The area method at the primary level with median line problems

In this work, it comes from solving the median line problem at primary level, we will develop and expand it to include new problems on the basis of changing the ratio of each dividing point on the two sides of the triangle and/or adding appropriate assumptions.

Vol No 4_October 2022 ISSN: 2354 - 1431 http://tckh.daihoctantrao.edu.vn/ TẠP CHÍ SCIENTIFIC JOURNAL OF TAN TRAO UNIVERSITY TẠP CHÍ KHOA HỌC ĐẠI HỌC TÂN TRÀO TẠP CHÍ KHOA HỌC ĐẠI HỌC TÂN TRÀO ISSN: 2354 - 1431 TẠP CHÍ KHOA HỌC ĐẠI HỌC TÂN TRÀO http://tckh.daihoctantrao.edu.vn/ KHOA HỌC GIÁO DỤC SCIENTIFIC EDUCATION Tập 8, Số - 10/2022 ISSN: 2354 - 1431 Tập 8, Số (Tháng 10/2022) Volume 8, Issue (October 2022) PHƯƠNG PHÁP DIỆN TÍCH Ở BẬC TIỂU HỌC VỚI BÀI TOÁN ĐƯỜNG TRUNG TUYẾN Trần Đình Tướng1 , Khổng Chí Nguyện2,∗ Trường Đại học Tài - Marketing, thành phố Hồ Chí Minh, Việt Nam Trường Đại học Tân Trào, Tuyên Quang, Việt Nam ∗ Email: nguyenkc69@gmail.com https://doi.org/10.51453/2354-1431/2022/5xx Thơng tin viết Tóm tắt: Ngày nhận bài: 21/9/2022 Ngày duyệt đăng: / /2022 Khi giải tốn nói chung tốn hình học nói riêng, ta thường có suy nghĩ để tìm thêm lời giải khác, làm để có tốn từ tốn Từ khóa: Tam giác, Diện tích, Tỷ số, Đường cao, Cạnh đáy gốc Đối với tốn hình học, để có thêm tốn mới, ta thường thay đổi, bổ sung thêm giả thiết/kết luận Trong báo xuất phát từ việc giải Bài toán đường trung tuyến cho học sinh tiểu học, ta phát triển, mở rộng thêm số toán sở thay đổi tỷ lệ điểm chia hai cạnh tam giác và/hoặc bổ sung thêm giả thiết phù hợp 166 184| Vol No 4_October 2022 TẠP CHÍ ISSN: 2354 - 1431 http://tckh.daihoctantrao.edu.vn/ Tập 8, Số - 10/2022 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx TẠP CHÍ KHOA HỌC ĐẠI HỌC TÂN TRÀO TẠP CHÍ KHOA HỌC ĐẠI HỌC TÂN TRÀO SCIENTIFIC JOURNAL OF TAN TRAO UNIVERSITY KHOA HỌC GIÁO DỤC SCIENTIFIC EDUCATION ISSN: 2354 - 1431 Tập 8, Số (Tháng 10/2022) Volume 8, Issue (October 2022) THE AREA METHOD AT THE PRIMARY LEVEL WITH MEDIAN LINE PROBLEMS Tran Dinh Tuong1 , Khong Chi Nguyen2,∗ University of Finance - Marketing, Ho Chi Minh City, Viet Nam Tan Trao University, Tuyen Quang, Viet Nam ∗ Email: nguyenkc69@gmail.com https://doi.org/10.51453/2354-1431/2022/5xx Article info Abtract: Received: 21/9/2022 Accepted: / /2021 When solving any problem or geometry problem, we often have thoughts and find the other solutions or how to discover new problems from the original problem For geometry problems, in order to have a new problem we often change or add assumptions/conclusions and etc In this work, it comes from solving the median line problem at primary level, we will develop and expand it to include new problems on the basis of changing the ratio of each dividing point on the two sides of the triangle and/or adding appropriate assumptions Keywords: Triangle, Area, Ratio, Altitude, Base side Introduction In the Primary Math program, geometry problems are always content of knowledge with many difficult math problems In particular, the problems of the area, and the ratio of measurements between geometric figures are challenges for most pupils and some teachers These problems require students and teachers to have a solid foundation of geometric knowledge, to have coherent thinking and, to be able to read and analyze geometric figures to determine the exact solution for the problem In the entrance exams to grade at key junior high schools, the problem of the geometrical area is always taken to increase the differentiation for candidates However, most pupils have no points in the problems of the area and are solved by the area method Finding the solution of a geometry problem by the area method and appropriately presenting the object, and helping pupils develop thinking capacity is always a requirement for every teacher Starting from the Median line Problem, Example 3.1, we will develop and expand many more problems by changing the ratio of each dividing point on the two sides of the triangle and/or adding make appropriate assumptions In the scope of this paper, we mainly study area problems related to triangles and |185 167 Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx special quadrilaterals such as parallelograms, trapezoids, rectangles The article is structured as follows: Following the Introduction section is the Preliminary knowledge section, which recalls some related knowledge that will be used in Section of this article Section is the research content Finally, the Conclusion section SABCD = (AB + CD) × AH If AB = CD, we get the parallelogram ABCD, and the area of parallelogram ABCD is calculated by the formula Preliminary 2.1 Area of a trapezoid, parallelogram, and rectangle: Given trapezoid ABCD, small base AB, longer base CD, and the altitude AH perpendicular to the bases at H on the side CD Then, the area of trapezoid ABCD is calculated by the formula Area formulas Triangle area: Given triangle ABC, altitude AH is perpendicular to side BC at point H We have, the area formula of triangle ABC is SABC = BC × AH BC is called the base side, and AH is the altitude corresponding to BC The formula for area is (verbally) stated as follows: The area of a triangle is the product of the base length with the corresponding height divided by SABCD = AB × AH If AB = CD and the altitude AH is the side AD, we get the rectangle ABCD, and the area of rectangle ABCD is calculated by the formula SABCD = AB × AD 2.2 Find two numbers when the sum and difference are known Now we will consider the ratio of two triangles through the base lengths and the heights Let’s say a > b Find two numbers a and b if the sum is T = a + b, the difference is H := a − b We have Consider triangle M N P , altitude M K is perpendicular to N P at point K We have, a = (T + H) ÷ 2, b = (T − H) ÷ 2, SM N P NP × MK = Then, BC AH BC × AH SABC = × = SM N P NP × MK NP MK If BC = N P then AH SABC = SM N P MK If AH = M K ten BC SABC = SM N P NP or b = T − a, or b = a − H 2.3 Find two numbers sum/difference and known when ratio the are Let’s say a > b Find two numbers a and b if the sum is T = a + b (or, difference is H := a − b) and the ratio is ab = M N It is easy to see that the sum (or, difference) of two numbers a and b is divided into M + N (or, M − N ) parts, where a occupies the M part, b occupies the N part Then each 168 186| Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx part has a value of p = T ÷ (M + N ) (or, p = H ÷ (M − N )), and we have a = M × p, b = N × p SP BC = × SP BM or b = T − a (or, b = a − H) Two triangles P BM and P BC share the same altitude drawn from P to BC and the equality (3.1), so we have (3.3) Two triangles BAN and BCN share the same altitude drawn from B to AC and N A = N C, so we have Contents We consider and solve the problem of the median line theorem in a triangle, in line with the thinking of primary students Example 3.1 Given triangle ABC Take the points M, N on the sides BC, CA respectively such that M C = M B, N A = N C Let P be the intersection of AM and BN Prove that P A = × P M, BP = × P N A SBAN = SBCN (3.4) Two triangles P AN and P CN share the same altitude drawn from P to AC and N A = N C, so we have SP AN = SP CN (3.5) By (3.4) and (3.5), we have SP AB = SP BC (3.6) Therefore, by (3.3) and (3.6), we get N SP BA = × SP BM ⇒ P B M C ANALYSIS By assumption, to prove P A = × P M , we will prove SCP A = × SCP M Draw a line segment P C, it is easy to see that SP BC = × SP CM , SP BC = SP AC So, SCP A = × SCP M Therefore, P A = × P M Similarly, we also prove the case BP = 2×P N SOLUTION Draw a line segment P C By assumption, we have BC = × BM, AC = × AN (3.1) (3.2) SP BA = SP BM (3.7) Two triangles P BA and P BM share the same altitude drawn from B to AM and the ratio (3.7) Then, the ratio of the bases is equal to the area ratio of the triangles We have PA SP BA = = ⇒ P A = × P M PM SP BM Next, we prove BP = × P N It is clear that the two triangles P CN and P AC share the same altitude drawn from P to AC and the ratio (3.2), so we have SP AC = × SP CN (3.8) Two triangles ABM and ACM share the same altitude drawn from A to BC and M B = M C, so we have SABM = SACM (3.9) 169 |187 Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx Two triangles P BM and P CM share the same altitude drawn from P to BC and M B = M C, so we have C N SP BM = SP CM (3.10) E By (3.9) and (3.10), we get SP AB = SP AC (3.11) A By (3.8) and (3.11), we get SAP B = × SAP N ⇒ SAP B = SAP N (3.12) Two triangles AP B and AP N share the same altitude drawn from A to BN and the ratio (3.12), so the ratio of the two bases is equal to the area ratio of the triangles We have PB SAP B = = ⇒ P B = × P N PN SAP N The proof is complete Example 3.1 is a basic problem for the requirement to calculate the ratio of the lengths of two line segments on a side connecting the vertex of the triangle to the midpoint of the opposite side by using the area method To solve the problem, detecting and drawing the subline P C play an important role To get more new problems, we will change the ratio of the points M, N on the two sides of the triangle ABC and/or add another, appropriate assumptions We consider some such problems through the examples presented as follows Example 3.2 Given triangle ABC Take the points M on AB, N on AC such that AM = M B, N A = × CN The notation E is the intersection of AM and BN Suppose, the area of triangle AEN is equal to 18cm2 Calculate the area of triangle ABC M B ANALYSIS Thank to the assumption N A = × N C, we can calculate the area of triangle EAC, the area ratio of the triangles EBC and EBA, namely SEAC = 27cm2 , SEBC = SEBA On the other hand,from the assumption AM = M B, then the area ratio of the triangles EBM and EBA is SEBM = SEBA We infer that two triangles EBC and EBM have equal areas and SEBC = 1, SEBM and then EC = EM Therefore, SAEM = SAEC = 27cm2 Then, triangle ABC has an area of 108cm2 SOLUTION By the assumption N A = × A = 23 Therefore, the N C, we have the ratio N AC area ratio of the triangles EN A and EAC is SACE ⇒ SACE = 27(cm2 ) = SEN A (3.13) It is also from the assumption N A = × N C, C = 12 Thus we have the we have the ratio N NA area ratio of the triangles EN C and EN A as SEN C = 12 Thus, two altitudes drawn from C, SEN A 170 188| Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx A to N B have the ratio 12 So the area ratio of the triangles EBC and EBA is 1 SEBC ⇒ SEBC = × SEBA = SEBA 2 (3.14) On the other hand, under the assumption AM = M B, we have AB = × M B Therefore, the area ratio of the triangles EM B and EBA is SEM B ⇒ SEAB = × SEM B = SEBA F is the intersection of CD and BE Suppose, the area of triangle DBF is equal to 100cm2 Calculate the area of triangle ABC B D K I F (3.15) By (3.14) and (3.15), we have SBEC = SBEM Since two triangles BEC and BEM share the same altitude drawn from B to M C, so two bases corresponding to B must be equal to each other, and EC = EM We get SEM A = SECA = 27(cm2 ) Thus, we have SCAM = × SEAM = 54(cm2 ) So, SABC = × SCAM = 108cm2 Answer: SABC = 108cm2 In Example 3.2, the "subline" AE is given Then, to get the solution of the problem, we EC = 1, or the rahave to calculate the ratio EM EB tio EN = This is the critical middle step to determine SABC Therefore, this problem may require proving EC = EM or EB = × EN to get another problem We can also replace the area of triangle EN A with the area of triangle EN C, EAM and etc Finally, F is the intersection of the extended AE that intersects the side BC It is easy to see that the ratio AE BE = BN = 34 Readers should prove these AF claims In the next example, we add more assumptions in addition to the ones about the dividing point on two sides of the triangle Example 3.3 Given triangle ABC Take the points D on AB, E on AC such that AB = × BD, AC = × AE The notation H A E C ANALYSIS By assumption, it is easy to see that the triangle DBF lies within the triangle ABE and SABC = × SABE Therefore, to compute SABC , we would have to find SABE through SDBF Two triangles DBF, ABE have two bases on the side BE, so the altitudes are parallel Draw the altitude AH, DI perpendicular to BE We will calculate the ratio of two altitudes and two bases of the triangles DBF, ABE Thereby, the ratio of the areas of the two triangles DBF, ABE is BE 15 AH SABE × = = SDBF DI BF 2 So SABE = 750cm Hence SABC = 3000cm2 SOLUTION Draw AF and the altitudes AH, DI perpendicular to BE, H, I are in AE We have BE × AH BF × DI SABE = , SDBF = 2 (3.16) i) Computing the ratio AH Draw AF , and DI F K perpendicular to AB The triangle F DB and the triangle F AB share the same altitude AB = (by F K, and the ratio of the bases DB SF AB assumption AB = × BD ) So SF DB = Hence, SF AB = × SF DB (3.17) 171 |189 Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx Now consider two triangles DF B and AF B with the base side F B Combined with (3.17) = We we get the ratio of two altitudes AH DI deduce, According to the results (3.22), the two triangles CF E and CF B share the altitude drawn from C to BE, so the ratio of two bases is FE = 32 Thus, we have FB AH = × DI BE FE + FB 3+2 = = = FB FB 2 (3.18) through the raii) Computing the ratio BE BF SCF E tio area SCF B Two triangles CAD and CBD share the same altitude drawn from C to AB AD = (since and the ratio of the two bases BD AB = × BD), so the area ratio is SCAD = SCBD We deduce We deduce BE = × F B (3.23) iii) Computing the area of triangle ABC By the (3.16), (3.18) and (3.23) we get 15 1 × BE × AH = × × BF × DI 2 15 15 = × SDBF = × 100 = 750(cm2 ) 2 SABE = SCAD = × SCBD (3.19) Now we consider two triangles CAD, CBD with base CD Combined with (3.19), we get the ratio of two altitudes drawn from A and B to CD equal to The triangles CF A and CF B share the base CF , and the ratio of two altitudes corresponding to the base CF is Hence the area ratio CF A = 2, so is SSCF B SCF B = × SCF A (3.20) Two triangles F AC, F EC share the altitude drawn from F to AC and the ratio of two bases AC = 43 (by assumption, AC = × AE) So, is EC AC = 43 the area ratio of two the triangles is SSFF EC We get SF AC = × SF EC SABC = × SABE = × 750 = 3000(cm2 ) Answer: SABC = 3.000cm2 Example 3.3 has the solution presented with a different approach, through finding the ratio between the altitudes, the bases of two triangles DBF and ABE So we determine the area ratio of these triangles and the area of the triangle ABF - an important calculation in computing the area of the triangle ABC Of course, we can also compute the area of triangle ABC through the area of triangle BCD (3.21) It thanks to (3.20), (3.21), we have SCF B = × 43 × SCF E = 23 × SCF E Hence the area ratio is SCF E = SCF B Two triangles ABE, ABC share the altitude drawn from B to AC, AC = × AE Thus, (3.22) Example 3.4 Given triangle ABC Take the point M in BC, the point N in AC such that, BM = M C, CN = × N A E is the intersection of AM and BN If the area of triangle ABC is equal to 420cm2 , calculate the area of AE triangle AEN and the ratio AM 172 190| Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx On the other hand, the triangles ABN and ABC share the altitude drawn from B to BC and the ratio of the bases is N A = 14 × AC, so A N E SABN = 1 × SABC = × 420 = 105(cm2 ) 4 Therefore SAEN = × 105 = 21(cm2 ) AE Finally, we calculate the ratio AM The triangles ABE and ABM share the altitude drawn B M C from B to AM Therefore, the area ratio of the triangles ABE and ABM is equal to the ratio ANALYSIS By the assumptions and condiof the bases AE, AM Therefore, tions about two triangles having equal area, SABE AE it is easy to see that = AM SABM 1 SAEN = × SAEC = × SAEB 4 We have SABE = SABN − SAEN = 105 − 21 = 84(cm2 ) So, the ratio to be found is Hence, SAEN = × SABN We deduce SAEN = 84 AE = = AM 210 × SABC = 21(cm2 ) 20 SOLUTION Draw the line segment EC By assumption, triangle ABM and triangle AM C share the altitude drawn from A to BC and BM = M C Therefore, SABM = SACM Similarly, triangle EBM and triangle EM C share the altitude drawn from E to BC and BM = M C Hence, SEBM = SECM Deduce SABE = SABM − SEBM = SACM − SECM = SACE (3.24) Now we will calculate SAEN in terms of SABN By the assumption CN = × N A, then N A = × AC The triangles AEN and AEC share the altitude drawn from E to AC, and the ratio of two bases is N A = 14 × AC Therefore SAEN = × SAEC Combined with (3.24), we have SAEN = SAEB , and then SAEN = × SABN × AE = 25 Answer: SAEN = 21cm2 ; AM The subline CE in Example 3.4 is still the deciding factor to get the correct answer The problem can be changed the assumptions, conclusions, and others to have new problems: - Change the conclusions about calculating areas and ratios, for example, the area of the AE BE or BN , triangle AEB or BEM , the ratios EM EN or EB , etc - Change the assumption about the position of the points M, N in BC, AC respectively, for example, BM = 12 × M C and EN = 12 × N C and etc These changes can make the more difficult problems We will consider an example for these cases Example 3.5 Given triangle ABC Take the point M in BC, the point N in AC such that, BM = 12 × M C, CN = 12 × N A, E is the intersection of AM and BN Supopose that the area of the triangle ABC is 420cm2 Calculate 173 |191 Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx the area of the triangle AEN and the ratio A AE AM Combined with (3.25), we have SAEN = SAEB We deduce SAEN = E SABN = M C ANALYSIS According to the assumptions and conditions about the area ratio of two triangles with equal altitudes, the area ratio is the ratio of two bases We have SAEN = × SAEC = × SAEB 3 Therefore, SAEN = 47 × SABN We deduce 8 SAEN = × SABC = × 420 = 160(cm2 ) 21 21 SOLUTION Draw the line segment EC By assumption, the triangles ABM and AM C share the altitude drawn from A to BC, and BM = 12 ×M C Therefore, SABM = 12 ×SACM × SABN 2 × SABC = × 420 = 280(cm2 ) 3 Therefore, SAEN = × 280 = 160(cm2 ) Finally, we calculate the ratio EM Triangles AE BEM and BEA share the altitude drawn from B to AM Thus, the area ratio of the triangles BEM and BEA is the ratio of the bases AE, EM Therefore, EM SBEM = AE SBEA We have SBEA = SABN −SAEN = 280−160 = 120(cm2 ), Similarly, two triangles EBM and EM C share the altitude from E to BC and BM = 12 ×M C Thus, SEBM = 12 × SECM We deduce so SABE = SABM − SEBM 1 = × SACM − × SECM = × SACE 2 So, the ratio to be found is Therefore, SACE = × SABE (3.25) Now we will calculate the area SAEN through SABN The triangles AEN and AEC share the altitude drawn from E to AC and N A = 23 × AC So we have SAEN = × SAEC × On the other hand, the triangles ABN and ABC share the altitude drawn from B to AC and N A = 23 × AC, then N B SBEM = SABM −SEBA = 140−120 = 20(cm2 ) 20 EM = = EA 120 Answer: SAEN = 160cm2 ; EM = 16 EA Example 3.6 Given triangle ABC Take the points M in BC and N in AC such that BM = M C, CN = × N A The line passing through M and N intersects the side BA extended at E Determine the area ratio of the triangles AN E, ABC and the length ratio of the sides EN , EM 174 192| Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx and ABC share the same altitude from B to AC and AC = × AN , so we have E A SABN = N × SABC (3.27) Combined with (3.26) we get the result SEAN = 15 × SABC , or SEAN = SABC 15 B M C ANALYSIS By the assumptions we have SABN = 51 Therefore, to determine the ratio SABC SAEN AEN we must calculate the ratio SSABN SABC It is clear that SEBM = SECM , SN BM = SN CM Therefore, SEBN = SECN On the other hand, SEAN EAN EAN = 14 Hence, SSEBN = 14 , so SSABN = 13 We SECN deduce SOLUTION We draw BN and CE Acording to the assumptions, the triangles EBM and ECM share the altitude drawn from E to BC and M B = M C, so SEBM = SECM Similarly, we also have SN BM = SN CM We deduce, SEBN = SECN Two triangles EBM and ECM share the altitude drawn from E to AC and CN = × AN , so we get SEAN = × SECN (3.28) Now we will determine SBN A through SBN M The triangles BN M , BN C share the altitude drawn from the vertex N and BM = 12 × BC, so we have × SBN C On the other hand, by the assumptions we have SBN C = 45 × SABC , so SBN M = 25 × SABC , or SABC = × SBN M Combining with (3.27), we get the result SBN A = × SBN M Therefore, by (3.28), we have SBN E = × SBN M Thus, the ratio of the bases EN , EM is the ratio of two areas SBEN , SBN M , and we have We deduce × SABN × SBN A SBEN = SBN M = × SEBN SEAN = According to The trangles BN E, BN M share the altitude drawn from B to EM and the area ratio Therefore, SEAN SBEN = SBN M = EN SAEN and = = SABC 15 EM EN EM Next, we calculate the ratio (3.26), we have (3.26) From the assumption CN = × AN , so AC = × AN Furthermore, the two triangles ABN SBEN EN 2 EN = ⇒ = = NM SBN M EM Answer: SEAN SABC = EN , 15 EM = 25 175 |193 Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx This is a difficult problem that is not given specific data, especially the requirement to find EN The problem will be more diffithe ratio EM EN cult if it only requires calculating the ratio EM or changing the assumption about the point M , for example, BM = × CM Calculatis an intermediate step and ing the ratio SSAEN ABC also requires a lesser degree of difficulty When adding specific data to simplify the problem, we should start from the area of triangle AEN , or the area of triangle ABC and calculate the area of the remaining triangles Note that, if we choose a specific figure for the area of triangle ABC, it should be divisible by 3, 4, so that the results are integers Example 3.7 Given triangle ABC Take the points M in BC and N in AC such that BM = 13 ×BC, AN = 14 ×AC The notation P is the intersection of AM and BN Calculate the area ratio of the triangles P BM , P AN and prove that P A = P M A SOLUTION Drawn the line segment P C By the assumptions, we have × N C (3.29) The triangles P BM , P CM share the altitude drawn from P to BC, and BM = 13 × BC, so we have (3.30) SP BM = × SP BC The triangles P AN , P CN share the altitude drawn from P to AC, and the ratio (3.29), so we have (3.31) SP AN = × SP CN The triangles BAN , BCN share the altitude drawn from P to AC, and the ratio (3.29), so we get AN = × SBCN (3.32) Acording to (3.31) and (3.32), we have SBAN = × SP BC (3.33) Combined with the (3.29), we get the result SP AB = N P B will prove P A = P M , thereby determining the ratio of two triangles SP BM , SP AN SP AB = SP M B M (3.34) The triangles BP A, BP M share the altitude drawn from B to AM Hence, we have immediately P A = P M C ANALYSIS Draw the line segment P C Thanks to the assumptions we have Next, we calculate the area ratio of the triangles P AN , P BM SP BM = × SP BC = SP AB ⇒ P A = P M On the other hand, it is easy to see that It is assumed that BM = we have BM = 12 × CM 1 × SP AC , SP AN = × SP AC Therefore, SP AB = × SP AN We deduce SP BM = × SP AN To solve this problem we SP AB = × BC Therefore, The triangles ABM , ACM share the altitude drawn from A to BC, and BM = 12 × CM , so we have (3.35) SABM = × SACM 176 194| Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx SOLUTION Draw the line segment AM Acording to the assumptions, we have From (3.29) and (3.35), we deduce SP AB = × SP AC (3.36) On the other hand, two triangles P AN , P AC share the altitude drawn from A to AC, and AN = 14 AC, so we have SP AN = × SP AC (3.37) × BC, P B = × AB MB = SP AN Answer: SP AN SP BM = 12 , P A = P M Example 3.8 Given triangle ABC Take the points M, N, P on the sides BC, CA, AB respectively such that M C = × M B, N A = × N C, P B = × P A Prove that SM N P = × SABC A (3.39) Two