相似三角形经典题(含答案)（Similar triangle classic questions (including answers)）

相似三角形经典 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 (含 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 )（Similar triangle classic questions (including answers)） 相似三角形经典题(含答案)(Similar triangle classic questions (including answers)) Similar triangle classical exercises Example 1. Choose a similar triangle from the following triangles Example 2 is known: as in figure ABCD, the ratio of the perimeter to the sum if... Figure 3 cases, known to, to prove that. Example 4 which of the following statements are true and which ones are wrong? (1) all right triangles are similar. (2) all isosceles triangles are alike (3) all isosceles right triangles are similar. (4) all equilateral triangles are alike Figure 5 example, D is a point on the AC, D DE E the dotted line, in the side, the small triangle and point D, point E and a vertex with similar composition. Draw as much as possible to meet the conditions of the graphics, and that of line DE painting. Figure 6 cases, a person holding a small scale paintings engraved with cm, standing about 30 meters away from the poles, the arm straight forward, small scale vertical ruler, see about 12 paintings just over the poles, the known arm length of about 60 cm high, for the wire rod. Figure 7 cases, in order to measure a high-rise MN Xiaoming, put a mirror in the A from N 20m, NA back to C along the Xiao Ming, just from the mirror to see the roof of M, if m, his eyes from the ground height of 1.6m, please help you calculate Xiaoming the height of the building (accurate to 0.1M). The two triangles in the 8 lattice diagram are similar triangles, and the reasons are given Example 9 determines whether the case is similar and explains the reasons for the following groups of conditions: (1) (2) (3) Example 10. In the following graph, there is no similar triangle. If it exists, show them in letters, and briefly explain the basis for identification Example 11 is known: as in Fig., in the case of angular bisector, try using a triangle similar relation description Example 12, the known three side length is 5, 12 and 13, and its similar maximum length is 26, the area of S. 13 cases in a mathematics activity class, the teacher let the students to the playground to measure the height of the flagpole, and then come back to AC measurement method for their measurement is. Xiaofang: take a 3.5 meter high pole upright in the 27 meters away from the flagpole at C (pictured), then walk along the BC direction D, the top of the flagpole and pole top A visual E is in the same line, C D, and measured the distance between two points is 3 meters, Xiaofang mesh is 1.5 meters high, so that you can know the high flagpole. Do you think this measurement method is feasible? Please explain the reason Figure 14. cases, in order to estimate the width of the river on the other side of the river, we can select a target as A, on this side of the river and then points B and C, so, then choosing E, BC and AE to determine the intersection point is D, measured in meters, meters, meters, you can find the distance between the two sides of AB roughly? Figure 15. cases, in order to find the island peak height of AB, DC and FE to establish a benchmark in D and F, the benchmark is 3 feet high, separated by 1000 step (step 1 is equal to 5 feet), and AB, CD and EF in the same plane, from the G DC benchmark back the 123 step, can see peaks A and C benchmark top end in a straight line, from the H FE benchmark 127 steps back, can see peaks A and E benchmark top on a straight line. How much is the horizontal distance BD AB and its peak height and benchmark CD? (ancient problems) Figure 16 example, known Delta ABC boundary AB = AD, AC = 2, BC = high on the side. (1) seeking the length of BC; (2) if there is a square edge on AB, the other two vertices are on AC, BC, respectively, and the area of this square is called Similar triangle classic Exercises answer 1. cases of the solution, five and six, and the similar, similar, three or four, and similar 2. solution is a parallelogram, so, l ~, Again, so, and the perimeter of the perylene ratio is 1:3. Again, dry. 3 cases analysis, so as to, if further proof, the problem must pass. To prove dreams, *. Again, l, Star. To dreams, *. In dreams, and in R ~ Case 4. analysis (1) is incorrect, because in the right triangle, the size of the two angles is uncertain, so the shape of the right triangle is different (2) not correct either, The vertices of an isosceles