ACCA_P3_关键模型汇总

ACCA_P3_关键模型汇总
ACCA_P3_关键模型汇总

ACCA P3 关键模型汇总

汇编:詹也

浙江财经大学

2014年

快速浏览法

Part A(战略位势)

1.1The strategy lenses(战略维度)

2.Strategy as design.

The design lens views strategy as the deliberate positioning of an organisation as the result of some ‘rational, analytical, structured and directive process’. It is the responsibility of top management to plan the destiny of the organisation. Lower levels of management carry out the operational actions required by the strategy.

The design lens is associated with objective setting and a plan for moving the organisation towards these objectives.

1Strategy as experience.

The experience lens views strategy development as the combination of individual and collective experience together with the taken-for-granted assumptions of cultural influences. Strategy as experience seems innately conservative. It could work well when a small incremental change is required within a stable environment. However, this view may become a major barrier to developing innovative strategies as experience may become rigid.

2Strategy as ideas.

It has a central role for innovation and new ideas. It sees strategy as emerging

from the variety and diversity in an organisation. It is as likely to come from the bottom of the organisation as from the top. Consequently, the organization should foster conditions that allow ideas to emerge and to be considered for inclusion in

a ‘mainstream strategy’.

1.2PESTEL

应用范围:当题目要求做“environmental analysis”、“analysis of the macro-environmental”或“analysis of the position of company,都可以用这个模型。

写作方式:

1.开头写一段“The PESTEL framework can be used to analyse the macro-environment. It considers political, economic, socio-cultural, technological, legal and environmental forces that affect the company”。

2. 根据案例内容,分别分析这6个宏观因素,以及各个因素如何影响企业的战略。

3. 根据分值,再加一个“summary”。

注意事项:

1.根据题目给出的信息量,不需要对这6个方面的因素面面俱到(有时候题目

并没有提及某一方面的信息)。当然,如果时间充裕,可以自己推测一下2.PESTEL模型只关注企业的外部情境,不适合做企业的内部资源和能力分析。

1.3Porter’s diamond(波特的“钻石模型”)

2010 June (Q 2,10分);2013 Dec (QA3,10分)

应用范围:适合分析国家或产业的竞争优势。

Porter’s diamond model identifies four main determinants of national advantage. The four main determinants are:

1.The nation’s position in factor conditions, such as skilled labour or infrastructure,

necessary for firms to compete in a given industry.

2.The nature of the home demand conditions for the industry’s product or service.

Home demand influences economies of scale, shapes the rate and character of improvement and innovation.

3.The presence of related and supporting industries can provide a good local

supply chain and hence quality and cost advantages that are internationally competitive.

4.The firm strategy, structure and rivalry concerns the conditions in the nation

governing how companies are created, organised and managed. It also considers the nature of domestic rivalry and the role of government.

1.4Scenarios planning(情境分析)

分析思路:

1. 解释什么是情境分析:A scenario is a detailed and consistent view of how the business environment of an organisation might develop in the future. Scenarios can relate to macro factors such as political changes, economic growth or interest rates, or industry factors such as changes in input prices or entry of a new competitor.

2. 根据题目给出的信息,分析宏观环境的变化将如何影响企业的战略。

1.55 Forces Model(波特的“5力模型”)

应用范围:适合分析企业的竞争地位,经常配合PESTEL模型一起用(取决于分值)。

The 5 Forces Model identifies four main determinants of market competition that the firm faces.

1.Threat of new entrants. Barriers to entry will reduce competition by deterring

companies from joining an industry.

2.Threat of substitutes. The presence of substitutes will reduce profits because

customers can switch if prices rise

3.Bargaining power of buyers. Factors such as relative size and dependence will

affect the ability of buyers to drive down prices.

4.Bargaining power of suppliers. similarly, if suppliers in an industry have strong

negotiating power, the price of inputs will be driven up and profits reduced

https://www.360docs.net/doc/7e17008678.html,petition and rivalry. Intense competition will reduce profits. This may result

from slow growing or declining markets, excess capacity, barriers to exit and other factors

1.6Forecasting method

linear regression:2011 Dec (Q4)

Time series:Pilot 2011 (Q3,12分)

两者结合:2013 Dec (Q2a,15分)

1.6.1 linear regression

linear regression defines the equation of a straight line that ‘best’fits the data by minimising the squares of deviations of actual values from the mean. In least squares analysis, one set of data is defined as the independent variable (x –in this case, time) and the other set of data, sales, is defined as y –the dependent variable.

一般要求解释方程以及指标的意义,如:

●The positive value of b suggests that the overall sales trend is upwards.

●The correlation coefficient ‘r’shows the strength of the statistical relationship

between the two variables. e.g. the value of r is 0.253, which suggests that the two variables are weakly connected.

●The coefficient of determination (r2) shows that xxx% of the variation in sales (y)

is due to the passage of time (x). Low coefficients of correlation (and

determination) are usual when data is widely scattered around the mean and/or are related in a non-linear fashion.

注意:linear regression has little practical use when the data pattern is caused by large

seasonal variations.

Time series analysis uses a moving average to define a trend.

1.7Resource audit

应用范围:当题目要求分析“current strategic position from an internal perspective”,用该模型。

分析思路:

3.解释定义:Resource audits identify human, financial and material resources and

how they are deployed within a company.

4.结合案例信息,分析企业的优势与劣势(类似与SWOT分析)

注意:如果案例给了大量的财务数据,应做数据分析并结合起来分析(e.g. 2011 Dec. Q1)。

1.8SWOT Analysis

应用范围:当题目要求分析企业的“strategic capability”、“evaluate the strengths and weaknesses of xxx”、“current strategic position (from an internal perspective)”的时候,用该模型。

分析思路:

1.解释什么是SWOT分析:“SWOT analysis summarises the key issues from the

business environment and the strategic capability of an organisation that are most likely to impact on strategy development.

2.根据题目给出的信息,分别分析该企业的“Strengths”(优势)、“Weakness”

(劣势)、“Opportunity”(机会)与“Threats”(威胁)。

注意:要看清楚题目的问题。例如,SWOT分析中的“优势和劣势”是针对企业内部因素,而“机会和威胁”是针对企业的外部环境,如果题目要求分析“strategic position”,需要全面的分析四个方面;如果题目要求分析“strategic capability”,则可以只分析“优势和劣势”。此外,当题目只要求分析“strengths and weaknesses”的时候,不需要分析机会与威胁。

1.9Marketing mix(营销组合)

Marketing mix is a set of strategies that used to attract and retain customers, also known as the 4Ps or, for service organisations, 7Ps. These are:

1. A product(or service) is anything that satisfies a need or want, including

design, features, quality and packaging.

2.Place deals with how the product is distributed. It concerns with channel and

logistics.

3.Promotion includes all marketing communications which let the public know

of the product or service, e.g. advertising, sales promotions, public relations

4.Price is the only one in the marketing mix which brings in revenue. It is

important to set an appropriate price with reference to factors such as cost, competitors’ prices, perceived quality, firm strategy etc.

5.People carry the interaction between customers and staff in person or over the

telephone

6.Processes. Fast and efficient processes (e.g. booking a service) may be a

significant marketing advantages

7.Physical evidence. Because services are intangible, it is sometimes important

to provide evidence of ownership, e.g. a ticket to travel or certificate of attainment for training.

