please solve this question using heron's formula...
The lengths of the sides of a triangle are 5cm, 12cm and 13cm.Find the length of perpendicular from the opposite vertex to the side whose length is 13cm.
This is a right triangle since $\displaystyle 13^2=5^2+12^2$
See diagram:
The side of 5 is the geometric mean between x and 13.
$\displaystyle \frac{x}{5}=\frac{5}{13}$
$\displaystyle 13x=25$
$\displaystyle x=\frac{25}{13}$
The side 12 is the geometric mean between y and 13.
$\displaystyle \frac{y}{12}=\frac{12}{13}$
$\displaystyle 13y=144$
$\displaystyle x=\frac{144}{13}$
The length of the perpendicular (h) to the hypotenuse is the geometric mean between x and y.
$\displaystyle \frac{\frac{25}{13}}{h}=\frac{h}{\frac{144}{13}}$
Can you finish up.
Sorry, I misread your instructions. I did not use Heron's formula to solve this.....I'll see if someone else does. Otherwise, I'll be back.
Okay, I'm back. I really don't know how Heron's formula fits into all this. Heron's formula is a formula for finding the area of any triangle knowing only the lengths of the sides.
$\displaystyle A=\sqrt{s(s-a)(s-b)(s-c)}$, where a, b, and c are the sides and s is the semi-perimeter. That is,
$\displaystyle s=\frac{a+b+c}{2}$
The perimeter of your triangle = 5 + 12 + 13 = 30.
The semi-perimeter (s) = 15
Substituting into Heron's formula to find Area:
$\displaystyle A=\sqrt{15(15-5)(15-12)(15-13)}=\sqrt{900}=30$
This could just as easily been found using $\displaystyle A=\frac{1}{2}bh$ using b = 12 and h = 5. But, I digress.
Now, since you just found what the area was, you can use this to solve for your missing perpendicular.
$\displaystyle A=\frac{1}{2}bh$
$\displaystyle 30=\frac{1}{2}(13)h$
$\displaystyle 60=13h$
$\displaystyle h=\frac{60}{13}$
All done!!