Question

Find the value of x of this question ​

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={\sqrt{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}}}$