Triangle 1 has vertices at (A, B), (C, D), and (E, F). Triangle 2 has vertices at (A+1,B+3), (C+1,D+3), and (E+1,F+3). What can you conclude about triangle 2?

Given:

Triangle 1 has vertices at (A,B), (C,D), and (E,F).

Triangle 2 has vertices at (A+1,B+3), (C+1,D+3), and (E+1,F+3).

To find:

The conclusion about the triangle 2.

Solution:

From the given information it is clear that the x-coordinates of the vertices of triangle 2 are 1 more than the x-coordinates of the vertices of triangle 1.

Similarly, the y-coordinates of the vertices of triangle 2 are 3 more than the y-coordinates of the vertices of triangle 1.

So, the rule of translation is:

It means the triangle 1 is shifted 1 unit right and 3 units up to get the triangle 2.

Therefore, triangle 2 is congruent to triangle 1 and triangle 2 is the translation of triangle 1 by 1 unit right and 3 units up.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

step-by-step explanation:

hihi so drawing these 2 triangles is fairly simple. all that's left is finding the height of each one. luckily, we have a side length and an angle, so there is a trig function we can use to solve for the height. the easiest one to use would be tangent, since we have 2/3 things needed to find and the third needed is what we are solving for. tan(73°) * 30 ~= 98.13 and tan(73°) * 35 ~= 114.45