In a ` DeltaA B C ,` prove the following :`\ If\ a=sqrt(2),\ b=sqrt(3)\ a n d\ c=sqrt(5)` , shos that its area is `1/2sqrt(6)` sq. units.

Explanation

Given : `a=sqrt2`, `b=sqrt3`, `c=sqrt5`

`therefore cos C=(a^2+b^2-c^2)/(2 a b)`

`Rightarrow cos C=(2+3-5)/(2 times sqrt{6})=0 `

`Rightarrow cos C=0`

`Rightarrow cos C=cos 90^{circ} `

`Rightarrow C=90^{circ}`

Thus, `sin C=sin 90^{circ}=1`

Hence, Area of `triangle A B C=frac{1}{2} a b sin C=frac{1}{2} sqrt{6} times 1=frac{sqrt{6}}{2} s q`. units .