In the quadrilateral learning content of junior high school mathematics, we will generally learn quadrilateral, parallelogram, trapezoid, rectangle, diamond, square and other related knowledge. Today, we will talk about how mathematics in the senior high school entrance examination examines the diamond shape.
What is a diamond?
A set of parallelograms with equal adjacent sides is called a diamond.
As a special parallelogram, rhombus not only has all the properties of parallelogram, but also has its own unique properties: for example, the four sides of rhombus are equal; Diagonal lines of the rhombus are vertically bisected, and each diagonal line bisects a set of diagonal lines of the rhombus; The diamond is not only a central symmetrical figure, but also an axisymmetric figure.
Mathematics in the senior high school entrance examination, diamond-related problems, typical example analysis 1:
As shown in the figure, it is known that BD bisects ABF and intersects AE at point d,
(1) solution: BAE's bisector AP (requirements: draw with a ruler, keep drawing traces, and don't write);
(2) Let AP intersect BD at point O and BF at point C to connect CD. When ACBD proves that quadrilateral ABCD is a diamond.
Test center analysis:
Determination of diamond shape; Drawing-basic drawing.
Stem analysis:
(1) Just make the bisector AP of BAE according to the method of angular bisector;
(2) Prove ABOCBO according to ASA, and get AO=CO and AB = CB; then prove ABOADO according to ASA, and get BO = DO. If the quadrilateral bisected by the diagonal is a parallelogram and a group of parallelograms with equal adjacent sides are diamonds, it can be proved that the quadrilateral ABCD is a diamond.
Thinking about solving problems:
This topic mainly examines the method of angle bisector, the judgment of diamond and the judgment and nature of congruent triangles. Mastering the judgment of diamonds is the key to solving the problem.
Mathematics in the senior high school entrance examination, diamond-related problems, analysis of typical examples II:
As shown in the figure, in ABC, ACB = 90°, D and E are the midpoint of BC and BA respectively, and d E and F are connected on the extension line of DE, and AF = AE.
(1) verification: quadrilateral ACEF is a parallelogram;
(2) If the quadrilateral ACEF is a diamond, find the degree of B. 。
The nature of diamonds; Determination of parallelogram.
(1) CE=AE=BE According to the fact that the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse, AF = CE 1=2 can be obtained according to the property that three lines of an isosceles triangle are in one line; Then F = 3； can be obtained according to the equilateral corner; Then you can get 2 = F；; Then according to the equivalent angle, two straight lines are parallel to each other, and then CEAF can be obtained by using a set of parallel opposite sides.
(2) According to the fact that all four sides of the rhombus are equal, AC=CE = AE can be obtained, so that AEC is an equilateral triangle, and CAE = 60 can be obtained according to the fact that each angle of the equilateral triangle is 60, and then the two acute angles of the right triangle are complementary angles.
This topic examines the nature of rhombus, the judgment of parallelogram and equilateral triangle. The midline on the hypotenuse of right triangle is equal to half of the hypotenuse, and the two acute angles of right triangle are complementary angles. Memorizing all the properties and judgment methods is the key to solving problems.
Judgement theorem of diamonds;
1, definition: a set of parallelograms with equal adjacent sides is a diamond;
2. Theorem 1: A quadrilateral with four equilateral sides is a diamond;
Theorem 2: Parallelograms with diagonal lines perpendicular to each other are rhombic.
The formula for calculating the area of rhombus is: S rhombus = length and height of base = half of the product of two diagonals.
In the mathematics of senior high school entrance examination, the diamond will be combined with other knowledge and closely linked to form more complicated comprehensive questions. Therefore, in the usual mathematics learning process, we must seriously master the relevant knowledge of the diamond and thoroughly understand every knowledge point, so that even if we encounter complicated problems, we don't have to be afraid.
Mathematics in the senior high school entrance examination, diamond-related problems, typical examples analysis 3:
As shown in the figure, it is known that the image of the quadratic function y=x2+bx+c intersects with the X axis at points A and B, intersects with the Y axis at point P, and the vertex is C( 1,? 2).
(1) Find the relation of this function;
(2) Make a symmetrical point D of point C about the X axis, and connect A, C, B and D in turn. If there is a point E on the parabola, make the straight line PE divide the quadrangle ABCD into two quadrangles with equal area, and find the coordinates of the point E;
(3) Under the condition of (2), is there a point F on the parabola that makes PEF a right triangle with P as its vertex? If it exists, find the coordinates of point F and the area of PEF; If it does not exist, please explain why.
Quadratic function synthesis problem; Algebraic geometry synthesis problem.
(1) Substitute vertex coordinate C( 1, 2) into y=x2+bx+c to get the relationship of this quadratic function;
(2) First, the analytical expression of the straight line PM is obtained, and then the coordinates of point E can be obtained immediately by connecting it with the quadratic function;
(3) According to the similar properties of triangles, first find GP=GF, then find the coordinates of point F, and then find the area of PEF.
This problem, called quadratic function synthesis, involves the solution of parabolic formula, the similarity of triangles and other knowledge points. It is a hot and difficult point in the senior high school entrance examination. Pay attention to the application of mathematical ideas such as the combination of numbers and shapes when solving problems, and students should strengthen training.
Mathematics for the senior high school entrance examination, problems related to diamonds, and analysis of typical examples 4:
As shown in the figure, parabola y=? 5x2/4+ 17x/4+ 1 intersects the Y axis at point A, the straight line passing through point A intersects the parabola at another point B, the BCx axis passing through point B and the vertical foot at point C (3,0).
(1) Find the functional relationship of straight line AB;
(2) The moving point P moves from the origin to C at a speed of one unit per second on the line segment o C, with the crossing point P as the PNx axis, the intersecting line AB at the M point, and the intersecting parabola at the N point. Let the moving time of the P point be t seconds and the length of MN be s units, find the functional relationship between S and T, and write the range of T;
(3) Connect CM and BN under the condition of (2) (regardless of the coincidence of point P with point O and point C). What is the value of t, and the quadrilateral BCMN is a parallelogram? Is the parallelogram BCMN a diamond for the value of t? Please explain the reason.
Quadratic function synthesis problem.
(1) The coordinates of A and B can be easily obtained from the meaning of the question, and then the functional relationship of straight line AB can be obtained by the undetermined coefficient method;
(2) from s=MN=NP? MP, you can get s=? 5t2/4+ 17t/4+ 1？ (t/2+ 1), the answer can be obtained by simplification;
(3) If the quadrilateral BCMN is a parallelogram, then MN=BC, equation:? 5t2/4+ 15t/4=5/2, the value of t can be obtained by solving the equation, and then the quadrilateral BCMN is a diamond when t is taken.
This topic examines the analytical expression of function by undetermined coefficient method, the relationship between line segment length and function expression, the nature and judgment of parallelogram and rhombus. This problem is very comprehensive and difficult, and the key to solve it is the application of the idea of combining numbers with shapes.