Is it possible to form a triangle with the given
side lengths? If not, explain why not.
1.5 cm, 7 cm, 10 cm
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5
Thus, you can form a triangle with side lengths 5 cm,
7 cm,and10cm.
ANSWER:
Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5
2.3 in., 4 in., 8 in.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, , 3 in., 4 in., 8 in. do not form a
triangle.
ANSWER:
No;
3.6 m, 14 m, 10 m
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6, you
can form a triangle with side lengths 6 m, 14 m, 10
m.
ANSWER:
Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6
4.MULTIPLE CHOICE If the measures of two
sides of a triangle are 5 yards and 9 yards, what is
the least possible measure of the third side if the
measure is an integer?
A 4 yd
B 5 yd
C 6 yd
D 14 yd
SOLUTION:
Let x represents the length of the third side. Next, set
up and solve each of the three triangle inequalities.
5 + 9 > x, 5 + x > 9, and 9 + x > 5
That is, 14 > x, x > 4, and x > –4.
Notice that x > –4 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 > 4 and x < 14, which can be written
as 4 < x < 14. So, the least possible measure of the
thirdsidecouldbe5yd.
The correct option is B.
ANSWER:
B
PROOF Write a two-column proof.
5.Given:
Prove:
SOLUTION:
Think backwards when considering this proof. Notice
that what you are trying to prove is an inequality
statement. However, it isn't exactly related to
, except for instead of side being used, it
is Since it is given that , you can
easilyusethisinasubstitutionstep.
Given:
Prove: YZ + ZW > XW
Proof:
Statements (Reasons)
1. (Given)
2. XW = YW (Def. of segments)
3. YZ + ZW > YW ( Inequal. Thm.)
4. YZ + ZW > XW (Substitution Property.)
ANSWER:
Given:
Prove: YZ + ZW > XW
Proof:
Statements (Reasons)
1. (Given)
2. XW = YW (Def. of segments)
3. YZ + ZW > YW ( Inequal. Thm.)
4. YZ + ZW > XW (Subst.)
Is it possible to form a triangle with the given
side lengths? If not, explain why not.
6.4 ft, 9 ft, 15 ft
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, , you can not form a triangle with
side lengths 4 ft, 9 ft, 15 ft.
ANSWER:
No;
7.11 mm, 21 mm, 16 mm
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, 11 + 21 > 16, 11 + 16 > 21, and 16 + 21 > 11,
you can form a triangle with side lengths 11 mm, 21
mm, and 16 mm.
ANSWER:
Yes; 11 + 21 > 16, 11 + 16 > 21, and 16 + 21 > 11
8.9.9 cm, 1.1 cm, 8.2 cm
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since , you can not form a triangle
with side lengths 9.9 cm, 1.1 cm,and8.2cm.
ANSWER:
No;
9.2.1 in., 4.2 in., 7.9 in.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, , you can not form a triangle
with side lengths 2.1 in., 4.2 in.,and7.9in.
ANSWER:
No;
10.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since , you can not form a triangle
with side lengths .
ANSWER:
No;
11.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since , you
can form a triangle with side lengths
.
ANSWER:
Yes;
Find the range for the measure of the third side
of a triangle given the measures of two sides.
12.4 ft, 8 ft
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than 4 +
8. Therefore, n < 12.
If n is not the largest side, then 8 is the largest and 8
must be less than 4 + n. Therefore, 4 < n.
Combining these two inequalities, we get 4 < n < 12.
ANSWER:
4 ft < n < 12 ft
13.5 m, 11 m
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
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.
ANSWER:
6 m < n < 16 m
14.2.7 cm, 4.2 cm
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than 2.7 +
4.2. Therefore, n < 6.9.
If n is not the largest side, then 4.2 is the largest and
4.2 must be less than 2.7 + n. Therefore, 1.5 < n.
Combining these two inequalities, we get 1.5 < n <
6.9.
ANSWER:
1.5 cm < n < 6.9 cm
15.3.8 in., 9.2 in.
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than 3.8 +
9.2. Therefore, n < 13.
If n is not the largest side, then 9.2 is the largest and
9.2 must be less than 3.8 + n. Therefore, 5.4 < n.
Combining these two inequalities, we get 5.4 < n <
13.
ANSWER:
5.4 in. < n < 13 in.
16.
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than
. Therefore, .
If n is not the largest side, then is the largest and
must be less than . Therefore, .
Combining these two inequalities, we get
.
ANSWER:
17.
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than
. Therefore, n < .
If n is not the largest side, then is the largest and
must be less than . Therefore, .
Combining these two inequalities, we get
.
ANSWER:
PROOF Write a two-column proof.
18.Given:
Prove:
SOLUTION:
The key to this proof is to figure out some way to get
BC=BD so that you can substitute one in for the
other using the Triangle Inequality Theorem.
Consider the given statement, if two angles of a
triangle are congruent, what kind of triangle is it and,
therefore, how do you know that BC must equal BD?
Proof:
Statements (Reasons)
1. (Given)
2. (Converse of Isosceles Thm.)
3. BC = BD (Def. of segments)
4. AB + AD > BD ( Inequality Thm.)
5. AB + AD > BC (Substitution Property.)
ANSWER:
Proof:
Statements (Reasons)
1. (Given)
2. (Conv.Isos. Thm.)
3. BC = BD (Def. of segments)
4. AB + AD > BD ( Inequal. Thm.)
5. AB + AD > BC (Subst.)
19.Given:
Prove: KJ +KL> LM
SOLUTION:
Think backwards when considering this proof. Notice
that what you are trying to prove is an inequality
statement. However, it isn't exactly related to
, except for instead of side being used, it
is Since it is given that , you can
easily use this in a substitution step
Proof:
Statements (Reasons)
1. (Given)
2. JL = LM (Def. of segments)
3. KJ + KL > JL ( Inequality Thm.)
4. KJ + KL > LM (Substitution Property)
ANSWER:
Proof:
Statements (Reasons)
1. (Given)
2. JL = LM (Def. of segments)
3. KJ + KL > JL ( Inequal. Thm.)
4. KJ + KL > LM (Subst.)
SENSE-MAKINGDetermine the possible
values of x.
20.
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Notice that isalwaystrueforanywhole
number measure for x.The range of values that
would be true for the other two inequalities is
and , which can be written as
.
ANSWER:
6 < x < 17
21.
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Notice that isalwaystrueforanywhole
number measure for x. Combining the two remaining
inequalities, the range of values that fit both
inequalities is and , which can be
written as .
ANSWER:
22.TRAVEL Keyan wants to take the most efficient
route from his hotel to the hockey game at The
Sportsplex. He can either take Highway 521 or
take Highway 3 and Route 11 from his hotel to the
arena.
a. Which of these two possible routes is the shorter?
Explain your reasoning.
b. Suppose Keyan always drives the speed limit. If
he chooses to take Highway 521, he is on it for 30
miles and the speed limit is 40 miles per hour. If he
chooses to take the other route, the speed limit is
60 miles per hour on both roads and he is on
Highway 3 for 22 miles and Route 11 for 25 miles.
Whichrouteisfaster?Explainyourreasoning.
SOLUTION:
a. Highway 521; sample answer: In a triangle, the
sum of two of the sides is always greater than the
third side, so the sum of the distance on Highway 3
and the distance on Route 11 is greater than the
distance on Highway 521. Or you can add the
distances using Highway 3 and Route 11 and
compare their sum to the 30 miles of Highway 521.
Since 47 miles is greater than 30 miles, Highway 521
istheshortestdistance.
b. Highway 521; sample answer: Since Keyan drives
the 40 mph speed limit on Highway 521 and the
distance is 30 miles, it will take him about 30/40 =
0.75 hour or 45 minutes to get to The Sportsplex. He
has to drive 47 miles on Highway 3 and Route 11,
and the speed limit is 60 miles per hour, so it will take
him 47/60 = 0.78 hour or about 47 minutes. The route
onHighway4521willtakelesstimethantheroute
on Highway 3 and Route 11.
ANSWER:
a. Highway 521; sample answer: In a triangle, the
sum of two of the sides is always greater than the
third side, so the sum of the distance on Highway 3
and the distance on Route 11 is greater than the
distance on Highway 521.
b. Highway 521; sample answer: It will take Keyan
45 minutes to drive on Highway 521 and 47 to
drive Highway 3 to Route 11.
23.PROOF Write a two-column proof.
Given: ΔABC
Prove: AC + BC > AB (Triangle Inequality Theorem)
(Hint: Draw auxiliary segment , so that C is
between B and D and .)