triangles ABM , ACM share the altitude drawn from A to BC, and the ratio (3.38) We have From (3.34), (3.36) and (3.37), we deduce SP AN = × SP BM ⇒ = SP BM (3.38) SABM = × SABC (3.40) Two triangles M P B, M AB share the altitude drawn from M to AB, and the ratio (3.39) We get SM P B = × SM AB (3.41) From (3.40) and (3.41) we get the result SM P B = × SABC (3.42) Similarly, we have P N × SABC , = × SABC SM N C = (3.43) SN P A (3.44) Combined the (3.42), (3.43) and (3.44), we deduce SM N P = × SABC B M C ANALYSIS Draw the line segment AM By The proof is complete the assumptions we have SABM = 13 × SABC Example 3.8 is also a basic problem, has a and SM P B = 23 × SM BA Therefore, simple solution, and is solved by the area method However, if the subline AM is not SM P B = × SABC detected, the problem will not be possible to solve Our problem will becomes more difficult Similarly, we get if the points M, N , and P are defined on the 2 SP N A = × SABC , SM N C = × SABC sides of the triangle with different ratios 9 We deduce, SM N P = × SABC Next, we will consider some area problems in the special quadrilaterals 177 |195 Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx Example 3.9 Given a parallelogram ABCD, height AH = 15cm, base AB = 20cm Take the point M on the side AB such that AM = × M B The line segment M C intersects the diagonal BD at K Calculate the area of triangle KM B A M BKM = 35 It is clear the ratio of the areas is SSDKM that we get the problem: Find two numbers when the sum and ratio of the two numbers are known, where the sum is SBKM + SDKM = BKM SM BD = 90(cm2 ), the ratio is SSDKM = 35 and SKM B is small number Therefore, we have SKM B = 90 ữ (3 + 5) ì = 33, 75(cm2 ) B Answer: 33,75cm2 K D H C ANALYSIS By assumption, it is easy to calculate the area of triangles M BD, M CD, CBD and CBM On the other hand, assumptively, B = 35 Therefore, to calwe have the ratio M CD culate SM BK , we have to calculate the ratio SM BK K = 38 , or the ratio SSBM = 38 Then, SM BD BM C SBM K Example 3.10 Given rectangle ABCD Take the point M in CD, the diagonal BD and the line segment AM intersect at I such that the area of triangle BM C is equal to 36cm2 and area of triangle IM D Calculate equal to 16 the area of rectangle ABCD D M C H I K 3 = × SBM D = × SBM C 8 SOLUTION Acording to the assumptions, we have M B = 35 × AB = 35 × 20 = 12(cm) Therefore, we get SDM B = 12 × M B × AH = × 12 × 15 = 90(cm2 ) = SCM B , By the same assumption, we have SM CD = × CD × AH = 150(cm2 ) Two triangles M CD, CM B share the altitude B AH = 15(cm), and the ratio of the bases M = CD SCM B 3 Thus, the ratio of the areas is SM CD = A B ANALYSIS Acording to the assumptions SBM C = 36cm2 = × SIM D 16 Thus, SIM D = 64cm2 , and we find the area of triangle IAB So, to calculate the area of rectangle ABCD we will compute the area of two triangles IAD, IBM On the other hand, triangles M CD, CM B share the bases M C Then, the altitude drawn from the vertices B, D to M C has the ratio 35 Since SIAD = 12 × ID × AH and SM ID = × ID × M K = 64cm2 , we would determine the relationship between the line segments ID, K =M IB or AH, M K Notice that ID IB AH The triangles BKM , DKM share the bases KM , and the ratio of the altitudes is 35 So SOLUTION By assumptions, we have immediately SIM D = 64cm2 Next, the triangles 178 196| Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 K.C Nguyen/Vol 8, No.4_Oct 2022|p.xxx – xxx DAB, M AB share the base side AB and the height DA, so SM AB = SDAB = SBCD (3.45) triangle AOB is 4cm2 Find the area of trapezoid ABCD A B K I On the other hand, the triangles BCD, M AB share the triangles IM B We deduce H SIAB = SIDM + SM BC = 100cm2 We determine the triangles IAD, IM B By (3.45) and the triangles DAB, M AB share the triangle IAB, so we have SIAD = SIBM , and then ID MK = IB AH 1 × ID × M K, SIAB = × IB × AH 2 Therefore, SIM D ID M K 64 = × = 100 SIAB IB AH ANALYSIS