Saturday
April 19, 2014

Homework Help: Geometry

Posted by Emily on Friday, January 22, 2010 at 4:10pm.

Hello, I hope that I got all these questions right, but is important that I do a good job on this for my grade. So, it would be great if someone would check my work for me- just to be sure. :)
Thank you for your help!
-em

(10 points)
Score

1. The coordinates of the vertices of parallelogram RMBS are R(4, 5), M(1, 4), B(2, 1), and S(3, 0).
Using the diagonals, prove that RMBS is a rhombus. Show all your work and state appropriate formulas and theorems used.
Answer:
slope of line rb={(5+1)/(-4-2)}=-1
slope of line ms={(4-0)/(1+3)}=1
m1*m2=-1 so both line are perpendicular.
if diagonals of a parallelogram s perpendicular then its a rhombus.





3. Explain the purposes of inductive and deductive reasoning in mathematics. Be sure to define both inductive reasoning and deductive reasoning and describe how each can help you develop and prove theorems.
Answer:

Deductive reasoning means you take a known principle or group of known principles and using those you go on to prove something more complex. For example,
Prove: The sum of two odds integers is an even integer.
Pf; Assume that n and m are odd integers. This means that n and m take the form 2(l)+1 where l is the set of all integers. Then n+m = 2(l)+1 + 2(l)+1
= 4(l)+2 = 2 (2(l)+1)
Which is clearly the form of an even integer.
Therefore, the sum of two odd integers is an even integer.

What we did is take past principles that we know, like the form of an odd and even integer, and use it to prove something more complex.

Inductive reasoning is when you prove something based on the assumption that if things behave this way now they will continue. In math induction it goes like this:

Suppose there is some proposition P(n) and we wanted to prove that P(n) is true where n >= 1 and is an integer. Then two things must be true:
(1) P(1) must be true, this means that the first number or proposition that you're trying to prove must be true. For instance, to prove this 1+2+3+...+n = n(n+1)/2 ; P(1) would be 1=1(1+1)/2 = 1, therefore P(1) is true.

(2) For each P(k); if P(k) is true than P(k+1) is also true.

So essentially induction says, if you can say the proposition is true for the first instance, and for some arbitrary instance and the one immediately following the arbitrary one, you can say that a particular theorem or assumption is true. This is beneficial because clearly there are things that cannot be handled by deductive reasoning alone.


(10 points)
Score


5. Perform these transformations on ∆TRL and compute the perimeter of the pre-image and final image.
a. Rotate ∆TRL 180 about the origin and label the resulting image ∆T R L.
b. Reflect ∆T R L across the y-axis and label the resulting image ∆T R L.
c. Translate ∆T R L according to (x, y) (x + 3, y 3) and label the image ∆T R L.

Answer:
The original triangle is a right-angled one with sides 2, 2, sqrt8.
The transformations are all transformations that preserve size and shape. This is also defined as an isometric translation, so the final triangle will also have perimeter 4 + sqrt8. (or 6.828)

NOTE: I am really good at graphing, so I am positive that I got the graphing right. I just need help with the above questions. Thanks!

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