SOLUTION:
Proof:
Statements (Reasons)
1. Construct sothatC is between B and D and
. (Ruler Postulate)
2. CD = AC (Definition of )
3. (Isosceles Theorem)
4. m ∠CAD = m ∠ADC (Definition. of
angles )
5. m ∠BAC + m ∠CAD = m ∠BAD (∠Addition
Postulate)
6. m ∠BAC + m ∠ADC = m ∠BAD (Substitution)
7. m ∠ADC < m ∠BAD (Definition of inequality)
8. AB < BD (Angle–Side Relationships in Triangles)
9. BD = BC + CD (Segment Addition Postulate)
10. AB < BC + CD (Substitution (Steps 8, 9))
11. AB < BC + AC (Substitution (Steps 2, 10))
ANSWER:
Proof:
Statements (Reasons)
1. Construct sothatC is between B and D and
.(Ruler Post.)
2. CD = AC (Def. of )
3. ∠CAD ADC(Isos. Thm)
4. m ∠CAD = m ∠ADC (Def. of ∠s )
5. m ∠BAC + m ∠CAD = m ∠BAD (∠Add. Post.)
6. m ∠BAC + m ∠ADC = m ∠BAD (Subst.)
7. m ∠ADC < m ∠BAD (Def. of inequality)
8. AB < BD (Angle–Side Relationships in Triangles)
9. BD = BC + CD (Seg. Add. Post.)
10. AB < BC + CD (Subst.(Steps 8, 9))
11. AB < BC + AC (Subst. (Steps 2, 10))
24.MULTI-STEP Toya rides her bike 3680 feet
down her street, Meadow Court, turns left onto
Holly Lane, and rides 2740 feet to get to the park.
One day she decides to try to find a shorter path to
the park by cutting straight through the field that is
next to the two roads.
However, if she decides to cut through the field, she
must walk instead of riding her bike.
a. If Toya is able to ride her bike 3 times as fast as
she can walk, how long should the path through the
field be for it to be a shortcut? How much time
would Toya save by taking the shortcut, if she can
ride her bike 9 miles per hour?
b. Describe your solution process.
c. What assumptions did you make?
SOLUTION:
a. - b. I drew a triangle formed by Meadow Court,
HollyLane,andthepaththroughthefield.
The Triangle Inequality Theorem to find the possible
sidelengths.
3680 + 2740 > x or 6420 > x
x + 2740 > 3680 or x > 940
x+3680>2740orx > –-940
Then,thepathisbetween940and6420feet.
Next, I found the time it would take Toya to ride her
bike to the park. She rides her bike 6420 feet. I
dividedby5280feettofinddistanceinmiles.Next,I
multiplied this number by 60 minutes and divided by 9
miles to get the number of minutes. This was
approximately8.1minutes,or0.135hours.
Since Toya can ride her bike 3 times as fast as she
canwalk,shecanwalk3milesperhour.Ifittakes
Toya longer than 8.1 minutes to walk through the
field,itisnolongerashortcut.
To find the maximum distance of the path, I
multiplied 3 miles per hour by 0.135 hours to get
0.405 miles or, after multiplying by 5280 feet, 2140
feet. So, the path can only be between 940 and 2140
feetinorderforittobeashortcut.Tofindthe
maximum amount of time Toya can save by taking
the shortcut, I took the shortest possible path, 940
feet,andconvertedittomiles.
Then, I multiplied the shortest possible distance in
miles by 60 minutes and divided by 3 mph to get the
minutes it took her to walk the shortest possible path.
Thiswasapproximately3.6minutes.
Next, I subtracted the two times and found that Toya
could save 4.5 minutes if the shortcut was 940 feet.
If the shortcut were any longer, she would save less
time. So Toya could save from no time to 4.5
minutes.
c. I assumed that the path through the field was a
straight line.
ANSWER:
a. The distance must be greater than 940 feet and
less than 2140 feet. Toya can save up to 4.5
minutes by taking the shortcut.
b. Sample answer: I drew a triangle formed by
Meadow Court, Holly Lane, and the path through
the field. The Triangle Inequality Theorem says that
the path is between 940 and 6420 feet. Since Toya
can ride her bike 3 times as fast as she can walk,
the path can only be between 940 and 2140 feet in
order for it to be a shortcut. Next, I found the time
it would take Toya to ride her bike to the park. She
rides her bike 6420 feet. I divided by 5280 feet to
find distance in miles. Next, I multiplied this number
by 60 minutes and divided by 9 miles to get the
number of minutes. This was approximately 8.1
minutes. To find the maximum amount of time Toya
can save by taking the shortcut, I took the shortest
possible path, 940 feet, and converted it to miles.
Since she can ride her bike 3 times as fast as she
can walk, Toya can walk 3 mph. Then, I multiplied
the shortest possible distance in miles by 60 minutes
and divided by 3 mph to get the minutes it took her
to walk the shortest possible path. This was
approximately 3.6 minutes. Next, I subtracted the
two times and found that Toya could save 4.5
minutes if the shortcut was 940 feet. If the shortcut
were any longer, she would save less time. So Toya
could save from no time to 4.5 minutes.
c. Sample answer: I assumed that the path through
the field was a straight line.
Find the range of possible measures of x if each
set of expressions represents measures of the
sides of a triangle.
25.x, 4, 6
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x is the largest side, then x must be less than 4 + 6.
Therefore, x < 4+6 or x <10.
If x is not the largest side, then 6 is the largest and 6
must be less than 4 + x. Therefore, 4 + x >6 or x >
2.
Combining these two inequalities, we get 2 < x<10.
ANSWER:
2 < x < 10
26.8, x, 12
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x is the largest side, then x must be less than 8 +
12. Therefore, x < 8 + 12 or x < 20.
If x is not the largest side, then 12 is the largest and
12 must be less than 8 + x. Therefore, 8 + x>12 or
x > 4.
Combining these two inequalities, we get 4 < x<20.
ANSWER:
4 < x < 20
27.x + 1, 5, 7
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x + 1 is the largest side, then x + 1 must be less
than 5 + 7. Therefore, x + 1 < 5 + 7 or x + 1 < 12 or
x <11.
If x + 1 is not the largest side, then 7 is the largest
and 7 must be less than 5 + (x + 1). Therefore, 5 + x
+ 1 > 7 or 6 + x > 7 or x > 1.
Combining these two inequalities, we get 1 < x<11.
ANSWER:
1 < x < 11
28.x – 2, 10, 12
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x – 2 is the largest side, then x – 2 must be less
than 10 + 12. Therefore, x – 2 < 10 + 12 or x – 2 <
22 or x < 24.
If x – 2 is not the largest side, then 12 is the largest
and 12 must be less than 10 + (x – 2). Therefore, 10
+ x – 2 > 12 or 8 + x > 12 or x > 4.
Combining these two inequalities, we get 4 < x<24.
ANSWER:
4 < x < 24
29.x + 2, x + 4, x + 6
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Noticethat and arealwaystruefor
any whole number measure for x. So, the only
required inequality is x > 0.
ANSWER:
x > 0
30.x, 2x + 1, x + 4
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Notice that isalwaystrueforanywhole
number measure for x and isalwaystrue.So,
the required inequality is .
ANSWER:
31.DRAMA CLUB Anthony and Catherine are
working on a ramp up to the stage for the drama
club's next production. Anthony's sketch of the ramp
is shown below. Catherine is concerned about the
measurements and thinks they should recheck the
measures before they start cutting the wood. Is
Catherine'sconcernvalid?Explainyourreasoning.
SOLUTION:
The measurements on the drawing do not form a
triangle. According to the Triangle Inequality
Theorem, the sum of the lengths of any two sides of
a triangle is greater than the length of the third side.
The lengths in the drawing are 1 ft, ft, and ft.
Since , the triangle is impossible. They
should recalculate their measurements before they
cutthewood.Therefore,Catherine'sconcernis
valid.
ANSWER:
Yes; sample answer: The measurements on the
drawing do not form a triangle. According to the
Triangle Inequality Theorem, the sum of the lengths
of any two sides of a triangle is greater than the
length of the third side. The lengths in the drawing
are 1 ft, ft, and ft. Since , the
triangle is impossible. They should recalculate their
measurementsbeforetheycutthewood.
32.MODELING Aisha is riding her bike to the park
and can take one of two routes. The most direct
route from her house is to take Main Street, but it is
safer to take Route 3 and then turn right onto Clay
Road as shown. The additional distance she will
travel if she takes Route 3 to Clay Road is between
how many miles?
