Name: ______________________ Class: _________________ Date: _________ ID: A
SSS
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____ 1. Given the lengths marked on the figure and that AD bisects BE, use SSS to explain why DABC @ DDEC.
a. AC@ CD, AB @ ED, BC @ CE c. AC@ CB, AB @ ED, CD @ CE b. AC@ CD, AB @ ED, BC @ BC d. The triangles are not congruent.
____ 2. The figure shows part of the roof structure of a house. Use SAS to explain why DRTS @ DRTU.
Complete the explanation.
It is given that [1]. Since –RTS and –RTU are right angles, [2] by the Right Angle Congruence Theorem. By the Reflexive Property of Congruence, [3]. Therefore, DRTS @ DRTU by SAS. a. [1] RT@ RT [2] –SRT @ –URT [3] ST@ UT c. [1] ST@ UT [2] –RTS @ –RTU [3] RT@ RT b. [1] ST@ UT [2] –SRT @ –URT [3] ST@ UT d. [1] ST@ UT [2] –RTS @ –RTU [3] SU@ SU
Name: ______________________ ID: A ____ 3. What additional information do you need to prove DABC @ DADC by the SAS Postulate?
a. AB@ AD c. –ABC @ –ADC
b. –ACB @ –ACD d. BC@ DC
____ 4. Determine if you can use ASA to prove DCBA @ DCED. Explain.
a. AC@ DC is given. –CAB @ –CDE because both are right angles. No other
congruence relationships can be determined, so ASA cannot be applied.
b. AC@ DC is given. –CAB @ –CDE because both are right angles. By the Adjacent
Angles Theorem, –ACB @ –DCE. Therefore, DCBA @ DCED by ASA.
c. AC@ DC is given. –CAB @ –CDE because both are right angles. By the Vertical
Angles Theorem, –ACB @ –DCE. Therefore, DCBA @ DCED by ASA.
d. AC@ DC is given. –CAB @ –CDE because both are right angles. By the Vertical
Name: ______________________ ID: A
____ 5. Determine if you can use the HL Congruence Theorem to prove DACD @ DDBA. If not, tell what else you need to know.
a. Yes.
b. No. You do not know that –C and –B are right angles. c. No. You do not know that AC@ BD.
d. No. You do not know that ABÄ CD.
____ 6. For these triangles, select the triangle congruence statement and the postulate or theorem that supports it.
a. DABC @ DJLK, HL c. DABC @ DJLK, SAS
Name: ______________________ ID: A ____ 7. A pilot uses triangles to find the angle of elevation –A from the ground to her plane. How can she
find m–A?
a. DABO @ DCDO by SAS and –A @ –C by CPCTC, so m–A = 40∞ by substitution. b. DABO @ DCDO by CPCTC and –A @ –C by SAS, so m–A = 40∞ by substitution. c. DABO @ DCDO by ASA and –A @ –C by CPCTC, so m–A = 40∞ by substitution. d. DABO @ DCDO by CPCTC and –A @ –C by ASA, so m–A = 40∞ by substitution. ____ 8. Find the value of x.
a. x = 6 c. x = 2
b. x = 4 d. x = 8
____ 9. Find m–Q.
a. m–Q = 30º c. m–Q = 70º
Name: ______________________ ID: A ____ 10. Find CA.
a. CA = 10
b. CA = 12
c. CA = 14
d. Not enough information. An equiangular triangle is not necessarily equilateral. ____ 11. Find the measure of each numbered angle.
a. m–1 = 54∞, m–2 = 117∞, m–3 = 63∞ b. m–1 = 117∞, m–2 = 63∞, m–3 = 63∞ c. m–1 = 54∞, m–2 = 63∞, m–3 = 63∞ d. m–1 = 54∞, m–2 = 63∞, m–3 = 117∞
ID: A
SSS
Answer Section
MULTIPLE CHOICE1. ANS: A
It is given that AB@ ED, BC @ CE, and AD bisects BE. By the definition of segment bisector,
AC@ CD. All three pairs of corresponding sides of the triangles are congruent. Therefore, DABC @
DDEC by SSS.
Feedback A Correct!
B Use the fact that segment AD bisects segment BE.
