The sides of a triangle are x , x + 1 and 2 x -1. It’s area is x √10 square units. Find the value of x. ​

Dear Student,

Here is solution:

s=(x+x+1+2x-1)/2

=4x/2

=2x

Now Area=√[s(s-x){s-(x+1)}{s-(2x-1)}]

=√[2x(2x-x){2x-(x+1)}{2x-(2x-1)}]

=√[2x^2(2x-x-1)(2x-2x+1)]

=√[2x^2(x-1)×1]

=√(2x^3–2x^2)

According to question

√(2x^3–2x^2)=x√10

Or, {√(2x^3–2x^2)}^2={x√10}^2

Or, 2x^3–2x^2=10x^2

Or, 2x^3–10x^2–2x^2=0

Or,2x^3–12x^2=0

Or, 2x^2(x-6)=0

Either 2x^2=0

Or, x=0

/OR x-6=0

Or,x=6

But x can not be 0

Therefore x=6

Thanks