Practice Test - Chapter 5
1. GARDENS Maggie wants to plant a circular flower bed within a triangular area set off by three pathways. Which
point of concurrency related to triangles would she use for the center of the largest circle that would fit inside the
triangle?
SOLUTION: The incenter is the center of a circle that intersects each side of the triangle at one point. For this reason, the
incenter always lies in the interior of a triangle. So, Maggie should use the incenter to find the center of the circle
inscribed in the triangle.
In
, K is the centroid and DK = 16. Find each length.
2. KH
SOLUTION: By the Centroid Theorem,
.
Substitute.
We know that DK + KH = DH.
3. CD
SOLUTION: Here, G is a midpoint of
So, CG = GD = 9.
4. FG
SOLUTION: By the Centroid Theorem,
.
Substitute.
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Practice Test - Chapter 5
4. FG
SOLUTION: By the Centroid Theorem,
.
Substitute.
PROOF Write an indirect proof.
5. Given:
Prove:
SOLUTION: Given:
Prove:
Step 1: Assume that x < 9.
Step 2: Make a table with several possibilities for x, assuming x < 9.
When x < 9, 5x + 7 < 52.
Step 3: The assumption leads to the contradiction of the given information that
assumption that x < 9 must be false, so the original conclusion that
must be true.
. Therefore the
Find each measure.
6. TQR
SOLUTION: Triangle QRS and triangle QRT are congruent by SAS. Therefore, angle SQR is congruent to angle TQR is 43º.
7. XZ
SOLUTION: By AAA,
Solve for x.
. By CPCTC,
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Substitute
in XZ.
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6. TQR
SOLUTION: Practice
TestQRS
- Chapter
5 QRT are congruent by SAS. Therefore, angle SQR is congruent to angle TQR is 43º.
Triangle
and triangle
7. XZ
SOLUTION: By AAA,
Solve for x.
Substitute
. By CPCTC,
in XZ.
8. GEOGRAPHY The distance from Tonopah to Round Mountain is equal to the distance from Tonopah to Warm
Springs. The distance from Tonopah to Hawthorne is the same as the distance from Tonopah to Beatty. Determine
which distance is greater, Round Mountain to Hawthorne or Warm Springs to Beatty.
SOLUTION: o
o
Since we have two pairs of congruent distances, as described in the given information, and we know that 95 >80 ,
we can use the Hinge Theorem to prove that the distance from Warm Springs to Beatty is greater than the distance
from Round Mountain to Hawthorne.
9. MULTIPLE CHOICE If the measures of two sides of a triangle are 3.1 feet and 4.6 feet, which is the least
possible whole number measure for the third side?
A 1.6 feet
B 2 feet
C 7.5 feet
D 8 feet
SOLUTION: Let n represent the length of the third side. According to the Triangle Inequality Theorem, the largest side cannot be greater than the sum of the other two
sides. If n is the largest side, then n must be less than 3.1 + 4.6. Therefore, n < 7.7.
If n is not the largest side, then 4.6 is the largest and 4.6 must be less than 3.1 + n. Therefore, 1.5 < n.
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Combining these two inequalities, we get 1.5 < n < 7.7. So, the least possible measure of the third side is the first
whole number greater than 1.5, or 2. The correct option is B.
SOLUTION: o
o
Since we have two pairs of congruent distances, as described in the given information, and we know that 95 >80 ,
we can
use- the
Hinge 5Theorem to prove that the distance from Warm Springs to Beatty is greater than the distance
Practice
Test
Chapter
from Round Mountain to Hawthorne.
9. MULTIPLE CHOICE If the measures of two sides of a triangle are 3.1 feet and 4.6 feet, which is the least
possible whole number measure for the third side?
A 1.6 feet
B 2 feet
C 7.5 feet
D 8 feet
SOLUTION: Let n represent the length of the third side. According to the Triangle Inequality Theorem, the largest side cannot be greater than the sum of the other two
sides. If n is the largest side, then n must be less than 3.1 + 4.6. Therefore, n < 7.7.
If n is not the largest side, then 4.6 is the largest and 4.6 must be less than 3.1 + n. Therefore, 1.5 < n.
Combining these two inequalities, we get 1.5 < n < 7.7. So, the least possible measure of the third side is the first
whole number greater than 1.5, or 2. The correct option is B.
Point H is the incenter of
. Find each measure.
10. DH
SOLUTION: Since H is equidistant from the sides of
Theorem.
, by the Incenter Theorem DH = FH. Find FH using the Pythagorean
Since length cannot be negative, use only the positive square root, 7.
Since DH = FH, DH = 7.
11. BD
SOLUTION: Since H is equidistant from the sides of Pythagorean Theorem.
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, by the Incenter Theorem DH = FH. Find FH using the
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Since length cannot be negative, use only the positive square root, 7.
Practice
SinceTest
DH =- Chapter
FH, DH =5 7.
11. BD
SOLUTION: Since H is equidistant from the sides of Pythagorean Theorem.
, by the Incenter Theorem DH = FH. Find FH using the
Since length cannot be negative, use only the positive square root, 7.
