Question

The diagonals of a parallelogram are 56 in and 34 in and intersect at angle of 120° find the length of the shorter side

Answer 1

Diagonals of a parallelogram bisect each other.

The opposite sides of a parallelogram are parallel and equal.

In a triangle, the larger angle has a longer opposite side and a smaller angle has a shorter opposite side.

Law of cosine: If a, b, c are three sides of a triangle and A is the angle opposite to the side a, then

`a^2=b^2+c^2-2bc\cos A`

The diagonals of a parallelogram are 56 inches and 34 inches. They bisect each other and form 4 triangles.

Let ABCD is a parallelogram and the diagonals AC and BD intersect each other at point O.

AB parallel to CD , AB=CD.

BC parallel to AD , BC=AD.

Diagonals intersect at an angle of 130 degrees.

m∠AOD=120 degree.

BD is a straight line. So,

m∠AOD+m∠AOB=180 degree

120+m∠AOB=180 degree

∠AOB =180-120=60 degree.

The opposite side of 130∘, (AD and BC) are the longer sides and the opposite side of 60∘, (AB and CD) are the shorter sides.

Use the law of cosine in triangle AOB,

`AB^2=OA^2+OB^2+2(OA)(OB)\cos 60^(\circ)`
`AB^2=28^2+17^2+2*28*17\cos 60^(\circ)`
`AB^2=784+289+476`
`AB^2=1549`
`AB=39.35\text{ in}`

The length of shorter side is AB =39.35 in.