People teaches printing plate eighth grade mathematics final examination questions.
A, multiple-choice questions * * * 3 points for each small question, ***30 points * * *
1. Known? If the circumference of ABCD is 32 and AB=4, then BC = * * * * *
A.4b 12 c . 24d . 28
2. If the value of the score is 0, * * * * *
A.x=﹣3 B. x= 3 C. x=3 D. x=0
3. In the following transformation from left to right, * * * * * is factorized.
A.x2﹣6x+9=x***x﹣6﹣9***·***a+2******a﹣2***=a2﹣4
C.2a***b﹣c***=2ab﹣2bc·y2﹣4y+4=***y﹣2***2
4. In the following statement, the wrong one is * * * * * *
A. inequality x
B.-2 is the inequality 2x- 1
C. the solution set of inequality -3x > 9 is x > 3.
D. inequality x
5. As shown in the figure, △ABC and △A 1B 1C 1 are symmetrical about point O, and make the following statement:
①∠BAC =∠b 1a 1c 1； ②AC = a 1c 1； ③OA = OA 1；
④ The area of △ ABC is equal to the area of △A 1B 1C 1, and the correct one is * * * *.
A. 1 B. 2 C. 3 D.4
6. As shown in the figure, △ABC is known, so that the distance from P to ∠A is equal, and PA=PB. The correct way to determine point P is * * * *.
A.p is the intersection of bisectors of a and b.
B.p is the intersection of the bisector of the angle ∠A and the perpendicular bisector of AB.
C.p is the intersection of the heights of AC and AB.
D.p is the intersection of the perpendicular lines on both sides of AC and AB.
7. The following deformation is correct * * * *
A.B.
C.D.
8. As shown in the figure, if ∠ A = 100, then ∠B+∠D is * * * *.
A.80 b . 100 c . 160d . 180
9. If the equation about X = has an increasing root, then the value of m is * * * * * *.
A. 1d ﹣ 1
10. As shown in the figure, in? In ABCD, BC=7, CD=5, ∠ D = 50, and divided into ∠ABC, the following conclusion is incorrect * * * *.
A.∠C = 130 b . AE = 5 C . ED = 2d .∠BED = 130
Second, fill in the blanks * * * 3 points for each small question, ***24 points * * *
1 1. The range of x that makes the expression 1+ meaningful is.
12. If 9x2+kx+ 16 is a completely flat mode, then the value of k is or.
13. If the sum of the inner angles of a polygon is half of the sum of its outer angles, then the polygon is a polygon.
14. As shown in the drawing, △ABC rotates counterclockwise around point A, and then translates the grid to the right to get △DEF.
15. The integer solution of the inequality group is.
16. As shown in the figure, it is known that in △ABC, AB=AC=8cm, AD bisects ∠BAC, and point E is the midpoint of AC, then DE=.
17. As shown in the figure, the diagonal of ABCD intersects with O, AB=6, and the circumference of △OCD is 23. The sum of the two diagonal lines of ABCD is.
18. Observe the following equations in turn: A1=1-,a2=, a3=, a4=… Try to guess that the nth equation ***n is a positive integer * * an =, and the simplified result is.
Third, answer questions.
19. Break down the following types:
*** 1***x2﹣9y2
***2***ab2﹣4ab+4a.
20. Simplified evaluation: * * * *, where a=3 and b=.
2 1. Solve the inequality group:, and represent the solution set on the number axis.
22. As shown in the figure, in the plane rectangular coordinate system, the coordinates of the three vertices of △ABC are known as A * * *-5, 1 * * *, B * * *-2, 2 * * *, C * *-1respectively.
* * * 1 * * Shift △ABC to the right by 4 unit lengths and then shift 1 unit length to get △ A1b1c, and draw △ A1b1c.
* * * 2 * * △A2B2C2 and △ABC are symmetrical about the origin O, and draw △A2B2C2.