注意:传统产业为4P,服务产业为7P。除非题目有特定的要求。

1.10CSI & KPI(关键成功因素分析法& 关键绩效指标)

分析思路:

1.(1)解释什么是CSI:Critical success factors (CSFs) are those product features

that are particularly valued by a group of customers and, therefore, where the organization must excel to outperform competition. (2)结合案例分析哪些因素是顾客最重视的要素(对企业而言就是关键的成功因素)。

2.(1)解释什么是KPI:CSFs are normally measured through key performance

indicators (KPIs). These are targets that the organisation has to achieve.(2)分析哪些指标可以用来考核上述的关键成功因素(可结合“the balance scorecard”,强调同时应用财务指标和非财务指标)。

1.11Value chain(价值链)

1.Value chain is developed to analyze a firm’s activities – ways in which they

add value.

2.具体内容根据题目提供的信息

Primary activities:

(a) Inbound logistics – receiving, storing and handling stocks of raw materials

(b) Operations – processing raw materials into finished goods

(c) Outbound logistics – storing finished goods and distributing them to

customers

(d) Marketing and sales – marketing and selling activities

(e) Service – after or during sales services separate from the product (eg

warranties)

Support activities:

(a) Procurement – purchasing function

(b) Human resources – all functions related to staff recruitment and

development

(c) Technology development – management of IT and R&D functions (d) Infrastructure – everything else! (eg senior managers and finance function) Inbound Logistics 内部后勤Operations 运营

Outbound

Logistics

外部后勤Marketing & Sales 营销Service 服务

Primary Activities Procurement

采购

Human Resource Management 人力资源管理

Technology Development 技术开发Firm Infrastructure 企业基础设施

The Value Chain

Support

Activities Primary Activities

注意:题目有时候只要求分析“primary activities ”或“support activities ”,要看清楚。

1.12 Stakeholder mapping (利益相关者矩阵)

1.解释什么是“stakeholder mapping”:An organization's stakeholder relationships

must be managed in accordance with their power and degree of interest. The stakeholder mapping analysis classifies stakeholders into different groups and use different strategies to manage relationship with them.

2.根据题目,把利益相关者分为四类,并提出分别的战略。

(1) “high power and high interest”: actively manage these key players

(2) “high power and low interest”: keep these people satisfied

(3) “low power and high interest”: keeping them informed about progress of the

company

(4) “low power and low interest”: minimal effort should be put into managing

them, sometimes even ignore them.

1.13Ethical stance(伦理立场)

Ethical stance is the extent to which an organization will exceed its minimum obligation to stakeholders and society at large.

1.Short-term shareholder interest(短期股东利益): only accept a duty of

obedience to the demands of the law.

2.Long-term shareholder interest(长期股东利益): taker a wider view of ethical

responsibilities and thus enhance corporate image

3.Multiple stakeholder obligations(多方利益相关者义务): accept and balance

the expectations of different stakeholders, including shareholders, suppliers, employers and customers etc.

4.Shaper of society(社会的标杆): largely the role of public sector organization

and charities.

1.14Culture web(文化网络)

应用范围:当题目要求分析“the culture of xxx”或“assess the underlying organisational cultural issues”时,应用该模型。

The cultural web is a representation of the taken-for-granted assumptions, or paradigm, of an organisation.

1.Symbols(符号)such as logos, offices, cars, titles, language and terminology are

a shorthand representation of the nature of the organisation.

2.Power structures(权力结构)represents the most powerful groupings within the

organization. These people have the greatest amount of influence on decisions, operations, and the strategic direct of an organization.

https://www.360docs.net/doc/7e17008678.html,anisational structure(组织结构)reflects power and shows important roles

and relationships.

4.Control systems(控制系统), measurements and reward systems emphasise what

is important to monitor in the organisation.

5.Routines and rituals(惯例与仪式)define the ‘way we do things around here’.

6.Stories(故事)are used by members of the organisation to tell people what is

important in the organization. They are usually concerned with success, disasters, heroes, villains and mavericks.

Finally, the company paradigm(范式)summarises and reinforces the other elements of the cultural web.

1.15Corporate governance(组织治理)

Corporate governance is the conduct of the organization's senior effort. The board of directors is the main governing body, others including an executive committee or the

board of trustees (unusually for not-for-profit organizations). The board must decides on a stewardship role and ensure that the organization’s activities is not directed to management’s own ends rather than those of stakeholders.

2. Strategic choices(战略选择)

2.1 corporate rationales for adding value(母公司创造价值的三大角色)

Portfolio managers, synergy managers and parental developers represent three corporate rationales for value creation in a multi-business organization.

1.Portfolio managers seek to acquire under-performing or under-valued companies

and to improve their performance so that they can later be sold at a premium.

Portfolio managers manage businesses with a low cost centre and do not intervene significantly in the running of each business in the portfolio. The value-added activities of a portfolio manager are usually restricted to investment, setting expectations and standards and for monitoring performance.

2.Synergy managers pursue economies of scope through the shared use of

competences and resources. Developing strategic capabilities, achieving synergies and transferring managerial capabilities are value-adding activities of a synergy manager. This will require greater intervention in the operation of the SBUs.

3.Parental developers use the competencies of the parent to add value to

businesses in the portfolio. In this instance, the parent company is confident about its resources and capabilities and wishes to use these to enhance the value of the businesses in the portfolio. For parental developers, achieving synergies between companies in the portfolio is not a priority. The focus is on providing the companies in the portfolio with the competencies of the parent.

解题思路:

1.解释三类母公司的角色和定位分别是什么。

2.分析本案例中,母公司扮演了什么样的角色。

2.2BCG Matrix(波士顿矩阵)

According to their position on the matrix, SBUs will be categorised as follows: 1.Stars offer good future returns so the parent needs to invest in and develop them.

Due to the industry life cycle, stars will become cash cows in time.

2.Cash cows do not need much investment so will generate cash income. Parents

can use this cash to invest in stars or simply provide a return to shareholders.

3.Question marks should be assessed to see whether they have the potential to

become stars. If so, the parent should invest in them, if not, they should be sold or run down.

4.Dogs can tie up funds and provide a poor return. In general, they should be sold

off although may be retained if they are a useful niche business.

2.3The public sector portfolio matrix(公共部门组合矩阵)

应用范围:Government or service agency

1.Public sector star is something that the system is doing well and should not

change. They are essential to the viability of the system.

2.Political hot boxes are services that the public want, or which are mandated, but

for which there are not adequate resources or competences.

3.Golden fleeces are services that are done well but for which there is low demand.

They are potential targets for cost cutting.

4.Back drawer issues are unappreciated and hve low priority for funding. They are

obvious candidates for cuts. But if managers perceive them as essential, they should attempt to increase support for them and move them into the political hot box category.

2.4The General Electric Model(通用电气公司模型)

The GE Model is based on the enterprise’s competitive capacity and the market’s attractiveness. This model attempts to match competence within the company to conditions within the market place.

