Question

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

Answer 1

Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

                                     AD+DB=3

                                       x+DB = 3

                                        DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

$BC^{2} = DB^{2} +CD^{2}$

$BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}$

$BC^{2} =(3-x)^{2} +2$

Since given AD=BC,let us plugin BC=x in above step.

$x^{2} =(3-x)^{2} +2$

$x^{2} =9-6x+x^{2} +2$

6x=11

x=$\frac{11}{6}$

Now we know AD=x=$\frac{11}{6}$ and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

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

$AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}$

$AC^{2} =\frac{193}{36}$

$AC=\sqrt{\frac{193}{36} }$ = 2.315cm