Question

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.

Answer 1

CD=14 cm and DE=21 cm

Explanation:

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

Triangles CDE and FNE are similar, thus,

`(CD)/(FN)=(CE)/(FE)=(DE)/(NE)`

or

`(CD)/(x)=(8+12)/(12)=(DE)/(DE-x).`

Hence,

`CD=(5x)/(3)`

and

`20(DE-x)=12DE,\\ \\8DE=20x,\\ \\DE=(5x)/(2).`

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

`(5x)/(3)+(5x)/(2)+20=55,\\ \\(25x)/(6)=35,\\ \\x=(6\cdot 35)/(25)=8.4.`

Therefore, CD=14 cm and DE=21 cm