Question

Perimeter of a triangle ABC is 72 cm. Find the perimeter of the triangle DEF with vertices D,E and F as teh mid-points of the sides of the given triangle.

Answer 1

D and E are the midpoints of AB and BC respectively
therefore DE=1/2AC {MIDPOINT THEOREM}… similarly FE =1/2AB and DF=1/2BC .... DE+EF+FD=1/2(AB+CD+EF)=1/2*72cm=36cm

Answer 2

Answer:

36 cm

Step-by-step explanation:

Perimeter of triangle ABC = 72 cm

D,E and F as teh mid-points of the sides of the given triangle.

So, by mid point theorem

DE=1/2AC

FE =1/2AB

DF=1/2BC

So, Perimeter of triangle DEF = DE+EF+DF = $\frac{1}{2}AC+\frac{1}{2}AB+\frac{1}{2}BC=\frac{1}{2}(AB+BC+AC)=\frac{1}{2}(72)=36$

Hence the perimeter of the triangle DEF is 36 cm