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The perimeter of △CDE is 55 cm. A rhombus DMFN is inscribed in this triangle so that vertices M, F, and N lie on the sides CD , CE , and DE respectively. Find CD and DE if CF=8 cm and EF=12 cm.

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Answer:

CD=14 cm and DE=21 cm

Step-by-step explanation:

Let the rhombus's side be x cm, DN=NF=FM=DM=x xm.

Triangles CDE and FNE are similar, thus,

[tex]\dfrac{CD}{FN}=\dfrac{CE}{FE}=\dfrac{DE}{NE}[/tex]

or

[tex]\dfrac{CD}{x}=\dfrac{8+12}{12}=\dfrac{DE}{DE-x}.[/tex]

Hence,

[tex]CD=\dfrac{5x}{3}[/tex]

and

[tex]20(DE-x)=12DE,\\ \\8DE=20x,\\ \\DE=\dfrac{5x}{2}.[/tex]

Since the perimeter of the triangle CDE is 55 cm, we have that

[tex]\dfrac{5x}{3}+\dfrac{5x}{2}+20=55,\\ \\\dfrac{25x}{6}=35,\\ \\x=\dfrac{6\cdot 35}{25}=8.4.[/tex]

Therefore, CD=14 cm and DE=21 cm

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