Question

find x and all sides of the triangle

Explanation:

Step1: Apply Pythagorean theorem for the first triangle

For the right - triangle \(ABC\) (left - hand side), by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = x\), \(b=x + 2\), and \(c=x + 4\).

$$x^{2}+(x + 2)^{2}=(x + 4)^{2}$$

Expand the equations:

$$x^{2}+x^{2}+4x + 4=x^{2}+8x + 16$$

Simplify the equation:

$$x^{2}+x^{2}+4x + 4-(x^{2}+8x + 16)=0$$

$$x^{2}-4x - 12 = 0$$

Factor the quadratic equation:

$$x^{2}-4x - 12=(x - 6)(x+2)=0$$

Since \(x>0\) (length cannot be negative), \(x = 6\)
The sides of the first triangle: \(x=6\), \(x + 2=8\), \(x + 4 = 10\)

Step2: Apply Pythagorean theorem for the second triangle

For the right - triangle \(ABC\) (right - hand side), by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a=x-4\), \(b\) (base) \(=x + 1\), and \(c=x\)

$$x^{2}=(x - 4)^{2}+(x + 1)^{2}$$

Expand the equations:

$$x^{2}=x^{2}-8x + 16+x^{2}+2x + 1$$

Simplify the equation:

$$x^{2}-x^{2}+8x - 16 - x^{2}-2x - 1=0$$

$$x^{2}-6x + 17 = 0$$

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-6\), \(c = 17\))
\(\Delta=b^{2}-4ac=(-6)^{2}-4\times1\times17=36 - 68=-32<0\) (This is wrong, let's re - check the right - hand side triangle. Maybe it's a mis - draw. Assume it's a right - triangle with legs \(x-4\) and \(x\) and hypotenuse \(x + 1\))

$$x^{2}+(x - 4)^{2}=(x + 1)^{2}$$

Expand:

$$x^{2}+x^{2}-8x + 16=x^{2}+2x + 1$$

$$x^{2}-10x + 15 = 0$$

Using the quadratic formula \(x=\frac{10\pm\sqrt{100 - 60}}{2}=\frac{10\pm\sqrt{40}}{2}=\frac{10\pm2\sqrt{10}}{2}=5\pm\sqrt{10}\)
Since \(x-4>0\) (length), \(x=5+\sqrt{10}\)
The sides: \(x=5+\sqrt{10}\), \(x-4=1+\sqrt{10}\), \(x + 1=6+\sqrt{10}\)

Answer:

For the first triangle: \(x = 6\), sides are \(6\), \(8\), \(10\). For the second (assuming the correct right - angle assignment) triangle: \(x=5+\sqrt{10}\), sides are \(1+\sqrt{10}\), \(5+\sqrt{10}\), \(6+\sqrt{10}\)