Question

In $$\triangle ABC$$, $$\angle ABC=90^0$$, $$BD \perp AC$$. If $$BD=8\ cm,AD=4\ cm.$$ Find $$CD$$ and $$AB.$$

Answer 1

Given, $$BD = 8 cm, AD = 4 cm, ∠ABC = 90^°$$ and $$BD⊥AC.$$

We know that, If a perpendicular is drawn from the vertex containing a right angle of a right triangle to the hypotenuse then the triangle on each side of the perpendicular are similar to each other and to the whole triangle.

So, $$ΔDBA∼ΔDCB $$ (A-A similarity)

$$BD/CD = AD/BD$$

[Since, triangles are similar, hence corresponding sides will be proportional]

$$BD^² = AD \times DC$$

$$(8)^² = 4 \times DC$$

$$64 = 4DC$$

$$DC = 64/4 $$

$$DC = 16 $$

In $$\triangle ADB $$, By Pythagoras theorem,

$$AB^2=BD^2+AD^2$$

$$AB=\sqrt{8^2+4^2}=\sqrt{100}=10 \ cm $$

Hence, the length of $$CD$$ is $$16$$ cm and $$AB$$ is $$10\ cm$$