11_17

Today in class we started off with returning to the Theorem //The area of a parallelogram is the product of the length of a base and the// //corresponding height.//

Alexia started us off by constructing a parallelogram with an altitude h and base a, and restating that the area of the parallelogram is A=ah=bg as stated above in our theorem. She then, proposed the question of adding the triangle formed in our parallelogram by our altitude, and adding it to the opposite side of the parallelogram. We would take our parallelogram ABCD, by proposing we create and altitude form point A to F we create line segment AF. Consider line segment AF but F could be any length therefore line segment BF may not form and exact right triangle we wanted to add on the side AB as Alexia proposed. However, we use ▲AFB congruent to ▲DEC. By angle C and angle E supplements to each other, then we know angle FBA and angle ECD are congruent so we know that sides FB and EC are congruent. __This would create a rectangle and we could then use the area of the rectangle to prove that the area of a parallelogram is ah. However, we ran into some difficulties proving those two triangles congruent. We first needed to know that the two sides of the parallelogram were congruent. We haven't proved that yet but we began with proving the angles of a parallelogram congruent.__ Next, we proposed that we prove proposition that angles of a parallelogram are congruent. __Brett's proof of this__ __thereom required that we had first proved the sides of a parallelogram were congruent, which we have not yet proved. Bob's proof used the properties of vertical angles and alternate interior angles to prove that the angles of a parallelogram were congruent.__We proposed the question… Can this be proved by contradiction? __Instead of proving by contradiction, Emina showed us a direct proof in which we can prove sides and angles of a parallelogram congruent at the same time.__ Consider parallelogram ABCD and a diagonal from angle D to B, transversal angles by ASA state ▲ DBA congruent to ▲BDC there def of congruent ▲s implies segment AD congruent to segment BC and segment DC congruent to AB and angle A congruent to angle C. By addition of angles prop 3.15 and definition of congruent triangles angle ADC congruent to angle ABC. While Emina was at the board discussing different views of the parallelogram and why we should not focus on just one view of the parallelogram, Dave commented on how is way of proving this theorem to Emina earlier was POOP. __We have now proven that the sides and angles of a parallelogram are congruent but we still have not proven that the area of a parallelogram is ah.__ We then broke up into our groups and discussed different ways to __prove this theorem.__ Finally we came to the conclusion of our Theorem. Consider the parallelogram ABDE, we construct an altitude from angle A to some point F, and an altitude from angle D to some point C to make ▲AFE and ▲DCB, we also constructed and an altitude from angle B to E’. By AIA we get angle BDE congruent to angle AEF and by SAS we get ▲AFE congruent to ▲BDE. Angle D is a supplement of ▲BDE and ▲BDC. Angle A is a supplement of ▲AEF and ▲ABE to conclude that ▲AFE congruent to ▲DCB = to ▲BE’D.

Conclusion- Area of a rectangle ACDF constructed, = area of ▲AFE + __area of ▲BCE__ + area of EDBA. = __area ▲BCD__ + area ▲E’BD + area of rectangle FE’BA… They are the same so the Area = ah. We then worked on Theorem: //The area of a triangle is half the product of the length of a base and the corresponding height.// //We constructed triangles ABC and DCB. Triangles ABC congruent to triangle DCB by definition of congruent triangles// //line CD parallel to line AB// //angle A congruent to angle DBC by def of congruent triangles.// //ABDC is a parallelogram// //Area ABDC = ah = 2A of triangle ABC.// //A3 ( Area or triangle ABC = area of triangle DCB)// //Area of triangle ABE = ½ ah.// //Square→ rectangle→ parallelogram→ triangle// //From a rectangle we get a right triangle→ triangle → parallelogram// //Area triangle = 1/2ah by ½xh+ ½ (a-x)h= ½ ah.// //We ended the class with the question// //Is it possible to express area of a triangle using sides only?//