23.*** 10 * * * * 2014? A school in Zaozhuang simulated grade seven is going to buy a batch of notebooks to reward outstanding students. When I bought it, I found that I could get a 10% discount on each notebook. After the discount, more notebooks were bought with 360 yuan money than before 10. What is the price of each notebook before the discount?
24. * * *11* * * * 2015 spring? Juancheng county was last known. As shown in the figure, in the parallelogram ABCD, AC and BD intersect at point O, and point E and point F are the midpoint of BO and DO respectively. Try to prove:
*** 1***OA=OC，OB = OD
* * * 2 * * Quadrilateral AECF is a parallelogram;
* * * 3 * * If point E and point F are on the extension lines of DB and BD respectively, and BE=DF is satisfied, is the above conclusion still valid? Please explain the reason.
25. * * *11* * * * 2015 spring? As shown in Juancheng county map at the end of the period, it is known that ∠ C = 90 in Rt△ABC, and this triangle is folded along a straight line passing through point B, so that point C and point D coincide on the AB side.
*** 1*** When ∣ ∣ ∣.
***2*** Under the condition of * *1* *, if DE= 1, find the area of △ABC.
Reference answer
A, multiple-choice questions * * * 3 points for each small question, ***30 points * * *
1. Known? If the circumference of ABCD is 32 and AB=4, then BC = * * * * *
A.4b 12 c . 24d . 28
Test site: the nature of parallelogram. property in copyright
Special topic: calculation problems.
Analysis: AB=CD and AD=BC are obtained according to the properties of parallelogram, and the answer can be obtained according to 2 * * * AB+BC * * = 32.
Solution: Solution: ∫ Quadrilateral ABCD is a parallelogram.
∴AB=CD,AD=BC，
The circumference of parallelogram ABCD is 32,
∴2***AB+BC***=32，
∴BC= 12.
So choose B.
Comments: This question mainly examines the understanding and mastery of the properties of parallelogram, and the key to solve this problem is to use the properties of parallelogram to calculate.
2. If the value of the score is 0, * * * * *
A.x=﹣3 B. x= 3 C. x=3 D. x=0
Test center: the condition that the score value is zero. property in copyright
Analysis: If the score is zero, two conditions must be met at the same time: * *1* * molecule is 0; ***2*** Denominator is not 0.
Solution: Solution: x2﹣9=0, and x+3≠0 is obtained by the condition that the value of the fraction is zero.
The solution is x = 3, x ≠ 3,
∴x=3，
So choose: C.
Comments: This question examines the condition that the score is 0. If the score is zero, two conditions must be met at the same time: ** 1*** molecule is 0; ***2*** The denominator is not 0, which is the key to solve the problem.
3. In the following transformation from left to right, * * * * * is factorized.
A.x2﹣6x+9=x***x﹣6﹣9***·***a+2******a﹣2***=a2﹣4
C.2a***b﹣c***=2ab﹣2bc·y2﹣4y+4=***y﹣2***2
Test center: the significance of factorization. property in copyright
Analysis: according to factorization, a polynomial is transformed into the product of several algebraic expressions, and the answer can be obtained.
Answer: Solution: A, x2-6x+9 = * * * x-3 * * 2, so A is wrong;
B is the multiplication of algebraic expressions, so B is wrong;
C is the multiplication of algebraic expressions, so C is wrong;
D, transform a polynomial into the product of several algebraic expressions, so d is correct;
Therefore, choose: d.
Comments: This question examines the significance of factorization. Factorization is to transform a polynomial into the product of several algebraic expressions. Pay attention to the difference between factorization and algebraic expression multiplication.
4. In the following statement, the wrong one is * * * * * *
A. inequality x
B.-2 is the inequality 2x- 1
C. the solution set of inequality -3x > 9 is x > 3.
D. inequality x
Test site: inequality solution set. property in copyright
Analysis: According to the nature of inequality, we can get the solution set of inequality.