triangle are not of definite size, so the shape of an isosceles triangle is also different (3) right. There are isosceles right triangle ABC and, among them, Then, The three sides are a, B, and C, and the edges are, Then, So, l ~. (4) is correct, and is an equilateral triangle, the corresponding angles are equal, the corresponding edge is proportional to it. Answer: (1) and (2) incorrect. (3) and (4) correct Example 5. solutions: Painting slightly. The analysis of 6. cases of the narrative can draw the geometry as shown below, the CM cm m, m, m, and BC. ~ ~ because, again, so, so you can find the BC long. So, l ~ solution. Hence. Again, l, So, l ~ *,. And cm cm meters, meters, meters, meters. The pole star is 6 meters high. Example 7. analysis according to the law of Physics: the incident angle of light is equal to the angle of reflection, so that the similarity relation is clear Because the solution, so so. So, that is. So (m) This shows that this is a practical application, the method seems simple, but in fact it is very clever, saving the use of instrumentation to measure the trouble Example 8.. It is impossible to judge these two graphs if they are not painted in the grid. In fact, the lattice virtually adds to the condition the length and the angle The solution is in the grid, so.., Again. So. So ~. Explain the problems encountered in the grid point, we must fully find the various conditions, do not make omissions In 9. cases (1) because the solution to it; (2) because the two triangles only, the other two are not equal, and not so similar; (3) because, so it is similar In 10. cases (1) and two equal solution; (2) to two equal; (3) to two equal; (4) to both sides proportionally equal angles; (5) to both sides proportionally equal angles; (6) to both sides proportionally equal angles. Analysis of 11. cases with a 65 degree angle of the isosceles triangle, the angle is 72 degrees, and BD is the bisector of the corner, so, you can launch to, and then by the similar triangle corresponding edge is proportional to the ratio between the line launched. That star. But equally, dry. And so, so, so, so, L. That (1) has two angles equal, then the two triangles are similar, this is the judgment of two triangles. The most commonly used method, and according to the equal angle position, can determine which side is the corresponding edge. (2) to explain the product of a line, or the square formula, usually to prove the scaling formula, or, again, to derive the product formula or the square formula according to the basic nature of the proportion By the analysis of 12 cases of the three sides can be judged as a right triangle, and because it is also a right triangle, so, then by the maximum edge length is 26, can calculate the similarity ratio, two right angle side to calculate, and obtain the area. The solution of a three side in order,,, L. And to dreams, *, Again, *. *. 13. cases analysis method to judge whether it is feasible, should consider the use of this method combined with our existing knowledge can be obtained according to the flagpole high. This measuring method, F to G, CE to H, so that, and GF, HF, EH and AG, this can be obtained, so the AB can be obtained. The flagpole The solution is feasible. The reasons are as follows: The flagpoles high. F for G, CE H (pictured). So ~. Because, so So, that is, by, so the solution (m) So the height of the flagpole is 21.5 meters It shows that the method should be practical and feasible in concrete measurement Example 14. solutions:, L ~, (m), a: between the two sides of AB is roughly 100 meters away. Example 15. answer: rice, step, (Note:.) 16. cases analysis: BC long, need to draw solution, because AB and AC are higher than AD, so there are two kinds of situations, namely D in BC or D in the BC extension line, so long for the BC to two to discuss the situation. For the area of a square key is the length of the side for a square. Solution: (1) as above, by the AD BC group, by the Pythagorean theorem BD = 3, DC = 1, BC = so BDDC = 3 + 1 = 4. As follows, BD = 3, DC = 1, so BC = BD = CD = 3-1 = 2. (2) as shown by the graph, BC = 4, and so is ABC. Hence, the right triangle. The AEGF is a square, set GF = x, FC = 2x, GF "AB dreams, so, that is. So, dry. As follows, when BC = 2, AC = 2, Delta ABC is an isosceles triangle, as an CP AB in P, AP = r, In Rt APC, by the Pythagorean theorem CP = 1, Dreams GH / / AB, R ~ Delta CGH Delta CBA, dreams, R Therefore, the square has an area of or Third (lower) similar triangle First