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 ,

Question

AleksAgata [21]

3 years ago

12

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.

Mathematics

1 answer:

ivanzaharov [21] · Answer 1 · 2021-12-08T23:04:50+03:00

Answer:

CD = 14 cm; DE = 21 cm

Step-by-step explanation:

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

<em>Check</em>

CD + DE = (14 +21) cm = 35 cm . . . . . matches requirements