Triangle Height Problem: Express AD When Area = 4X+16 cm²

The area of triangle ABC is:

$\frac{AB\times AC}{2}=S$

Into this formula, we insert the given data:

$\frac{AB\times(x+4)}{2}=4x+16$

$\frac{AB\times(x+4)}{2}=4(x+4)$

Notice that X plus 4 on both sides is reduced, and we are left with the equation:

$\frac{AB}{2}=4$

We then multiply by 2 and obtain the following:

$AB=4\times2=8$

If we now observe the triangle ABC we are able to find side BC using the Pythagorean Theorem:

$AB^2+AC^2=BC^2$

We first insert the existing data into the formula:

$8^2+(x+4)^2=BC^2$

We extract the root:

$BC=\sqrt{64+x^2+2\times4\times x+4^2}=\sqrt{x^2+8x+64+8}=\sqrt{x^2+8x+72}$

We can now calculate AD by using the formula to calculate the area of triangle ABC:

$S_{\text{ABC}}=\frac{AD\times BC}{2}$

We then insert the data:

$4x+16=\frac{AD\times\sqrt{x^2+8x+80}}{2}$

$AD=\frac{(4x+16)\times2}{\sqrt{x^2+8x+80}}=\frac{8x+32}{\sqrt{x^2+8x+80}}$