SOLUTION:
The distance from Aisha's house to the park via
MainSt.representsthethirdsideofatriangle.
From the Triangle Inequality Theorem the length of
this side must also be greater than 7.5 – 6 or 1.5
miles and must be less than 6 + 7.5 or 13.5 miles.
Therefore, the distance d from her house to the park
via Main St. can be represented by 1.5 < d < 13.5.
The distance to the park by taking Route 3 to Clay
Roadis7.5+6or13.5miles.
The least additional number of miles she would travel
would be greater than 13.5 –13.5 or 0. The greatest
number of additional miles she would travel would be
less than 13.5 – 1.5 or 12. Therefore, the additional
distance she will travel if she takes Route 3 to Clay
Roadisbetween0and12miles.
ANSWER:
0 and 12
33.DESIGN Carlota designed an awning that she and
her friends could take to the beach. Carlota decides
to cover the top of the awning with material that will
drape 6 inches over the front. What length of
material should she buy to use with her design so that
it covers the top of the awning, including the drape,
when the supports are open as far as possible?
Assume that the width of the material is sufficient to
cover the awning.
SOLUTION:
Let x be the length of the material needed.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x is the largest side, then x must be less than 4 + 3.
Therefore, n < 7. Since the material will drape 6
inches or 0.5 feet over the front, the minimum length
of material she should buy is 7 + 0.5 or 7.5 feet at the
most.
ANSWER:
She should buy no more than 7.5 ft.
ESTIMATION Without using a calculator,
determine if it is possible to form a triangle with
the given side lengths. Explain.
34.
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , .
is between and . Since = 1 and =
2, .
Since , .
5.9 > 2.9 + 1.5, so it is not possible to form a
triangle with the given side lengths.
ANSWER:
No; since , since or 2
and or 1, and since . So,
.
35.
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , then .
Also, since , then .
And since , then .
9.9 < 6.9 + 8.1, so yes, it is possible to form a
triangle with the given side lengths.
ANSWER:
Yes. since , since
, and since .6.9+8.1>
9.9, so it is possible.
36.
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , then .
Also, since , then .
And since , then .
4.9< 3.9 + 1.9, so yes, it is possible to form a
triangle with the given side lengths.
ANSWER:
Yes. since , since
, and since .
1.9 + 3.9 > 4.9, so it is possible.
37. .
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , then .
Also, since ,then .
And since , then .
11.1> 2.1 + 5.1 so no, it is not possible to form a
triangle with the given side lengths.
ANSWER:
No; since ,
since , and
since . So, 2.1 + 5.1 11.1.
REASONING Determinewhetherthegiven
coordinates are the vertices of a triangle.
Explain.
38.X(1, –3), Y(6, 1), Z(2, 2)
SOLUTION:
We can graphically show that given coordinates form
atrianglebygraphingthem,asshownbelow.
We can algebraically prove that the given
coordinates form a triangle by proving that the
length of the longest side is greater than the sum of
thetwoshortersides.
Use the distance formula. has endpoints X(1,–3)
and Y(6, 1).
has endpoints Y(6, 1) and Z(2, 2).
has endpoints Z(2, 2) and X(1, –3).
Here, XY + YZ > XZ, XY + XZ > YZ, and XZ + YZ >
XY.
Thus,thegivencoordinatesformatriangle.
ANSWER:
Yes; XY + YZ > XZ, XY + XZ > YZ, and XZ + YZ >
XY
39.F(–4, 3), G(3, –3), H(4, 6)
SOLUTION:
We can graphically show that given coordinates form
atrianglebygraphingthem,asshownbelow.
We can algebraically prove that the given
coordinates form a triangle by proving that the
length of the longest side is greater than the sum of
thetwoshortersides.
Use the distance formula. has endpoints F(–4,3)
and G(3, 3).
has endpoints G(3, 3) and H(4, 6).
has endpoints H(4, 6) and F(–4, 3).
Here, FG + GH > FH, FG + FH > GH, and GH +
FH > FG .
Thus, the coordinated form a triangle.
ANSWER:
Yes; FG + GH > FH, FG + FH > GH, and GH +
FH > FG
40.J(–7, –1), K(9, –5), L(21, –8)
SOLUTION:
Wecangraphicallydetermineifthegiven
coordinates form a triangle by graphing them, as
shownbelow.
We can algebraically prove that the these three
points are collinear and therefore, by showing that
the sum of the two shorter segments is equal to the
longestsegment.
Use the distance formula. has endpoints J(–7,–
1) and K(9, –5).
has endpoints K(9, –5) and L(21, –8).
has endpoints L(21, –8) and J(–7, –1).
Here JK + KL = JL. You can also confirm this by
using your calculator. Compute
toconfirmthatitequals0.
Thus the given coordinates do not form a triangle.
ANSWER:
No; JK + KL = JL
41.Q(2, 6), R(6, 5), S(1, 2)
SOLUTION:
We can graphically determine if the given
coordinates form a triangle by graphing them, as
shownbelow.
We can algebraically prove that the given
coordinates form a triangle by proving that the
lengthofoneofthesidesequalszero.
Use the distance formula.
has endpoints Q(2,6) and R(6, 5).
has endpoints R(6, 5) and S(1, 2).
has endpoints S(1, 2) and Q(2, 6).
Then and
.
Thus the coordinates form a triangle.
ANSWER:
Yes;
42.MULTIPLE REPRESENTATIONS In this
problem, you will use inequalities to make
comparisons between the sides and angles of two
triangles.
a. GEOMETRIC Draw three pairs of triangles that
have two pairs of congruent sides and one pair of
sides that is not congruent. Mark each pair of
congruent sides. Label each triangle pair ABC and
DEF, where and .
b. TABULAR Copy the table below. Measure and
record the values of BC, m A, EF, and m D for
each triangle pair.
c. VERBAL Make a conjecture about the
relationship between the angles opposite the
noncongruent sides of a pair of triangles that have
two pairs of congruent legs.
SOLUTION:
a. Using a ruler, compass, or drawing tool, make sure
that and , in each of the
trianglepairsmade.
b. Use a protractor and ruler to carefully measure
the indicated lengths and angle measures in the table
below. Look for a pattern when comparing to
.
c. Sample answer: The angle opposite the longer of
the two noncongruent sides is greater than the angle
opposite the shorter of the two noncongruent sides.
ANSWER:
a.
b.
c. Sample answer: The angle opposite the longer of
the two noncongruent sides is greater than the angle
opposite the shorter of the two noncongruent sides.
43.REASONING What is the range of possible
perimeters for figure ABCDE if AC = 7 and DC = 9?
Explain your reasoning.
SOLUTION:
The perimeter is greater than 36 and less than 64.
Sample answer: From the diagram we know that
and , and
because vertical angles are
congruent, so .
Using the Triangle Inequality Theorem, if 9 is the
longest length of the triangle, then the minimum
length of or is 9 – 7=2.If or is
the longest length of the triangle, then the maximum
value is 9+7=16. Therefore, the minimum value of
the total perimeter, p, of the two triangles is greater
than 2(2 + 7 + 9) or 36, and the maximum value of
the perimeter is less than 2(16 + 7 + 9) or 64 or,
expressed as an inequality, .
ANSWER:
The perimeter is greater than 36 and less than 64.
Sample answer: From the diagram we know that
and , and
because vertical angles are
congruent, so . Using the Triangle
Inequality Theorem, the minimum value of AB and
ED is 2 and the maximum value is 16. Therefore, the
minimum value of the perimeter is greater than 2(2 +
7 + 9) or 36, and the maximum value of the perimeter
is less than 2(16 + 7 + 9) or 64.
44.CHALLENGE What is the range of lengths of
each leg of an isosceles triangle if the measure of the
base is 6 inches? Explain.
SOLUTION:
Each leg must be greater than 3 inches. According to
the Triangle Inequality Theorem, the sum of any two
sides of a triangle must be greater than the sum of
the third side. Therefore. if you consider an isosceles
triangle with lengths x, x, and 6, we know three
inequalities must hold true: .
Since the last two inequalities are
the same, we will only consider the solutions of the
firsttwo.
Since 6 >0 is always true, the solution for the lengths
of the legs of the isosceles triangle is greater than 3.
Thereisnomaximumvalue.
ANSWER:
Each leg must be greater than 3 inches. Sample
answer: When you use the Triangle Inequality
Theorem to find the minimum leg length, the solution
is greater than 3 inches. When you use it to find the
maximum leg length, the inequality is 0 < 6, which is
always true. Therefore, there is no maximum length.
45.WRITING IN MATH What can you tell about a
triangle when given three side lengths? Include at
least two items.
SOLUTION:
When given three side lengths, you can determine
whether or not the side lengths actually form a
triangle, what the smallest and largest angles are,
whether the triangle is equilateral, isosceles, or
scalene.
ANSWER:
Sample answers: whether or not the side lengths
actually form a triangle, what the smallest and largest
angles are, whether the triangle is equilateral,
isosceles, or scalene
46.CHALLENGE The sides of an isosceles triangle
are whole numbers, and its perimeter is 30 units.
What is the probability that the triangle is equilateral?
SOLUTION:
Let x be the length of the congruent sides of an
isosceles triangle. Based on the Triangle Inequality
Theorem and properties of isosceles triangles, we
know that the following inequality can be written and
solved:
Therefore, based on the given information that the
two congruent sides are whole numbers greater than
7.5 and the perimeter of the triangle is 30 units, we
can create a list of possible side lengths for this
triangle:
* 10,10 10 is equilateral so the probability of the
triangle being equilateral is .
ANSWER:
47.OPEN-ENDED The length of one side of a triangle
is 2 inches. Draw a triangle in which the 2-inch side
is the shortest side and one in which the 2-inch side is
the longest side. Include side and angle measures on
your drawing.
SOLUTION:
When drawing your triangles, be sure to choose side
lengths that follow the conditions of the Triangle
Inequality Theorem. For the triangle where 2 is the
longest side length, the other two sides must each be
less than 2, however, their sum must be greater than
2. For the triangle where 2 is the shortest side, one of
the other sides plus 2 must have a greater sum than
the length of the third side. Sample sketches are
providedbelow.
ANSWER:
48.WRITING IN MATH Suppose your house is
mile from a park and the park is 1.5 miles from a
shopping center.
a. If your house, the park, and the shopping center
are noncollinear, what do you know about the
distance from your house to the shopping center?
Explain your reasoning.
b. If the three locations are collinear, what do you
know about the distance from your house to the
shopping center? Explain your reasoning.
SOLUTION:
a.BytheTriangleInequalityTheorem,thedistance
from my house to the shopping center is greater than
mile and less than miles.
b. The park (P) can be between my house (H1) and
the shopping center (S), which means that the
distance from my house to the shopping center is
miles, or my house (H2) can be between the park
(P) and the shopping center (S), which means that
the distance from my house to the shopping center is
mile.
ANSWER:
a. Sample answer: By the Triangle Inequality
Theorem, the distance from my house to the
shopping center is greater than mileandless
than miles.
b. Sample answer: The park can be between my
house and the shopping center, which means that the
distance from my house to the shopping center is
miles, or my house can be between the park and
the shopping center, which means that the distance
from my house to the shopping center is mile.
49.Which of the following could not be the length of
?
A 0.4 cm
B 3.3 cm
C 6.2 cm
D 7.5 cm
SOLUTION:
Since LM = 3.9 cm and LN = 3.6 cm, the possible
values of MN can be found using the Triangle
Inequality Theorem
LM + LN > MN
3.9 + 3.6 > MN
7.5 > MN
MN + LN > LM
MN + 3.6 > 3.9
MN > 0.3
0.3 < MN <7.5,ThecorrectanswerischoiceD.
ANSWER:
D
50.Three towns are connected by straight roads as
shown in the figure. Nick is driving from Crestfield to
Seaview.
What is the greatest whole number of miles Nick
may have to drive?
A2mi
B7mi
C 8 mi
D 15 mi
SOLUTION:
By the Triangle Inequality Theorem, the distance
between Elmwood and Seaview plus the distance
between Elmwood and Crestfield must be greater
than the distance from Crestfield to Seaview. So, 4.5
+ 3.2 = 7.7 must be greater than the distance from
Crestfield to Seaview. The exercise asks for the
greatest whole number of miles Nick may have to
drive,soBisthecorrectanswer.
ANSWER:
B
51.A mountain is 2.8 miles at the base and has one slope
that is 1.6 miles. Which is the best description for the
length of the other slope?
A between 1.2 mi and 4.4 mi
B between 1.6 mi and 2.8 mi
C less than 4.4 mi
D greater than 1.2 mi
SOLUTION:
By the Triangle Inequality Theorem, the base of the
mountain plus the length of one slope must be greater
than the length of the other slope. The Triangle
Inequality Theorem also tells us that the sum of the
lengths of the two slopes must be greater than the
base of the mountain. The following inequalities
describe the relationships between the three sides of
themountain.
2.8 + 1.6 > unknown slope
2.8 + unknown slope > 1.6
1.6 + unknown slope > 2.8
Solving these inequalities will determine the best
descriptionoftheunknownslope.
2.8 + 1.6 = 4.4 > unknown slope
The second inequality is meaningless as it results in a
negativelength.
unknown slope > 2.8 – 1.6 = 1.2
So the correct answer is A, the length of the other
slope is between 1.2 miles and 4.4 miles.
ANSWER:
A
52.In ΔABC, AB = 6.5 m and BC = 8.1 m. What is the
greatest possible whole-number perimeter of ΔABC,
in meters?
SOLUTION:
By the Triangle Inequality Theorem, the sum of the
lengths of any two sides of a triangle must be greater
than the length of the third side. The relationship
between the sides of the triangle are given by the
followinginequalities.
6.5 + 8.1 > x
6.5 + x > 8.1
8.1 + x > 6.5
Thus, 1.6 < x<14.6.
Since we are looking for the greatest possible whole-
number perimeter, assume that x is as large as it can
possible be. The perimeter would be less that the
sumof6.5+8.1+14.6or29.2.
The greatest possible whole-number perimeter is 29
meters.
ANSWER:
29
53.The figure shows five straight hiking trails. Selena
plans to hike on the trail that goes from point T to
point R.
Which of the following is a possible distance Selena
mighthike,assumingshecompletesthetrail?Select
all that apply.
A 1.2 km
B 4.2 km
C 5.1 km
D 6.7 km
E 9.1 km
F 9.4 km
SOLUTION:
is part of ΔTQR and ΔSRT. The following
inequalities describe the relationships between
andtheotherlegsofthetriangles.
5.3 + 4.1 > TR→TR< 9.4
5.3 + TR >4.1→TR> – 1.2
4.1 + TR >5.3→TR> 1.2
2.9 + 3.8 > TR→TR< 6.7
2.9 + TR >3.8→TR> 0.9
3.8 + TR >2.9→TR< – 0.9
So the range of possible values for TR is 1.2 < TR <
6.7.SothecorrectanswersareBandC.
ANSWER:
B, C
54.MULTI-STEP Naomi has some wooden dowels
that she wants to glue together to form a triangular
picture frame. The lengths of some of the dowels are
showninthetable.
a. Can Naomi make the frame using dowels Q, R,
andS?Explain.
b. Can Naomi make the frame using dowels P, Q,
andS?Explain.
c. Naomi decides to use dowels Q and R and a third
dowel that is not shown in the table. What is the
greatest possible whole number length for the third
dowel in centimeters?
d. How many different frames can Naomi make with
the dowels in the table?
Dowel Length (cm)
P18.5
Q20.6
R31.1
S40.8
SOLUTION:
a. Yes, 20.6 + 31.1 > 40.8, 20.6 + 40.8 > 31.1, 31.1 +
40.8>20.6.
b. No, 18.5 + 20.6 < 40.8
c. The third dowel must be less than the sum of the
lengths of the two dowels.
Q + R = 20.6 + 31.1 = 51.7
The dowel could be at most 51 cm as a whole
number length
d. There are three different possibilities for a
triangular frame, because P, Q, and R and P, R, and
S also make triangles according to the Triangle
Inequality Theorem.
Dowel Length (cm)
P18.5
Q20.6
R31.1
S40.8
ANSWER:
a. Yes, 20.6 + 31.1 > 40.8, 20.6 + 40.8 > 31.1, 31.1 +
40.8>20.6.
b. No, 18.5 + 20.6 < 40.8
c. 51
d. 3 triangular frames, because P, Q, and R; P, R,
and S; and Q, R, and S make triangles.
congruence
Triangle
congruence
Is it possible to form a triangle with the given
side lengths? If not, explain why not.
1.5 cm, 7 cm, 10 cm
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5
Thus, you can form a triangle with side lengths 5 cm,
7 cm,and10cm.