Solution: solution: a, inequality x
B, ﹣2 is unequal 2x ﹣ 1
C, the solution set of inequality -3x > 9 is X.
D, inequality x
So choose: C.
Comments: This question examines the solution set of inequality, and using the nature of inequality is the key to solving the problem.
5. As shown in the figure, △ABC and △A 1B 1C 1 are symmetrical about point O, and make the following statement:
①∠BAC =∠b 1a 1c 1； ②AC = a 1c 1； ③OA = OA 1；
④ The area of △ ABC is equal to the area of △A 1B 1C 1, and the correct one is * * * *.
A. 1 B. 2 C. 3 D.4
Test center: the center is symmetrical. property in copyright
Analysis: it can be judged according to the graphic properties of central symmetry.
Solution: Solution: If two centrally symmetric figures are congruent, then 124 is correct;
The distance from the symmetry point to the symmetry center is equal, so ③ is correct;
So 12344 are all correct.
So choose D.
Comments: This question mainly examines the nature of the central symmetric figure, and a correct understanding of the nature is the key to solving the problem.
6. As shown in the figure, △ABC is known, so that the distance from P to ∠A is equal, and PA=PB. The correct way to determine point P is * * * *.
A.p is the intersection of bisectors of a and b.
B.p is the intersection of the bisector of the angle ∠A and the perpendicular bisector of AB.
C.p is the intersection of the heights of AC and AB.
D.p is the intersection of the perpendicular lines on both sides of AC and AB.
Test center: the nature of the angular bisector; The nature of perpendicular bisector. property in copyright
Topic: the finale.
Analysis: Answer according to the judgment theorem of angle bisector and line segment perpendicular bisector.
Solution: Solution: The distance from point P to point A is equal.
Point p is on the bisector of point a;
And ∵PA=PB,
∴ Point P is on the perpendicular of line segment AB.
That is, the intersection of the bisector of P ∠A and the perpendicular bisector of AB.
So choose B.
Comments: This question examines the judgment theorem of the bisector of the angle and the vertical line in the line segment.
Points with equal distances to both sides of the angle are on the bisector of the angle; The point with equal distance to both ends of a line segment is on the middle vertical line of this line segment.
7. The following deformation is correct * * * *
A.B.
C.D.
Test center: the basic nature of scores. property in copyright
Analysis: According to the fact that both the numerator and denominator of the fraction are multiplied or divided by the same non-zero algebraic expression, and the value of the fraction remains unchanged, the answer can be obtained.
Solution: Solution: A, the numerator denominator is divided by different algebraic expressions, so A is wrong;
B, numerator and denominator are multiplied by different algebraic expressions, so b is wrong;
C and a are equal to zero, which makes no sense, so C is wrong;
D, the numerator and denominator of the fraction are multiplied or divided by the same non-zero algebraic expression, so D is correct;
Therefore, choose: d.
Comments: This question examines the basic nature of the score. Both the numerator and denominator of a fraction are multiplied or divided by the same non-zero algebraic expression, and the value of the fraction remains unchanged.
8. As shown in the figure, if ∠ A = 100, then ∠B+∠D is * * * *.
A.80 b . 100 c . 160d . 180
Test site: the nature of parallelogram. property in copyright
Analysis: according to the diagonal equality of parallelogram and the complementary solution of adjacent internal angles.
Solution: solution: ∵ parallelogram ABCD
∴∠B=∠D= 180 ﹣∠A
∴∠B=∠D=80
∴∠B+∠D= 160
So choose C.
Comments: This question examines the nature of parallelogram and must be mastered skillfully.
9. If the equation about X = has an increasing root, then the value of m is * * * * * *.
A. 1d ﹣ 1
Test center: increase the root of the fractional equation. property in copyright
Special topic: calculation problems.
Analysis: the denominator of the fractional equation is transformed into an integral equation, the value of X can be obtained by adding roots to the fractional equation, and the value of M can be obtained by substituting into the integral equation.
