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

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

$AD^{2} + CD^{2} = AC^{2}$

$AD^{2}+(\sqrt{2})^2= AC^{2}$

$AD^{2}+2= AC^{2}$

$AD^{2}=AC^{2} -2$ ...(ii)

Plug value (ii) into (i)

$AC^{2} + AC^{2}-2 = 9$

$2AC^{2} -2=9$

$2AC^{2} =11$

$AC^{2} =\frac{11}{2}$

$AC=\sqrt{\frac{11}{2} }$

Hence final answer is $AC=\sqrt{\frac{11}{2} }$

Answer 2

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

$2AC^{2} -2=9$

$2AC^{2} =11$

$AC^{2} =\frac{11}{2}$

$AC=\sqrt{\frac{11}{2} } cm$