ANSWER:
Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5
2.3 in., 4 in., 8 in.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, , 3 in., 4 in., 8 in. do not form a
triangle.
ANSWER:
No;
3.6 m, 14 m, 10 m
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6, you
can form a triangle with side lengths 6 m, 14 m, 10
m.
ANSWER:
Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6
4.MULTIPLE CHOICE If the measures of two
sides of a triangle are 5 yards and 9 yards, what is
the least possible measure of the third side if the
measure is an integer?
A 4 yd
B 5 yd
C 6 yd
D 14 yd
SOLUTION:
Let x represents the length of the third side. Next, set
up and solve each of the three triangle inequalities.
5 + 9 > x, 5 + x > 9, and 9 + x > 5
That is, 14 > x, x > 4, and x > –4.
Notice that x > –4 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 > 4 and x < 14, which can be written
as 4 < x < 14. So, the least possible measure of the
thirdsidecouldbe5yd.
The correct option is B.
ANSWER:
B
PROOF Write a two-column proof.
5.Given:
Prove:
SOLUTION:
Think backwards when considering this proof. Notice
that what you are trying to prove is an inequality
statement. However, it isn't exactly related to
, except for instead of side being used, it
is Since it is given that , you can
easilyusethisinasubstitutionstep.
Given:
Prove: YZ + ZW > XW
Proof:
Statements (Reasons)
1. (Given)
2. XW = YW (Def. of segments)
3. YZ + ZW > YW ( Inequal. Thm.)
4. YZ + ZW > XW (Substitution Property.)
ANSWER:
Given:
Prove: YZ + ZW > XW
Proof:
Statements (Reasons)
1. (Given)
2. XW = YW (Def. of segments)
3. YZ + ZW > YW ( Inequal. Thm.)
4. YZ + ZW > XW (Subst.)
Is it possible to form a triangle with the given
side lengths? If not, explain why not.
6.4 ft, 9 ft, 15 ft
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, , you can not form a triangle with
side lengths 4 ft, 9 ft, 15 ft.
ANSWER:
No;
7.11 mm, 21 mm, 16 mm
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, 11 + 21 > 16, 11 + 16 > 21, and 16 + 21 > 11,
you can form a triangle with side lengths 11 mm, 21
mm, and 16 mm.
ANSWER:
Yes; 11 + 21 > 16, 11 + 16 > 21, and 16 + 21 > 11
8.9.9 cm, 1.1 cm, 8.2 cm
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since , you can not form a triangle
with side lengths 9.9 cm, 1.1 cm,and8.2cm.
ANSWER:
No;
9.2.1 in., 4.2 in., 7.9 in.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since, , you can not form a triangle
with side lengths 2.1 in., 4.2 in.,and7.9in.
ANSWER:
No;
10.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since , you can not form a triangle
with side lengths .
ANSWER:
No;
11.
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Since , you
can form a triangle with side lengths
.
ANSWER:
Yes;
Find the range for the measure of the third side
of a triangle given the measures of two sides.
12.4 ft, 8 ft
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than 4 +
8. Therefore, n < 12.
If n is not the largest side, then 8 is the largest and 8
must be less than 4 + n. Therefore, 4 < n.
Combining these two inequalities, we get 4 < n < 12.
ANSWER:
4 ft < n < 12 ft
13.5 m, 11 m
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
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.
ANSWER:
6 m < n < 16 m
14.2.7 cm, 4.2 cm
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than 2.7 +
4.2. Therefore, n < 6.9.
If n is not the largest side, then 4.2 is the largest and
4.2 must be less than 2.7 + n. Therefore, 1.5 < n.
Combining these two inequalities, we get 1.5 < n <
6.9.
ANSWER:
1.5 cm < n < 6.9 cm
15.3.8 in., 9.2 in.
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than 3.8 +
9.2. Therefore, n < 13.
If n is not the largest side, then 9.2 is the largest and
9.2 must be less than 3.8 + n. Therefore, 5.4 < n.
Combining these two inequalities, we get 5.4 < n <
13.
ANSWER:
5.4 in. < n < 13 in.
16.
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than
. Therefore, .
If n is not the largest side, then is the largest and
must be less than . Therefore, .
Combining these two inequalities, we get
.
ANSWER:
17.
SOLUTION:
Let nrepresentthelengthofthethirdside.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If n is the largest side, then n must be less than
. Therefore, n < .
If n is not the largest side, then is the largest and
must be less than . Therefore, .
Combining these two inequalities, we get
.
ANSWER:
PROOF Write a two-column proof.
18.Given:
Prove:
SOLUTION:
The key to this proof is to figure out some way to get
BC=BD so that you can substitute one in for the
other using the Triangle Inequality Theorem.
Consider the given statement, if two angles of a
triangle are congruent, what kind of triangle is it and,
therefore, how do you know that BC must equal BD?
Proof:
Statements (Reasons)
1. (Given)
2. (Converse of Isosceles Thm.)
3. BC = BD (Def. of segments)
4. AB + AD > BD ( Inequality Thm.)
5. AB + AD > BC (Substitution Property.)
ANSWER:
Proof:
Statements (Reasons)
1. (Given)
2. (Conv.Isos. Thm.)
3. BC = BD (Def. of segments)
4. AB + AD > BD ( Inequal. Thm.)
5. AB + AD > BC (Subst.)
19.Given:
Prove: KJ +KL> LM
SOLUTION:
Think backwards when considering this proof. Notice
that what you are trying to prove is an inequality
statement. However, it isn't exactly related to
, except for instead of side being used, it
is Since it is given that , you can
easily use this in a substitution step
Proof:
Statements (Reasons)
1. (Given)
2. JL = LM (Def. of segments)
3. KJ + KL > JL ( Inequality Thm.)
4. KJ + KL > LM (Substitution Property)
ANSWER:
Proof:
Statements (Reasons)
1. (Given)
2. JL = LM (Def. of segments)
3. KJ + KL > JL ( Inequal. Thm.)
4. KJ + KL > LM (Subst.)
SENSE-MAKINGDetermine the possible
values of x.
20.
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Notice that isalwaystrueforanywhole
number measure for x.The range of values that
would be true for the other two inequalities is
and , which can be written as
.
ANSWER:
6 < x < 17
21.
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Notice that isalwaystrueforanywhole
number measure for x. Combining the two remaining
inequalities, the range of values that fit both
inequalities is and , which can be
written as .
ANSWER:
22.TRAVEL Keyan wants to take the most efficient
route from his hotel to the hockey game at The
Sportsplex. He can either take Highway 521 or
take Highway 3 and Route 11 from his hotel to the
arena.
a. Which of these two possible routes is the shorter?
Explain your reasoning.
b. Suppose Keyan always drives the speed limit. If
he chooses to take Highway 521, he is on it for 30
miles and the speed limit is 40 miles per hour. If he
chooses to take the other route, the speed limit is
60 miles per hour on both roads and he is on
Highway 3 for 22 miles and Route 11 for 25 miles.
Whichrouteisfaster?Explainyourreasoning.
SOLUTION:
a. Highway 521; sample answer: In a triangle, the
sum of two of the sides is always greater than the
third side, so the sum of the distance on Highway 3
and the distance on Route 11 is greater than the
distance on Highway 521. Or you can add the
distances using Highway 3 and Route 11 and
compare their sum to the 30 miles of Highway 521.
Since 47 miles is greater than 30 miles, Highway 521
istheshortestdistance.
b. Highway 521; sample answer: Since Keyan drives
the 40 mph speed limit on Highway 521 and the
distance is 30 miles, it will take him about 30/40 =
0.75 hour or 45 minutes to get to The Sportsplex. He
has to drive 47 miles on Highway 3 and Route 11,
and the speed limit is 60 miles per hour, so it will take
him 47/60 = 0.78 hour or about 47 minutes. The route
onHighway4521willtakelesstimethantheroute
on Highway 3 and Route 11.
ANSWER:
a. Highway 521; sample answer: In a triangle, the
sum of two of the sides is always greater than the
third side, so the sum of the distance on Highway 3
and the distance on Route 11 is greater than the
distance on Highway 521.
b. Highway 521; sample answer: It will take Keyan
45 minutes to drive on Highway 521 and 47 to
drive Highway 3 to Route 11.
23.PROOF Write a two-column proof.
Given: ΔABC
Prove: AC + BC > AB (Triangle Inequality Theorem)
(Hint: Draw auxiliary segment , so that C is
between B and D and .)