Solution: after naming: m- 1 =-x,
Starting from the fact that the fractional equation has an increasing root, it is concluded that x-2 = 0, that is, x-2 = 0 2,
Substituting x=2 into the whole equation gives m =- 1,
So choose D.
Comments: This question examines the increasing roots of fractional equations. The problem of increasing roots can be carried out as follows: ① Let the simplest common denominator be 0, and determine the increasing roots; (2) transforming the fractional equation into an integral equation; ③ Substituting the added root into the whole equation can get the value of relevant letters.
10. As shown in the figure, in? In ABCD, BC=7, CD=5, ∠ D = 50, and divided into ∠ABC, the following conclusion is incorrect * * * *.
A.∠C = 130 b . AE = 5 C . ED = 2d .∠BED = 130
Test site: the nature of parallelogram. property in copyright
Analysis: According to the nature of parallelogram and the definition of angular bisector, AB=AE, so AE=AB=CD=5, DE=2, C and D are adjacent, so they are complementary, so C = 130, so the answer can be determined.
Solution: Solution: ∫ parallelogram
∴∠ABC=∠D=50，∠C= 130
Once again ∵ evenly distributed ∠ABC
∴∠EBC=25
∴∠BED= 180 ﹣25 = 155
∴ The incorrect one is D,
So choose D.
Comments: This question mainly investigates the properties of parallelogram. In parallelogram, when the bisector of the angle appears, an isosceles triangle can be constructed, and then the problem can be solved by using the properties of isosceles triangle.
Second, fill in the blanks * * * 3 points for each small question, ***24 points * * *
1 1. The range of X that makes the expression 1+ meaningful is x≠ 1.
Test center: conditions for meaningful scores. property in copyright
Analysis: the score is meaningful and the denominator is not equal to zero.
Answer: solution: from the meaning of the question, the denominator x- 1 ≠ 0,
That is, when x≠ 1, the formula 1+ is meaningful.
So the answer is: x≠ 1
Comments: This question examines the conditions for the significance of scores. Understand the concept of score from the following three aspects:
*** 1*** score is meaningless? Denominator is zero;
***2*** Is the score meaningful? Denominator is not zero;
***3*** fractional value is zero? The numerator is zero and the denominator is not zero.
12. If 9x2+kx+ 16 is a completely flat mode, then the value of k is 24 or -24.
Test center: completely flat. property in copyright
Analysis: The first two terms and the last two terms here are the squares of 3x and 4, so the middle term is twice the product of 3x and 4, so k = 24.
Solution: The median term is twice the positive and negative product of 3x and 4.
So k = 24.
So fill in 24; ﹣24.
Comments: This question examines the completely flat way. The sum of the squares of two numbers, plus or MINUS twice their product, constitutes a completely flat way. Pay attention to the sign of the double product so as not to miss the solution.
13. If the sum of the inner angles of a polygon is half of the sum of its outer angles, then the polygon is a triangle.
Test center: polygon inner corner and outer corner. property in copyright
Analysis: use the theorem of polygon external angle sum to find its internal angle sum, and then get.
Solution: Solution: The sum of the inner angles of a polygon is half of the sum of its outer angles, and the sum of the outer angles of any polygon is 360.
The sum of the internal angles of this polygon is 180, so this polygon is a triangle.
So the answer is: three.
Comments: This question mainly investigates the inner and outer angles of polygons, and concludes that the sum of the inner angles of polygons is the key to solving problems.
14. As shown in the drawing, △ABC rotates 90 degrees counterclockwise around point A, and then translates 6 squares to the right to get △DEF.
Test center: the nature of rotation; The essence of translation. property in copyright
Analysis: By observing the image, we can see that △ABC rotates 90 counterclockwise around point A, and then translates to the right.
Solution: According to the image, Δ △ABC rotates 90 counterclockwise around point A, which is the same as Δ def, and it can coincide with Δ def by moving 6 squares to the right.
