no ai please
2.
(05.01 LC)
Look at the right triangle ABC:
Right triangle ABC has a right angle at B. Segment BD meets segment AC at a right angle.
A student made the following chart to prove that AB2 + BC2 = AC2:
Statement Justification
1. Triangle ABC is similar to triangle BDC 1. Angle ABC = Angle BDC and Angle BCA = Angle BCD
2. BC2 = AC ⋅ DC 2. BC ÷ DC = AC ÷ BC because triangle ABC is similar to triangle BDC
3. Triangle ABC is similar to triangle ABD 3. Angle ABC = Angle BAD and Angle BAC = Angle ABD
4. AB2 = AC ⋅ AD 4. AB ÷ AD = AC ÷ AB because triangle ABC is similar to triangle ADB
5. AB2 + BC2 = AC ⋅ AD + AC ⋅ DC = AC (AD + DC) 5. Adding Statement 1 and Statement 2
6. AB2 + BC2 = AC2 6. AD + DC = AC
What is the first incorrect justification? (1 point)
Justification 4
Justification 1
Justification 2
Justification 3
3.
(05.01 LC)
Look at the figure shown below:
A triangle RPQ is shown. S is a point on side PR and T is a point on side PQ. Points S and T are joined using a straight line.
Nora is writing statements as shown to prove that if segment ST is parallel to segment RQ, then x = 45.
Statement Reason
1. Segment ST is parallel to segment RQ Given
2. Angle QRS is congruent to angle TSP Corresponding angles formed by parallel lines and their transversal are congruent
3. Angle SPT is congruent to angle RPQ Reflexive property of angles
4. Triangle SPT is similar to triangle RPQ Angle-Angle Similarity Postulate
5. 60: (60+x) = Corresponding sides of similar triangles are in proportion
Which of the following can she use to complete statement 5? (1 point)
60:(48 + 36)
60:36
48:36
48:(48 + 36)
4.
(05.01 MC)
Gary is using an indirect method to prove that segment DE is not parallel to segment BC in the triangle ABC shown below:
A triangle ABC is shown. D is a point on side AB and E is a point on side AC. Points D and E are joined using a straight line.
He starts with the assumption that segment DE is parallel to segment BC.
Which inequality will he use to contradict the assumption? (1 point)
4:10 ≠ 6:14
4:6 ≠ 6:14
4:10 ≠ 6:8
4:14 ≠ 6:10
5.
(05.01 MC)
The figure shows three right triangles. Triangles PQS, QRS, and PRQ are similar.
Theorem: If two triangles are similar, the corresponding sides are in proportion.
Figure shows triangle PQR with right angle at Q. Segment PQ is 4 and segment QR is 9. Point S is on segment PR and angles QSP a
Using the given theorem, which two statements help to prove that if segment PR is x, then x2 = 97? (1 point)
Segment PR ⋅ segment PS = 16
Segment PR ⋅ segment SR = 36
Segment PR ⋅ segment PS = 36
Segment PR ⋅ segment SR = 81
Segment PR ⋅ segment PS = 16
Segment PR ⋅ segment SR = 81
Segment PR ⋅ segment PS = 81
Segment PR ⋅ segment SR = 16
6.
(05.03 LC)
The figure below shows segments AC and EF which intersect at point B. Segment A F is parallel to segment EC.
Two line segments AC and FE intersect at B. Segments A F and EC are parallel to each other.
Which of these facts is used to prove that triangle AB F is similar to triangle CBE? (1 point)
Angle A FB is congruent to angle CBE.
Angle ABF is congruent to angle CBE.
Line segment BC is congruent to line segment BE.
Line segment AB is congruent to line segment FB.
7.
(05.03 MC)
Look at the figure below:
Triangles ABC and BDC have a common base BC. E is the point of intersection of BD and AC. AE equals EC and ED equals BE.
Based on the figure, which pair of triangles is congruent by the Side Angle Side Postulate? (1 point)
Triangle ABE and triangle CBE
Triangle AEB and triangle CED
Triangle BEC and triangle DEC
Triangle ABC and triangle DCE
8.
(05.03 MC)
The figure below shows a square ABCD and an equilateral triangle DPC:
ABCD is a square. P is a point inside the square. Straight lines join points A and P, B and P, D and P, and C and P. Triangle D
Jake makes the chart shown below to prove that triangle APD is congruent to triangle BPC:
Statements Justifications
In triangles APD and BPC; DP = PC Sides of equilateral triangle DPC are equal
Sides of square ABCD are equal
In triangles APD and BPC; angle ADP = angle BCP Angle ADC = angle BCD = 90° and angle ADP = angle BCP = 90° − 60° = 30°
Triangles APD and BPC are congruent SAS postulate
Which of the following completes Jake's proof? (1 point)
In triangles APD and BPC; AD = BC
In triangles APD and BPC; AP = PB
In triangles APB and DPC; AD = BC
In triangles APB and DPC; AP = PB
9.
(05.03 HC)
The figure below shows a trapezoid, ABCD, having side AB parallel to side DC. The diagonals AC and BD intersect at point O.
ABCD is a trapezoid with DC parallel to AB. Diagonals AC and DB intersect at point O.
If the length of AO is three times the length of CO, the length of DC is (1 point)
three times the length of OC
three times the length of DO
one-third of the length of AC
one-third of the length of AB