2.5The Shell directional policy matrix(指导性政策矩阵)

The directional policy matrix model is based on the enterprise’s competitive capacity

and the prospects for sector profitability. This model is deepens upon managerial judgment rather than simple numerical scores.

2.6The Ashridge portfolio Model

The Ashridge model assesses the benefit SBUs can derive from a corporate parent playing the parental developer role. The Ashridge model has two variables.

(1) 'Feel': The degree of fit between the parent's skills and resources, and the critical success factors of the SBU. This measures how well the parent understands what the SBU must do well to succeed. Low fit in this area means that the parent may inadvertently damage the business.

(2) 'Benefit': the degree of fit between the parent’s skills and resources and the SBU opportunities.

According to this model, the SBUs can be classified into 4 categories:

(1)Heartland businesses have opportunities to improve that the parent knows how

to address and are well understood

(2)Ballast businesses are well understood by the parent, however, the parent has no

real opportunities to add value. Ballast businesses can provide steady earnings but can also be a distraction for management. If they cannot be moved into heartland businesses, managers should consider divesting them.

(3)Value trap businesses present opportunities for parents to add value, perhaps by

exploiting synergies or transferable skills. However, the parents have a limited understanding of what is critical for the SBU to succeed and so there is a high risk that they will make decisions that reduce value. If they cannot be converted to heartland businesses they should be divested.

(4)Alien businesses are poorly understood and offer no opportunities for adding

value. It is likely that value is being destroyed and they should be divested as soon as possible

2.7The strategy clock(战略钟模型)

The strategy clock suggests 8 approaches to creating value for the customer. A customer will buy from the provider whose approach to price and perceived benefits matches their own. Each strategy has its on critical success factor.

1. A no frills approach seeks to deliver the lowest possible price. It is most

appropriate where customers are price-sensitive, switching costs are low and there is little opportunity to compete on product features.

2. A low price strategy seeks to provide a similar value of product or service to

competitors, but at a lower price. This is relatively easy for competitors to copy so will only be sustainable if the company has a cost advantage over its competitors for a given level of quality.

3.The hybrid approach attempts to simultaneously price lower than competitors

while delivering enhanced value to customers. This may be achieved by producing higher volumes than competitors, or focusing very clearly on one aspect of added value or a particular market segment.

4.Differentiation is a strategy aiming to provide services which are different or

unique in terms of value provided to customers. This may be based on factors such as product quality, marketing or innovation. Differentiation allows customers to earn a higher margin by charging higher prices, or gain market share by offering more value at the same price as competitors.

5.Focused differentiation means providing high perceived value to justify charging

a significantly higher price than other products. This usually means targeting a

specific market segment.

(6), (7) & (8) which combine high price and low perceived value will most likely fail.

2.8Porter’s generic strategies model

Porter developed 3 generic strategies for competitive advantages.

1.Cost leadership: being the lowest cost producer

2.Differentiation: making the product different from competitors’ products

3.Focus: specializing on a segment of the market

a)Cost-focus: provide goods and/or services at lower cost

b)Differentiation-focus: provide a differentiated product or service

2.9The TOWS Matrix

TOWS analysis provides four potential strategic directions:

1. WT (weakness-threat) strategy aims to minimize weaknesses and threats.

2. WO (weakness-opportunity) strategy aims to minimise weaknesses and exploit opportunities.

3. ST (strength-threat) strategy aims to use a firm’s strengths to deal with threats.

4. SO (strength-opportunity)use a firm’s strengths to exploit opportunities identified.

2.10The Ansoff’s Matrix

This model attempts to determine which products should be sold in which markets. 1.Market Penetration means increasing market share of existing products via

promotions, price reductions, increasing usage etc. It represents a relatively low risk strategy. Alternatively, a company may simply aim to maintain or even reduce its position in a market.

2.Market Development means seeking new customers for existing products, e.g.

exporting or selling via new distribution channels. Risk here is still reasonably low.

3.Product Development is selling new products to existing customers

(“cross-selling”). This is slightly riskier as it may involve investment in new products and its success may depend on the relationship a firm has with its customers.

4.Diversification, selling new products to new customers, may offer significant

growth potential but it is risky as it may require significant investment and new competencies.

2.11Methods of development(发展模式)

2013 Dec.(QA1,18分)

Internal development takes place when strategies are developed by building on or developing the organisation’s own capabilities. It is often termed organic growth. It is suited for organization with risk-averse attitude and cautious culture. The organic approach spreads cost and risk over time and growth is much easier to control and manage. However, growth can be slow and restricted by the breadth of the organisation’s capabilities.

Acquisition strategy is one where one organisation takes ownership of other existing organisations in the target countries. It allows an organisation to enter a new product or market area. However, acquisitions usually require considerable expenditure at some point in time and evidence suggests that there is a high risk that they will not deliver the returns that they promised. Furthermore, acquisitions also bring political and cultural issues, and acquisitions often results from problems of cultural fit.

A strategic alliance takes place when two or more organisations share resources and activities to pursue a particular strategy. This approach allow the organization to enter into a marketplace without the large financial outlay of acquiring a local organisation. Furthermore, it would avoid the cultural dislocation of either acquiring or merging with another organisation. However, loss of control and difficult to trust alliance partners are always challenges for alliance management.

A joint venture is an arrangement where a newly created organisation is jointly owned by the parents.

2.12Success criteria(战略成功标准)

2013 Dec.(QA2,18分)

应用范围:问某方案或目标收购对象的可行性/吸引力/成功可能性?

回答思路:1. 解释战略成功的三大标准:适宜性、可接受性、可行性(suitability, acceptability and feasibility);2. 结合案例分析。

1.Suitability. Suitability is concerned with whether a proposed strategy addresses

the circumstances in which an organisation is operating - its strategic position.

如果是收购案例,评价suitability的标准是“an acquisition makes particular sense if speed to market is vital”,可以结合案例资料从目标市场的competitors、自身的capability(如economies of scale)、cultural problems等来分析。要说明该项议案是否合乎企业的战略需求。

2.Acceptability. The acceptability of a proposed strategy is concerned with the

expected performance outcomes of a strategy in terms of return, risk and stakeholder reactions.

有可能涉及财务指标计算,如gross profit margin, ROCE, gearing ratio等。3.Feasibility. Feasibility is concerned with whether an organisation has the

resources and competences to deliver a strategy. E.g. financial feasibility

3. Strategic action(战略行为)

3.1 7 types of structure

1. Functional

2. Multidivisional

3. Holding company

4. Matrix

5. Transnational

6. Team

7. Project

重点:分析每一种结构的优点和缺点。其中,最关键的三种结构是functional、matrix和project。

3.2Stereotypical configuration(六大组织结构模型)

An organisation’s configuration considers how the structure, processes and relationships of an organisation work together. Henry Mintzberg has identified six configuration stereotypes.

1.Simple/Entrepreneurial(简单结构)

Strategic apex is dominant. There is no formal structure and little planning.

Control is exercised by direct supervision. It can be very effective in small entrepreneurial organisations where flexibility is important.

2.Machine bureaucracy(机械型组织)

Technostructure is dominant.. Control is by rules and procedures and work is highly standardised. Likely to have a hierarchical structure. This may be appropriate in a stable environment where the focus is on improving performance rather than solving problems.

