Question

In an isosceles right angled triangle one of the congruent sides is of length 10 cm. Find the length of an altitude on the hypotenuse.

Answer 1

Let the given isosceles right angled triangle be $△$ABC.
Here, AB = BC = 10 cm and $\angle$ABC = 90^o
BD is the altitude on the hypotenuse AC.


In right triangle ABC, we have:

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2(\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {\mathrm{AC}}^2={10}^2+{10}^2 \\ \Rightarrow {\mathrm{AC}}^2=200 \\ \Rightarrow \mathrm{AC}=10\sqrt{2}\mathrm{cm} \end{gathered}$$

Length of the hypotenuse AC = $10\sqrt{2}$cm.
Now, height BD will bisect the base AC, because if a point is equidistant from two given points,
then it lies on the perpendicular bisector of the segment joining the points.
AD = DC = $\frac{1}{2}\mathrm{AC}\hspace{0.17em}=\frac{1}{2}\times 10\sqrt{2}=5\sqrt{2}\mathrm{cm}$

Now, in right triangle BDC, we have:

$$\begin{gathered} {\mathrm{BC}}^2={\mathrm{DC}}^2+{\mathrm{BD}}^2(\mathrm{by}\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {10}^2=(5{\sqrt{2)}}^2+{\mathrm{BD}}^2 \\ \Rightarrow {\mathrm{BD}}^2=100-50 \\ \Rightarrow {\mathrm{BD}}^2=50 \\ \Rightarrow \mathrm{BD}=5\sqrt{2}\mathrm{cm} \end{gathered}$$

∴ The length of the altitude on the hypotenuse is $5\sqrt{2}$ cm.