Question

the perimeter of triangle is 50 cm.one side of triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. find the areas of triangle.

Answer 1

Given,

Let smaller side = a           --------------- (i)
one side = a + 4                 --------------- (ii)
third side = 2a - 6              --------------- (iii)

On solving (i),(ii),(iii) we get

= a + a + 4 + 2a - 6 =50

= 4a - 2 = 50

a = 13

Substituting a value in (i),(ii),(iii)

smaller side = 13

one side = 17

third side = 20

Perimeter  = 20 + 17 + 13/2

                   = 50/2

                   = 25

Area = s(s-a)(s-b)(s-c)^1/2

         = (25(25-13)(25-17)(25-20))^1/2

         = 20(30)^1/2

         = 20 root 30


Hope it helps!

Answer 2

Perimeter of triangle = 50 cm
Let the length of the smaller side be x cm. 
Length of the second side = (x + 4) cm
Length of the third side = (2x-6) cm

Sum of lengths of triangle = Perimeter
x + (x + 4) + (2x-6) = 50
x + x + 4 + 2x - 6 = 50
x + x + 2x + 4 - 6 = 50
4x - 2 = 50
4x = 50 + 2
x = 52/4 = 13
x = 13
Length of the first side = x cm = 13 cm
Length of the second side = (x + 4) cm = (13 + 4) cm = 17 cm
Length of the third side = (2x - 6) cm = (2 * 13 - 6) cm = 20 cm

FINDING OUT THE AREA OF THE TRIANGLE USING HERON'S FORMULA
Let a = 13, b = 17, c = 20
$s = \frac{a+b+c}{2}$
$s = \frac{13 + 17 + 20}{2}$
$s = \frac{50}{2}$
$s = 25$

$\sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{25(25-13)(25-17)(25-20)}$
$= \sqrt{25 * 12 * 8 * 5}$
$= \sqrt{12000}$
$= 20\sqrt{30}$
I have just simplified the square root as it is a surd. 
∴ $20 \sqrt{30}$ $cm^{2}$ or 109.54451 $cm^{2}$ is the area of the triangle. 

Hope this may help you.