Part A
Answer: Choice D) [tex]\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}[/tex]
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Explanation:
Let E be the intersection point of the diagonals. For any parallelogram, the diagonals bisect each other. This means that AE = EC = 2n+5 and it also means EB = DE = 4n-3
Focus on triangle AED. This triangle is a right triangle with angle AED as a 90 degree angle (any rhombus has its diagonals that are perpendicular to one another). The legs of this triangle are AE = 2n+5 and ED = 4n-3
Apply the pythagorean theorem to find hypotenuse AD
[tex](\text{leg1})^2+(\text{leg2})^2 = (\text{hypotenuse})^2\\\\(\text{AE})^2+(\text{ED})^2 = (\text{AD})^2\\\\(\text{AD})^2 = (\text{AE})^2+(\text{ED})^2\\\\\text{AD} = \sqrt{(\text{AE})^2+(\text{ED})^2}\\\\\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}\\\\[/tex]
This points to choice D as the final answer.
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Part B
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Explanation:
If AC = 14, then half of this is AE = 7. The diagonals bisect each other, so they cut each other in half with E at the midpoint.
Since AE = 2n+5, we can solve for n like so
2n+5 = 7
2n = 7-5
2n = 2
n = 2/2
n = 1
Then we can find the length of AD
[tex]\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}\\\\\text{AD} = \sqrt{(2*1+5)^2+(4*1-3)^2}\\\\\text{AD} = \sqrt{(2+5)^2+(4-3)^2}\\\\\text{AD} = \sqrt{7^2+1^2}\\\\\text{AD} = \sqrt{49+1}\\\\\text{AD} = \sqrt{50}\\\\\text{AD} \approx 7.0710678\\\\\text{AD} \approx 7.07\\\\[/tex]
Segment AD is roughly 7.07 units long. The other external sides of the rhombus (AB, BC and CD) are also this length because the four sides of any rhombus are the same length.