Question

if the sodes of the triangle are x,(x+1),(2x-1) and its area is x root 10. than find x. ​

Answer 1

$s = \frac{(x) + (x + 1) + (2x + 1)}{2} \\ = x + 1$

$area = x \sqrt{10} \\ \sqrt{(x + 1)x + (x + 1)(x + 1) + (x + 1)(2x + 1)} = x \sqrt{10} \\ squaring \: on \: both \: sides \: \\ (x + 1)(x) + (x + 1)(x + 1) + (x + 1)(2x + 1) = 10{x}^{2} \\ {x}^{2} + x + {x}^{2} + 2x + 1 + 2 {x}^{2} + 3x + 1 = 10 {x}^{2} \\ 4 {x}^{2} + 6x + 2 = 10 {x }^{2} \\ 6 {x}^{2} - 6x - 2 = 0 \\ dividing \: throughout \: by \: 2 \\ 3 {x}^{2} - 3x - 1 = 0$

$x = \frac{ 3 \frac{ + }{} \sqrt{17} }{6}$