Tenth Grade (Grade 10) Similar and Congruent Figures Questions for Tests and Worksheets

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Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.B.4

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What is the missing reason in step 9?

Addition Property of Equality
Transitive Property of Equality
Subtraction Property of Equality
Corresponding angles of congruent triangles are congruent

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

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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

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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

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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.2

Matt is flying a kite and measures the angle of his string to the ground as 30 degrees. Samantha is also flying a kite using a string which is 150 feet long. Her kite is at an altitude of 75 feet. Are the two triangles formed by Matt's and Samantha's kite strings to the ground similar? Show your work.

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.B.5

Angle C is a right angle. Given that side AC = 15 and hypotenuse AB = 30, find the length of segment AD. Show your work and round to the nearest tenth.

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.B.4

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What is the missing statement in step 15?

$Delta ABC \ ~ \ Delta ANM$
$Delta ABC \ ~ \ Delta AMN$
$Delta ABC \ ~ \ Delta MAN$
$Delta ABC \ ~ \ Delta MNA$

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.B.4

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What is the missing statement in step 30?

$(CD)/(AB) = (AB)/(AC)$
$(BC)/(AD) = (AB)/(AC)$
$(AC)/(AD) = (AC)/(AB)$
$(AC)/(AD) = (AB)/(AC)$

Grade 10 Similar and Congruent Figures CCSS: HSG-SRT.B.4

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What is the missing statement in step 14?

$Delta ACD \ ~ \ Delta CAB$
$Delta ACD \ ~ \ Delta BCA$
$Delta ACD \ ~ \ Delta ABC$
$Delta ACD \ ~ \ Delta ACB$

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

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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

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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

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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.