Question

ABCD is a Trapezium with AB parallel to DC. M is the midpoint of BC.prove that ar(AMD)=1/2ar(ABCD)​

Answer 1

ABCD is a trapizium in which AB II CD, and M is the midpoint of BC,

we have to prove:--

triangle AMD = 1/2 ABCD

AB II MN II DC,

now in triangle DBC, OM II DC, and M is the midpoint of BC,

hence, O is the midpoint of DB,

MO = 1/2 DC ( by the converse of midpoint theory),

In triangle BAD,

O is the mid point of BD,

MN II BA

hence, N is the midpoint of AD,

therefore:---

MN = 1/2 BA,( by converse of midpoint theory),

MO + ON = 1/2 DC + 1/2 BA=1/2 ( DC+BA),

from MO + ON = MN.

hence MN = 1/2 (DC + BA),

now:----

area of triangle AMD ÷ area of trapizium ABCD

= 1/2 AD × MN÷ 1/2 (BA+CD) × AD

=MN ÷ (BA +CD),

= 1/2 ( DC + BA) ÷ ( BA+CD) = 1/2

=triangle AMD = 1/2 ABCD,

Ans:-Area of triangle AMD = area of 1/2 trapizium ABCD.

Answer 2

ar(AMD)=1/2ar(ABCD)​.

Proved below.        

Step-by-step explanation:

Given: ABCD is a trapezium in which AD║BC and M is the midpoint of CD.

To prove: area ∆AMB is half the area of trapezium ABCD

​Construction: ML parallel to DA​ and CB. Join BD and let BD intersect ML at O.

DA ∥ML∥CB.

Now in ΔDBC,

OM∥CB and M is the midpoint of DC.

Hence O is the midpoint of DB and also MO =$\frac{1}{2}$ BC (by converse of midpoint theorem)

In ΔDAB,

O is the midpoint of DB and OL ∥DA.

Hence L is the midpoint of AB and also OL=$\frac{1}{2}$ DA (by converse of midpoint theorem)

∴MO+OL = $\frac{1}{2}$ BC+$\frac{1}{2}$ DA

MO+OL = $\frac{1}{2}$ (BC+DA)

And form the figure, MO+OL = ML

Hence ML= $\frac{1}{2}$ (BC+DA)               [1]

Now, $\frac{area \triangle AMB}{area trapezium ABCD}=\frac{\frac{1}{2}\times AB \times ML }{\frac{1}{2}\times (DA+CB)\times AB }$

$\frac{area \triangle AMB}{area trapezium ABCD}={\frac{ML}{DA+CB}$

$\frac{area \triangle AMB}{area trapezium ABCD}={\frac{(\frac{1}{2} )(BC+DA)}{DA+CB}$              [from 1]

$\frac{area \triangle AMB}{area trapezium ABCD}=\frac{1}{2}$

Hence proved.