Question

part a
in rhombus abcd, diagonals \\( \overline{bd} \\) and \\( \overline{ac} \\) intersect at point e. if \\( be = 4n - 3 \\) and \\( ec = 2n + 5 \\), which expression can be used to represent \\( ab \\)?
image of rhombus abcd with diagonals intersecting at e
\\( \bigcirc \\) a \\( (2n + 5)^2 + (4n - 3)^2 \\)
\\( \bigcirc \\) b \\( 2(2n + 5) + 2(4n - 3) \\)
\\( \bigcirc \\) c \\( \sqrt{(2n + 5) + (4n - 3)} \\)
\\( \bigcirc \\) d \\( \sqrt{(2n + 5)^2 + (4n - 3)^2} \\)

part b
if \\( ac = 14 \\), what is the measure of \\( bd \\)?

Explanation:

Part A

Step1: Recall properties of a rhombus

In a rhombus, the diagonals are perpendicular bisectors of each other. So, \( AE = EC=2n + 5\) and \( BE=ED = 4n-3\), and \(\triangle AED\) (or \(\triangle AEB\), \(\triangle BEC\), \(\triangle CED\)) is a right triangle with legs \(AE\) and \(BE\) and hypotenuse \(AB\) (since all sides of a rhombus are equal, \(AB = AD\)).

Step2: Apply the Pythagorean theorem

For a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\), the Pythagorean theorem states that \(c=\sqrt{a^{2}+b^{2}}\). Here, \(a = AE=2n + 5\), \(b=BE = 4n-3\) and \(c = AD\). So, \(AD=\sqrt{(2n + 5)^{2}+(4n-3)^{2}}\)

Step1: Find the value of \(n\)

We know that \(AC=AE + EC\) and \(AE = EC=2n + 5\) (diagonals bisect each other in a rhombus). Given \(AC = 14\), so \(AE+EC=(2n + 5)+(2n + 5)=14\)

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Step2: Find the length of \(BE\)

We know that \(BE = 4n-3\). Substitute \(n = 1\) into the expression for \(BE\):
\(BE=4(1)-3=1\)

Step3: Find the length of \(BD\)

Since the diagonals bisect each other, \(BD=2\times BE\) (because \(BE = ED\)). So, \(BD = 2\times1=2\)? Wait, no, wait. Wait, \(BE=4n - 3\), when \(n = 1\), \(BE=4(1)-3 = 1\), and \(BD=2\times BE\) (since diagonals bisect each other, \(BD = BE+ED\) and \(BE = ED\)). But wait, let's re - check.

Wait, from the property of rhombus diagonals: \(AC = 2\times EC\), we used \(AC=14\), so \(EC=\frac{AC}{2}=7\). And \(EC = 2n+5\), so \(2n + 5=7\), \(2n=2\), \(n = 1\). Then \(BE=4n-3=4(1)-3 = 1\). Since diagonals are perpendicular bisectors, \(BD = 2\times BE\) (because \(BE=ED\)). So \(BD=2\times1 = 2\)? Wait, no, that seems too short. Wait, maybe I made a mistake. Wait, \(BE\) is a segment of the other diagonal. Wait, the diagonals are \(AC\) and \(BD\). \(AC\) is split into \(AE\) and \(EC\) each of length \(7\) (since \(AC = 14\)). Then \(BE=4n-3\), with \(n = 1\), \(BE = 1\), so \(BD=BE + ED=1 + 1=2\)? Wait, but let's check the formula again.

Wait, in the rhombus, diagonals bisect each other at right angles. So \(BD = 2\times BE\), and \(BE=4n - 3\). We found \(n = 1\), so \(BE=1\), so \(BD = 2\times1=2\)? Wait, that seems correct based on the calculations. But let's verify with the value of \(n\).

Wait, \(EC=2n + 5\), \(AC = 14\), so \(EC=\frac{AC}{2}=7\), so \(2n+5 = 7\), \(n = 1\). Then \(BE=4n-3=1\), and since \(BD = 2\times BE\) (because diagonals bisect each other), \(BD = 2\times1=2\). Wait, but maybe I messed up the direction. Wait, \(BD\) is composed of \(BE\) and \(ED\), and \(BE = ED\), so \(BD=2\times BE\). So with \(n = 1\), \(BE = 1\), so \(BD = 2\).

Wait, but let's re - check the calculation for \(n\):

\(AC=AE + EC\), and \(AE = EC\) (diagonals bisect each other). So \(AC = 2\times EC\), \(EC=2n + 5\), \(AC = 14\), so \(2(2n + 5)=14\), \(2n+5 = 7\), \(2n=2\), \(n = 1\). Then \(BE=4n-3=4(1)-3 = 1\). Then \(BD=2\times BE=2\times1 = 2\).

Answer:

D. \(\sqrt{(2n + 5)^{2}+(4n - 3)^{2}}\)

Part B