# In each of the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle.

Question:

In each of the figures given below, an altitude is drawn to the hypotenuse by a right-angled triangle. The length of different line-segment are marked in each figure. Determine xyz in each case.

Solution:

(i) $\triangle A B C$ is right angled triangle right angled at $B$

$A B^{2}+B C^{2}=A C^{2}$

$x^{2}+z^{2}=(4+5)^{2}$

$x^{2}+z^{2}=9^{2}$

$x^{2}+z^{2}=81$......$(i)$

$\triangle B D A$ is right triangle right angled at $\mathrm{D}$

$B D^{2}+A D^{2}=A B^{2}$

$y^{2}+4^{2}=x^{2}$

$y^{2}+16=x^{2}$

$16=x^{2}-y^{2}$...(ii)

$\triangle B D C$ is right triangle right angled at $D$

$B D^{2}+D C^{2}=B C^{2}$

$y^{2}+25=z^{2}$

$25=z^{2}-y^{2}$$\ldots \ldots( iii ) By canceling equation(i) and(ii) by elimination method, we get y canceling and by elimination method we get z^{2}=45 z=\sqrt{45} z=\sqrt{3 \times 3 \times 5} z=3 \sqrt{5} Now, substituting z^{2}=45 in equation (iv) we get y^{2}+z^{2}=65 y^{2}+45=65 y^{2}=65-45 y^{2}=20 y=\sqrt{20} y=\sqrt{2 \times 2 \times 5} y=2 \sqrt{5} Now, substituting y^{2}=20 in equation (ii) we get x^{2}-y^{2}=16 x^{2}-20=16 x^{2}=16+20 x^{2}=36 x=\sqrt{36} x=\sqrt{6 \times 6} x=6 Hence the values of x, y, z is 6,2 \sqrt{5}, 3 \sqrt{5} (ii) \triangle P Q R is a right triangle, right angled at Q 6+z^{2}=(4+x)^{2} 36+z^{2}=16+x^{2}+8 x z^{2}-x^{2}-8 x=16-36 z^{2}-x^{2}-8 x=-20 \ldots \ldots (i) \Delta Q S P is a right triangle right angled at S Q S^{2}+P S^{2}=P Q^{2} y^{2}+4^{2}=6^{2} y^{2}+16=36 y^{2}=36-16 y^{2}=20 y=\sqrt{20} y=\sqrt{2 \times 2 \times 5} y=2 \sqrt{5} \triangle Q S R is a right triangle right angled at S Q S^{2}+R S^{2}=Q R^{2} y^{2}+x^{2}=z^{2}.....(2) Now substituting y^{2}+x^{2}=z^{2} in equation (i) we get y^{2}+x^{2}-x^{2}-8 x=-20 y^{2}+x^{2}-x^{2}-8 x=-20 y^{2}-8 x=-20$$\ldots \ldots(i i i)$

Now substituting $y^{2}=20$ in equation (iii) we get

$y^{2}-8 x=-20$

$20-8 x=-20$

$-8 x=-20-20$

$-8 x=-40$

$x=\frac{40}{8}$

$x=5$

Now substituting $x=5$ and $y^{2}=20$ in equation (ii) we get

$y^{2}+x^{2}=z^{2}$

$20+5^{2}=z^{2}$

$20+25=z^{2}$

$45=z^{2}$

$\sqrt{3 \times 3 \times 5}=z^{2}$

$3 \sqrt{5}=z$

Hence the value of $x, y$ and $z$ are $5,2 \sqrt{5}, 3 \sqrt{5}$