pages, 6 pages (similarity triangle's nature and application) practice roll Fill in the blanks 1. When the similarity ratio between two similar triangles is 3, their perimeter ratio is..; 2, if the delta delta A to ABC 'B' C ', and the perimeter of delta ABC is 12cm, then the perimeter of delta A' B 'C' for; 3, as shown in Figure 1, in ABC, BE, CD line intersect at point G, then the delta GED:S Delta GBC= = S; 4, as shown in Figure 2, the ABC / B= / AED, AB=5, AD=3, CE=6, AE=; 5, as shown in Figure 3, ABC, M AB is the midpoint of the N on BC, BC=2AB / BMN= / C, is a ~ Delta, similarity ratio =; 6, as shown in Figure 4, the trapezoidal ABCD, AD / / BC S, Delta ADE:S Delta BCE=4:9, Delta ABD:S Delta ABC= S; The perimeter of 7 and two similar triangles are 5cm and 16cm, respectively, and the ratio of the bisector of their corresponding angles is; 8, as shown in Figure 5, the BC=12cm in ABC, D, and F are three points AB, E, G is three points AC, DE+FG+BC=; The ratio of the area of the two and the 9 triangles is 2:3, and the ratio of them to the angle is equal to the ratio of the height of the opposite side; 10, it is known that there are two triangles similar, one side length is 2, 3 and 4 respectively, and the other side length is x, y and 12 respectively. Then the values of X and y are respectively; Two, multiple-choice questions 11, the following polygon must be similar to (), A, two rectangles, B, two diamond, C, two squares, D, two parallelogram In 12, ABC, BC=15cm, CA=45cm, AB=63cm, the shortest edge of another and it is similar to the triangle is 5cm, is the longest side (18cm) is A, B, 21cm C, 24cm D, 19.5cm 13, as shown in ABC, BD, CE to the high point of O, the following conclusion is wrong () A, CO, CE=CD, CA, B, OE, OC=OD, OB C, AD, AC=AE, AB, D, CO, DO=BO, EO 14, known in ABC / ACB=900, CD, AB in group D, if BC=5, CD=3, AD (long) A, 2.25 B, 2.5 C, 2.75 D, 3 15, as shown in figure ABCD, the edge of square BC is on the bottom QR of the isosceles right triangle PQR, The other two vertices, A and D, are on PQ and PR, and PA:PQ equals () A, 1:B, 1:2, C, 1:3, D, 2:3 16, as shown in figure D, and E are Delta ABC edge AB and AC point, ==3, And / AED= / B, Delta AED and delta ABC is the area ratio is () A, 1:2, B, 1:3, C, 1:4, D, 4:9 Three, answer questions 17, figure, known in the delta ABC, CD=CE / A= / ECB, CD2=AD - BE test. 18, known as shown in ABC, DE, BC, AD=5, BD=3, S and delta ADE:S Delta ABC value. 19, known square ABCD, C straight line, respectively, AD, AB extension line at points E, F, and AE=15, AF=10, square ABCD for the length of the side. 20, known as shown in the equilateral Delta CDE and B respectively, A ED, DE extension line, DE2=AD and EB, and the degree of angle ACB. 21, known as shown in ABC / C=600, AD, BC in D group, BE group AC E, Delta CDE Delta CBA to explain. 22, known, as shown in figure F, ABCD edge, DC extension of the line point, link AF, pay BC at G, hand in BD at E, try to explain AE2=EG EF 24. ABC, D, E / C=900, respectively AB, AC on AD, AB=AE AC, ED AB (13) to verify the above In 25, ABC, M and AC is the midpoint, side of the AE=BA connection EM, and extend the BC line to D, verify the BC=2CD AB=AC, the 26 known isosceles triangle ABC, AD, BC in group D, CG, AB, AD, AC BG respectively in E, F, BE2=EF and EG prove: 27, known in ABC, AD / BAC=900 BC in D P group, AD midpoint, BP extension line AC to E EF BC in F, an EF2=AE AC confirmation: 28., as shown in the parallelogram, 1. APD ~ CDQ Two Map your own painting, with a triangle of 30 degrees can be drawn out Dreams of an isosceles triangle ABC / ABC = 120 DEG L / DAP= / DCQ=30 / CDQ / PDA=150 ~ * ~ / ADP / APD=150 degrees and dreams L / CDQ= / APD / DAP= / QCD and dreams Star delta APD Delta CDQ ~ AP/CD=PD/DQ frequency D is the midpoint of AC AD=DC dreams AP/DP=AD/DQ AP/AD=PD/QD perylene perylene perylene / PDQ= / PAD dreams Star delta APD to DPQ 3. a triangle has 1 angles of 30, and the other has 2 30 degrees angles, in favor of the 155| review (6) (1) dreams / ABC=120 / A= / L degrees, C=30 degrees, Dreams / ADP+ / APD=150 / ADP+ / QDC=150 degrees degrees, L / APD= / CDQ, Star delta APD to CQD (2) set up; as shown Dreams / ADP+ / APD=150 / ADP+ / QDC=150 degrees, degrees, R / APD= / CDQ / A= / C, and Star delta APD to CQD / A= / C only, the other corresponding angle are not equal, therefore, Delta APD and delta DPQ is similar; (3), two triangle into a more general condition, but the ABC must be an isosceles triangle, and / EDF= / A, otherwise it is not established.