Finding the Area of a Parallelogram

Question

I usually know how to find it, but I can't seem to find an equation or anything that works for this one. I have asked multiple questions about this that have gotten mixed answers so I'm confused.

The question is: "Find the area of a parallelogram with sides of length 6" and 8" and the diagonals intersect at an angle of 114 degrees."

I can't find a way to get the area given the angle that the diagonals hit. I've done it with diagonal lengths and with an angle at the corner, but not like this. Any help would be appreciated.

Answer 1

I saw an expert answer with video, and his solution was correct. However, I will try to re-explain the same solution here:

When you draw the diagonals, you will see 4 different triangles formed.

We can use the law of cosines twice. Once using angle 114 degrees and side 8", and again using angle 66 degrees and side 6":

Let a represent half of one of the diagonals, and b represent half of the other diagonals. We know that the diagonals of a parallelogram bisect each other.

Law of cosines:

c² = a²+b² - 2abcos(C)

So

8² = a²+b² - 2abcos(114)

6² = a²+b² - 2abcos(66)

Subtract one equation from the other to get

28 = -2abcos(114) + 2abcos(66)

14 = ab(cos(66) - cos(114))

ab = 14 / (cos(66) - cos(114))

We can simplify the denominator using a sum-to-product trigonometric identity:

cos(x) - cos(y) = -2sin((x+y)/2)*sin((x-y)/2)

So

cos(66) - cos(114) = -2sin((66+114)/2)*sin((66-114)/2)

= -2sin(90)*sin(-24)

= -2sin(-24)

also, sin(-x) = -sin(x)

so

cos(66) - cos(114) = 2sin(24)

We can plug this in to simplify the value of ab:

ab = 14/(2sin(24))

ab = 7/sin(24)

Now we can find the area of each triangle with

A = 1/2*ab*sin(angle) where the angle is between a and b.

Since we solved for ab, we can plug it in. And we have 2 sets of equal triangles, so we calculate the area for each triangle, and then multiply by 2.

A_{parallelogram} = 2A_triangle1 + 2A_triangle2

A_triangle1 = 1/2*a*b*sin(66)

A_triangle2 = 1/2*a*b*sin(114)

We know a*b is 7/sin(24), so you can do the rest to find the area of the parallelogram.

Answer 2

The parallelogram has dimensions 8" on top, 6" down one side, 8" on bottom, and 6" up other side.

The diagonals intersect at an angle of 114^o, so there are NO right angles.

Divide the parallelogram into 2 triangles, each with sides 6, 8 and unknown. The sum of the angles of a triangle is 180^o. Both of our 2 triangles has one angle > 90^o with the other two less than 90^o.

The larger angle = 114^o due to symmetry.

We have the situation where we have a side, angle and side (SAS) and the area of the triangle is

1/2*6*8*sin(114^o) = 1/2*6*6*0.91355 = 21.92509 in².

Thus the area of the parallelogram is 2*21.92509 = 43.85018 in² to 5 decimal places.