oh, and did I forget to mention THE QUIZ ON TUESDAY!!
OPTIONAL Quiz Review
hw #5-4A Due Tues Dec 14
1) Pg 182 #1-8
2) Pg 197-98 #1-16
hw #5-4B Due Tues Dec 14
1) Read pgs 184-85
2) Pg 186 #1-11
3) Copy the chart on pg 187 #1-10 into your GB (you can leave off the problem numbers)
4) Pg 187-89 11-41 Odd
EXTRA CREDIT for the first accurate paragraph proof posted to the blog for #41 on pg 189!!
Nick said...
ReplyDelete7 d WHA?
BTW rhombuses or rhombi?
Moy & Chisu
7d WHA WHA?
ReplyDeleteYou say to-MAY-toe I say to-MA-toe
on 37 how is the bisectors form a square?
ReplyDeletekevin
nevermind i got it
ReplyDeletekevin
ACCURATE PROOF OF PG 189 #41
ReplyDeleteGiven: ABCDE is a regular pentagon; X is the intersection of lines AC and BD
Prove: Quadrilateral AXDE is a rhombus.
Paragraph proof (the book asked for a paragraph proof):
Draw an auxiliary line between points A and D. Each angle in this shape is equal because this is a regular pentagon. The measure of the total inside degrees of the pentagon is 540 based on Theorem #3-13. Each angle is equal to 108 degrees by the division property of equality. Triangle BCA is an isosceles triangle. This is true, because this is a regular pentagon and sides BC and AB are congruent. Angle BAC is congruent to angle BCA by Theorem #4-1. Using the same steps, it can also be proved that triangle EDA is an isosceles triangle.
Angle B is congruent to angle E because this is a regular pentagon. Lines BA, CB, EA and DE are all congruent because this is a regular pentagon. By the SAS Postulate, triangle BAC is congruent to triangle EAD. Angle BAC is congruent to angle EAD because CPCTC. Angles B and E equal 108 degrees each. By Theorem #3-13, the sum of total interior degrees of a triangle is 180. By the subtraction property of equality, angles BAC and ACB must sum to a total of 72 degrees. By the division property of equality, angles BAC and ACB equal 36 degrees each. Angles EAD and ADE equal 36 degrees each because CPCTC. By the subtraction property of equality, angle CAD equals 36 degrees. Using these steps, it can also be proved that triangle CDB is congruent to triangle EDA; therefore, angle ADB is equal to 36 degrees.
Both triangles XAD and EAD are congruent because of the ASA Postulate. Because CPCTC, triangle XAD is an isosceles triangle based on Theorem #4-1. Lines XA, ED, DX and AE are congruent because triangles XAD and EAD are congruent and isosceles. Because all sides of quadrilateral AXDE are congruent, quadrilateral AXDE is a rhombus by the definition of a rhombus.
AWESOME! Napolean RULES!!
ReplyDeleteI don't believe we have a theorem for #15.
ReplyDeleteMoy & Chisu
#33, tonight's HW.
ReplyDeleteJKLM is a rhombus.
How?
Not too bad, don't need to get out the shottys for the zombies.
Sorry to spoil the fun.
JM and LK are half of FG by theorem 5-11 part 2, talking about lines going through 2 midpts being half of the base.
By the same token, JK and ML are half of EV by the same theorem.
Last, all the smaller segments that I just named are congruent because EV is congruent to FG, which is given. This makes it a rhombus.
Furthermore, we can't make it a square because we have nothing that depicts angles. No perpendiculars or measures or otherwise.
Moy & Chisu
For the test I'm going to miss can I take it tomorrow, and when?
ReplyDelete16 and 17 i had issues with
ReplyDeletewe aren't doing number 16 anonymous.
ReplyDeletefor number 17 prove triangles ACO and CBO and ADO and ABO congruent.
k bye!
MERRY CHRISTMAS TO X AND TO Y A GOODNIGHT!
ReplyDelete