Question

One side other the hypotenuse of right angle isoceles triangle is 6 cm. The length of the perpendicular on the hypotenuse from the opposite vertex is

• A. 6 cm
• B. $\frac{6}{\sqrt{2}}$ cm
• C. 4 cm
• D. $3\sqrt{2}$

Answer 1

The correct option is B $\frac{6}{\sqrt{2}}$ cm

Let AD be the perpendicular on CB of length p.

Since ABC is an isosceles $\Delta$, there $AC=AB$ & $\angle ACB=\angle ABC={45}^o$

In $\Delta ABC\Rightarrow B{C}^2=6^2+6^2\Rightarrow BC=6\sqrt{2}$

In $\Delta ACD$ & $\Delta ABD$:-

$AD=AD$(common side)

$\angle CAD=\angle OAB={45}^o$

$AC=BC$

By SAS, $\Delta ACD\sim \Delta ABD$ so $BD=CD$

So $BD=CD=\frac{BC}{2}=\frac{6\sqrt{2}}{2}=\frac{6}{\sqrt{2}}$

In $\Delta ABD$

$\Rightarrow A{B}^2=B{D}^2+P^2$

$P^2=A{B}^2-B{D}^2=6^2-\frac{6^2}{2}=\frac{6^2}{2}$

So $P=\frac{6}{\sqrt{2}}$.