SOLUTION:
Proof:
Statements (Reasons)
1. Construct sothatC is between B and D and
. (Ruler Postulate)
2. CD = AC (Definition of )
3. (Isosceles Theorem)
4. m ∠CAD = m ∠ADC (Definition. of
angles )
5. m ∠BAC + m ∠CAD = m ∠BAD (∠Addition
Postulate)
6. m ∠BAC + m ∠ADC = m ∠BAD (Substitution)
7. m ∠ADC < m ∠BAD (Definition of inequality)
8. AB < BD (Angle–Side Relationships in Triangles)
9. BD = BC + CD (Segment Addition Postulate)
10. AB < BC + CD (Substitution (Steps 8, 9))
11. AB < BC + AC (Substitution (Steps 2, 10))
ANSWER:
Proof:
Statements (Reasons)
1. Construct sothatC is between B and D and
.(Ruler Post.)
2. CD = AC (Def. of )
3. ∠CAD ADC(Isos. Thm)
4. m ∠CAD = m ∠ADC (Def. of ∠s )
5. m ∠BAC + m ∠CAD = m ∠BAD (∠Add. Post.)
6. m ∠BAC + m ∠ADC = m ∠BAD (Subst.)
7. m ∠ADC < m ∠BAD (Def. of inequality)
8. AB < BD (Angle–Side Relationships in Triangles)
9. BD = BC + CD (Seg. Add. Post.)
10. AB < BC + CD (Subst.(Steps 8, 9))
11. AB < BC + AC (Subst. (Steps 2, 10))
24.MULTI-STEP Toya rides her bike 3680 feet
down her street, Meadow Court, turns left onto
Holly Lane, and rides 2740 feet to get to the park.
One day she decides to try to find a shorter path to
the park by cutting straight through the field that is
next to the two roads.
However, if she decides to cut through the field, she
must walk instead of riding her bike.
a. If Toya is able to ride her bike 3 times as fast as
she can walk, how long should the path through the
field be for it to be a shortcut? How much time
would Toya save by taking the shortcut, if she can
ride her bike 9 miles per hour?
b. Describe your solution process.
c. What assumptions did you make?
SOLUTION:
a. - b. I drew a triangle formed by Meadow Court,
HollyLane,andthepaththroughthefield.
The Triangle Inequality Theorem to find the possible
sidelengths.
3680 + 2740 > x or 6420 > x
x + 2740 > 3680 or x > 940
x+3680>2740orx > –-940
Then,thepathisbetween940and6420feet.
Next, I found the time it would take Toya to ride her
bike to the park. She rides her bike 6420 feet. I
dividedby5280feettofinddistanceinmiles.Next,I
multiplied this number by 60 minutes and divided by 9
miles to get the number of minutes. This was
approximately8.1minutes,or0.135hours.
Since Toya can ride her bike 3 times as fast as she
canwalk,shecanwalk3milesperhour.Ifittakes
Toya longer than 8.1 minutes to walk through the
field,itisnolongerashortcut.
To find the maximum distance of the path, I
multiplied 3 miles per hour by 0.135 hours to get
0.405 miles or, after multiplying by 5280 feet, 2140
feet. So, the path can only be between 940 and 2140
feetinorderforittobeashortcut.Tofindthe
maximum amount of time Toya can save by taking
the shortcut, I took the shortest possible path, 940
feet,andconvertedittomiles.
Then, I multiplied the shortest possible distance in
miles by 60 minutes and divided by 3 mph to get the
minutes it took her to walk the shortest possible path.
Thiswasapproximately3.6minutes.
Next, I subtracted the two times and found that Toya
could save 4.5 minutes if the shortcut was 940 feet.
If the shortcut were any longer, she would save less
time. So Toya could save from no time to 4.5
minutes.
c. I assumed that the path through the field was a
straight line.
ANSWER:
a. The distance must be greater than 940 feet and
less than 2140 feet. Toya can save up to 4.5
minutes by taking the shortcut.
b. Sample answer: I drew a triangle formed by
Meadow Court, Holly Lane, and the path through
the field. The Triangle Inequality Theorem says that
the path is between 940 and 6420 feet. Since Toya
can ride her bike 3 times as fast as she can walk,
the path can only be between 940 and 2140 feet in
order for it to be a shortcut. Next, I found the time
it would take Toya to ride her bike to the park. She
rides her bike 6420 feet. I divided by 5280 feet to
find distance in miles. Next, I multiplied this number
by 60 minutes and divided by 9 miles to get the
number of minutes. This was approximately 8.1
minutes. To find the maximum amount of time Toya
can save by taking the shortcut, I took the shortest
possible path, 940 feet, and converted it to miles.
Since she can ride her bike 3 times as fast as she
can walk, Toya can walk 3 mph. Then, I multiplied
the shortest possible distance in miles by 60 minutes
and divided by 3 mph to get the minutes it took her
to walk the shortest possible path. This was
approximately 3.6 minutes. Next, I subtracted the
two times and found that Toya could save 4.5
minutes if the shortcut was 940 feet. If the shortcut
were any longer, she would save less time. So Toya
could save from no time to 4.5 minutes.
c. Sample answer: I assumed that the path through
the field was a straight line.
Find the range of possible measures of x if each
set of expressions represents measures of the
sides of a triangle.
25.x, 4, 6
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x is the largest side, then x must be less than 4 + 6.
Therefore, x < 4+6 or x <10.
If x is not the largest side, then 6 is the largest and 6
must be less than 4 + x. Therefore, 4 + x >6 or x >
2.
Combining these two inequalities, we get 2 < x<10.
ANSWER:
2 < x < 10
26.8, x, 12
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x is the largest side, then x must be less than 8 +
12. Therefore, x < 8 + 12 or x < 20.
If x is not the largest side, then 12 is the largest and
12 must be less than 8 + x. Therefore, 8 + x>12 or
x > 4.
Combining these two inequalities, we get 4 < x<20.
ANSWER:
4 < x < 20
27.x + 1, 5, 7
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x + 1 is the largest side, then x + 1 must be less
than 5 + 7. Therefore, x + 1 < 5 + 7 or x + 1 < 12 or
x <11.
If x + 1 is not the largest side, then 7 is the largest
and 7 must be less than 5 + (x + 1). Therefore, 5 + x
+ 1 > 7 or 6 + x > 7 or x > 1.
Combining these two inequalities, we get 1 < x<11.
ANSWER:
1 < x < 11
28.x – 2, 10, 12
SOLUTION:
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x – 2 is the largest side, then x – 2 must be less
than 10 + 12. Therefore, x – 2 < 10 + 12 or x – 2 <
22 or x < 24.
If x – 2 is not the largest side, then 12 is the largest
and 12 must be less than 10 + (x – 2). Therefore, 10
+ x – 2 > 12 or 8 + x > 12 or x > 4.
Combining these two inequalities, we get 4 < x<24.
ANSWER:
4 < x < 24
29.x + 2, x + 4, x + 6
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Noticethat and arealwaystruefor
any whole number measure for x. So, the only
required inequality is x > 0.
ANSWER:
x > 0
30.x, 2x + 1, x + 4
SOLUTION:
Set up and solve each of the three triangle
inequalities.
Notice that isalwaystrueforanywhole
number measure for x and isalwaystrue.So,
the required inequality is .
ANSWER:
31.DRAMA CLUB Anthony and Catherine are
working on a ramp up to the stage for the drama
club's next production. Anthony's sketch of the ramp
is shown below. Catherine is concerned about the
measurements and thinks they should recheck the
measures before they start cutting the wood. Is
Catherine'sconcernvalid?Explainyourreasoning.
SOLUTION:
The measurements on the drawing do not form a
triangle. According to the Triangle Inequality
Theorem, the sum of the lengths of any two sides of
a triangle is greater than the length of the third side.
The lengths in the drawing are 1 ft, ft, and ft.
Since , the triangle is impossible. They
should recalculate their measurements before they
cutthewood.Therefore,Catherine'sconcernis
valid.
ANSWER:
Yes; sample answer: The measurements on the
drawing do not form a triangle. According to the
Triangle Inequality Theorem, the sum of the lengths
of any two sides of a triangle is greater than the
length of the third side. The lengths in the drawing
are 1 ft, ft, and ft. Since , the
triangle is impossible. They should recalculate their
measurementsbeforetheycutthewood.
32.MODELING Aisha is riding her bike to the park
and can take one of two routes. The most direct
route from her house is to take Main Street, but it is
safer to take Route 3 and then turn right onto Clay
Road as shown. The additional distance she will
travel if she takes Route 3 to Clay Road is between
how many miles?
