dHello, reiward!

HallsofIvy gave you an excellent game plan for Problem #3.

Did you follow it?

3. One of the legs of a right triangle has a length of 4cm.

Express the length of the altitude perpendicular to the hypotenuse

as a function of the length of the hypotenuse. Code:

C
o
*| *
* | *
* | *
4 * |y *
* | *
* | *
* | *
A o * * * o * * * * * * * o B
: x D h-x :
: - - - - h - - - - - - :

We have right triangle $\displaystyle ABC,\;C = 90^o,\;AC = 4.$

Let $\displaystyle h = AB$, the hypotenuse.

Let $\displaystyle y = CD$, atitude to hypotenuse.

Let $\displaystyle x = AD,\;\text{then: }h-x \:=\:DB$

We find that: .$\displaystyle \Delta ADC \:\sim \: \Delta CDB \:\sim\:\Delta ACB$

Hence: .$\displaystyle \dfrac{y}{x} \:=\:\dfrac{h-x}{y} \quad\Rightarrow\quad y^2 \:=\:hx-x^2$ .[1]

In $\displaystyle \Delta ADC\!:\;x^2+y^2 \:=\:4^2 \quad\Rightarrow\quad x \:=\:\sqrt{16-x^2}$

Substitute into [1]: .$\displaystyle y^2 \:=\:h\sqrt{16-y^2} - (16-y^2)$

. . $\displaystyle y^2 \:=\:h\sqrt{16-y^2} - 16 + y^2 \quad\Rightarrow\quad h\sqrt{16-y^2} \:=\: 16 $

. . $\displaystyle \sqrt{16-y^2} \:=\:\dfrac{16}{h} \quad\Rightarrow\quad 16 - y^2 \:=\:\dfrac{256}{h^2}$

. . $\displaystyle y^2 \:=\:16 - \dfrac{256}{h^2} \:=\:\dfrac{16h^2 - 256}{h^2} \:=\:\dfrac{16(h^2-16)}{h^2} $

Therefore: .$\displaystyle y \;=\;\dfrac{4\sqrt{h^2-16}}{h}$