Let's understand the concept:-
- Here angle B is 90°
- So [tex]\triangle ABC[/tex] and [tex]\triangle ABD [/tex] Are right angled triangle
- So we use Pythagoras thereon for solution
Required Answer:-
perpendicular=p=8cm
Hypontenuse =h =10cm
According to Pythagoras thereon
[tex]{\boxed{\sf b^2=h^2-p^2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf b^2=10^2-p^2[/tex]
[tex]\longrightarrow[/tex][tex]\sf b={\sqrt {10^2-8^2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf b={\sqrt{100-64}}[/tex]
[tex]\longrightarrow[/tex][tex]\bf b={\sqrt {36}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf b=6[/tex]
[tex]\therefore[/tex][tex]\overline{BC}=6cm[/tex]
BD=BC+CD
[tex]\longrightarrow[/tex][tex]BD=9+6[/tex]
[tex]\longrightarrow[/tex][tex]BD=15cm [/tex]
- Now in [tex]\triangle ABD [/tex]
Perpendicular=p=8cm
Base =b=15cm
- We need to find Hypontenuse =AD(x)
According to Pythagoras thereon
[tex]{\boxed {\sf h^2=p^2+b^2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf h^2=8^2+15^2 [/tex]
[tex]\longrightarrow[/tex][tex]\sf h={\sqrt {8^2+15^2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf h={\sqrt {64+225}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf h={\sqrt {289}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf h=17cm [/tex]
[tex]\therefore[/tex][tex]{\underline{\boxed{\bf x=17cm}}}[/tex]
