Question

Please answer me fast​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer:

Given:

• $AB \perp BD$
• $AC = BD$
• $AB = 4 \ cm$
• $BC = 2 \ cm$

Solution:

$In \ \triangle ABC, \ using \ pythagoras \ theorem$

$\boxed{AB^2 + BC^2 = AC^2}$

$\implies 4^2 + 2^2 = AC^2$

$\implies AC^2 = 16 + 4$

$\implies AC ^2 = 20$

$\implies AC = 2\sqrt5$

$In \ \triangle ABD , \ using \ pythagoras \ theorem$

$\boxed{AB^2 + BD^2 = AD^2}$

$\implies 4^2 + AC^2 = AD^2 \ \ \ (\because BD = AC)$

$\implies 4^2 + (2\sqrt5)^2 = AD^2$

$\implies AD^2 = 16 + 20$

$\implies AD^2 = 36$

$\implies AD = \sqrt{36}$

$\boxed{\implies{\boxed{AD = 6 \ cm}}}$