A line segment (DE) joining the midpoints of two sides of a triangle is parallel
to the third sid

This is  called a mid segment theorem in which the al ine segment (DE) joining the midpoints of two sides of a triangle is parallel to the third side.

Step-by-step explanation:

Suppose we have a triangle ABC. Then the midpoints can be located as D, E and F. If we join D , E and F another triangle is formed.

From the figure we can see that

AE≅ CE

AD≅DB

BF≅CF  

BECAUSE  all the given points are the midpoints which divide the lines into two equal halves.

If we increase the line DE to a point L we find out that DL is parallel to BC i.e. it does not meet at any point with BC. ( the two lines do not meet)

(1)

If we join C with L we find out that the the line DE is half in length to the line BC.

AS

AE= CE  (midpoints dividing into equal line segements.)

LE= DE

Triangle CEL= Triangle DEF

so

DL= BC

But DE = 1/2 DL

therefore

DE= 1/2 BC (2)

Therefore from 1 and 2 we find that a line segment (DE) joining the midpoints of two sides of a triangle is parallel to the third side

rectangle: 9 1/8 x 4 1/9 = 37 37/72in

parallelogram: b times h= 10.3 x 6.2 = 63.86

trapezoid: a+b/2 times h= 42.5

volume of prism: l times w times h= 36

im alright what do you need

step-by-step explanation: