Question

the sides of a trianglenare x/y+ y/z and y/z+z/x and z/x+x/y then area of a triangle​

Answer 1

Answer:

Step-by-step explanation:

The given sides of a triangle are;

a = y/z+z/x

b = z/x+x/y

c = x/y+y/z

We can find the area of the triangle by using Heron's formula;

Area = Sqr (s x (s-a) x (s-b) x (s-c))

where a, b, c are the sides of the triangle and s = (a + b c)/2

s = (a + b c)/2

 = (y/z + z/x + z/x + x/y + x/y + y/z)/2

 = 2 x (x/y + y/z + z/x) / 2

 = x/y + y/z + z/x

Area = Sqr (s x (s-a) x (s-b) x (s-c))

       = Sqr ((x/y + y/z + z/x) x (x/y) x (y/z) x (z/x))

       = Sqr (/ + xy/yz + xz/xy + xy/yz + / + yz/xz + xz/xy + yz/xz + /)

       = Sqr [/ +  / + / + 2(xy/yz + xz/xy + yz/xz)]

Answer 2

Refer to the attachment!!^^

Area = √((x/y + y/z + z/x) x (x/y) x (y/z) x (z/x))

= √ (x^{2}/y^{2} + xy/yz + xz/xy + xy/yz + y^{2}/z^{2} + yz/xz + xz/xy + yz/xz + z^{2}/x^{2})

= √ [x^{2}/y^{2} + y^{2}/z^{2} + z^{2}/x^{2} + 2(xy/yz + xz/xy + yz/xz)]