A Math Forum Project

Geometry Forum - Problem of the Week Solutions - Breaking Up a Triangle, Feb. 20-24, 1995 Annie says: This is another example of a problem that can be solved in several ways. None of the solutions included are exactly like the others - they all break the original triangle into different areas to compare. Read each of the problems carefully to see how they did it. One thing I like to see is that all but one of the solutions stated their conclusion right off the bat, then went on to show why. David and Sarah McMenomy I hope this is not too late to post a solution to the problem of the week. The answer still has not appeared on the board. This is the solution generated by David and Sarah McMenomy (home schooled) last week, assuming that the problem as written was not exactly correct: properly speaking, F should not be able to move along seg BC, since it is defined as a point on seg AC. We assumed a typo, and that it should be able to be moved anywhere on seg AC. If this is wrong, of course, the following proof is meaningless. ----- The area of the quadrilateral is exactly equal to the excluded triangles in all cases, without regard to the position of point F (assuming it remains on AC). ----- Given: E the midpoint of AB; D the midpoint of CB. 1. CA = 2DE. Segment connecting the midpoints of two sides of a triangle is half the length of the third side. 2. Area of tri ABC = 4 * area of tri DBE. Tri DBE has half the height of tri ABC and a base half its length. 3. Area of tri FED = area of tri BED. Base and height are equal. 4. Area of quad FEBD = half area tri ABC 1/4 ABC + 1/4 ABC = 1/2 ABC 5. Area tri ABC - area quad FEBD = 1/2 area tri ABC Area ABC - 1/2 area ABC = 1/2 area ABC 6. Area tri ABC - area quad FEBD = area FCD + area FAE Tri's FCD and FAE are what is left over when you take quad FEBD away from tri ABC. 7. Area tri AFD + area tri FAE = 1/2 area tri ABC Area tri ABC - 1/2 area tri ABC = 1/2 area tri ABC 8. Area tri FCD + area tri FAE = area quad FEBD Substitution: 1/2 area tri ABC = 1/2 area tri ABC Emillia Ortiz From: Emillia Ortiz Grade:9 Mr. Delaney's Geometry Class, Westmont High School. Problem of the Week Feb. 20-24 Answer: The areas of the triangle when added together, equal the area of the quadrilateral, no matter where F is located along AC. The area of the two triangle always change when F is moved along AC, but the area of the quadrilateral always remains the same. The area of the quadrilateral always remains the same because the triangle EBD within the quadrilateral always remains the same no matter where F is because the two midpoints D & E are stationary. The other part of the quadrilateral DEF which changes position, always has the same area because as F moves along AC, the base, DE, remains the same length. The height of the same triangle is always the same because AC and ED are parallel and as F moves along AC it remains the same height away from its base ED. The reason that the sum of the areas of the two triangles equal the area of the quadrilateral is: Draw a line from F to B, the area of triangle FDB now equals the area of triangle FDC because the length of the bases are the same and the height is the same because it is the same line. Triangle EBF's area equals the area of triangle EAF because the bases equal each other and the height is the same line. Now from this you know that the sum of the area's of the two triangles FCD and FAE are equal to the area of the quadrilateral FEBD. Kate Devine, Jen Hinkel, and Jordan Walsh Mount St. Joseph Academy Kate Devine (9th), Jen Hinkel (9th), Jordan Walsh (9th) [this was submitted as a sketch, but I just pulled the text out] As point F moves along AC, the area of the two triangles change but their sum remains constant and equal to the area of the quadrilateral. Triangle BED ~Triangle BAC and ED=!/2 AC. Since the scale factor of the 2 triangles is 1:2, the ratio of their areas is 1:4. If the area of triangle BAC is 4 square units and the area of triangle Bed is 1 square unit, then the area of trapezoid AEDC is 3 square units. Triangle AEF, triangle EFD, and triangle FDC all have the same height, so the ratio of their areas is equal to the ratio of their bases. The ratio of triangle EFD to triangles (AEF and FDC added together) is 1:2. Thus , the area of the quadrilateral (BEFD) is 2 square units because it is made up of triangles BED and EFD and the sum of the areas of the other 2 triangles is also 2 square units. Joan Scott and Kate Devine [this was submitted as a sketch, but I just pulled the text out] 1) ED is parallel to AC.(any segment joining the midpoints of two sides of a triangle is parallel to the third side) 2) Triangle BAC is similar to triangle BED (AA Similarity Prop.) 3) ED is equal to one half AC.( A segment joining the midpoints of two sides of a triangle and parallel to the third side is one half the length of the third side.) 4) The scale factor of triangle BED to triangle BAC is 1/2. 5) The ratio of the area of triangle BED to the area of triangle BAC is 1/4( If two triangles are similar, then the ratio of their areas equals the square of their scale factors.) 6) The ratio of the triangles ABC and DEF is 4 to 1. By: Joan Scott and Kate Devine Mount St. Joseph Academy grades 10 and 9 Susanna Puntel and Maria Szczesniak [this was submitted as a sketch, but I just pulled the text out] Susanna Puntel Maria Szczesniak Mount St. Joseph Academy Grade 10 We discovered that the sum of the areas of the triangles is equal to the area of the quadrilateral. We also discovered that the ratio of the areas of the triangles are equal to the ratio of the bases of the triangle. Also, when point F is in the middle of segment AC, then the ratio of the areas of the triangles is equal to the ratio of the heights of the triangles. Construct segment FB to divide the quadrilateral into 2 triangles. Construct the perpendicular from F to AB. This segment (FH) is the height of both Triangle BFE and EFA. Their bases are equal because of the definition of a midpoint. Therefore their Areas (1/2 base * height) Are equal. Now construct the perpendicular from F to CB. This segment forms the height of both Triangle DFC and BFD. Their bases are equal because of the definition of a midpoint. Therefore Their areas (1/2 base* height) are equal. Use the addition property from algebra to prove that triangle BDF and BFE (the quadrilateral) together have the same area as triangles AEF and DFC. 9th Grade Class, Wheatland Jr. High Assume that triangle ABC is an acute scalene triangle (we determined that the kind of triangle does not affect the answer.) Construct the lines described in the problem, but make F the midpoint of CA. By calculating the areas of the two triangles, they can be shown to be the same. If you construct line DE you will find that triangles DEB and DEF are congruent and are also congruent to CDF. Also, if line DE was constructed, the four triangles could be proved congruent using the mid-point theorem of triangles. Quadrilateral BDFE (which is comprised of triangles DEF and DEB) has the same area as the sum of the areas of triangles CDF and FEA. When F is not the midpoint of CA, the sum of the areas of CDF and FEA does not change. The total base does not change because as one base gets larger the other gets smaller, and the altitude remains the constant. Wheatland Jr. High 9th grade Geometry class Lancaster PA Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page