1. ## please solve this question

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.

2. Originally Posted by rickymylv
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 $13^2=5^2+12^2$

See diagram:

The side of 5 is the geometric mean between x and 13.

$\frac{x}{5}=\frac{5}{13}$

$13x=25$

$x=\frac{25}{13}$

The side 12 is the geometric mean between y and 13.

$\frac{y}{12}=\frac{12}{13}$

$13y=144$

$x=\frac{144}{13}$

The length of the perpendicular (h) to the hypotenuse is the geometric mean between x and y.

$\frac{\frac{25}{13}}{h}=\frac{h}{\frac{144}{13}}$

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.

$A=\sqrt{s(s-a)(s-b)(s-c)}$, where a, b, and c are the sides and s is the semi-perimeter. That is,

$s=\frac{a+b+c}{2}$

The perimeter of your triangle = 5 + 12 + 13 = 30.
The semi-perimeter (s) = 15
Substituting into Heron's formula to find Area:

$A=\sqrt{15(15-5)(15-12)(15-13)}=\sqrt{900}=30$

This could just as easily been found using $A=\frac{1}{2}bh$ 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.

$A=\frac{1}{2}bh$

$30=\frac{1}{2}(13)h$

$60=13h$

$h=\frac{60}{13}$

All done!!

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# find length of perpendicular to vertex

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