C The corresponding sides need to belong to different triangles. Use the fact that segment AD bisects segment BE.
D The corresponding sides of the triangles are congruent. Use the fact that segment
AD bisects segment BE.
PTS: 1 DIF: Basic REF: Page 242
OBJ: 4-4.1 Using SSS to Prove Triangle Congruence NAT: 12.3.5.a TOP: 4-4 Triangle Congruence: SSS and SAS
2. ANS: C
It is given that ST@ UT. Since –RTS and –RTU are right angles, –RTS @ –RTU by the Right Angle Congruence Theorem. By the Reflexive Property of Congruence, RT@ RT. Therefore, DRTS @ DRTU by SAS.
Feedback
A Check the figure to see what is given.
B Angle SRT and angle URT are not right angles.
C Correct!
D Segment SU being congruent to itself does not help in proving the triangles congruent.
PTS: 1 DIF: Average REF: Page 243 OBJ: 4-4.2 Application NAT: 12.3.5.a TOP: 4-4 Triangle Congruence: SSS and SAS
ID: A 3. ANS: B
The SAS Postulate is used when two sides and an included angle of one triangle are congruent to the corresponding sides and included angle of a second triangle.
From the given, BC@ DC.
From the figure, AC@ AC by the Reflexive Property of Congruence.
You have two pair of congruent sides, so you need information about the included angles. Use these pairs of sides to determine the included angles.
The angle between sides AC and BC is –ACB. The angle between sides AC and DC is –ACD.
You need to know –ACB @ –ACD to prove DABC @ DADC by the SAS Postulate.
Feedback
A This information is needed to use the SSS Postulate.
B Correct!
C You need the included angle between the two sides.
D This information is already given. Find information that you need that is not given or true in the figure.
PTS: 1 DIF: Advanced NAT: 12.3.5.a TOP: 4-4 Triangle Congruence: SSS and SAS
4. ANS: C
AC@ DC is given. –CAB @ –CDE because both are right angles. By the Vertical Angles Theorem,
–ACB @ –DCE. Therefore, DCBA @ DCED by ASA.
Feedback
A Look for vertical angles.
B Adjacent angles are angles in a plane that have their vertex and one side in common but have no interior points in common. Angle ACB and angle DCE are not adjacent angles.
C Correct!
D Use ASA, not SAS, to prove the triangles congruent.
PTS: 1 DIF: Basic REF: Page 253
OBJ: 4-5.2 Applying ASA Congruence NAT: 12.3.2.e TOP: 4-5 Triangle Congruence: ASA AAS and HL
ID: A 5. ANS: A
AB Ä CD is given. In addition, by the Reflexive Property of Congruence, AD @ AD. Since AC Ä BD and AC^PB, by the Perpendicular Transversal Theorem BD ^PB. By the definition of right angle, –ABD is a right angle. Similarly, –DCA is a right angle. Therefore, DABD @ DDCA by the HL Congruence Theorem.
Feedback A Correct!
B Since line segment AC is parallel to line segment BD, what does the Perpendicular Transversal Theorem tell you about line segment BD and line segment PB?
C What do you know about the other pair of legs of the right triangles ABD and
DCA?
D What do you know about line segments AB and CD? PTS: 1 DIF: Average REF: Page 255 OBJ: 4-5.4 Applying HL Congruence NAT: 12.3.2.e TOP: 4-5 Triangle Congruence: ASA AAS and HL
6. ANS: B
Because –BAC and –KJL are right angles, DABC and DJKL are right triangles.
You are given a pair of congruent legs AC@ JL and a pair of congruent hypotenuses CB @ LK. So a hypotenuse and a leg of DABC are congruent to the corresponding hypotenuse and leg of DJKL. DABC @ DJKL by HL.
Feedback
A Segment AC is congruent to segment JL. Make sure the triangle vertices correspond accordingly.
B Correct!
C Segment AC is congruent to segment JL. Make sure the triangle vertices correspond accordingly. For SAS, the angle is included between the sides.
D For SAS, the angle is included between the sides.
