You can find the interior angle measures in a regular polygon and then use what you know about triangles and special quadrilaterals to analyze complex figures.

Question

The first time I submitted this assignment, my teacher said that this was wrong:

 for task 1 you needed to find the measure of the angles in the shape and show all workings. For task 2 you needed to identify what the shape will be using theorems/postulates. For task 3 you need to prove congruence using one of the methods that you gave. Please make the necessary adjustments and resubmit the entire project. 

here is the link: https://intervisualtechnology.us/pxudpothrb/167/9129/Unit_6_Project.pdf

i would appreciate any help i can get!

Answer 1

Task 1

To start, the question tells us that we are dealing with a regular hexagon, so right off the bat, we know that all sides are the same length and each interior angle is the same 120 degrees. This is known because all regular hexagons have these properties.

Image: https://i.imgur.com/VZm6vTI.jpg

Next, the diagonals given to us by the problem go from corner to corner within the hexagon and make four congruent isosceles triangles. Namely, triangles FAB, EFA, CDE, and DCB. We can also deduce that the two smaller base angles for each of these triangles are all 30 degrees. This is done by knowing that if you add up all the angles inside a triangle, they will always add to 180 degrees; we also know that the two base angles are equivalent to each other.

The algebra to find this would be:

[the big angle] + [the left little angle] + [the right little angle] = 180

120 + x + x = 180 (Both little base angles are equal to each other, so let's call them x)

120 + 2x = 180 (Combine like terms)

--subtract 120 from both sides--

2x = 60 (Result from subtracting)

--divide both sides by 2--

x = 30 (Result from dividing)

Image: https://i.imgur.com/nZpv7Sd.jpg

The diagonals also make two smaller congruent isosceles triangles. These two are similar to the four triangles mentioned before by the Angle-Angle-Angle theorem (or AAA for short). Notice how the base angles for the two isosceles triangles are actually the base angles for the bigger isosceles triangles at the same time! Then, because all angles in a triangle must add to 180 degrees, the missing big angle in these two triangles must be 120 degrees.

Image: https://i.imgur.com/5Oq0kIS.jpg

Now we can determine the interior angles of quadrilateral GBHE. We can find angles EHB and BGE by noticing that they are vertical angles to angles FHA and CGD respectively, thus both angles FHA and CGD must equal 120 degrees. As for angles HEG and HBG, notice how they are part of the hexagon's interior angles (which are all 120 degrees). The angles we are looking for (HEG and HBG) are both equal to one of the hexagon's interior angles minus the two base angles of the isosceles triangles (which are all 30 degrees).

The math you would do to find HEG and HBG is:

[hexagon's interior angle] - [base angle] - [base angle] = ??

120 - 30 - 30 = 60 degrees

Image: https://i.imgur.com/ABvfx8n.jpg

Now to find which shape is GBHE. Fortunately, this is much more straightforward than finding the interior angles. With all of the information on the angles we have found so far, we have all the information we need to see that triangles AHB, CGB, DGE, and FHE are all right triangles. Not only that, but they are all congruent by the Angle-Side-Angle theorem (or ASA for short). If each of these triangles are congruent, then all the sides of the quadrilateral GBHE, must all be congruent because it is made up of all the hypotenuses of these four right triangles. The only shape that has four congruent sides but different angles is a rhombus.

Image: https://i.imgur.com/mS23F40.jpg

Task 2

In this task, parallelogram JKLM is extended by x. The strategy I would recommend for this problem would be to show that triangles PKQ and RMS are congruent, then show that triangles SJP and QLR are congruent. We can show all of that with the Side-Angle-Side theorem (or SAS for short), but we are missing the "angle" part.

All parallelograms have a property that stats that all opposite angles are congruent. This means, within the context of our given problem, the pair of interior angles M and K are congruent to each other. Also, the pair of interior angles J and L are congruent to each other. Every interior angle has an exterior angle that is supplementary to it (they both add to 180 degrees). If interior angles M and K are congruent, then their exterior angles (angles RMS and PKQ) must also be congruent; the same goes for interior angles J and L and their exterior angles (angles SJP and QLR). Now we have enough information to conclude that triangles PKQ and RMS are congruent to each other, and that triangles SJP and QLR are congruent to each other; we do this by using the Side-Angle-Side theorem.

Now, because congruent parts of congruent polygons are congruent, we can say that line segments PQ and RS are congruent. We can also say that line segments SP and QR are congruent. A quadrilateral whose opposite sides are congruent is a parallelogram.

Task 3

This is the most straightforward task out of the three. Start by finding the coordinates of each corner of both triangles. If you use the distance formula to find the length of each side, using the coordinates you found, you should find that both triangles have matching sides and thus must be congruent by the Side-Side-Side theorem.

Another way is to find the slope of line segments AB and AC and notice that the two slopes are the opposite-reciprocals of each other (you change the slope's sign and flip the fraction); this means that the two line segments are perpendicular to each other which means that angle BAC is a right angle. You can do the same thing with line segments EC and EF to find the same result. Now all you have to do is use the distance formula to find the length of line segments AB, AC, EC, and EF. Now you have all the information you need to claim that both triangles are congruent by the Side-Angle-Side theorem.

Hope this helps!

-Hunter

Answer 2

1. Rhombus
2. Parallelogram
3. By 3 sides