Question

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 BCD and Angle BCA = Angle DBC
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 ADB and Angle BAC = Angle BAD
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

Which justification is incorrect?

Justification 4
Justification 1
Justification 2
Justification 3

Question 3(Multiple Choice Worth 1 points)

(05.01 LC)

The figure below shows two triangles EFG and KLM:

Two triangles EFG and KLM are drawn. Angle KML is a right angle. The measures of the sides of the triangles are, In triangle EFG, EG measures a, GF measure b, and EF measures c. In triangle KLM, KM measures a and ML measures b.

Which of the following can be used to prove that triangle EFG is also a right triangle?
Use Pythagorean Theorem to prove that KL is equal to c.
Prove that the ratio of EF and KL is greater than 1 and hence, the triangles are similar by AA postulate.
Prove that the sum of the squares of a and c is greater than square of b.
Prove that the sum of the squares of a and b is greater than square of c.

Question 4(Multiple Choice Worth 1 points)

(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 DPC is an equilateral triangle.

Jim 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
In triangles APD and BPC; AD = BC 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

Which of the following completes Jim's proof?

ASA postulate
HL postulate
SAS postulate
SSS postulate

Question 5(Multiple Choice Worth 1 points)

(05.03 HC)

The figure below shows a trapezoid, ABCD, having side AB parallel to side DC. The diagonals AC and B

Answer 1

Don't post all your questions in one post

Also include the images

Answer 2

This was an accident. I was trying to post just one part and it put all.

Answer 3

Ah okay