3.Professional bureaucracy(专业型组织)

Operating core is dominant. Processes are too complex to be standardised and so control is exercised via training, individuals’ expertise and their professional ethos.

Common in professional firms and hospitals.

4.Divisionalised(分部型组织)

Middle line is dominant. The organisation is too large or complex to be managed as one unit so is split into divisions. Control is exercised via performance measures such as profit. Each division may well be configured as a machine bureaucracy.

5.Adhocracy(创新型组织)

Support staff are dominant, as they are the only group providing continuity.

Innovation is critical and work tends to be project-based. Teams with a mix of skills will form and disperse as required. There are few formal controls and work is complex and ambiguous.

6.Missionary(使命型组织)

Missionary organisations have little structure or formal control but are held together by a shared set of values, reinforced in a strong culture. May be effective in start-up companies or campaigning groups

3.3Types of change(变革的种类)

Adaptation is change that can be accommodated within the current paradigm and can be introduced incrementally in the organisation.

Reconstruction is change that may be rapid and create upheaval in the organization but which does not fundamentally change the underlying paradigm.

Evolution is a change that does require a paradigm change but one that can be introduced over time

Revolution is a change that requires rapid change associated with a change in

小学数学常见几何模型典型例题及解题思路

* 小学数学常见几何模型典型例题及解题思路(1) 巧求面积 常用方法:直接求;整体减空白;不规则转规则(平移、旋转等);模型(鸟头、蝴蝶、漏斗等模型);差不变 1、ABCG 是边长为12厘米的正方形,右上角是一个边长为6厘米的正方形FGDE ,求阴影部分的面积。答案:72 A H F E C B I D G 思路:1)直接求,但是阴影部分的三角形和四边形面积都无法直接求;2)整体减空白。关键在于如何找到整体,发现梯形BCEF 可求,且空白分别两个矩形面积的一半。 2、在长方形ABCD 中,BE=5,EC=4,CF=4,FD=1。△AEF 的面积是多少答案:20 |

A D B F C E 思路:1)直接求,无法直接求;2)由于知道了各个边的数据,因此空白部分的面积都可求 3、如图所示的长方形中,E 、F 分别是AD 和DC 的中点。 (1)如果已知AB=10厘米,BC=6厘米,那么阴影部分面积是多少平方厘米答案: (2)如果已知长方形ABCD 的面积是64平方厘米,那么阴影部分的面积是多少平方厘米答案:24 B C D F E 思路(1)直接求,无法直接求;2)已经知道了各个边的数据,因此可以求出空白的位置;3)也可以利用鸟头模型 4、正方形ABCD 边长是6厘米,△AFD (甲)是正方形的一部分,△CEF (乙)的面积比△AFD (甲)大6平方厘米。请问CE 的长是多少厘米。答案:8 @

A B D C F 思路:差不变 5、把长为15厘米,宽为12厘米的长方形,分割成4个三角形,其面积分别为S 1、S 2、S 3、S 4,且S 1=S 2=S 3+S 4。求S 4。答案:10 D C E F S 1 S 2 S 3 S 4 思路:求S4需要知道FC 和EC 的长度;FC 不能直接求,但是DF 可求,DF 可以由三分之一矩形面积S1÷AD ×2得到,同理EC 也求。最后一句三角形面积公式得到结果。 6、长方形ABCD 内的阴影部分面积之和为70,AB=8,AD=15。求四边形EFGO 的面积。答案10。 A B C D F O E G 思路:看到长方形和平行四边形,只要有对角线,就知道里面四个三

初中数学9大几何模型

初中数学九大几何模型 一、手拉手模型----旋转型全等 (1)等边三角形 【条件】:△OAB 和△OCD 均为等边三角形; 【结论】:①△OAC ≌△OBD ;②∠AEB=60°;③OE 平分∠AED (2)等腰直角三角形 【条件】:△OAB 和△OCD 均为等腰直角三角形; 【结论】:①△OAC ≌△OBD ;②∠AEB=90°;③OE 平分∠AED (3)顶角相等的两任意等腰三角形 【条件】:△OAB 和△OCD 均为等腰三角形; 且∠COD=∠AOB 【结论】:①△OAC ≌△OBD ; ②∠AEB=∠AOB ; ③OE 平分∠AED O D E 图 1 O A B C D E 图 2 O A B C D E 图 1 O A C D E 图 2 O A B C D E O C D E 图 1 图 2

二、模型二:手拉手模型----旋转型相似 (1)一般情况 【条件】:CD ∥AB , 将△OCD 旋转至右图的位置 【结论】:①右图中△OCD ∽△OAB →→→△OAC ∽△OBD ; ②延长AC 交BD 于点E ,必有∠BEA=∠BOA (2)特殊情况 【条件】:CD ∥AB ,∠AOB=90° 将△OCD 旋转至右图的位置 【结论】:①右图中△OCD ∽△OAB →→→△OAC ∽△OBD ; ②延长AC 交BD 于点E ,必有∠BE=∠BOA ; ③ ===OA OB OC OD AC BD tan ∠OCD ;④BD ⊥AC ; ⑤连接AD 、BC ,必有22 22CD AB B C AD +=+;⑥BD AC 2 1 S △BCD ?= 三、模型三、对角互补模型 (1)全等型-90° 【条件】:①∠AOB=∠DCE=90°;②OC 平分∠AOB 【结论】:①CD=CE ;②OD+OE=2OC ;③2△OCE △OCD △DCE OC 2 1 S S S =+= 证明提示: ①作垂直,如图2,证明△CDM ≌△CEN ②过点C 作CF ⊥OC ,如图3,证明△ODC ≌△FEC ※当∠DCE 的一边交AO 的延长线于D 时(如图4): 以上三个结论:①CD=CE ;②OE-OD=2OC ; ③2△OCD △OCE OC 21 S S =- O C O C D E O B C D E O C D A O B C D E 图 1 A O B C D E M N 图 2 A O B C D E F 图 3 A O B C D E M N 图 4

小学奥数 几何五大模型(等高模型)

模型一 三角形等高模型 已经知道三角形面积的计算公式: 三角形面积=底?高2÷ 从这个公式我们可以发现:三角形面积的大小,取决于三角形底和高的乘积. 如果三角形的底不变,高越大(小),三角形面积也就越大(小); 如果三角形的高不变,底越大(小),三角形面积也就越大(小); 这说明当三角形的面积变化时,它的底和高之中至少有一个要发生变化.但是,当三角形的底和高同时发生变化时,三角形的面积不一定变化.比如当高变为原来的3倍,底变为原来的13 ,则三角形面积与原来的一样.这 就是说:一个三角形的面积变化与否取决于它的高和底的乘积,而不仅仅取决于高或底的变化.同时也告诉我们:一个三角形在面积不改变的情况下,可以有无数多个不同的形状. 在实际问题的研究中,我们还会常常用到以下结论: ①等底等高的两个三角形面积相等; ②两个三角形高相等,面积比等于它们的底之比; 两个三角形底相等,面积比等于它们的高之比; 如图 12::S S a b = b a S 2S 1 D C B A ③夹在一组平行线之间的等积变形,如右上图ACD BCD S S =△△; 三角形等高模型与鸟头模型

反之,如果ACD BCD S S △△,则可知直线AB 平行于CD . ④等底等高的两个平行四边形面积相等(长方形和正方形可以看作特殊的平行四边形); ⑤三角形面积等于与它等底等高的平行四边形面积的一半; ⑥两个平行四边形高相等,面积比等于它们的底之比; 两个平行四边形底相等,面积比等于它们的高之比.

