For k:(j + k)/3 = 5
2 answers:
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Answer:
Please find attached the required plot accomplished with an online tool
Part A:
1/4
Part B:
P''(-1, 0), Q''(0, -1), and R''(2, -1)
Part C:
Triangle PQR is similar to triangle P''Q''R'' but they are not congruent
Step-by-step explanation:
Part A:
Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)
Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)
The dimensions of the sides of the triangle are given by the relation;
Where;
(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment
For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);
By substitution into the length equation, we get;
The length of segment PQ = 4·√2
The length of segment PR = 4·√10
The length of segment RQ = 8
The length of segment P'Q' = √2
The length of segment P'R' = √10
The length of segment R'Q' = 2
Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4
Part B:
Reflection of (x, y) across the y-axis gives;
(x, y) image after reflection across the y-axis = (-x, y)
The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;
P'(1, 0) image after reflection across the y-axis = P''(-1, 0)
Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)
R'(-2, -1) image after reflection across the y-axis = R''(2, -1)
Part C:
Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.
Answer:
8 and 12
Step-by-step explanation:
Sides on one side of the angle bisector are proportional to those on the other side. In the attached figure, that means
AC/AB = CD/BD = 2/3
The perimeter is the sum of the side lengths, so is ...
25 = AB + BC + AC
25 = AB + 5 + (2/3)AB . . . . . . substituting AC = 2/3·AB. BC = 2+3 = 5.
20 = 5/3·AB
12 = AB
AC = 2/3·12 = 8
_____
<em>Alternate solution</em>
The sum of ratio units is 2+3 = 5, so each one must stand for 25/5 = 5 units of length.
That is, the total of lengths on one side of the angle bisector (AC+CD) is 2·5 = 10 units, and the total of lengths on the other side (AB+BD) is 3·5 = 15 units. Since 2 of the 10 units are in the segment being divided (CD), the other 8 must be in that side of the triangle (AC).
Likewise, 3 of the 15 units are in the segment being divided (BD), so the other 12 units are in that side of the triangle (AB).
The remaining sides of the triangle are AB=12 and AC=8.
