in a triangle ABC De is parallel to BC and D is the midpoint of side a b . find the perimeter of triangle ABC when AE= 4.5 cm,DE= 5 cm and DB = to 3.5 CM.
Answers
Answer:
It is assumed that the point D lies on AB and the point E lies on AC.
Draw a figure with the given data.
Now increase the measure of angle B a little bit. Move the vertex A towards point D so that the length of side AC remains the same. In this process the lengths of AD and DE become smaller - however, the lengths of BD, BC and AC remain the same.
This shows that the data given is insufficient and, with the given data, you can have an infinite number of solutions.
Since DE∥BC,△ADE∼△ABC.
⇒ADAB=AEAC=DEBC
⇒ADAD+3=AE6.4=DE3.5
We can now assume a value for AD (subject to the condition that AB is not greater that AC + BC) and get the corresponding values of DE (and, if you want, AE also).
Let AD = 2 units. Then, 22+3=DE3.5⇒DE=25×3.5=1.4 units.
Let AD = 4 units. Then, 44+3=DE3.5⇒DE=47×3.5=2 units.
Answer:
Here , BC | | DE and D is mid point of AB , So
DB = AD = 3.5 cm ( As given DB = 3.5 cm )
From converse of mid point theorem we get
ADDE = ABBC⇒ADDE = AD + DBBC⇒3.55 = 3.5 + 3.5BC⇒3.55 = 7BC⇒15 = 2BC⇒BC= 10
And if D is mid point of AB and DE | | BC , so from converse of mid point theorem we get that E is mid point of AC , So
AE = CE = 4.5 cm ( As given AE = 4.5 cm )
Then,
AC = AE + CE = 4.5 + 4.5 = 9 cm
Therefore,
Perimeter of triangle ABC = AB + BC + AC = 10 + 7 + 9 = 26 cm ( Ans )
Step-by-step explanation: