Question

the lengths of the side of the triangle are 7cm ,13cm and 12cm find the lengths of perpendicular from the
opposite vertex to the side whose lengths is 12cm ​

Answer 1

Area=square root of (S)(s-a)(s-b)(s-c)
Here s=(7+13+12)/2=32/2=16
So area = sq root 16*9*3*4 =24/3
1/2*12h=24/3
So h=6*3^1/2

Answer 2

Given:

Length of sides of triangle are:

7cm

13cm

12cm

━━━━━━━━━━━━━━━

Need to find:

The length of perpendicular from the opposite vertex to the side 12cm

━━━━━━━━━━━━━━━

Solution:

${ s = \dfrac{Perimeter}{2}}$

$\implies{ s = \dfrac{ a + b + c}{2}}$

$\implies{ s = \dfrac{ 7 + 13 + 12}{2}}$

$\implies{ s = \dfrac{ 32}{2} = 16cm}$

=

━━━━━━━━━━━━━━━

We know

$\bold{\red{Area\:of\:triangle = \sqrt{ (s) ( s - a ) ( s - b ) ( s - c)}}}$

$\implies{\sqrt{ (16) ( 16 - 7) ( 16 - 13) ( 16 - 12 )}}</p><p>$

$=\sqrt{ (16) (9) (3) (4)}= </p><p></p><p> </p><p>$

$</p><p>=\sqrt{1728}$

$= 41.569cm^2</p><p>$

━━━━━━━━━━━━━━━

${ Area\:of\:triangle = \dfrac{1}{2}\times{Base}\times{height}}$

$\implies{ 41.569 = \dfrac{1}{2}\times{12}\times{height}}$

$\implies{ 41.569 = 6h}$

\implies{\red{\large{\bold{\boxed{h = 6.9282cm}}}}}⟹

h=6.9282cm