Question

in the figure 13.82the line segment joining the midpoint of two non parallel side of a Trapezium is equal to half of their sum​

Answer 1

Proved that line segment joining the midpoint of two non parallel side of a Trapezium is equal to half of their sum​

Step-by-step explanation:

Let say ABCD is Trapezium

M & N are mid points of AD & BC

=> MN ║ AB ║ CD

Join AC which intersect MN at P

=> MP ║ AB ║ CD   & NP ║ AB ║ CD  as P is point on MN

Now in ΔABC

PN ║ AB

=> PN/AB  = CN/BC

CN = BC/2  ( as N is mid point of BC)

=> CN/BC = 1/2

=> PN/AB = 1/2

=> PN = AB/2

Similarly  in ΔACD

MP ║ CD

=> AM/AD = MP/CD

AM = AD/2   ( as M is mid point of AD)

=> AM/AD = 1/2

=> MP/CD = 1/2

=> MP = CD/2

MP + PN = AB/2 + CD/2

=> MN = (AB + CD)/2

Hence Proved that line segment joining the midpoint of two non parallel side of a Trapezium is equal to half of their sum​

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Answer 2

Answer:

Step-by-step explanation:

See the attached diagram.

ABCD is a trapezium with two non-parallel sides BC and DA.

Now, E and F are the midpoints of DA and BC respectively.

Then, EF will be parallel to AB and CD.

We have to prove that the length of the line segment EF is equal to half of the sum of lengths of its parallel sides i.e. AB and CD.

Draw a diagonal of the trapezium AC and it divides the trapezium into two triangles Δ ABC and Δ ACD.

From, Δ ABC, F is the midpoint of BC and FG ║ AB, hence, G will be the midpoint of AC and $FG = \frac{1}{2} AB$ ........... (1)

Now, from Δ ACD, E and G are the midpoints of DA and AC respectively, and hence $EG = \frac{1}{2}CD$ ............. (2)

Therefore, EF = EG + GF = $\frac{1}{2}(AB + CD)$ (Proved)