Question

In each of the figure given bellow, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segment is marked in each figure. Determine x, y, z in each case.

Answer 1

In a right triangle, the altitude that’s perpendicular to the hypotenuse has a special property: it creates two smaller right triangles that are both similar to the original right triangle.

$\left(i\right)$ In $△ABC$, $BD$ is altitude.

If an altitude is drawn to the hypotenuse of a right triangle as shown in the above figure, then

$(AD{)}^2=AD\times AC$

$\Rightarrow$ $(x{)}^2=4\times 9$

$\Rightarrow$ $x^2=36$

$\therefore$ $x=6$

Now, $(BC{)}^2=DC\times AC$

$\Rightarrow$ $z^2=5\times 9$

$\Rightarrow$ $z^2=45$

$\therefore$ $z=3\sqrt{5}$

Next, $(BD{)}^2=AD\times DC$

$\Rightarrow$ $y^2=4\times 5$

$\Rightarrow$ $y^2=20$

$\therefore$ $y=2\sqrt{5}$

$\left(ii\right)$ In $△PQR$, $QS$ is an altitude.

If an altitude is drawn to the hypotenuse of a right triangle as shown in the above figure, then

$(PQ{)}^2=PS\times PR$

$(6{)}^2=4\times (4+x)$

$36=16+4x$

$4x=20$

$\therefore$ $x=5$

$\therefore$ $SR=5$

Now, $Q{R}^2=SR\times PR$

$Z^2=5\times (4+5)$

$z^2=5\times 9$

$z^2=45$

$\therefore$ $z=3\sqrt{5}$

Next, $Q{S}^2=PS\times SR$

$y^2=4\times 5$

$y^2=20$

$\therefore$ $y=2\sqrt{5}$