So the answer is: 90, 6.
Comments: This question examines the types of geometric transformation. Geometric transformation only changes the position of the graph, not the shape and size of the graph. This topic uses rotation transformations and translation transformations.
15. The integer solution of the inequality group is 0, 1, 2.
Test site: integer solution of linear inequality. property in copyright
Special topic: calculation problems.
Analysis: First, we can get the range of X according to the unary linear inequality group, and get the integer solution of the inequality group according to the fact that X is an integer solution.
Solution: solution: inequality group,
Solution,
Integer solutions of inequality groups are 0, 1 and 2;
So the answer is 0, 1, 2.
Comments: This question examines the solution of linear inequality and linear equation of one variable. According to the range of x, we can get the integer solution of x, and then we can get the value of a by substituting it into the equation. To find the solution set of inequality groups, we should follow the following principles: take the largest with the same size, take the smallest with the same size, find the middle with the smallest, and the largest cannot solve the largest solution.
16. As shown in the figure, it is known that in △ABC, AB=AC=8cm, AD bisects ∠BAC, and point E is the midpoint of AC, then DE= 4cm.
Test center: the center line on the hypotenuse of the right triangle; Properties of isosceles triangle. property in copyright
Analysis: AD⊥BC can be obtained according to the nature of isosceles triangle, and then the answer can be obtained according to the fact that the median line on the hypotenuse in a right triangle is equal to half of the hypotenuse.
Solution: solution: AB = AC, AD equally divided ∠BAC,
∴AD⊥BC，
∴∠ADC=90，
Point e is the midpoint of AC,
∴DE=AC=4cm.
So the answer is: 4cm.
Comments: This question mainly investigates the properties of isosceles triangle and right triangle. The key is to master the right triangle, and the median line on the hypotenuse is equal to half of the hypotenuse.
17. As shown in the figure, the diagonal of ABCD intersects with O, AB=6, and the circumference of △OCD is 23. The sum of the two diagonal lines of ABCD is 34.
Test site: the nature of parallelogram. property in copyright
Analysis: firstly, the length of CD can be obtained from the properties of parallelogram, and the length of OD+OC can be obtained from the condition that the circumference of △OCD is 23. Then, the diagonal lines of the four parallel sides of the parallelogram are divided into two, and the sum of the two diagonal lines of the parallelogram can be obtained.
Solution: Solution: ∫ Quadrilateral ABCD is a parallelogram.
∴AB=CD=6，
∫△OCD has a circumference of 23,
∴OD+OC=23﹣6= 17，
BD = 2DO，AC=2OC，
∴ and ABCD = BD+AC = 2 * * * Do+OC * * = 34,
So the answer is: 34.
Comments: This question mainly examines the basic properties of parallelogram and uses properties to solve problems. The basic properties of parallelogram are as follows: ① two groups of opposite sides of parallelogram are parallel respectively; ② Two groups of opposite sides of parallelogram are equal respectively; ③ The two diagonal lines of parallelogram are equal respectively; Diagonal bisection of parallelogram.
18. Observe the following equations in turn: a 1= 1﹣, a2=, a3=, a4=… Try to guess that the nth equation ***n is a positive integer ***an= ﹣, and the simplified result is.
Test center: conventional type: digital variation type. property in copyright
Analysis: According to the meaning of the question, we can know a 1= 1﹣, a2=﹣, a3=﹣, ... and then we can get the answer by simplifying it.
Answer: solution: ∫a 1 = 1,
a2=﹣，
a3=﹣，
…
∴ the nth equation an =,
As a result of simplification.
So the answer is:.
Comments: This question examines the changing law of numbers, and requires students to analyze the meaning of the question first, find the law and deduce the answer.
Third, answer questions.
19. Break down the following types:
*** 1***x2﹣9y2
***2***ab2﹣4ab+4a.
Test center: the comprehensive application of common factor method and formula method. property in copyright
Special topic: calculation problems.
