Question

Given that DE, DF, and EF are midsegments of △ABC, and DE=3.2 feet, EF=4 feet, and DF=2.4 feet, the perimeter of △ABC is ___feet.

Answer 1

Perimeter of ΔABC = 19.2 ft

Explanation:

Given that DE, DF and EF are midsegments of ΔABC and DE=3.2 feet,

EF=4 feet, and DF=2.4 feet

The value of the midsegment is half value of the corresponding side of the triangle, which means:

EF = 1/2 AB => AB = 2 · EF = 2 · 4 = 8 ft

AB = 8 ft

DE = 1/2 BC => BC = 2 · DE = 2 · 3.2 = 6.4 ft

BC = 6.4 ft

DF = 1/2 AC => AC = 2· 2.4 = 4.8 ft

AC = 4.8 ft

Perimeter of the ΔABC is:

P = AB + BC + AC = 8 + 6.4 + 4.8 = 19.2 ft

P = 19.2 ft

God with you!!!

Answer 2

Answer: P=19.2 ft

Explanation:

By the Triangle midsegment theorem, you have:

EF=1/2AB

DE=1/2BC

DF=1/2AC

Then, the lengths of the sides of the triangle are:

AB=2EF=8 ft

BC=2DE=6.4 ft

AC=2DF=4.8 ft

Add them to obtain the perimeter, then you obtain the following result:

P=8 ft+6.4 ft+ 4.8 ft

P=19.2 ft