【例 1】 你有多少种方法将任意一个三角形分成:⑴ 3 个面积相等的三角形; ⑵ 4个面积相等的三角形;⑶6个面积相等的三角形。 【解析】 ⑴ 如下图,D 、E 是BC 的三等分点,F 、G 分别是对应线段的中点, 答案不唯一: C E D B A F C D B A G D B A ⑵ 如下图,答案不唯一,以下仅供参考: ⑸ ⑷⑶⑵⑴ ⑶如下图,答案不唯一,以下仅供参考: 【例 2】 如图,BD 长12厘米,DC 长4厘米,B 、C 和D 在同一条直线上。 ⑴ 求三角形ABC 的面积是三角形ABD 面积的多少倍? ⑵ 求三角形ABD 的面积是三角形ADC 面积的多少倍? 【解析】 因为三角形ABD 、三角形ABC 和三角形ADC 在分别以BD 、BC 和DC 为底时,它们的高都是从A 点向BC 边上所作的垂线,也就是说三个三角形的高相等。 于是:三角形ABD 的面积12=?高26÷=?高 三角形ABC 的面积124=+?()高28÷=?高 三角形ADC 的面积4=?高22÷=?高 所以,三角形ABC 的面积是三角形ABD 面积的43 倍; 三角形ABD 的面积是三角形ADC 面积的3倍。 【例 3】 如右图,ABFE 和CDEF 都是矩形,AB 的长是4厘米,BC 的长是3厘米, 那么图中阴影部分的面积是 平方厘米。 C D B A

全等三角形常见的几何模型

1、绕点型(手拉手模型) (1)自旋转:?????? ?,造中心对称遇中点旋 全等遇等腰旋顶角,造旋转 ,造等腰直角 旋遇,造等边三角形旋遇自旋转构造方法00 00018090906060 (2 )共旋转(典型的手拉手模型) 例1、在直线ABC 的同一侧作两个等边三角形△ABD 和△BCE ,连接AE 与CD ,证明: (1) △ABE ≌△DBC (2) AE=DC ( 3) AE 与DC 的夹角为60。 (4) △AGB ≌△DFB (5) △ EGB ≌△CFB (6) BH 平分∠AHC (7) GF ∥AC 变式练习1、如果两个等边三角形△ABD 和△BCE ,连接 AE 与CD ,证明: (1) △ABE ≌△DBC (2) AE=DC (3) AE 与DC 的夹角为60。 (4) AE 与DC 的交点设为H,BH 平分∠AHC 变式练习2、如果两个等边三角形△ABD 和△BCE ,连接AE 与CD ,证明: (1)△ABE ≌△DBC (2)AE=DC (3)AE 与DC 的夹角为60。 (4)AE 与DC 的交点设为H,BH 平分∠AHC

3、(1)如图1,点C 是线段AB 上一点,分别以AC ,BC 为边在AB 的同侧作等边△ACM 和△CBN ,连接AN ,BM .分别取BM ,AN 的中点E ,F ,连接CE ,CF ,EF .观察并猜想△CEF 的形状,并说明理由. (2)若将(1)中的“以AC ,BC 为边作等边△ACM 和△CBN ”改为“以AC ,BC 为腰在AB 的同侧作等腰△ACM 和△CBN ,”如图2,其他条件不变,那么(1)中的结论还成立吗?若成立,加以证明;若不成立,请说明理由. 例4、例题讲解: 1. 已知△ABC 为等边三角形,点D 为直线BC 上的一动点(点D 不与B,C 重合),以AD 为边作菱形ADEF(按A,D,E,F 逆时针排列),使∠DAF=60°,连接CF. (1) 如图1,当点D 在边BC 上时,求证:① BD=CF ? ②AC=CF+CD. (2)如图2,当点D 在边BC 的延长线上且其他条件不变时,结论AC=CF+CD 是否成立?若不成立,请写出AC 、CF 、CD 之间存在的数量关系,并说明理由; (3)如图3,当点D 在边BC 的延长线上且其他条件不变时,补全图形,并直接写出AC 、CF 、CD 之间存在的数量关系。 2、半角模型 说明:旋转半角的特征是相邻等线段所成角含一个二分之一角,通过旋转将另外两个和为二分之一的角拼接在一起,成对称全等。 例1、如图,正方形ABCD 的边长为1,AB,AD 上各存在一点P 、Q ,若△APQ 的周长为2, 求PCQ 的度数。 Q

六年级奥数专题-4几何五大模型——鸟头模型

几何五大模型——鸟头模型 本讲要点 一两点都在边上:鸟头定理: (现出“鸟头模型” 。然后按一下出现一个鸟头,勾勒出鸟头的轮廓,出现如图的鸟头几何模型。最后真实的鸟头隐去,只留下几何模型。最后按一下,出公式。) S AD×AE △ADE = S AB×AC △ABC A E D B C 二一点在边上,一点在边的延长线上: S CD×CE △CDE = S BC×AC △ABC A E D B C

例 1 如图, AD=DB,AE=EF=FC,已知阴影部分面积为 5 平方厘米,△ABC 的面积是平方厘米. 例 2 例 2 ( 1)如图在△ ABC中, D、E 分别是 AB,AC上的点,且 AD:AB=2:5, AE:AC=4:7,△ ABC 的面积是 16 平方厘米,求△ ABC的面积。 (2)如图在△ ABC中, D 在 BA 的延长线上, E 在 AC上,且 AB:AD=5:2, AE:EC=3:2,△ ADE 的面积是12 平方厘米,求△ABC的面积。

例3 已知△ DEF的面积为12 平方厘米, BE=CE,AD=2BD,CF=3AF,求△ ABC的面积。 例4 三角形 ABC面积为 1, AB 边延长一倍到 D, BC 延长 2 倍到 E, CA延长 3 倍到 F,问三角形DEF的面积为多少? F A E C B D

例5 长方形 ABCD面积为 120, EF 为 AD上的三等分点, G、 H、 I 为 DC上的四等分点,阴影面积是多大? 例 6 如图,过平行四边形 ABCD内的一点 P 作边AD、BC的平行线 EF 、GH,若 PBD 的面积为 8 平方分米,求平行四边形PHCF 的面积比平行四边形PGAE 的面积大多少平方分米? AG D P E F B H C

初中几何常见九大模型解析(完美版)

初中几何常见九大模型解析(完美版) -CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN

初中几何常见九大模型解析 模型一:手拉手模型-旋转型全等 (1)等边三角形 ?条件:均为等边三角形 ?结论:①;②;③平分。 (2)等腰 ?条件:均为等腰直角三角形 ?结论:①;②; ?③平分。 (3)任意等腰三角形 ?条件:均为等腰三角形 ?结论:①;②; ?③平分 模型二:手拉手模型-旋转型相似 (1)一般情况 ?条件:,将旋转至右图位置 ?结论: ?右图中①; ?②延长AC交BD于点E,必有

