A triangle has sides 5cm, 12cm, 13cm. The length of the perpendicular from the opposite vertex to the side whose length is 13cm is?

Jul 21, 2018

Below.

Explanation:

It is clear that the triangle is a right angled triangle.

Let AB=12 cm BC=5 cm and AC=13 cm

Now a perpendicular is drawn from A to AC.

ar(ABC) = ½ AB · BC = ½ CA · BD.

 BD = ( AB · BC ) / (CA

Putting the respective values of the sides we get,

$B D = \frac{60}{13}$ cm

You can also apply the heron's formula to get the answer to the question.

$\frac{60}{13}$

Explanation:

The given side of triangle are $5 , 12$ & $13$.

The semi-perimeter $s$ of given triangle with sides $a = 5 , b = 12 , c = 13$ is given as

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

$= \setminus \frac{5 + 12 + 13}{2}$

$= 15$

Now, using Heron's formula, the area $\setminus \Delta$ of given triangle is given as

$\setminus \Delta = \setminus \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

$= \setminus \sqrt{15 \left(15 - 5\right) \left(15 - 12\right) \left(15 - 13\right)}$

$= 30$

If $h$ is the length of perpendicular to the side $c = 13$ drawn from opposite vertex then the area of given triangle

$\setminus \Delta = \frac{1}{2} \left(13\right) \left(h\right)$

$30 = \frac{13}{2} h$

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