# Length of diagonals of parallelogram r 8 & 10 cm, angle included between diagonals is 60 degree. Find the area of parallelogram.

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The area of parallelogram could be found in various ways. One formula says that area of parallelogram is the sum of the areas of the 4 triangles which are formed by diagonals, inside of parallelogram.

We name the diagonals: AC and BD. The intersection between them is the point O. We know that the 2 segments (AO=OC and BO=OD) which are the result of the intersection of diagonals are equal, so AO=OC=5cm and BO=OD=4cm.

The triangle formed is AOD, in which we know the following:

2 sides and the angle between:

AO=5cm, DO=4cm and <AOD=60

AREA OF AOD TRIANGLE= (AO*OD*sin60)/2=(5*4*0.86)/2=8.6cm^2

AREA OF AOB TRIANGLE = (AO*OB*sin 120)/2= (5*4* 0*86)/2=8.6cm^2

AREA OF PARALLELOGRAM ABCD= 4*8.6= 34.4cm^2

Let's name the pallellogram ABCD. The diagonals are called AC and the other one BD. The angle of intersection is 60 degrees. Let's name the intersection O. AO=CO=10/2=5cm as intersection of diagonals are equal, as well as BO=DO=8/2=4cm.

To find the triangle AOD, we use 1/2 ab sin60

= 1/2(AO)(DO)sin60

= 1/2*4*5*sin60

= 10sin60

= 8.66 cm square

To find area of parellogram, knowing that each of the triangle equals to each other, we:

area of parallelogram= 8.66*4

= 34.64 cm square.

We know that the diagonals bisect each other in a parallelogram.

We konow that a diagonal divides a parallelogram into two congruent triangles.

Thus the area of 4 triangles in the parallelogram ABCD with its diagonals bisecting at E have to be equal in area and their sum is the area of the parallelogram. Let the angle BEC =60 deg.

Then the area of the triangle BEC=(1/2)EB*ECsin60

Area of the parallelogram ABCD= 4*{(1/2)EB*EC*sin60=

=4(1/2)(8/2)(10/2)sin60=40sin60=34.6411 sqcm