Question

1. in the cube abcdefgh with opposite vertices c and e, j and i are the midpoints of edges (overline{fb}) and (overline{hd}), respectively. the edge length of the cube is 2. let r be the ratio of the perimeter of the cross - section ejci to the perimeter of one of the faces of the cube. what is r? (adapted from 2018 amc 8 problem, question #24)

a. (\frac{5}{2}) b. (\frac{sqrt{5}}{4}) c. (\frac{sqrt{5}}{2}) d. (\frac{5}{4}) e. (\frac{2sqrt{5}}{5})

Explanation:

Step1: Calculate the length of \( EJ\)

In right - triangle \(EFJ\), \(EF = 2\), \(FJ=\frac{2}{2}=1\). By the Pythagorean theorem \(EJ=\sqrt{EF^{2}+FJ^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{4 + 1}=\sqrt{5}\).

Step2: Calculate the length of \(JC\)

In right - triangle \(JBC\), \(JB = 1\), \(BC=2\). By the Pythagorean theorem \(JC=\sqrt{JB^{2}+BC^{2}}=\sqrt{1^{2}+2^{2}}=\sqrt{1 + 4}=\sqrt{5}\).

Step3: Calculate the length of \(CI\)

In right - triangle \(CDI\), \(CD = 2\), \(DI = 1\). By the Pythagorean theorem \(CI=\sqrt{CD^{2}+DI^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{4+1}=\sqrt{5}\).

Step4: Calculate the length of \(IE\)

In right - triangle \(IEH\), \(EH = 2\), \(HI = 1\). By the Pythagorean theorem \(IE=\sqrt{EH^{2}+HI^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{4 + 1}=\sqrt{5}\).

Step5: Calculate the perimeter of cross - section \(EJCI\)

The perimeter of \(EJCI\) is \(P_{1}=EJ + JC+CI + IE=4\sqrt{5}\).

Step6: Calculate the perimeter of one face of the cube

The perimeter of one face of the cube (a square with side length \(a = 2\)) is \(P_{2}=4\times2=8\).

Step7: Calculate the ratio \(R\)

\(R=\frac{P_{1}}{P_{2}}=\frac{4\sqrt{5}}{8}=\frac{\sqrt{5}}{2}\).

Answer:

C. \(\frac{\sqrt{5}}{2}\)