How do you do those geometry problems? Math and Geometry Questions?

Updated on educate 2024-04-01
12 answers
  1. Anonymous users2024-02-07

    Geometry... This question is king, first of all, you can consolidate the knowledge points by doing the problem.

    For example, if you want to prove that the angles are equal, you can react quickly, by parallelism, or by similar congruence, or by relation to common angles, etc.

    If you want to prove a relation, such as ae+cf=ef, you have to learn to transform, for example, find a congruent shape, and correspond three variables together, such as a straight line segment, or a triangle, so that the relationship is very provable.

    I just liked geometry back then, so I did a book of exercises during the winter vacation, and I was familiar with geometry at that time, basically except for super difficult problems, many of the problems even knew where to start at a glance, and there were no dead ends in the process of confidently writing.

    Doing problems is a royal way, which allows you to see different problems and different solutions. If you really don't know how to do the general proof question, you can try the method of reverse extrapolation. Push backwards, from the conditions of the proof, as long as the known conditions can be introduced, and then write the other way around.

    But don't take it for granted, some drawings are very precise and tend to give you preconceived conditions... For example, if you don't explicitly tell you that it's a right triangle of 30, 60 and 90, you can't take this condition for granted.

    The writing process should be careful and the steps should be detailed, so that you can check the problem in time when you find something wrong. Writing carefully will help you clear your mind.

    There are also some variables that ask you to find the maximum and minimum value, this is to apply the function to the geometry problem, or no matter how it changes, a certain side, a certain angle is a fixed value... This requires you to find congruence with known fixed angles and fixed edges, similarly.

    The most important !! It is the knowledge points that you have to gnaw through, so that as soon as the question gives you conditions, you can quickly react to the implied conditions. Here's an example:

    Tell you that the two sides are parallel, you have to think that the inner wrong angle, the same angle is equal, the same side of the inner angle complementary, and then farther away, when there is an angle bisector, think of the angle bisector of the same side of the inner angle perpendicular to each other.

    This is my impression when I was learning geometry in junior high school, and I want to give you a reference.

  2. Anonymous users2024-02-06

    Find a few examples that are thoroughly studied.

  3. Anonymous users2024-02-05

    Question 2: The sum of the internal angles of the quadrilateral = 360 degrees, because the angle cfa=bea=90 degrees, because vertical, because the angle a=50 degrees, so the angle eof=130 degrees, because the angle eof=angle boc, the angle to the apex is equal, so the angle boc=130 degrees.

  4. Anonymous users2024-02-04

    The square side length is 5

    The solution yields a diamond with a high pf = 4

    From the Pythagorean law, bf=3

    So pe=2

    So the area of PEBC = (5+2)*4 2=14 shaded part area = square area - area of PEBC = 5x5-14=11

  5. Anonymous users2024-02-03

    GE eb=df fb=ad ab=ad 2ebSo:ge=ad 2=dm, the distance from the point d to both sides is equal, and it divides the angle equally. ∠dce=20°/2=10°

    dcb=10°+20°=30°

  6. Anonymous users2024-02-02

    The area of the triangle BCD = the area of the three-closed bright angle ABC (the same base and the same height) to obtain the area of the triangle AOB = 3

    Area of BCD = 12

    The ratio of BOC = 9:12=3 4 indicates that the height of AOD is 1 3 of BOC

    Area of AOD: Area of ACD = 1:4

    The area of AOD = 1

    Area of ABCD = 9 + 3 + 3 + 1 = 16

  7. Anonymous users2024-02-01

    It is proved that D and F are the midpoints of AD and BG respectively, so DF is the median line of the triangle bag, so DF is parallel to AG, that is, FH is parallel to AG

    In the same way, it can be shown that GH is parallel to AF

    So the quadrilateral AFHG is a parallelogram.

    If you connect AH and cross BC at the point M, then M is the center of GF, and G, F are the third points of BC, so M is the midpoint of BC, and DM is parallel to AC and DM is parallel to BH

    So AC is parallel to BH

    The same goes for connecting EM

    CH is parallel to AB

    So the quadrilateral abhc is a parallelogram.

  8. Anonymous users2024-01-31

    Do AE first, DF is perpendicular to the points E, F. After the AEDF is removed, the remaining part is combined into the triangle A(D)BC, thus.

    Because: angle ABC+angle DCB=90 degrees, then angle BA(D)C=180 degrees-90 degrees=90 degrees.

    Because: M is the midpoint of AD, N is the midpoint of Benfc, then Mn is the midline of BC (after resection)!

    Then: the midline theorem with a triangle gets: mn=1 2

    bc-ad)

  9. Anonymous users2024-01-30

    Proof: After D is used as DEAB to BC to E, and DFMN is passed to BC to F, BE=AD, MN=DF, NF=MD, and DEC=B can be obtained

    cf=nc-nf=nc-md=bc/2-ad/2=1/2(bc-ad)

    Whereas ec=bc-be=bc-ad

    Therefore, CF=1 2EC, then F is the midpoint of EC.

    From dec=b and b+c=90, dec+c=90, edc=90, dec is rt, then df=1 2ec, so mn=df=1 2ec=1 2(bc-ad).

  10. Anonymous users2024-01-29

    The question is wrong, the median line of the trapezoid should be equal to the sum of the upper bottom plus the lower bottom divided by 2.

  11. Anonymous users2024-01-28

    First of all, it's a matter of course, when I was in junior high school, I did a book of circle practice questions just to learn circles, practice makes perfect.

    Again, as everyone said, the picture is very important, the hand must not be lazy, and when there is no picture, you must draw the picture yourself.

    In fact, there are many geometry questions, at least at first glance you should be able to know which type of geometric figure this question is tested, whether it is a parallelogram, a triangle, or a circle, and then analyze it down. This is the most basic first step.

    Then, when you determine what kind of graph is mainly examined, you have to think about it, what are the theorems of this graph, in fact, there are very few in junior high school, there are only a few in each category, one by one, not blind, there are skills.

    To combine the topic, see what it asks, for example, ask for the edge, you have to see, what kind of edge is this, are there other sides in the topic that have a special relationship with it, if so, what is the connection between them? At this time, it is necessary to pass the theorem, see which theorem can connect them, and do more questions, naturally at a glance.

  12. Anonymous users2024-01-27

    Review the question first, and then mark the conditions given by the question on the diagram, it is best to draw a diagram on the scratch paper now, and try to draw on the scratch paper if you need to do auxiliary lines, and then think about the process to see if you can draw a conclusion.

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