PTS: 1 DIF: Advanced NAT: 12.3.5.a TOP: 4-5 Triangle Congruence: ASA AAS and HL
ID: A 7. ANS: A
From the figure, CO@ AO,and DO @ BO. –AOB @ –COD by the Vertical Angles Theorem. Therefore, DABO @ DCDO by SAS and –A @ –C by CPCTC. m–A = 40∞ by substitution.
Feedback A Correct!
B First, show that the triangles are congruent. Then, show that their corresponding parts are congruent.
C First, show that the triangles are congruent. Then, show that their corresponding parts are congruent.
D First, show that the triangles are congruent. Then, show that their corresponding parts are congruent.
PTS: 1 DIF: Average REF: Page 260 OBJ: 4-6.1 Application NAT: 12.3.2.e TOP: 4-6 Triangle Congruence: CPCTC
8. ANS: A
The triangles can be proved congruent by the SAS Postulate. By CPCTC, 3x- 5 = 2x+ 1.
Solve the equation for x. 3x- 5 = 2x+ 1
3x= 2x+ 6
x= 6
Feedback A Correct!
B When solving, you can either add 5 or subtract 1 from each side.
C Remember to combine the like terms when solving.
D These two triangles have SAS congruence, so the two expressions are equal by CPCTC.
PTS: 1 DIF: Advanced NAT: 12.3.2.e TOP: 4-6 Triangle Congruence: CPCTC
ID: A 9. ANS: D
m–Q = m–R = 2x+ 15( )∞ Isosceles Triangle Theorem m–P+ m–Q + m–R = 180∞ Triangle Sum Theorem
x+ 2x+ 15( )+ 2x+ 15( ) = 180 Substitute x for mand m–R. –P and substitute 2x+ 15 for m–Q 5x= 150 Simplify and subtract 30 from both sides.
x= 30 Divide both sides by 5.
Thus m–Q = 2x+ 15( )∞ = [2 30( )+ 15]∞ = 75∞.
Feedback
A This is x. The measure of angle Q is 2x + 15.
B By the Isosceles Triangle Theorem, the measure of angle Q equals the measure of angle R. Use the Triangle Sum Theorem and solve for x.
C By the Isosceles Triangle Theorem, the measure of angle Q equals the measure of angle R. Use the Triangle Sum Theorem and solve for x.
D Correct!
PTS: 1 DIF: Average REF: Page 274
OBJ: 4-8.2 Finding the Measure of an Angle NAT: 12.3.3.f TOP: 4-8 Isosceles and Equilateral Triangles
10. ANS: C
DABC is equilateral. Equiangular triangles are equilateral. 2s- 10 = s + 2 Definition of equilateral triangle.
s= 12 Subtract s and add 10 to both sides of the equation.
AB= 2s - 10
AB= 2 12( )- 10 Substitute 12 for s in the equation for AB.
AB= 14 Simplify.
CA= AB Definition of equilateral triangle.
CA= 14 Substitute 14 for AB.
Feedback
A Equiangular triangles are equilateral. Use AB = BC to solve for s, and then use AC = AB or AC = BC to find AC.
B This is s. Substitute s in the original equation to find AC.
C Correct!
D By a corollary to the Isosceles Triangle Theorem, equiangular triangles are
ID: A 11. ANS: C
Step 1: –2 is supplementary to the angle that is 117∞. 117∞ + m–2 = 180∞. So m–2 = 63∞.
Step 2: By the Alternate Interior Angles Theorem, –2 @ –3. So m–2 = m–3 = 63∞.
Step 3: By the Isosceles Triangle Theorem, –2 and the angle opposite the other side of the isosceles triangle are congruent. Let –4 be that unknown angle.
Then, –2 @ –4 and m–2 = m–4 = 63∞.
m–1 + m–2 + m–4 = 180∞ by the Triangle Sum Theorem. m–1 + 63∞ + 63∞ = 180∞. So m–1 = 54∞.
Feedback
A Angle 2 is supplementary to the angle that measures 117 degrees.
B To find the measure of angle 1, use the Isosceles Triangle Theorem.
C Correct!
D By the Alternate Interior Angles Theorem, angle 2 is congruent to angle 3. PTS: 1 DIF: Advanced NAT: 12.3.2.e
