oubaas
AB = BD = CD = 16√2 / 2 = 8√2
area = AB^2 = 64*2 = 128 square units
or
area = BC^2/2 = 256/2 = 128 square units
Engr. Ronald
Given BC = 16
solving the height of the parallelogram.
sin(45) = BD/16
BD = 16sin(45)
BD = 11.31
Apply the Pythagorean theorem to solve for DC.
DC = 16^2 - (11.31)^2
DC = 11.31
Solving it's Area
A = Bh
A = (11.31)(11.31)
A = 127.91 square units or 128 square units...Answer//
lenpol7
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
Puzzling
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
llaffer
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