(2)特殊情况 ?条件:,,将旋转至右图位置 ?结论:右图中①;②延长AC交BD于点E,必有;③; ④; ⑤连接AD、BC,必有; ⑥(对角线互相垂直的四边形) 模型三:对角互补模型 (1)全等型-90° ?条件:①;②OC平分 ?结论:①CD=CE;②; ③ ?证明提示: ①作垂直,如图,证明; ②过点C作,如上图(右),证明; ?当的一边交AO的延长线于点D时: 以上三个结论:①CD=CE(不变); ②;③ 此结论证明方法与前一种情况一致,可自行尝试。 (2)全等型-120° ?条件:①; ?②平分; ?结论:①;②; ?③

?证明提示:①可参考“全等型-90°”证法一; ②如图:在OB上取一点F,使OF=OC,证明为等边三角形。 (3)全等型-任意角 ?条件:①;②; ?结论:①平分;②; ?③. ?当的一边交AO的延长线于点D时(如右上图): 原结论变成:①;②;③; 可参考上述第②种方法进行证明。请思考初始条件的变化对模型的影响。 ?对角互补模型总结: ①常见初始条件:四边形对角互补;注意两点:四点共圆及直角三角形斜边中线; ②初始条件“角平分线”与“两边相等”的区别; ③两种常见的辅助线作法; ④注意平分时,相等如何推导? 模型四:角含半角模型90° (1)角含半角模型90°-1 ?条件:①正方形;②; ?结论:①;②的周长为正方形周长的一半; 也可以这样: ?条件:①正方形;② ?结论:

中考数学常见几何模型简介精编版

几何问题 初中几何常见模型解析 (1)等边三角形 ?条件:均为等边三角形 ?结论:①;②;③平分。(2)等腰 ?条件:均为等腰直角三角形 ?结论:①;②;③平分。(3)任意等腰三角形 ?条件:均为等腰三角形 ?结论:①;②;③平分。?

(1)一般情况 ?条件:,将旋转至右图位置 ?结论:右图中①;②延长AC交BD于点E,必有 (2)特殊情况 ?条件:,,将旋转至右图位置 ?结论:右图中①;②延长AC交BD于点E,必有; ③;④;⑤连接AD、BC,必有 ; ⑥(对角线互相垂直的四边形) ?

(1)全等型-90° ?条件:①;②OC平分 ?结论:①CD=CE; ②;③ ?证明提示: ①作垂直,如图,证明; ②过点C作,如上图(右),证明;?当的一边交AO的延长线于点D时: 以上三个结论:①CD=CE(不变);②;③此结论证明方法与前一种情况一致,可自行尝试。

(2)全等型-120° ?条件:①;②平分; ?结论:①;②;③ ?证明提示:①可参考“全等型-90°”证法一; ②如图:在OB上取一点F,使OF=OC,证明为等边三角形。 ?当的一边交AO的延长线于点D时(如上图右): 原结论变成:①; ②; ③; 可参考上述第②种方法进行证明。 (3)全等型-任意角 ?条件:①;②; ?结论:①平分;②;③ . ?当的一边交AO的延长线于点D时(如右上图): 原结论变成:①; ②; ③; 可参考上述第②种方法进行证明。 ◇请思考初始条件的变化对模型的影响。

? 如图所示,若将条件“平分”去掉,条件①不变,平分,结论变化如下: 结论:①;②;③.

盘点小升初平面几何常考五大模型

盘点小升初平面几何常考五大模型 (一)等积变换模型性质与应用简介 导读:平面几何问题,是历年小升初的必考题目,也在各大杯赛中占有很大比例,这些题目都是以等积变形为主导思想,结合五大模型的变化应用交织而成的,这一期我们讲解了解一下五大模型第一块——等积变换模型。 等积变换模型例题讲解与课后练习题 (一)例题讲解与分析 ?【例1】:如右图,在△ABC中,BE=3AE,CD=2AD.若△ADE的面积是1平方厘米,那么三角形ABC的面积是多少 【解答】连接BD,S△ABD和S△ AED同高,面积比等于底边比,所以三角形ABD的面积是4, S△ABD和S△ABC同高面积比等于底边比,三角形ABC的面积是ABD的3倍,是12. 【总结】要找准那两个三角形的高相同。 【例2】:如图,四边形ABCD中,AC和BD相交于O点,三角形ADO的面积=5,三角形DOC的面积=4,三角形AOB的面积=15,求三角形BOC的面积是多少

【解答】S△ADO=5,S△DOC=4根据结论2,△ADO与△DOC同高所以面积比等于底的比,即AO/OC=5:4同理S△AOB/S△BOC=AO/OC=5:4,因为S△AOB=15所以S△BOC=12。 【总结】从这个题目我们可以发现,题目的条件和结论都是三角形的面积比,我们在解题过程中借助结论2,先把面积比转化成线段比,再把线段比用结论2转化成面积比,解决了问题。事实上,这2次转化的过程就相当于在条件和结论中搭了一座“桥梁”,请同学们体会 一下。 (二)课后练习题讲解与分析 (二)鸟头定理(共角定理)模型 导语:平面几何问题,是历年小升初的必考题目,也在各大杯赛中占有很大比例,这些题目都是以等积变形为主导思想,结合五大模型的变化应用交织而成的,第二期我们讲解了解一下五大模型第二块——鸟头定理(共角定理)模型。

(完整版)中考数学常见几何模型简介

初中几何常见模型解析 模型一:手拉手模型-旋转型全等 (1)等边三角形 ?条件:均为等边三角形 ?结论:①;②;③平分。 (2)等腰 ?条件:均为等腰直角三角形 ?结论:①;②; ?③平分。 (3)任意等腰三角形 ?条件:均为等腰三角形 ?结论:①;②; ?③平分 模型二:手拉手模型-旋转型相似 (1)一般情况 ?条件:,将旋转至右图位置 ?结论: ?右图中①; ?②延长AC交BD于点E,必有 (2)特殊情况 ?条件:,,将旋转至右图位置 ?结论:右图中①;②延长AC交BD于点E,必有; ③; ④; ⑤连接AD、BC,必有; ⑥(对角线互相垂直的四边形)

模型三:对角互补模型 (1)全等型-90° ?条件:①;②OC平分 ?结论:①CD=CE; ②;③ ?证明提示: ①作垂直,如图,证明; ②过点C作,如上图(右),证明; ?当的一边交AO的延长线于点D时: 以上三个结论:①CD=CE(不变); ②;③ 此结论证明方法与前一种情况一致,可自行尝试。 (2)全等型-120° ?条件:①; ?②平分; ?结论:①;②; ?③ ?证明提示:①可参考“全等型-90°”证法一; ②如图:在OB上取一点F,使OF=OC,证明 为等边三角形。 (3)全等型-任意角 ?条件:①;②; ?结论:①平分;②; ?③. ?当的一边交AO的延长线于点D时(如右上图): 原结论变成:①; ②; ③; 可参考上述第②种方法进行证明。请思考初始条件的变化对模型的影响。 ?对角互补模型总结: ①常见初始条件:四边形对角互补;注意两点:四点共圆及直角三角形斜边中线; ②初始条件“角平分线”与“两边相等”的区别; ③两种常见的辅助线作法; ④注意平分时,相等如何推导?

