PLEEAAASE HELP!! Thx!
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 20 and 15!!!!!!
Answers
draw same picture and mark points as given
A rhombus DMFN is inscribed in this triangle so that vertices M, F, and N lie on the sides CD , CE , and DE respectively.
Since, in rhombus, the opposite edges are parallel,
MF ║ DE
=> cm/md=cf/ef
since given that CF=8cm and EF=12cm , we have
=>cm/md=8/12
That is, CM=8 and MD=+12
Therefore, CD=md+cm=8+12=20
But According to the question,
CD+DE+CE=55
=>20+DE+20=55
=> DE=15
Answer: CD=20 cm and DE= 15 cm
Answer:
In Rhombus DNFM, diagonal DF is also the angle bisector of Angle D. Therefore, the two triagles created (DFC and DFE) are similar.
Since EF:FC = 12:8, then DE:DC is also equal to 12:8 (For similar triangles, the sides should have the same proportion).
We know perimiter of CDE is 55 and Side CE = 20 (8+12). Therefore the other two sides should be 35 (55-20).
Applying the ratio of 12:8, we get DE = 35*12/20 = 21 and DC = 35*8/20 = 14
Answer = DE is 20 and DC is 14
Step-by-step explanation: