Question

The sides of a triangle are x , x+1 , 2x-1 and its area is x√10 what is the value of x

Answer 1

Sides of triangle are x, x+1, 2x-1
Area of triangle is $x \sqrt{10}$
To find out:- Value of x
We Know That,
Semi-Perimeter(S)= Perimeter/2
                               =x+x+1+2x-1/2
                               =4x/2
                              =2x
Area Of Triangle= $\sqrt{s(s-a)(s-b)(s-c)}$
    or, $x \sqrt{10}$  =$\sqrt{2x(2x-x)(2x-x-1)(2x-2x+1)}$
     or, $x \sqrt{10}        =[tex] \sqrt{2x.x(x-1).1}$
      or, $x \sqrt{10}       =[tex] \sqrt{2x^2(x-1)}$
       or, $x \sqrt{10}      =<strong></strong>[tex]x\sqrt{2(x-1)}$
        Squaring Both Sides, We get
      or, 10=2(x-1)    
      or, 10=2x-2
     or, 10+2=2x
   or, 12/2 = x
   or, 6 = x
Therefore, x = 6
hence, The value of X is 6 Ans..

Answer 2

Answer:

Step-by-step explanation:

Sides are x, x+1, 2x-1

S= x+x+1+2x-1 the whole divided by 2

= 4x÷2

=2x

Area=x^10

=^s (s-a) (s-b) (s-c)

=^2x (2x-x) (2x-x-1) (2x-2x+1)

=^2x x (x-1)

= x^2(x-1)

=x^2(x-1)

Both sides = 2(x-1) = 10

= x-1 = 5

= x =5+1 =6