Analysis: * * *1* * * The original formula can be decomposed by the square difference formula;
***2*** Extract A from the original formula, and then decompose it with the complete square formula.
Solution: Solution: * * 1 * * * Original formula = * * * x+3y * * * * * x ﹣ 3y * * *;
***2*** Original formula = A * * B2-4B+4 * * * = A * * * B-2 * * 2.
Comments: This topic examines the comprehensive application of common factor method and formula method, and mastering factorization is the key to solve this problem.
20. Simplified evaluation: * * * *, where a=3 and b=.
Test center: simplified evaluation of scores. property in copyright
Special topic: calculation problems.
Analysis: the two items in brackets in the original form are divided by the subtraction rule of fractions with the same denominator. At the same time, the simplest result is obtained through the deformation of the division rule, and the value of a and b can be obtained by substituting it into the calculation.
Solution: Solution: Original formula =? ***a+b***=，
When a=3 and b=, the original formula =.
Comments: The key to solve this problem is to examine the simplified evaluation of scores and master the laws of operational calculus.
2 1. Solve the inequality group:, and represent the solution set on the number axis.
Test site: solving a set of linear inequalities; Represents the solution set of inequality on the number axis. property in copyright
Analysis: Find the solution set of each inequality separately, and then find its common solution set and express it on the number axis.
Answer: Solution:,
From ①, x ≤ 3;
From ②, x >1,
Therefore, the solution set of the inequality group is:-1
On the number axis, it is expressed as:
Comments: This question examines the solution of a group of linear inequalities. Knowing the basic steps of solving a linear inequality is the key to solving this question.
22. As shown in the figure, in the plane rectangular coordinate system, the coordinates of the three vertices of △ABC are known as A * * *-5, 1 * * *, B * * *-2, 2 * * *, C * *-1respectively.
* * * 1 * * Shift △ABC to the right by 4 unit lengths and then shift 1 unit length to get △ A1b1c, and draw △ A1b1c.
* * * 2 * * △A2B2C2 and △ABC are symmetrical about the origin O, and draw △A2B2C2.
Test center: drawing-rotation transformation; Drawing-translation conversion. property in copyright
Special topic: geometric transformation.
Analysis: * * 1 * * According to the law of point translation, we get a1* *-1,0***, B 1** 2,1* *, C65438+.
* * * 2 * * A2 * * 5,-1* *, B2 * * 2,-2 * * *, C2*** 1,-4 * * are obtained according to the coordinate characteristics of points symmetrical about the origin.
Solution: solution: * * 1 * * * as shown in the figure:
***2*** as shown:
Comments: This topic examines the drawing-rotation transformation: according to the nature of rotation, the corresponding angles are all equal and equal to the rotation angle, and the corresponding line segments are also equal, so we can extract the equal line segments on the edge of the angle by making equal angles, find the corresponding points, and then connect them in turn to get the rotated figure. We also check the translation transformation.
23.*** 10 * * * * 2014? A school in Zaozhuang simulated grade seven is going to buy a batch of notebooks to reward outstanding students. When I bought it, I found that I could get a 10% discount on each notebook. After the discount, more notebooks were bought with 360 yuan money than before 10. What is the price of each notebook before the discount?
Test center: the application of fractional equation. property in copyright
Analysis: If the price before the discount is X yuan and the price after the discount is 0.9x yuan, indicating the purchase quantity before the discount and the purchase quantity after the discount, then the purchase quantity after the discount is more than 10, and the equation can be obtained and solved.
Solution: Solution: If the price before discount is X yuan, the price after discount is 0.9x yuan.
Judging from the meaning of the question,+10=,
Solution: x=4,
It is proved that x=4 is the root of the original equation.
A: Before the discount, the price of each notebook was 4 yuan.
Comments: This topic mainly investigates the application of fractional equation. The key is to correctly understand the meaning of the question, find out the equivalence relationship, and then list the equations. Pay attention to the test after solving the equations.