几何五大模型汇总

小学平面几何五大模型 一、 共角定理 两个三角形中有一个角相等或互补,这两个三角形叫做共角三角形. 共角三角形的面积比等于对应角(相等角或互补角)两夹边的乘积之比.如图在ABC △中,,D E分别是, AB AC上的点如图⑴(或D在BA的延长线上,E在AC上),则:():() S S AB AC AD AE =?? △△ 证明:由三角形面积公式S=1/2*a*b*sinC可推导出 若△ABC和△ADE中, ∠BAC=∠DAE 或∠BAC+∠DAE=180°, 则 ADE ABC S S ? ? = AE AD AC AB ? ? 二、等积模型 ①等底等高的两个三角形面积相等; ②两个三角形高相等,面积比等于它们的底之比; 两个三角形底相等,面积比等于它们的高之比; 如下图 12 :: S S a b = ③夹在一组平行线之间的等积变形,如右图 ACD BCD S S= △△ ; 反之,如果 ACD BCD S S = △△ ,则可知直线AB平行于CD. ④等底等高的两个平行四边形面积相等(长方形和正方形可以看作特殊的平行四边形); ⑤三角形面积等于与它等底等高的平行四边形面积的一半; ⑥两个平行四边形高相等,面积比等于它们的底之比;两个平行四边形底相等,面积比等于它们的高之比. b a S2 S1 D C B A

三、蝶形定理 1、任意四边形中的比例关系(“蝶形定理”): ①1243::S S S S =或者1324S S S S ?=? ②()()1243::AO OC S S S S =++ 速记:上×下=左×右 蝶形定理为我们提供了解决不规则四边形的面积问题的一个途径.通过构造模型,一方面 可以使不规则四边形的面积关系与四边形内的三角形相联系;另一方面,也可以得到与面积对应的对角线的比例关系. 2、梯形中比例关系(“梯形蝶形定理”): ①2213::S S a b = ②221324::::::S S S S a b ab ab =; ③S 的对应份数为()2a b +. 四、相似模型 (一)金字塔模型 (二) 沙漏模型 G F E A B C D A B C D E F G ①AD AE DE AF AB AC BC AG ===; ②22:ADE ABC S S AF AG =△△:. 相似三角形,就是形状相同,大小不同的三角形(只要其形状不改变,不论大小怎样改变它们都相似),与相似三角形相关的常用的性质及定理如下: ⑴相似三角形的一切对应线段的长度成比例,并且这个比例等于它们的相似比; ⑵相似三角形的面积比等于它们相似比的平方; A B C D O b a S 3 S 2 S 1S 4 S 4 S 3 S 2 S 1O D C B A

小学数学常见几何模型典型例题及解题思路

小学数学常见几何模型典型例题及解题思路(1) 巧求面积 常用方法:直接求;整体减空白;不规则转规则(平移、旋转等);模型(鸟头、蝴蝶、漏斗等模型);差不变 1、ABC G是边长为12厘米的正方形,右上角是一个边长为6厘米的正方形FGDE,求阴影部分的面积。答案:72 A H F E C B I D G 思路:1)直接求,但是阴影部分的三角形和四边形面积都无法直接求;2)整体减空白。关键在于如何找到整体,发现梯形BCEF 可求,且空白分别两个矩形面积的一半. 2、在长方形ABCD 中,B E=5,EC =4,CF=4,FD =1。△AEF 的面积是多少?答案:20

A D B F C E 思路:1)直接求,无法直接求;2)由于知道了各个边的数据,因此空白部分的面积都可求 3、如图所示的长方形中,E、F 分别是AD 和DC 的中点。 (1)如果已知AB=10厘米,BC=6厘米,那么阴影部分面积是多少平方厘米?答案:22。5 (2)如果已知长方形ABC D的面积是64平方厘米,那么阴影部分的面积是多少平方厘米?答案:24 B C D F E 思路(1)直接求,无法直接求;2)已经知道了各个边的数据,因此可以求出空白的位置;3)也可以利用鸟头模型 4、正方形A BCD 边长是6厘米,△AF D(甲)是正方形的一部分,△CEF(乙)的面积比△AFD (甲)大6平方厘米。请问C E的长是多少厘米.答案:8

A B D C F 思路:差不变 5、把长为15厘米,宽为12厘米的长方形,分割成4个三角形,其面积分别为S 1、S2、S 3、S4,且S1=S 2=S 3+S 4。求S4.答案:10 D C E F S 1S 2 S 3S 4 思路:求S4需要知道FC 和EC 的长度;FC不能直接求,但是DF 可求,DF 可以由三分之一矩形面积S 1÷AD ×2得到,同理EC 也求.最后一句三角形面积公式得到结果。 6、长方形ABCD 内的阴影部分面积之和为70,AB=8,A D=15。求四边形E FGO 的面积.答案10。 A B C D F O E G 思路:看到长方形和平行四边形,只要有对角线,就知道里面四个三

小学数学几何五大模型教师版

几何五大模型 一、五大模型简介 (1)等积变换模型 1、等底等高的两个三角形面积相等; 2、两个三角形高相等,面积之比等于底之比,如图①所示,S1:S2=a:b; 3、两个三角形底相等,面积在之比等于高之比,如图②所示,S1:S2=a:b; 4、在一组平行线之间的等积变形,如图③所示,S△ACD=S△BCD;反之,如果S△ACD=S△BCD,则可知直线AB平行于CD。 例、如图,三角形ABC的面积是24,D、E、F分别是BC、AC、AD的中点,求三角形DEF的面积。

(2)鸟头(共角)定理模型 1、两个三角形中有一个角相等或互补,这两个三角形叫共角三角形; 2、共角三角形的面积之比等于对应角(相等角或互补角)两夹边的乘积之比。 如图下图三角形ABC中,D、E分别是AB、AC上或AB、AC延长线上的点 则有:S△ABC:S△ADE=(AB×AC):(AD×AE) 我们现在以互补为例来简单证明一下共角定理! 如图连接BE,根据等积变化模型知,S△ADE:S△ABE=AD:AB、S△ABE:S△CBE=AE:CE,所以S△ABE:S△ABC=S△ABE:(S△ABE+S△CBE)=AE:AC,因此S△ADE:S△ABC=(S△ADE:S△ABE)×(S△ABE:S△ABC)=(AD:AB)×(AE:AC)。 例、如图在ΔABC中,D在BA的延长线上,E在AC上,且AB:AD=5:2,AE:EC=3:2,△ADE的面积为12平方厘米,求ΔABC的面积。

(3)蝴蝶模型 1、梯形中比例关系(“梯形蝴蝶定理”) 例、如图,梯形ABCD,AB与CD平行,对角线AC、BD交于点O,已知△AOB、△BOC 的面积分别为25平方厘米、35平方厘米,求梯形ABCD的面积。 2、任意四边形中的比例关系(“蝴蝶定理”):

