The perimeter of a rhombus is the sum of its side lengths.
The perimeter of the rhombus is [tex]\mathbf{ 4x\sqrt{17}}[/tex]
The given parameters are:
[tex]\mathbf{d_1 = 2x}[/tex]
[tex]\mathbf{d_2 = 8x}[/tex]
The perimeter is calculated as:
[tex]\mathbf{P = 4 \times \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}}[/tex]
So, we have:
[tex]\mathbf{P = 4\sqrt{(\frac{2x}{2})^2 + (\frac{8x}{2})^2}}[/tex]
[tex]\mathbf{P =4 \sqrt{(x)^2 + (4x)^2}}[/tex]
Evaluate all exponents
[tex]\mathbf{P = 4\sqrt{x^2 + 16x^2}}[/tex]
[tex]\mathbf{P = 4\sqrt{17x^2}}[/tex]
Take the square root of x^2
[tex]\mathbf{P = 4x\sqrt{17}}[/tex]
Hence, the perimeter of the rhombus is [tex]\mathbf{ 4x\sqrt{17}}[/tex]
Read more about perimeters at:
https://brainly.com/question/6465134
