Question

the sides of a triangle are x, x+1, 2x-1 and its area is √10x² units. find the value of x.

Answer 1

s=x+x+1+2x-1÷2=2x
area of triangl =√2x(2x-1)(2x-x+1)(2x-2x-1)
√10x^2 = √2x^2×(x+1)×(-1)
10x^2=-2x^3-2x^2
x=10x^2+2x^2÷2x^2
x=6

Answer 2

S =(x+x+1+2x-1)÷2 =2x (a,b,c are side)
area of triangle =
$\sqrt{s(s - a)(s - b)(s - c) }$
$\sqrt{2x(2x - x)(2x - x - 1)(2x - 2x + 1}$
=
$\sqrt{2x(x)(x - 1)(1) }$
=

$\sqrt{ {x}^{2} } ( \sqrt{2x - 2)}$
now this equal to
$\sqrt{10 {x }^{2} }$
x^2 (2x-2)=10x^2
2x=12
x=6