Find X

You know that the area of a triangle, by Heron's Formula, is $\sqrt { s(s-a)(s-b)(s-c) }$ , where $s$ is the semi-perimeter and $a,b,c$ are the side lengths of the triangle. Adding the side lengths and dividing by $2$ , you know that the semi-perimeter is $2x$ . Plugging into Heron's Formula, the area of the triangle is $\sqrt { 2x\times (2x-(x))\times (2x-(x+1))\times (2x-(2x-1)) } =\sqrt { 2{ x }^{ 3 }-2{ x }^{ 2 } }$ . But you also know that the area of the triangle is $x\sqrt { 10 }$ Setting those equal to each other, you have $x\sqrt { 10 } =\sqrt { 2{ x }^{ 3 }-2{ x }^{ 2 } } \Longrightarrow 10{ x }^{ 2 }=2{ x }^{ 3 }-2{ x }^{ 2 }\Longrightarrow 12{ x }^{ 2 }={ 2x }^{ 3 }\Longrightarrow 6{ x }^{ 2 }={ x }^{ 3 }\Longrightarrow x=6\quad or\quad 0$
We must reject zero, so we don't have a negative side length. $\boxed { x\quad =\quad 6 }$

Heron's Formula:

$A = \sqrt {p(p-a)(p-b)(p-c)}$

$A = \sqrt {2x \cdot (2x-x) \cdot (2x-(x + 1)) \cdot (2x-(2x-1))} = \sqrt {2x^3-2x^2}$

$= \sqrt {x^2(2x-2)} = x\sqrt {2x-2}$

But the area is also $x\sqrt {10}$ , so:

$x\sqrt {2x-2} = x\sqrt {10} \implies 2x-2 = 10 \implies 2x = 12 \implies x = \boxed{6}$

Please forgive me, because my solution is quite complicated. I didn't know Heron's formula until after I looked at the solutions page, so my solution was much longer.

You may view an illustration here.

Let triangle ABC have sides AC of length x+1 , CB of length 2x-1 , and AB of length x .

Let AB be the base of triangle ABC .

Let CD have length d , which would also be the height of triangle ABC . CD is perpendicular to AB .

Let CD intersect the triangle at point D so that n is the length of AD , and x-n is the length of BD .

Let y=x+1 , and z=2x-1 .

Visualize that AC of length y is the hypotenuse of both CD of length d and BD of length n .

$d^{2} + n^{2} = y^{2}$ $d^{2} + n^{2} = (x + 1)^{2}$ $d^{2} = (x + 1)^{2} - n^{2}$

Visualize that CB of length z is the hypotenuse of both CD of length d and BD of length x-n .

$d^{2} + (x - n)^{2} = z^{2}$ $d^{2} + (x - n)^{2} = (2x - 1)^{2}$ $d^{2} = (2x - 1)^{2} - (x - n)^{2}$

$(x + 1)^{2} - n^{2} = (2x - 1)^{2} - (x - n)^{2}$

$x^{2} + 2x + 1 - n^{2} = 4x^{2} - 4x + 1 - (x^{2} - 2xn + n^{2})$ $x^{2} + 2x + 1 - n^{2} = 4x^{2} - 4x + 1 - x^{2} + 2xn - n^{2}$ $x^{2} + 2x + 1 - n^{2} = 3x^{2} - 4x + 1 + 2xn - n^{2}$ $x^{2} + 2x + 1 = 3x^{2} - 4x + 1 + 2xn$ $x^{2} = 3x^{2} - 6x + 2xn$ $0 = 2x^{2} - 6x + 2xn$ $-2xn = 2x^{2} - 6x$ $n = (2x^{2} - 6x)/(-2x)$ $n = 3 - x$

Now, let us get d using the newfound value of n . In the Remember that AC of length y , or x+1 , is the hypotenuse of CD of length d and AD of length n .

$d^{2} + n^{2} = (x + 1)^{2}$ $d^{2} + (3 - x)^{2} = (x + 1)^{2}$ $d^{2} = (x + 1)^{2} - (3 - x)^{2}$ $d^{2} = (x^{2} + 2x + 1) - (9 - 6x + x^{2})$ $d^{2} = x^{2} + 2x + 1 - 9 + 6x - x^{2}$ $d^{2} = 2x + 1 - 9 + 6x$ $d^{2} = 1 - 9 + 8x$ $d^{2} = -8 + 8x$ $d^{2} = 8x - 8$

$d = \sqrt{8x - 8} = 2\sqrt{2x-2}$

Using the value of d, which is the length of the line CD that is perpendicular to the line AB, we can get the area of triangle ACD.

$aACD = \frac{(3-x) 2\sqrt{2x - 2}}{2}$ $aACD = (3-x)\sqrt{2x-2}$

We use the same solution to get the area of triangle CBD .

$aCBD = \frac{(x - (3-x))2\sqrt{2x-2}}{2}$ $aCBD = \frac{(2x - 3)2\sqrt{2x-2}}{2}$ $aCBD = (2x-3)\sqrt{2x-2}$

Getting the sum of the areas of triangles ACD and CBD would give us the area of triangle ABC . It is given that the area of triangle ABC is equal to x√10 .

$aACD + aCBD = aABC$ $(3-x)\sqrt{2x-2} + (2x-3)\sqrt{2x-2} = x√10$ $((3-x)+(2x-3))\sqrt{2x-2}=x\sqrt{10}$ $x\sqrt{2x-2}=x\sqrt{10}$

Ah! 2x-2 = 10 .

$2x = 12$

Therefore:

$\boxed{x = 6}$