AA for Similar Triangles and Transformations

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Let

$Delta ABC$ and

$Delta PQR$ be triangles such that

$ang A ~= ang P$ and

$ang B ~= ang Q$ . Assume that

$AB < PQ$ . The following questions will show, using similarity transformations, that if two angles of one triangle are congruent with two angles of another triangle, the triangles are similar.

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

Which of the following shows that

$ang C ~= ang R ?$

$m ang A + m ang B + m ang C = 180°$ and $m ang P + m ang Q + m ang R = 180°$
$m ang C = 180° - m ang A - m ang B = 180° - m ang P - m ang Q = m ang R$
$m ang C = m ang A + m ang B = m ang P + m ang Q = m ang R$
$m ang C = -m ang A - m ang B = -m ang P - m ang Q = m ang R$

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

If the coordinates of points B and Q are given by

$B(x_B, y_B)$ and

$Q(x_Q, y_Q)$ , which of the following gives the correct translation such that point B is translated to point Q? Let

$T_x$ represent the horizontal translation and

$T_y$ the vertical.

$T_x = x_Q - x_B$ and $T_y = y_Q - y_B$
$T_x = x_Q$ and $T_y = y_Q$
$T_x = x_P$ and $T_y = y_P$
$T_x = x_Q + x_B$ and $T_y = y_Q + y_B$

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

Using the translation from the previous question, translate

$Delta ABC$ by

$T_x$ units horizontally and

$T_y$ units vertically, resulting in

$Delta A_2B_2C_2$ . The transformed point

$B_2$ is coincident with point Q. If we want point

$A_2$ to lie on line segment

$bar{PQ}$ , and it does not already, which of the following transformations would ensure that it does?

A reflection over the line $stackrel{leftrightarrow}{PR}$ .
A reflection about the y-axis.
A rotation by some angle $theta$ about the point Q.
A rotation by some angle $theta$ about the origin.

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

After applying the transformation in the previous question to

$Delta A_2 B_2 C_2$ , the newly transformed triangle is

$Delta A_3B_3C_3$ and point

$A_3$ lies on

$bar{PQ}$ . It may be that point

$C_3$ lies on

$bar{PQ}$ . If not, a reflection over the line

$stackrel{leftrightarrow}{PQ}$ , applied to

$Delta A_3 B_3 C_3$ will ensure that it does. Why is it certain that point

$C_4$ (or

$C_3$ if the transformation is unnecessary) will lie on

$bar{QR} ?$

Congruent angles $ang B$ and $ang Q$ must have congruent arms.
Since a translation and rotation have already been applied, a reflection must transform $C$ to $bar{QR}$ .
For two congruent angles, $ang B$ and $ang Q$ , if the vertices are coincident, then the arms must be coincident.
For two congruent angles, $ang B$ and $ang Q$ , if the initial arms are coincident and both angles are measured in the same direction, then the terminal arms must be coincident.

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

Since only rigid transformations have been applied,

$Delta ABC ~= Delta A_4B_4C_4$ . Given this, and all given information, which of the following is/are correct? There may be more than one correct answer.

$ang B_4A_4C_4 ~= ang A ~= ang P$
$ang B_4 ~= ang B ~= ang Q$
$ang A_4C_4B_4 ~= ang C ~= ang R$
$bar{AC} ~= bar{A_4C_4} ~= bar{PR}$

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

Given the previous result, it can be concluded that

$bar{A_4 C_4} \ "||" \ bar{PR}$ . Which of the following is the reason why?

If two lines are cut by a transversal and corresponding angles are congruent, then the lines are parallel.
If two lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel
If two lines are cut by a transversal and alternate exterior angles are congruent, then the lines are parallel
If two lines are cut by a transversal and the sum of the measures of consecutive interior angles is 180°, then the lines are parallel

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

Which of the following conclusions can be reached, using the information above and the triangle theorem which states that if a line is parallel to one side of a triangle, and intersects the other two sides, then the line divides these two sides proportionally. Reminder: points

$B_4$ and

$Q$ are coincident.

$(A_4 B_4) / (QP) = (A_4 C_4) / (PR)$
$(B_4 C_4) / (QR) = (A_4 C_4) / (PR)$
$(A_4 P) / (A_4 B_4) = (C_4 R) / (B_4 C_4)$
$(QA_4)/(QC_4) = (A_4C_4) / (PR)$

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

Using the previous result, it can be shown that

$(PQ) / (A_4B_4) = (QR)/(B_4C_4)$ . Which of the following dilations would transform point

$A_4$ to point

$P ?$ Note: this dilation, applied to

$Delta A_4B_4C_4$ , would also transform point

$C_4$ to

$R$ .

A dilation of factor $PQ$ centered at point Q.
A dilation of factor $(PQ)/(A_4B_4)$ centered at point Q.
A dilation of factor $(PQ)/(A_4B_4)$ centered at the origin.
A dilation of factor $A_4B_4$ centered at the origin.

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

Using the information from the dilation in the previous question, what is the relationship between

$bar{A_4C_4}$ and

$bar{PR} ?$

$A_4C_4 = PR$
$(A_4C_4)/(PR) = 1$
$(PR)/(A_4C_4) = (PQ)/(A_4B_4)$
$(PR)/(A_4C_4) = (PQ)/(QR)$

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.A.3

The previous questions have shown that, if two angles of one triangle are congruent to two angles of another triangle, then all the conditions of similar triangles are met (all angles are congruent and the ratios of the corresponding sides are equal). What would change if the assumption

$AB < PQ$ was changed? Choose all correct answers.

If $AB = PQ$ , then the two triangles would be congruent.
If $AB = PQ$ , then the two triangles must be coincident before any transformations are performed.
If $AB > PQ$ , the two triangles would not be similar or congruent.
If $AB > PQ$ , the triangles would still be similar, and only minor changes to the math in showing them to be similar would be required.

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AA for Similar Triangles and Transformations

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