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. the answer is not cd=20 and de=15. i have tried that one, it does not work.
CD=20 cm and DE=15 cm. CD=20 cm and DE=15 cm.
A rhombus DMFN is inscribed in such a way that it is inscribed in triangle CDE with the vertices M, F, and N l on the sides CD , CE , and DE respectively.
Now, in rhombus, the parallel sides are always parallel, therefore
MF║DE, thus, by the basic proportionality condition.
It is given that CF=8 cm and EF=12 cm, therefore, .
That is CM=8 cm and MD=12 cm. Therefore, CD=CM+MD=8+12=20.
According to question, Perimeter of △CDE =55 cm
Hence, CD=20 cm and DE=15 cm.
CD = 14 cm; DE = 21 cm
The perimeter is the sum of side lengths (in centimeters), so ...
CD + DE + CF + EF = 55
CD + DE + 8 + 12 = 55 . . . . . . . substittute for CF and EF
CD + DE = 35 . . . . . . . . . . . . . . subtract 20
The segment DF is a diagonal of the rhombus, so bisects angle D. That angle bisector divides ΔCDE into segments that are proportional. That is, ...
CD/DE = CF/EF = 8/12 = 2/3
So, we have two segments whose sum is 35 (cm) and whose ratio is 2 : 3. The total of "ratio units is 2+3=5, so each must stand for a length unit of 35/5 = 7 (cm). The sides are ...
CD = 2·7 cm = 14 cm
DE = 3·7 cm = 21 cm
CD + DE = (14 +21) cm = 35 cm . . . . . matches requirements
CD=14 cm and DE=21 cm
Let the rhombus's side be x cm, DN=NF=FM=DM=x xm.
Triangles CDE and FNE are similar, thus,
Since the perimeter of the triangle CDE is 55 cm, we have that
Therefore, CD=14 cm and DE=21 cm