Question

Prove that area of parallelogram is 1/2(d1xd2)

Answer 1

here is ur answer buddy:)

In a parallelogram diagonals bisect each other.

Here we have 4 small triangles due to the 2 diagonals.

Now,by symmetry, there will be 2 set of triagnles of eql. area.

Let the angle between the diagonals be = k.

The supplimentry angle will be = 180 - k.

Area of a triangle = (1/2)abSin(C).

Thus the area of 1 set of trigls. will be = (1/2)*(d1/2 *d2/2)* Sin(k) = L(say).

As Sin(180-k) = Sin(k).

We will have the other two areas also = (1/2)*(d1/2 *d2/2)* Sin(k) = M(say).

Thus the total area is = 2L+2M = (1/2)*(d1*d2)*Sin(k).

You Cannot find the area without the angle between the diagonals. As the above formula depends on value of "k".

Area will be eql to (1/2)*(d1*d2) only when it is a Rhombus,

Square or a Rectangle , where angle between d1 d2 = 90 deg. Sin(k=90) = 1.

hope it helps you :)pls mark it as BRAINLIEST:::)

Answer 2

Answer:

Step-by-step explanation:

In a parallelogram diagonals bisect each other.

Here we have 4 small triangles due to the 2 diagonals.

Now,by symmetry, there will be 2 set of triagnles of eql. area.

Let the angle between the diagonals be = k.

The supplimentry angle will be = 180 - k.

Area of a triangle = (1/2)abSin(C).

Thus the area of 1 set of trigls. will be = (1/2)*(d1/2 *d2/2)* Sin(k) = L(say).

As Sin(180-k) = Sin(k).

We will have the other two areas also = (1/2)*(d1/2 *d2/2)* Sin(k) = M(say).

Thus the total area is = 2L+2M = (1/2)*(d1*d2)*Sin(k).

You Cannot find the area without the angle between the diagonals. As the above formula depends on value of "k".

Area will be eql to (1/2)*(d1*d2) only when it is a Rhombus,

Square or a Rectangle , where angle between d1 d2 = 90 deg. Sin(k=90) = 1.

hope it helps you :)pls mark it as BRAINLIEST:::)