem , as ,
[tex]\longrightarrow[/tex] base² + perpendicular² = hypotenuse ² .
• In ∆ABD , we have ;
[tex]\longrightarrow[/tex] AB² + BD² = AD²
[tex]\longrightarrow[/tex] AB² + x² = 10²
[tex]\longrightarrow[/tex] AB² = 10² - x²
[tex]\longrightarrow[/tex] AB² = 100 - x²
[tex]\rule{200}2[/tex]
• Again in ∆ADC , we have ;
[tex]\longrightarrow[/tex] AC² + AD² = CD²
[tex]\longrightarrow[/tex] AC² = 20² - 10²
[tex]\longrightarrow[/tex] AC² = 400 - 100
[tex]\longrightarrow[/tex] AC² = 300
[tex]\rule{200}2[/tex]
Again in ∆ABC , we have ,
[tex]\longrightarrow[/tex] AC² = AB² + BC²
Substituting the values from above ,
[tex]\longrightarrow[/tex] 300 = 100-x² + (20-x)²
[tex]\longrightarrow[/tex] 300 = 100 - x² + 400 + x² - 40x
[tex]\longrightarrow[/tex] 40x = 500 - 300
[tex]\longrightarrow[/tex] 40x = 200
[tex]\longrightarrow[/tex] x = 200/40
[tex]\longrightarrow[/tex] x = 5