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 ,
2 answers:
Answer:
CD=20 cm and DE=15 cm. CD=20 cm and DE=15 cm.
Step-by-step explanation:
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
⇒CD+DE+CE=55
⇒20+DE+20=55
⇒DE=15 cm
Hence, CD=20 cm and DE=15 cm.
You might be interested in
To solve for the longest side, the hypotenuse, you have to use the pythagorean theorem. It will be 10^2 + 9^2 = c^2. 100 + 81 =c^2.
c^2 = 181 so c = sqrt(181).
to find sin of A do opposite/hypotenuse which gives you 9/sqrt(181)
to find cos of A do adjacent/hypotenuse which gives you 10/sqrt(181)
c^2 = 181 so c = sqrt(181).
to find sin of A do opposite/hypotenuse which gives you 9/sqrt(181)
to find cos of A do adjacent/hypotenuse which gives you 10/sqrt(181)