初中几何模型及常见结论的总结归纳

初中几何模型及常见结论的总结归纳 三角形的概念 三角形边、角之间的关系:①任意两边之和大于第三边(任意两边之差小于第三边);②三角形内角和为0180(外角和为0 360);③三角形的外角等于不相邻的两内角和。 三角形的三线:(1)中线(三角形的顶点和对边中点的连线);三角形三边中线交于一点(重心) 如);DE 之到?S 如图,已知AB ,AC 的长,求AF 的取值范围时。我们可以通过倍长 中线。利用三角形边的关系在三角形ABD 中构建不等关系。(AC AB AF AC AB +- 2). (2)角平分线(三角形三内角的角平分线);三角形的三条内角平分线交于一点(内心)

如等 OE ; r = 2

(3)垂线(三角形顶点到对边的垂线);三角形三条边上的高交于一点(垂心) 如图,O为三角形ABC的垂心,我们可以得到比较多的锐角相等如 COD ABC ACO ABO∠ = ∠ ∠ = ∠;等。因此垂线(或高)这样的条件在题目中出现,我们往往可以得出比较多的锐角相等。(等角或同角的余角相等),此外,如果要求垂线段的长度或与垂线段有关的长度问题,我们通常用面积法求解。在上图中,若已知CE AC AB, ,的长度,求BE的长。 特别注意:在等腰三角形中,我们通常所指的三线合一就是指中线、角平分线、高线。三线合一:已知三角形三线中的任意两个条件是重合的,那么就可以得出第三条线也是重合的。在具体运用时,我们往往时把三线合一的等腰三角形补充完整再加以运用。 三角形全等 三角形全等我们要牢记住它的五个判定方法。(SSS,SAS,ASA,AAS,HL) 在具体运用时,我们需要找出判定三角形全等的各种条件,不外乎是关于边相等或相等的问题。 对于寻找角相等:常有四种方法:①两条平行线被第三条直线所截得出的“三线八角”的结论;②对顶角相等;③锐角互余;④三角形的外角等于不相邻的两内角和。 对于寻找边相等:常有三种方法:①特殊图形中隐含的条件(如等腰三角形、等边三角形、菱形、正方形。。。。。);②利用三线合一的正逆定理;③通过已有的全等三角形性质得出。对于证明角相等,证明边相等,我们都要优先考虑边或角所在的三角形全等。(一定要注意对应)如果不能直接通过全等证明,我们就要转化角或转化边(用上面的几种方法)然后再考虑全等。 全等三角形的基本图形: 平移类全等;对称类全等;旋转类全等;

几何五大模型一

几何五大模型 一、等积变换模型 1、等底等高的两个三角形面积相等。 2、两个三角形高相等,面积比等于它们的底之比。 3、两个三角形底相等,面积比等于它的的高之比。 二、共角定理模型 两个三角形中有一个角相等或互补,这两个三角形叫做共角三角形。共角三角形的面积比等到于对应角(相等角或互补角)两夹边的乘积之比。 三、蝴蝶定理模型 (说明:任意四边形与四边形、长方形、梯形,连接对角线所成四部的比例关系是一样的。) 四、相似三角形模型 相似三角形:是形状相同,但大小不同的三角形叫相似三角形。 相似三角形的一切对应线段的长度成比例,并且这个比例等于它们的相似比。相似三角形的面积比等于它们相似比的平方。 五、燕尾定理模 等积变形: 等积变形是小学几何里面一个非常重要的思想,小学所以的几何题,或多或少的都会用到等积变形的思想,几何五大模型也都是依托等积变形思想变化而成的。

一半模型 平行四边形、梯形、任意四边形中的一些一半模型。 一、 模型归纳总结 1、等面积变换模型 (1)直线AB 平行于CD ,可知BCD ACD S S ??=; 反之,如果BCD ACD S S ??=,则可知直线AB 平行于CD .如图A (2)两个三角形高相等,面积比等于它们的底之比; 两个三角形底相等,面积比等于它们的高之比; ::ABD ACD S S BD CD =△△如图 B D C B A D C B A 图A 图B (3)一半面积关系 S 4 S 3 S 2S 1 A B C D D C A 1 2 S S =阴影长方形 1324 S S S S +=+

【例1】、如图,每一个正方形四边中点的连线构成另一接小正方形,则阴影部分面积为原正方形面积的几分之几? 第8题 【例2】、如右图,过平行四边形ABCD 的一点P 作边的平行线EF 、GH ,若PBD 的面积为8平方分米,求平行四边形PHCF 的面积比平行四边形PGAE 的面积大多少平方分米? B C G H

全等三角形常见的几何模型

1绕点型(手拉手模型) 遇600旋60°,造等边三角形 遇90°旋90°,造等腰直角遇等腰旋 顶角,造旋转全等遇中点旋1800,造中 心对称 (2)共旋转(典型的手拉手模型) 例1、在直线ABC的同一侧作两个等边三角形△ (1)△ ABE ◎△ DBC (2)AE=DC (3)AE与DC的夹角为60。 (4)△ AGB ◎△ DFB (5)△ EGB ◎△ CFB (6)BH 平分/ AHC (7)GF // AC 变式练习2、如果两个等边三角形△ ABD和厶BCE,连接AE与CD,证明: ("△ ABE ◎△ DBC (2)AE=DC (3)AE与DC的夹角为60。 (4) AE与DC的交点设为H,BH平分/ AHC [D山3 Vi壮-U (I) ? 变式练习1、如果两个等边三角形△ABD和厶BCE,连接AE与CD,证明 (1) △ ABE ◎△ DBC (2) AE=DC (3) AE与DC的夹角为60。 (4) AE与DC的交点设为H,BH 平分/ AHC (1自旋转:自旋转构造方法 ABD和厶BCE,连接AE与CD,证明:

3、(1)如图1,点C是线段AB上一点,分别以AC, BC为边在AB的同侧作等边△ ACM和厶CBN ,连接AN , BM .分别取BM, AN的中点E, F,连接CE, CF, EF.观察并猜想△ CEF的形状,并说明理由. (2)若将(1)中的“以AC , BC为边作等边△ ACM和厶CBN”改为“以AC, BC为腰在AB的同侧作等腰△ ACM和△ CBN,”如图2,其他条件不变,那么(1)中的结论还成立吗?若成立,加以证明;若不成立,请说明理由. B 例4、例题讲解: 1.已知△ ABC为等边三角形,点D为直线BC上的一动点(点D不与B,C重合),以AD为边作菱形ADEF(按A,D,E,F 逆时针排列),使/ DAF=60 ° ,连接CF. (1)如图1,当点D在边BC上时,求证:① BD=CF 宓AC=CF+CD. (2)如图2,当点D在边BC的延长线上且其他条件不变时,结论AC=CF+CD是否成立?若不成立,请写出AC、CF、 CD之间存在的数量关系,并说明理由; ⑶如图3,当点D在边BC的延长线上且其他条件不变时,补全图形,并直接写出AC、CF、CD之间存在的数量关系。 2、半角模型 说明:旋转半角的特征是相邻等线段所成角含一个二分之一角,通过旋转将另外两个和为二分之一的角拼接在一起, 成对称全等。 D A D A M x N rt B D 例1、如图,正方形ABCD的边长为1, AB,AD上各存在一点P、0,若厶APQ的周长为2, A P

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