24. * * *11* * * * 2015 spring? Juancheng county was last known. As shown in the figure, in the parallelogram ABCD, AC and BD intersect at point O, and point E and point F are the midpoint of BO and DO respectively. Try to prove:
*** 1***OA=OC，OB = OD
* * * 2 * * Quadrilateral AECF is a parallelogram;
* * * 3 * * If point E and point F are on the extension lines of DB and BD respectively, and BE=DF is satisfied, is the above conclusion still valid? Please explain the reason.
Test center: the judgment and nature of parallelogram; Congruent triangles's judgment and nature. property in copyright
Analysis: * *1* * * The diagonal of the parallelogram is equally divided, and a conclusion can be drawn.
* * * 2 * * The quadrilateral whose diagonals bisect each other is a parallelogram, which can be proved according to this judgment theorem.
***3*** still holds, which proves that the quadrilateral still bisected by diagonal is a parallelogram.
Answer: Proof: * *1* * ∫ AC, BD is the diagonal line in the parallelogram ABCD, and O is the intersection point.
∴OA=OC,OB=OD.
* * * 2 * * *∫OB = OD, and points E and F are the midpoint of BO and DO respectively.
∴OE=OF，
OA = OC，
∴ Quadrilateral AECF is a parallelogram.
The conclusion of * * * 3 * * still holds.
Reason: BE = DF, OB=OD,
∴OE=OF，
OA = OC，
∴ Quadrilateral AECF is a parallelogram.
So the conclusion is still valid.
Comments: This topic examines the judgment and nature of parallelogram and congruent triangles.
25. * * *11* * * * 2015 spring? As shown in Juancheng county map at the end of the period, it is known that ∠ C = 90 in Rt△ABC, and this triangle is folded along a straight line passing through point B, so that point C and point D coincide on the AB side.
*** 1*** When ∣ ∣ ∣.
***2*** Under the condition of * *1* *, if DE= 1, find the area of △ABC.
Test center: folding transformation * * * folding problem * * *; Pythagorean theorem property in copyright
Special topic: proof questions; Open.
Analysis: * * *1* * According to the nature of folding: △ BCE △ BDE, BC=BD, when point D is just the midpoint of AB, AB=2BD=2BC, and ∠ C = 90, so ∠ A = 30; When the new condition ∠ A = 30, we can know from the folding properties that ∠ EBD = ∠ EBC = 30, and ∠ A = 30 and ED⊥AB, which proves that D is the midpoint of AB;
* * * 2 * * In Rt△ADE, the values of AE and AD can be found according to the values of ∠A and ED, and D is the midpoint of AB, so the length of AB can be found. In Rt△ABC, the values of AC and BC can be calculated according to the values of AB and ∠A, and substituted into S△ABC=AC×BC.
Answer: Solution: * * 1 * * * The new condition is ∠ A = 30.
Prove that ≈A = 30 and ∠ C = 90, so ∠ CBA = 60,
The point ∵C is folded and coincides with the point d on the side of AB.
∴BE equally divided ∠ ∠CBD, ∠ BDE = 90,
∴∠EBD=30，
∴∠EBD=∠EAB, so eb = ea.
∫ED is the high line of △EAB, so ED is also the center line of isosceles △EBA.
∴D is the midpoint of AB
***2***∵de= 1,ed⊥ab,∠a=30 ,∴ae=2.
In Rt△ADE, according to pythagorean theorem, AD==,
∴AB=2,∵∠A=30，∠c = 90°，
∴BC=AB=.
In Rt△ABC, AC==3,
∴S△ABC=×AC×BC=.
Comments: This topic examines the folding transformation of graphics. In the process of solving problems, we should pay attention to the fact that folding is a symmetrical transformation and belongs to axial symmetry. According to the nature of axial symmetry, the shape and size of the figure remain unchanged before and after folding.