Find the area of the parallelogram if BC=16.?

The area of the parallelogram = 2*area of one triangle .

C^2=256

BD=256/2=128/2

DC=256/2=128/2

Area tri=((128/2*64/2)*2)

area parallelogram= 256

Given BC = 16

solving the height of the parallelogram.

sin(45) = BD/16

BD = 16sin(45)

BD = 11.31

now, solving the area of the parallelogram

A = Bh

A = (16)(11.31)

A = 181 unit^2  Answer//

It's the same area a square would have if the length of a diagonal of the square was 16. 

That's not true of all parallelograms. It's true in this case because of the 45 degree and 135 degree angles here. 

So see that, move △ABD so segment AD aligns with segment BC. (Because it is a parallelogram, they have the same length.)

Apply Pythagoras 

BC^2 = BD^2 + CD^2 

DB = CD because the triangle is Isoscles ( 2 x 45 degree angles0 

Hence substituting 

16^2 = 2BD^2 

256 = 2BC^2 

128 = BC^2 

BD  = CD = sqrt(128)  = 2^3sqrt(2) 

The area of a parallelogram is the base CD X perpendicular height BD 

Hence A = (8sqrt(2))^2 = 64x2 = 128  units^2  

Using the Pythagorean Theorem, or your knowledge of the 45-45-90 triangle, you should be able to figure out that the base and the height of each triangle is 16/√2 = 8√2.

In a 45-45-90 triangle, the legs are congruent.

s² + s² = 16²

2s² = 256

s² = 128

But now notice that the area of a parallelogram is base times height (which are both s).

A = s²

A = 128 sq. units

2x^2 = 16^2

x^2 = 128

x = 11.313708498984761

The area of the parallelogram ABCD: 128 units^2

These are two 45-45-90 triangles so the two non-hypotenuse sides are the same length.

One length is shared in both triangles so both triangles are the same.

In 45-45-90 triangles, the hypotenuse is √2 times the length of one of the other two sides.  So we can use this to solve for that side:

x√2 = 16

x = 16 / √2

x = 16√2 / 2

x = 8√2

The height and width are the same, so the area of one triangle is:

A = bh/2

A = (8√2)(8√2) / 2

A = 64 * 2  / 2

A = 64

And finally, add in the second triangle and the total area is:

64 + 64 = 128 unit²

The surface area of the triangle BDC is:

a₁ = [BD * DC]/2 → where: BD = BC.cos(45)

a₁ = [BC.cos(45) * DC]/2 → where: DC = BC.cos(45)

a₁ = [BC.cos(45) * BC.cos(45)]/2

a₁ = BC².cos²(45) → given that: BC = 16

a₁ = 16².cos²(45) → recall: cos(45) = (√2)/2

a₁ = 16².[(√2)/2]²

a₁ = 256 * [2/4]

a₁ = 128

The surface area of the triangle ABD is: (similar to the triangle BDC)

a₂ = 128

The surface area of the parallelogram ABCD is:

= a₁ + a₂

= 256