Question

the side of a triangle are x, x + 1 and 2x-1 and its area is X√10. find the side of a triangle

Answer 1

First of all we are going to find semi perimeter


semi perimeter = sum of sides/2

=> semi perimeter = (x + x + 1 + 2x - 1)/2

=> semi perimeter = 4x/2

=> semi perimeter = 2x



Now given area = x√10

Using Heron's Formula we can compare with the area,

Heron's Formula =
$\sqrt{s(s - a)(s - b)(s - c)}$
where s = semi perimeter and a,b an c are sides

We know that Heron's Formula give area and area is given as x√10.

So the equation is,

$\sqrt{2x(2x - x)(2x - (x + 1))(2x - (2x - 1))}$
$= > \sqrt{2x(x)(2x - x - 1)(2x - 2x + 1)} \\ = > \sqrt{2x(x)(x - 1)(1)} \\ = > \sqrt{2 {x}^{2} (x - 1)}$
Area of given Triangle =
$\sqrt{2 {x}^{2}(x - 1) }$
but the given area is x√10

=>
$\sqrt{2 {x}^{2} (x - 1)} = x \sqrt{10} \\ = > x \sqrt{2(x - 1)} = x \sqrt{10} \\ \\ = > \sqrt{2(x - 1) } = \frac{x \sqrt{10} }{x} \\ \\ = > \sqrt{2(x - 1)} = \sqrt{10}$
now squaring both sides,

2(x - 1) = 10

=> x - 1 = 10/2

=> x - 1 = 5

=> x = 6


First side = x = 6


Second side = x + 1 = 6 + 1 = 7

Third side = 2x - 1 = 2(6) - 1 = 12 - 1 = 11



So the sides of the triangle are

6, 7 and 11

Hope it helps dear friend ☺️✌️✌️