Question

Part I: Midsegments
1. If D, E, and F are midpoints of the sides
of AABC, find the perimeter of AABC.
E
16

Answer 1

Answer:

The perimeter of triangle ABC is 70 units

Step-by-step explanation:

we know that

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side

so

$DF=\frac{1}{2}AB$

$ED=\frac{1}{2}BC$

$EF=\frac{1}{2}AC$

step 1

Find the length side AB

$DF=\frac{1}{2}AB$

we have

$DF=16\ units$

substitute

$16=\frac{1}{2}AB$

$AB=32\ units$

step 2

Find the length side BC

$ED=\frac{1}{2}BC$

we have

$ED=7\ units$

substitute

$7=\frac{1}{2}BC$

$BC=14\ units$

step 3

Find the length side AC

$EF=\frac{1}{2}AC$

we have

$EF=12\ units$

substitute

$12=\frac{1}{2}AC$

$AC=24\ units$

step 4

Find the perimeter of triangle ABC

$P=AB+BC+AC$

substitute values

$P=32+14+24=70\ units$

Answer 2

The perimeter of triangle ABC is 70 units

Midpoint theorem:

When the segment joining two sides of a triangle at the midpoints of those sides so it should be parallel to the third side and is half the length of the third side.

based on this,

DF = 0.5AB

ED = 0.5BC

EF = 0.5AC

Now the length of AB is

Since DF is 16 units

So, AB is 32 units

Likewise, the length of BC is

Since ED is 7 units

So, BC should be 14 units

Likewise, the length of AC is

Since EF is 12 units

So, BC should be 24 units

Now the perimeter of the triangle ABC is

= AB + BC + AC

= 32 + 14 + 24

= 70 units

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