From the assumptions, it is easy to calculate the area ratio of the triangles DAB and BCD, SDAB = SBCD We are also easy to calculate the area ratio of the triangles ABI and CBI, SABI = ⇒ SCBI = 6cm2 SCBI (3.47) Hence, SABD = 10cm2 So, SBCD = 15cm2 , and thereby SABCD = 25cm2 (3.48) SOLUTION The triangles DAB, BCD have the same height as the height of the trapezoid AB = 23 So the area and the ratio of the bases CD ratio of two triangles DAB, BCD is By (3.46) and (3.47) we get the result 64 ID ID ID × = ⇒ = IB IB 100 IB C (3.46) We drawn the altitudes AH, M K with H, K in BD Then, we have SIM D = D We deduce, × ID × AH = × × IB × AH 2 2 = × SIAB = × 100 = 40(cm2 ) 5 SIAD = From the above results, it follows that SABCD = × SABD = × (SIAB + SIAD ) = 280cm2 Answer: SABCD = 280cm2 Example 3.11 (Junior high school Le Quy Đon, 2020) In a trapezoid ABCD, the ratio between the base sides AB, CD is 23 Two diagonals AC, BD intersect at I The area of SDAB = , hay SBCD = × SDAB (3.49) SBCD The triangles ABD, CBD share the base BD and the ratio (3.49) Therefore, the ratio of the altitudes drawn from A and C to BD is 23 The triangles ABI, CBI share the base BI and the ratio of the altitudes drawn from A and C to BI is 23 We deduce SCBI = × SABI = 6cm2 (3.50) On the other hand, SIAD = SIBC = 6cm2 We deduce, SABD = SIAB + SIAD = + = 10(cm2 ) 179 |197 Tran Dinh Tuong/Vol No 4_October 2022| p.184-198 No.21_Jun 2021|p.166–176 According to (3.49), we get [3] Đỗ Đình Hoan (Chủ biên), Nguyễn Áng, Đỗ Tiến Đạt, Phan Thanh Tâm (2020), Bài tập Toán 4, Nhà xuất Giáo dục Việt Nam × 10 = 15(cm2 ) So, the area of trapezoid ABCD is SCBD = SABCD = SABD + SCBD = 10 + 15 = 25(cm2 ) Answer: SABCD = 25cm Conclusion The article has presented some development directions to detect new problems such as: changing the assumptions and/or adding new assumptions, changing the role between the assumptions and conclusion, and etc Improving the solutions to these examples is left to the reader Otherwise, these techniques that we proposed in this work can be applied to higher levels such as high school and undergraduate programs for some fields, such as geometry in algebra and geometric data analysis These problems arise in these fields, and some applications of geometry will be studied in future research REFERENCES [1] Bộ Giáo dục Đào tạo (2020), Toán lớp (Tập 1, 2), Nhà xuất Giáo dục Việt Nam [2] Bộ Giáo dục Đào tạo (2020), Toán lớp (Tập 1, 2), Nhà xuất Giáo dục Việt Nam [4] Đỗ Đình Hoan (Chủ biên), Nguyễn Áng, Đỗ Tiến Đạt, Phan Thanh Tâm (2020), Bài tập Toán 5, Nhà xuất Giáo dục Việt Nam [5] Trần Phương, Nguyễn Đức Tấn, Phạm Xuân Tiến (2011), Toán Chọn lọc Tiểu học (Tập 2: Các đề toán), Nhà xuất Giáo dục Việt Nam [1] Ministry of Education and Training, Grade Math (Volume 1, 2), Vietnam Education Publishing House [2] Ministry of Education and Training, Grade Math (Volume 1, 2), Vietnam Education Publishing House [3] Do Dinh Hoan (Editor), Nguyen Ang, Do Tien Dat, Phan Thanh Tam (2020), Math Exercise 4, Vietnam Education Publishing House [4] Do Dinh Hoan (Editor), Nguyen Ang, Do Tien Dat, Phan Thanh Tam (2020), Math Exercise 5, Vietnam Education Publishing House [5] Tran Phuong, Nguyen Duc Tan, Pham Xuan Tien (2011), Primary Selected Mathematics (Volume 2: Math problems), Vietnam Education Publishing House 180 198| ... calculate the area of triangle EAC, the area ratio of the triangles EBC and EBA, namely SEAC = 27cm2 , SEBC = SEBA On the other hand,from the assumption AM = M B, then the area ratio of the triangles... Take the point M on the side AB such that AM = × M B The line segment M C intersects the diagonal BD at K Calculate the area of triangle KM B A M BKM = 35 It is clear the ratio of the areas... the problem of the geometrical area is always taken to increase the differentiation for candidates However, most pupils have no points in the problems of the area and are solved by the area method