2.05 transformations and congruence activity on FLVS

The steps given in the question are;

Step 1: The type of triangle chosen is the equilateral equiangular triangle with rotation to prove SAS congruence

Step 2

The chosen type of triangle, the equilateral, equiangular triangle, ΔABC, is constructed on MS Excel using the points A(0, 0), B(6, 0), and C(3, 3·√3)

The lengths of the sides of the triangle are verified as follows;

AB = 6 - 0 = 6

AC = √(3² + (3·√3)²) = 6

CB = √((6 - 3)² + (3·√3)²) = √(3² + (3·√3)²) = 6

Therefore, ΔABC is an equilateral triangle

An equilateral triangle = An equiangular triangle

∴ ∠A  = ∠B = ∠C = 60°

The coordinates of the image of the preimage (x, y), following a rotation transformation of 180° clockwise is given as follows;

(x, y) 180°CW → (-x, -y)

Therefore, the coordinates of the image ΔA'B'C' of the preimage triangle ΔABC after a 180° rotation transformation are;

A'(0, 0), B'(-6, 0), and C'(-3, -3·√3)

The lengths of the sides of the triangle ΔA'B'C' are;

A'B' = 0 - (-6) = 6

A'C' = √((-3)² + (-3·√3)²) = 6

C'B' = √((-6 - (-3))² + (-3·√3)²) = √(3² + (3·√3)²) = 6

Therefore, ΔA'B'C' is an equilateral triangle and therefore an equiangular triangle

∴ ∠A' = ∠B' = ∠C' = 60°

Therefore, the lengths of the sides of the triangle ΔABC are equal to the lengths of the sides of the triangle ΔA'B'C', and we have;

AB = A'B' therefore, AB ≅ A'B'

AC = A'C', therefore, AC ≅ A'C'

∠A = ∠A' = 60°, therefore, ∠A ≅ ∠A'

Therefore, triangle ΔABC is congruent to ΔA'B'C', by Side-Angle-Side, also known as, SAS rule of congruency

Learn more about rigid transformations here;

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step-by-step explanation:

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.