a translation 2 units to the right and 1 unit up followed by a counter-clockwise rotation 90° about the origin
Pentagon A B C D E is reflected across the X-axis to form pentagon A B C D.
First plot points A, B, C and D on the coordinate plane and connect them to get quadrilateral ABCD. Then plot the vertices E, F, G, H of the quadrilateral EFGH in each case.
For each of these cases there are attached diagrams
1 case - reflection acraoss the x-axis
2 case - translation 5 units down
3 case - reflection across the y-axis
4 case - translation 7 units right
1. Coordinates of the vertices of the figure I (preimage):(10, - 5)(15, -5)(10, - 10)(15, -10)
2. Coordinates of the vertices of the figure II (image):(0,10)(0,15)(-5,10)(-5, 15)
Since many different rigid transformations can map the figure I into the figue II, you will need to use trial and error.
The most important is to do it in an educated way.
I will start by eliminating some options.
The first option, a reflection across the x-axis and 15 units left, does not work, because the reflection across the x-axis would shift the figure to a lower position than what you need.
The third option, a 90º counterclokwise rotation about the origin then a translation 10 units up, would move the figure to the second quadrant, and we need it in the third quadrant.
I will try now with the second choice, a 90º clockwise rotation and then 25 units up:
First, a 90º clockwise rotation, which is the same that a 270º counterclockwise rotation, follows the rule (x, y) → (y, -x)
Then, that results in:(10, - 5) → (-5, -10)(15, -5) → (-5, -15)(10, - 10) → (-10, -10)(15, -10) → (-10, -15)
Now, you can see that shifting 25 units up will not work, because you need that two x-coordinates become 0 (zero). So, this is not the correct set of transformations either.
A 180º rotation about the origin and a translation 10 units right follow this chain of rules:(x, y) → (-x, -y) → (-x + 10, -y)
That means:(10, - 5) → (-10,5) → (-10 + 10, 5) = (0, 5)(15, -5) → (-15, 5) → (-15 + 10, 5) = (-5, 5)(10, - 10) → (-10, 10) → (-10 + 10, 10) = (0, 10)(15, -10) → (-15, 10) → (-15 + 10, 10) = (-5, 10)
These last points do not coincide either with the vertices of the figure II.
In conclusion, neither of the choices gives the correct answer to the question.
a. (SSA) is not a valid mean for establishing triangle congruence. In this case we know the measure of two adjacent sides and the angle opposite to one of them. Since we don't know anything about the measure of the third side, the second side of the triangle can intercept the third side in more than one way, so the third side can has more than one length; therefore, the triangles may or may not be congruent. In our example (picture 1) we have a triangle with tow congruent adjacent sides: AC is congruent to DF and CB is congruent to FE, and a congruent adjacent angle: ∠CAB is congruent to ∠FDE, yet triangles ABC and DEF are not congruent.
b. (AAA) is not a valid mean for establishing triangle congruence. In this case we know the measures of the three interior sides of the triangles. Since the measure of the angles don't affect the lengths of the sides, we can have tow triangles with 3 congruent angles and three different sides. In our example (picture 2) the three angles of triangle ABC and triangle DEF are congruent, yet the length of their sides are different.
c. (SAA) is a valid means for establishing triangle congruence. In this case we know the measure of a side, an adjacent angle, and the angle opposite to the side; in other words we have the measures of two angles and the measure of the non-included side, which is the AAS postulate. Remember that the AAS postulate states that if two angles and the non-included side of one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent. Since SAA = AAS, we can conclude that SAA is a valid mean for establishing triangle congruence.
Task 2: geometric constructions
a. Step 1. Take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw another circle with radius AB but this time with center at B.
Step 3. Mark the two points, C and D, of intersection of both circles.
Step 4. Use the points C and D to mark a point E in the circle with center at A.
Step 5. Join the points C, D, and E to create the equilateral triangle CDE inscribed in the circle with center at A (picture 3).
b. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. The point B is the first vertex of the inscribed square.
Step 3. Draw a diameter from point B to point C.
Step 4. Set a radius form point B to point D passing trough A, and draw a circle.
Step 5. Use the same radius form point C to point E using the same measure of the radius BD from the previous step.
Step 6. Draw a line FG trough were the two circles cross passing trough point A.
Step 7. Join the points B, F, C, and G, to create the inscribed square BFCG (picture 4).
c. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw the diameter of the circle BC.
Step 3. Use radius AB to create another circle with center at C.
Step 4. Use radius AB to create another circle with center at B.
Step 5. Mark the points D, E, F, and G where two circles cross.
Step 6. Join the points C, D, E, B, F, and G to create the inscribed regular hexagon (picture 5).
In a dilation, the shapes are not always congruent, however they are similar.