find the value of x of this question ​

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]

Perpendicular=p=8cm

Base =b=15cm

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]

04.04 match the term with the definition

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I'll mark brainleiest whoever answers.​

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7th grade math need help please will give brainliest. ​

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