Since DH = FH, DH = 7.
Use the Pythagorean Theorem in
.
Since length cannot be negative, use only the positive square root, 8.5.
12. HAC
SOLUTION: Here,
. Also,
.
So by the Triangle Angle Sum Theorem,
.
Substitute.
Since
is an angle bisector of angle A,
Simplify.
13. DHG
SOLUTION: Here,
Substitute.
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. Also, .
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14. MULTIPLE CHOICE If the lengths of two sides of a triangle are 5 and 11, what is the range of possible lengths
Since
is an angle bisector of angle A,
Simplify.
Practice
Test - Chapter 5
13. DHG
SOLUTION: Here,
Substitute.
. Also, .
14. MULTIPLE CHOICE If the lengths of two sides of a triangle are 5 and 11, what is the range of possible lengths
for the third side?
F 6 < x < 10
G 5 < x < 11
H 6 < x < 16
J x < 5 or x > 11
SOLUTION: Let n represent the length of the third side. According to the Triangle Inequality Theorem, the largest side cannot be greater than the sum of the other two
sides. If n is the largest side, then n must be less than 5 + 11. Therefore, n < 16.
If n is not the largest side, then 11 is the largest and 11 must be less than 5 + n. Therefore, 6 < n.
Combining these two inequalities, we get 6 < n < 16. So, the correct option is H.
Compare the given measures.
15. AB and BC
SOLUTION: In
and
, we know that
we can state that AB < BC.
16. RST and
and
. By the Hinge Theorem, since
,
JKL
SOLUTION: In Manual -and
, we know that
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, we know that
and
.
. By the Converse of the Hinge Theorem, since Page 6
State the assumption necessary to start an indirect proof of each statement.
SOLUTION: In
and
, we know that
Practice
Test
- Chapter
we can
state
that AB <5 BC.
16. RST and
and
. By the Hinge Theorem, since
,
JKL
SOLUTION: In
and
, we know that
, we know that
and
. By the Converse of the Hinge Theorem, since
.
State the assumption necessary to start an indirect proof of each statement.
17. If 8 is a factor of n, then 4 is a factor of n.
SOLUTION: The first step in writing an indirect proof is to assume that the conclusion is false. 4 is not a factor of n.
18. m
M >m
N
SOLUTION: The first step in writing an indirect proof is to assume that the conclusion is false.
19. If
, then
.
SOLUTION: The first step in writing an indirect proof is to assume that the conclusion is false. a > 7
Use the figure to determine which angle has the greatest measure.
20. 1,
5,
6
SOLUTION: In the figure,
is an obtuse angle , angles
has the greatest measure.
21. 9,
8,
4,
3,
is also an acute angle. So, among these three
3
SOLUTION: In the figure,
is an obtuse angle , has the greatest measure.
22. is an acute angle, and is a right angle, and is an acute angle. So, among these three angles 2
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SOLUTION: In the figure,
is a right angle , Page 7
is an acute angle, and is also an acute angle. So, among these three
21. 9,
8,
3
SOLUTION: In theTest
figure,
is an obtuse angle , Practice
- Chapter
5
has the greatest measure.
22. 4,
3,
is a right angle, and is an acute angle. So, among these three angles 2
SOLUTION: In the figure,
is a right angle , is an acute angle, and angles
has the greatest measure.
is also an acute angle. So, among these three
PROOF Write a two-column proof.
bisects SRT.
23. Given:
Prove: m SQR > m SRQ
SOLUTION: Given:
Prove:
bisects .
Proof:
Statements (Reasons)
1.
. (Given)
bisects 2.
(Def. of bisector)
3.
(Def. of angles)
4.
(Exterior Angle Theorem)
5.
(Def. of Inequality)
6.
(Substitution)
Find the range for the measure of the third side of a triangle given the measures of the two sides.
24. 10 ft, 16 ft
SOLUTION: Let x represent the length of the third side. Next, set up and solve each of the three triangle inequalities.
Notice that x > –6 is always true for any whole number measure for x. Combining the two remaining inequalities, the
range of values that fit both inequalities is x > 6 and x < 26, which can be written as 6 ft < x < 26 ft.
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25. 23 m, 39 m
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3.
(Def. of angles)
4.
(Exterior Angle Theorem)
5.
(Def. of Inequality)
Practice
Test
Chapter
5
6.
(Substitution)
Find the range for the measure of the third side of a triangle given the measures of the two sides.
24. 10 ft, 16 ft
SOLUTION: Let x represent the length of the third side. Next, set up and solve each of the three triangle inequalities.
Notice that x > –6 is always true for any whole number measure for x. Combining the two remaining inequalities, the
range of values that fit both inequalities is x > 6 and x < 26, which can be written as 6 ft < x < 26 ft.
25. 23 m, 39 m
SOLUTION: Let x represent the length of the third side. Next, set up and solve each of the three triangle inequalities.
Notice that x > –16 is always true for any whole number measure for x. Combining the two remaining inequalities,
the range of values that fit both inequalities is x > 16 and x < 62, which can be written as 16 m < x < 62 m.
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