Question

with the diagram I want

Answer 1

In Parallelogram ABCD,

4 Triangles are there; AOB, BOC, COD, DOA.

Ar. (DOA/AOD) =
${4cm}^{2}$
Now, Let the Ar. of the Paralleogram be x.

So, Ar. of triangle ADC =
$\frac{1}{2} \: of \: ar(abcd) \\ = \frac{1}{2} \times x$
[Since, The diagonal of the parallelogram divides the parallelogram into 2 congruent triangles]

Now, In Triangles AOD & COD;

AO = OC [Since, the diagonals of a parallelogram bisects each other]

And The height will be same for both the triangles if the base we take is AO and OC respectively for both the Triangles.

So, their areas will be equal as ar of a triangle =
$\frac{1}{2} \times base \times height$
Therefore, their areas will be equal.

So, If the area of ABC =
$\frac{1}{2} x$
So, Area of AOD = Area of COD

$2 \times ar(aod) = \frac{1}{2} x \\ = > ar(aod) = \frac{1}{4} x$
Now, We know area of AOD =
${4cm}^{2}$
Therefore,
${4cm}^{2} = \frac{1}{4} x \\ = > x = {16cm}^{2}$
Thus, Area of ABCD = x =
${16cm}^{2}$