A rectangle intersects B rhombus

The condition that a set intersects represents that both sets should be satisfied.

So your assumption should be:

a=rhomboid={a quadrilateral satisfies 1 pair of sides parallel 2 sides equal 3 diagonals bisect and perpendicular}

b=rectangle={a quadrilateral satisfies 1 pair of sides parallel 2 diagonals bisect 3 four angles are equal} so. a intersection b = {a quadrilateral satisfies 1 pair of sides parallel 2 sides are equal 3 diagonals bisect and perpendicular and satisfies a parallel sides b bisects diagonals c four angles are equal}

That is, {a quadrilateral satisfies 1 pair of sides parallel 2 sides are equal 3 diagonals bisected and perpendicular 4 diagonals bisected 5 four angles are equal}=

a={rectangle} intersects b={rhombus}=

A rectangle is a parallelogram with a right angle, a diamond is a parallelogram with equal adjacent sides, and their intersection is a parallelogram with a right angle and a set of parallelograms with equal adjacent sides, then it can only be a square.

The square has the characteristics of both a rectangle and a diamond and is their common element.

Your problem is getting the elements wrong. In the set of all rectangles, there are marginal unequals, the kind of rectangle you speak; There are also marginal equals, i.e., squares. In the set of all diamonds, there are no right angles, that is, the kind of diamonds you say; There are also those with right angles, i.e., squares.

The intersection of the set of diamonds and the set of rectangles is the square. Your understanding is that the determination conditions of diamonds and rectangles are considered elements, and the actual elements are all rectangles and diamonds.

x|x is a square} a=, the geometric feature of its elements is a parallelogram with a large rubber fiber at right angles, b=, the geometric feature of its elements is a parallelogram with equal adjacent sides, by the nature of the intersection, a such as taking the characteristics of the elements in b is that there is a parallelism with a right angle and equal adjacent edges.

<> quadrilateral ABCD is a diamond, ab=bc, bac= dac, Zheng Bang ab=ac=10, Qiyan ab=bc=ac=10, abc is equilateral triangle shouting pure shape, bac=60°, dac=60°, bad= bac+ dac=120°

So the answer is: 60°

Solution:1The elements in set A B have both the characteristic properties of set A and the characteristic properties of set B. That is, the four sides are equal, and the four corners are right-angled, so it is a square. So.

The intersection of A and B is the set of Hui in the form of a square annihilation.

Answer: a=, b=, then a b=

The common feature of a diamond shape and a rectangular shape is the square.