SOLUTION:
The distance from Aisha's house to the park via
MainSt.representsthethirdsideofatriangle.
From the Triangle Inequality Theorem the length of
this side must also be greater than 7.5 – 6 or 1.5
miles and must be less than 6 + 7.5 or 13.5 miles.
Therefore, the distance d from her house to the park
via Main St. can be represented by 1.5 < d < 13.5.
The distance to the park by taking Route 3 to Clay
Roadis7.5+6or13.5miles.
The least additional number of miles she would travel
would be greater than 13.5 –13.5 or 0. The greatest
number of additional miles she would travel would be
less than 13.5 – 1.5 or 12. Therefore, the additional
distance she will travel if she takes Route 3 to Clay
Roadisbetween0and12miles.
ANSWER:
0 and 12
33.DESIGN Carlota designed an awning that she and
her friends could take to the beach. Carlota decides
to cover the top of the awning with material that will
drape 6 inches over the front. What length of
material should she buy to use with her design so that
it covers the top of the awning, including the drape,
when the supports are open as far as possible?
Assume that the width of the material is sufficient to
cover the awning.
SOLUTION:
Let x be the length of the material needed.
According to the Triangle Inequality Theorem, the
largest side cannot be greater than the sum of the
othertwosides.
If x is the largest side, then x must be less than 4 + 3.
Therefore, n < 7. Since the material will drape 6
inches or 0.5 feet over the front, the minimum length
of material she should buy is 7 + 0.5 or 7.5 feet at the
most.
ANSWER:
She should buy no more than 7.5 ft.
ESTIMATION Without using a calculator,
determine if it is possible to form a triangle with
the given side lengths. Explain.
34.
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , .
is between and . Since = 1 and =
2, .
Since , .
5.9 > 2.9 + 1.5, so it is not possible to form a
triangle with the given side lengths.
ANSWER:
No; since , since or 2
and or 1, and since . So,
.
35.
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , then .
Also, since , then .
And since , then .
9.9 < 6.9 + 8.1, so yes, it is possible to form a
triangle with the given side lengths.
ANSWER:
Yes. since , since
, and since .6.9+8.1>
9.9, so it is possible.
36.
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , then .
Also, since , then .
And since , then .
4.9< 3.9 + 1.9, so yes, it is possible to form a
triangle with the given side lengths.
ANSWER:
Yes. since , since
, and since .
1.9 + 3.9 > 4.9, so it is possible.
37. .
SOLUTION:
Estimate each side length by comparing the values to
perfect squares.
Since , then .
Also, since ,then .
And since , then .
11.1> 2.1 + 5.1 so no, it is not possible to form a
triangle with the given side lengths.
ANSWER:
No; since ,
since , and
since . So, 2.1 + 5.1 11.1.
REASONING Determinewhetherthegiven
coordinates are the vertices of a triangle.
Explain.
38.X(1, –3), Y(6, 1), Z(2, 2)
SOLUTION:
We can graphically show that given coordinates form
atrianglebygraphingthem,asshownbelow.
We can algebraically prove that the given
coordinates form a triangle by proving that the
length of the longest side is greater than the sum of
thetwoshortersides.
Use the distance formula. has endpoints X(1,–3)
and Y(6, 1).
has endpoints Y(6, 1) and Z(2, 2).
has endpoints Z(2, 2) and X(1, –3).
Here, XY + YZ > XZ, XY + XZ > YZ, and XZ + YZ >
XY.
Thus,thegivencoordinatesformatriangle.
ANSWER:
Yes; XY + YZ > XZ, XY + XZ > YZ, and XZ + YZ >
XY
39.F(–4, 3), G(3, –3), H(4, 6)
SOLUTION:
We can graphically show that given coordinates form
atrianglebygraphingthem,asshownbelow.
We can algebraically prove that the given
coordinates form a triangle by proving that the
length of the longest side is greater than the sum of
thetwoshortersides.
Use the distance formula. has endpoints F(–4,3)
and G(3, 3).
has endpoints G(3, 3) and H(4, 6).
has endpoints H(4, 6) and F(–4, 3).
Here, FG + GH > FH, FG + FH > GH, and GH +
FH > FG .
Thus, the coordinated form a triangle.
ANSWER:
Yes; FG + GH > FH, FG + FH > GH, and GH +
FH > FG
40.J(–7, –1), K(9, –5), L(21, –8)
SOLUTION:
Wecangraphicallydetermineifthegiven
coordinates form a triangle by graphing them, as
shownbelow.
We can algebraically prove that the these three
points are collinear and therefore, by showing that
the sum of the two shorter segments is equal to the
longestsegment.
Use the distance formula. has endpoints J(–7,–
1) and K(9, –5).
has endpoints K(9, –5) and L(21, –8).
has endpoints L(21, –8) and J(–7, –1).
Here JK + KL = JL. You can also confirm this by
using your calculator. Compute
toconfirmthatitequals0.
Thus the given coordinates do not form a triangle.
ANSWER:
No; JK + KL = JL
41.Q(2, 6), R(6, 5), S(1, 2)
SOLUTION:
We can graphically determine if the given
coordinates form a triangle by graphing them, as
shownbelow.
We can algebraically prove that the given
coordinates form a triangle by proving that the
lengthofoneofthesidesequalszero.
Use the distance formula.
has endpoints Q(2,6) and R(6, 5).
has endpoints R(6, 5) and S(1, 2).
has endpoints S(1, 2) and Q(2, 6).
Then and
.
Thus the coordinates form a triangle.
ANSWER:
Yes;
42.MULTIPLE REPRESENTATIONS In this
problem, you will use inequalities to make
comparisons between the sides and angles of two
triangles.
a. GEOMETRIC Draw three pairs of triangles that
have two pairs of congruent sides and one pair of
sides that is not congruent. Mark each pair of
congruent sides. Label each triangle pair ABC and
DEF, where and .
b. TABULAR Copy the table below. Measure and
record the values of BC, m A, EF, and m D for
each triangle pair.
c. VERBAL Make a conjecture about the
relationship between the angles opposite the
noncongruent sides of a pair of triangles that have
two pairs of congruent legs.
SOLUTION:
a. Using a ruler, compass, or drawing tool, make sure
that and , in each of the
trianglepairsmade.
b. Use a protractor and ruler to carefully measure
the indicated lengths and angle measures in the table
below. Look for a pattern when comparing to
.
c. Sample answer: The angle opposite the longer of
the two noncongruent sides is greater than the angle
opposite the shorter of the two noncongruent sides.
ANSWER:
a.
b.
c. Sample answer: The angle opposite the longer of
the two noncongruent sides is greater than the angle
opposite the shorter of the two noncongruent sides.
43.REASONING What is the range of possible
perimeters for figure ABCDE if AC = 7 and DC = 9?
Explain your reasoning.
SOLUTION:
The perimeter is greater than 36 and less than 64.
Sample answer: From the diagram we know that
and , and
because vertical angles are
congruent, so .
Using the Triangle Inequality Theorem, if 9 is the
longest length of the triangle, then the minimum
length of or is 9 – 7=2.If or is
the longest length of the triangle, then the maximum
value is 9+7=16. Therefore, the minimum value of
the total perimeter, p, of the two triangles is greater
than 2(2 + 7 + 9) or 36, and the maximum value of
the perimeter is less than 2(16 + 7 + 9) or 64 or,
expressed as an inequality, .
ANSWER:
The perimeter is greater than 36 and less than 64.
Sample answer: From the diagram we know that
and , and
because vertical angles are
congruent, so . Using the Triangle
Inequality Theorem, the minimum value of AB and
ED is 2 and the maximum value is 16. Therefore, the
minimum value of the perimeter is greater than 2(2 +
7 + 9) or 36, and the maximum value of the perimeter
is less than 2(16 + 7 + 9) or 64.
44.CHALLENGE What is the range of lengths of
each leg of an isosceles triangle if the measure of the
base is 6 inches? Explain.
SOLUTION:
Each leg must be greater than 3 inches. According to
the Triangle Inequality Theorem, the sum of any two
sides of a triangle must be greater than the sum of
the third side. Therefore. if you consider an isosceles
triangle with lengths x, x, and 6, we know three
inequalities must hold true: .
Since the last two inequalities are
the same, we will only consider the solutions of the
firsttwo.
Since 6 >0 is always true, the solution for the lengths
of the legs of the isosceles triangle is greater than 3.
Thereisnomaximumvalue.
