Question

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD=√2 cm.

Answer 1

In Δ ADC,

`AD^(2) + CD^(2) = AC^(2)`

`AD^(2)+2= AC^(2)`

`AD^(2)=AC^(2) -2` (1)

In Δ ABC,

`AC^(2) + BC^(2) = AB^(2)`

`AC^(2) + AD^(2) = AB^(2)` (Given BC = AD)

`AC^(2) + AC^(2)-2 = 9` from (1) and given that AB = 3 cm

`2AC^(2) -2=9`

`2AC^(2) =11`

`AC^(2) =(11)/(2)`

`AC=\sqrt{(11)/(2) } cm`

Answer 2

Given that triangle ABC is right angle triangle. CD is altitude such that AD=BC

ABC is a right angle triangle so apply Pythogorean theorem

`AC^(2) + BC^(2) = AB^(2)`

`AC^(2) + AD^(2) = AB^(2)` (Given that AD = BC)

`AC^(2) + AD^(2) = 3^(2)` (Given that AB=3)

`AC^(2) + AD^(2) = 9` ...(i)

ADC is a right angle triangle so apply Pythogorean theorem

`AD^(2) + CD^(2) = AC^(2)`

`AD^(2)+(√(2))^2= AC^(2)`

`AD^(2)+2= AC^(2)`

`AD^(2)=AC^(2) -2` ...(ii)

Plug value (ii) into (i)

`AC^(2) + AC^(2)-2 = 9`

`2AC^(2) -2=9`

`2AC^(2) =11`

`AC^(2) =(11)/(2)`

`AC=\sqrt{(11)/(2) }`

Hence final answer is
`AC=\sqrt{(11)/(2) }`