Hotel shooting bd

Solution: 1 In quadrilateral ABCD, AD//DC, AB=DC,

So the quadrilateral ABCD is an isosceles trapezoid, so the diagonal AC=BD.

That is, AC=BD= 10, and let the area of the trapezoid be s, then

S=(AC*BD)/2×sin 120 degrees (sin 120 degrees = root 3/2), that is.

S = [( 10× 10)/2× (root number 3)/2 = 25× root number 3 (area unit).

2. If the quadrilateral is a parallelogram, the diagonal AC and BD are equally divided, that is, AO = OC, BO = OD = 10/2 = 5.

If the vertical line passing through point B is AC at point E, then △BEO is a right triangle with angle E = 90 degrees and angle BOE=60 degrees (/kloc-complementary angle of 0/20 degrees). BE=BO×sin60 degrees = 5× (root number 3)/2

OE=BOcos60 degrees = 5× 1/2.

OE:BE=BE:AE, so AE = Be 2/OE = 75/ 10.

AO = AE+EO =(75/ 10)+5/2 = 10，

AC=2×AO=2× 10=20，

Let the area of the parallelogram be s,

Then s = 2× triangle ABC = 2× (1/2× AC× Be).

= (2× 1/2× 20× 10/4) radical number 3

= 50× root number 3

Answer: 1. When the quadrilateral ABCD is isosceles trapezoid, its area is 25× root number 3 (area unit);

2. When the quadrilateral ABCD is a parallelogram. Its area is 50× root number 3 (area unit).