ANSWER:
Each leg must be greater than 3 inches. Sample
answer: When you use the Triangle Inequality
Theorem to find the minimum leg length, the solution
is greater than 3 inches. When you use it to find the
maximum leg length, the inequality is 0 < 6, which is
always true. Therefore, there is no maximum length.
45.WRITING IN MATH What can you tell about a
triangle when given three side lengths? Include at
least two items.
SOLUTION:
When given three side lengths, you can determine
whether or not the side lengths actually form a
triangle, what the smallest and largest angles are,
whether the triangle is equilateral, isosceles, or
scalene.
ANSWER:
Sample answers: whether or not the side lengths
actually form a triangle, what the smallest and largest
angles are, whether the triangle is equilateral,
isosceles, or scalene
46.CHALLENGE The sides of an isosceles triangle
are whole numbers, and its perimeter is 30 units.
What is the probability that the triangle is equilateral?
SOLUTION:
Let x be the length of the congruent sides of an
isosceles triangle. Based on the Triangle Inequality
Theorem and properties of isosceles triangles, we
know that the following inequality can be written and
solved:
Therefore, based on the given information that the
two congruent sides are whole numbers greater than
7.5 and the perimeter of the triangle is 30 units, we
can create a list of possible side lengths for this
triangle:
* 10,10 10 is equilateral so the probability of the
triangle being equilateral is .
ANSWER:
47.OPEN-ENDED The length of one side of a triangle
is 2 inches. Draw a triangle in which the 2-inch side
is the shortest side and one in which the 2-inch side is
the longest side. Include side and angle measures on
your drawing.
SOLUTION:
When drawing your triangles, be sure to choose side
lengths that follow the conditions of the Triangle
Inequality Theorem. For the triangle where 2 is the
longest side length, the other two sides must each be
less than 2, however, their sum must be greater than
2. For the triangle where 2 is the shortest side, one of
the other sides plus 2 must have a greater sum than
the length of the third side. Sample sketches are
providedbelow.
ANSWER:
48.WRITING IN MATH Suppose your house is
mile from a park and the park is 1.5 miles from a
shopping center.
a. If your house, the park, and the shopping center
are noncollinear, what do you know about the
distance from your house to the shopping center?
Explain your reasoning.
b. If the three locations are collinear, what do you
know about the distance from your house to the
shopping center? Explain your reasoning.
SOLUTION:
a.BytheTriangleInequalityTheorem,thedistance
from my house to the shopping center is greater than
mile and less than miles.
b. The park (P) can be between my house (H1) and
the shopping center (S), which means that the
distance from my house to the shopping center is
miles, or my house (H2) can be between the park
(P) and the shopping center (S), which means that
the distance from my house to the shopping center is
mile.
ANSWER:
a. Sample answer: By the Triangle Inequality
Theorem, the distance from my house to the
shopping center is greater than mileandless
than miles.
b. Sample answer: The park can be between my
house and the shopping center, which means that the
distance from my house to the shopping center is
miles, or my house can be between the park and
the shopping center, which means that the distance
from my house to the shopping center is mile.
49.Which of the following could not be the length of
?
A 0.4 cm
B 3.3 cm
C 6.2 cm
D 7.5 cm
SOLUTION:
Since LM = 3.9 cm and LN = 3.6 cm, the possible
values of MN can be found using the Triangle
Inequality Theorem
LM + LN > MN
3.9 + 3.6 > MN
7.5 > MN
MN + LN > LM
MN + 3.6 > 3.9
MN > 0.3
0.3 < MN <7.5,ThecorrectanswerischoiceD.
ANSWER:
D
50.Three towns are connected by straight roads as
shown in the figure. Nick is driving from Crestfield to
Seaview.
What is the greatest whole number of miles Nick
may have to drive?
A2mi
B7mi
C 8 mi
D 15 mi
SOLUTION:
By the Triangle Inequality Theorem, the distance
between Elmwood and Seaview plus the distance
between Elmwood and Crestfield must be greater
than the distance from Crestfield to Seaview. So, 4.5
+ 3.2 = 7.7 must be greater than the distance from
Crestfield to Seaview. The exercise asks for the
greatest whole number of miles Nick may have to
drive,soBisthecorrectanswer.
ANSWER:
B
51.A mountain is 2.8 miles at the base and has one slope
that is 1.6 miles. Which is the best description for the
length of the other slope?
A between 1.2 mi and 4.4 mi
B between 1.6 mi and 2.8 mi
C less than 4.4 mi
D greater than 1.2 mi
SOLUTION:
By the Triangle Inequality Theorem, the base of the
mountain plus the length of one slope must be greater
than the length of the other slope. The Triangle
Inequality Theorem also tells us that the sum of the
lengths of the two slopes must be greater than the
base of the mountain. The following inequalities
describe the relationships between the three sides of
themountain.
2.8 + 1.6 > unknown slope
2.8 + unknown slope > 1.6
1.6 + unknown slope > 2.8
Solving these inequalities will determine the best
descriptionoftheunknownslope.
2.8 + 1.6 = 4.4 > unknown slope
The second inequality is meaningless as it results in a
negativelength.
unknown slope > 2.8 – 1.6 = 1.2
So the correct answer is A, the length of the other
slope is between 1.2 miles and 4.4 miles.
ANSWER:
A
52.In ΔABC, AB = 6.5 m and BC = 8.1 m. What is the
greatest possible whole-number perimeter of ΔABC,
in meters?
SOLUTION:
By the Triangle Inequality Theorem, the sum of the
lengths of any two sides of a triangle must be greater
than the length of the third side. The relationship
between the sides of the triangle are given by the
followinginequalities.
6.5 + 8.1 > x
6.5 + x > 8.1
8.1 + x > 6.5
Thus, 1.6 < x<14.6.
Since we are looking for the greatest possible whole-
number perimeter, assume that x is as large as it can
possible be. The perimeter would be less that the
sumof6.5+8.1+14.6or29.2.
The greatest possible whole-number perimeter is 29
meters.
ANSWER:
29
53.The figure shows five straight hiking trails. Selena
plans to hike on the trail that goes from point T to
point R.
Which of the following is a possible distance Selena
mighthike,assumingshecompletesthetrail?Select
all that apply.
A 1.2 km
B 4.2 km
C 5.1 km
D 6.7 km
E 9.1 km
F 9.4 km
SOLUTION:
is part of ΔTQR and ΔSRT. The following
inequalities describe the relationships between
andtheotherlegsofthetriangles.
5.3 + 4.1 > TR→TR< 9.4
5.3 + TR >4.1→TR> – 1.2
4.1 + TR >5.3→TR> 1.2
2.9 + 3.8 > TR→TR< 6.7
2.9 + TR >3.8→TR> 0.9
3.8 + TR >2.9→TR< – 0.9
So the range of possible values for TR is 1.2 < TR <
6.7.SothecorrectanswersareBandC.
ANSWER:
B, C
54.MULTI-STEP Naomi has some wooden dowels
that she wants to glue together to form a triangular
picture frame. The lengths of some of the dowels are
showninthetable.
a. Can Naomi make the frame using dowels Q, R,
andS?Explain.
b. Can Naomi make the frame using dowels P, Q,
andS?Explain.
c. Naomi decides to use dowels Q and R and a third
dowel that is not shown in the table. What is the
greatest possible whole number length for the third
dowel in centimeters?
d. How many different frames can Naomi make with
the dowels in the table?
Dowel Length (cm)
P18.5
Q20.6
R31.1
S40.8
SOLUTION:
a. Yes, 20.6 + 31.1 > 40.8, 20.6 + 40.8 > 31.1, 31.1 +
40.8>20.6.
b. No, 18.5 + 20.6 < 40.8
c. The third dowel must be less than the sum of the
lengths of the two dowels.
Q + R = 20.6 + 31.1 = 51.7
The dowel could be at most 51 cm as a whole
number length
d. There are three different possibilities for a
triangular frame, because P, Q, and R and P, R, and
S also make triangles according to the Triangle
Inequality Theorem.
Dowel Length (cm)
P18.5
Q20.6
R31.1
S40.8
ANSWER:
a. Yes, 20.6 + 31.1 > 40.8, 20.6 + 40.8 > 31.1, 31.1 +
40.8>20.6.
b. No, 18.5 + 20.6 < 40.8
c. 51
d. 3 triangular frames, because P, Q, and R; P, R,
and S; and Q, R, and S make triangles.
congruence
Triangle
congruence
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5-5 The Triangle Inequality
