Question

Solve step by step solution then only i can do it plxx ​

Answer 1

Let's understand the concept:-

Here angle B is 90°

So
`\triangle ABC` and
`\triangle ABD` Are right angled triangle

So we use Pythagoras thereon for solution

Required Answer:-

• First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

• We need to find base=b

According to Pythagoras thereon

`{\boxed{\sf b^2=h^2-p^2}}`

• Substitutethe values

`\longrightarrow`
`\sf b^2=10^2-p^2`

`\longrightarrow`
`\sf b={\sqrt {10^2-8^2}}`

`\longrightarrow`
`\sf b={√(100-64)}`

`\longrightarrow`
`\bf b={\sqrt {36}}`

`\longrightarrow`
`\sf b=6`

`\therefore`
`\overline{BC}=6cm`

• BD=BC+CD

`\longrightarrow`
`BD=9+6`

`\longrightarrow`
`BD=15cm`

• Now in
  `\triangle ABD`

Perpendicular=p=8cm

Base =b=15cm

• We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

`{\boxed {\sf h^2=p^2+b^2}}`

• Substitute the values

`\longrightarrow`
`\sf h^2=8^2+15^2`

`\longrightarrow`
`\sf h={\sqrt {8^2+15^2}}`

`\longrightarrow`
`\sf h={\sqrt {64+225}}`

`\longrightarrow`
`\sf h={\sqrt {289}}`

`\longrightarrow`
`\sf h=17cm`

`\therefore`
`{\underline{\boxed{\bf x=17cm}}}`