Savvas Learning Co Geometry Student Edition Chapter 5 Relationships Within Triangles Exercise 5.1 Midsegments Of Triangles
Savvas Learning Co Geometry Student Edition Chapter 5 Exercise 5.1 Midsegments Of Triangles Solutions Page 288 Exercise 1 Problem 1
From the given figure we can see that \(\overline{N O}\) is parallel to \(\overline{J K} \)
The segment \(\overline{N O}\) is parallel to \(\overline{J K} \).
Exercise 5.1 Midsegments Of Triangles Savvas Geometry Answers Page 288 Exercise 2 Problem 2
Given: Length of the side of the triangle.
To Find – Length of NM.
Use the relationship between the length of a midsegment and the length of the side of a triangle.
We have,LK = 46
Using the relationship
Length of a midsegment = \(\frac{1}{2}\) (length of the third side)
⇒ NM = \(\frac{1}{2}\)(46)
⇒ NM = 23
The length of the midsegment NM is 23.
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Exercise 5.1 Midsegments Of Triangles Savvas Geometry Answers Page 288 Exercise 3 Problem 3
A line segment that connects two midpoints of the sides of a triangle is called a midsegment.
It is a line that is drawn from one side to another, parallel to the other side.
A midsegment connects two mid points of the side of a triangle.
Exercise 5.1 Midsegments Of Triangles Savvas Geometry Answers Page 288 Exercise 4 Problem 4
The two noncollinear segments in the coordinate plane have the same slope, which means that the segments are neither perpendicular nor intersecting.
Thus, the two segments are parallel to each other which is the reason that the slopes of both the noncollinear segments are the same.
The two noncollinear segments are parallel to each other in the coordinate plane.
Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 288 Exercise 5 Problem 5
In the given figure, we know that NP = PT.
Which means that P is the midpoint of the side NT.
On the other hand, there is no information if OL is equal to LT, so we can say that L is not the midpoint of the side OT.
Thus, we can say that the conclusion given by the student is not correct and PL is not parallel to NO.
The error in the student’s reasoning is that L is not the midpoint of the side OT.
Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 288 Exercise 6 Problem 6
Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 288 Exercise 7 Problem 7
Given: A segment of the triangle.
To Find – The segment that is parallel to the given segment.
Use the triangle midsegment theorem.
We have the given segment,BC of the triangle
Using the triangle midsegment theorem, BC ∥ GF
The segment that is parallel to the segment BC is the segment GF.
Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 288 Exercise 8 Problem 8
Given: A segment of the triangle.
To Find – The segment that is parallel to the given segment.
Use the triangle midsegment theorem.
We have the given segment,EF of the triangle ABC.
Using the triangle midsegment theorem, EF ∥ AB
The segment parallel to the segment EF is the segment AB.
Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 288 Exercise 9 Problem 9
Given: A segment of the triangle ABC.
To Find – The segment parallel to the given segment.
Use the triangle midsegment theorem.
We have the given segment,CA of the triangle ABC.
Using the triangle midsegment theorem, CA ∥ EG
The segment parallel to the segment CA is the segment EG.
Solutions For Midsegments Of Triangles Exercise 5.1 In Savvas Geometry Chapter 5 Student Edition Page 288 Exercise 10 Problem 10
Given: A segment of the triangle ABC.
To Find – The segment parallel to the given segment.
Use the triangle midsegment theorem.
We have the segment,EG of the triangle FGE
Using the triangle midsegment theorem, EG ∥ CA
The segment parallel to the segment EG is the segment CA.
Solutions For Midsegments Of Triangles Exercise 5.1 In Savvas Geometry Chapter 5 Student Edition Page 288 Exercise 11 Problem 11
Given: Midpoints and lengths of the sides of the ΔTUV.
To Find – The length of HE.
Use the triangle midsegment theorem.
We have
⇒ UV = 80
⇒ TV = 100 and HD = 80 also,E,D and H are the midpoints of ΔTUV
Using the triangle midsegment theorem
⇒ HE = \(\frac{1}{2}\) VU
⇒ HE = \(\frac{1}{2}\) (80)
⇒ HE = 40
The length of HE is 40 units.
Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Detailed Answers Page 288 Exercise 12 Problem 12
Given: Midpoints and lengths of the sides of the ΔTUV.
To Find – The length of ED.
Use the triangle midsegment theorem.
We have
UV = 80, TV = 100 and HD = 80 also E,D and H are the midpoints of the sides of the ΔTUV
Using the triangle midsegment theorem
⇒ ED = \(\frac{1}{2}\) TV
⇒ ED = \(\frac{1}{2}\) (100)
⇒ ED = 50
The length of ED is 50 units.
Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Detailed Answers Page 288 Exercise 13 Problem 13
Given: HD = 80
To Find – The length of TU
In triangle TUV by midpoint theorem.
⇒ HD = \(\frac{1}{2}\) TU
⇒ 80 = \(\frac{1}{2}\) TU
⇒ TU = 160
The value of TU = 160
Geometry Chapter 5 Midsegments of Triangles Savvas Learning Co explanation guide Page 288 Exercise 14 Problem 14
Given: HD = 80
To Find – The length of TE
In triangle TUV by mid point theorem.
HD = \(\frac{1}{2}\) TU
80 = \(\frac{1}{2}\) TU
⇒ TU = 160
⇒ TU = 2 TE
⇒ TE = \(\frac{TU}{2}\)
⇒ TE = \(\frac{160}{2}\)
⇒ TE = 80
The value of TE = 80
Geometry Chapter 5 Midsegments Of Triangles Savvas Learning Co Explanation Guide Page 288 Exercise 15 Problem 15
Given: A diagram
To Find – The value of x
In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side
SO, x = \(\frac{1}{2}\) × 26
⇒ x = 13
The value of x = 13
Geometry Chapter 5 Midsegments of Triangles Savvas Learning Co explanation guide Page 288 Exercise 16 Problem 16
Given: A diagram
To Find – The value of x
In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side
SO, 5x = \(\frac{1}{2}\) × 45
⇒ x = 4.5
The value of x = 4.5
Savvas Learning Co Geometry Student Edition Chapter 5 Page 288 Exercise 17 Problem 17
Given: A diagram
To Find – The value of x
In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side
So, 6x = \(\frac{72}{2}\)
⇒ x = 6
The value of x = 6
Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289 Exercise 18 Problem 18
Given: A diagram
To Find – The value of x
In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side
So x − 4 = \(\frac{1}{2}\) × 17
⇒ x = \(\frac{17}{2}\) + 4
⇒ x = 12.5
The value of x = 12.5
Savvas Learning Co Geometry Student Edition Chapter 5 Page 289 Exercise 19 Problem 19
Given: A diagram
To Find – The value of x
In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side
So x + 2 = \(\frac{1}{2}\) × 38
⇒ x = 19 − 2
⇒ x = 17
The value of x = 17
Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289 Exercise 20 Problem 20
Given: A diagram
To Find – The value of x
In the triangle by midpoint theorem, segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
So, 5x − 4 = \(\frac{1}{2}\) × 8
⇒ 5x = 8
⇒ x = \(\frac{8}{5}\)
⇒ x = 1.6
The value of x = 1.6.
Savvas Learning Co Geometry Student Edition Chapter 5 Page 289 Exercise 21 Problem 21
Given: The number of strides is shown in the triangle and the average stride is 3.5 ft
To Find – The length of the longest side of the triangle.
In the triangle, the longest side will be in which maximum strides are paddled so the maximum number of strides is 250
The length of the longest side = 3.5 × 250
⇒ 875ft
The length of the longest side is 875ft
Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289 Exercise 21 Problem 22
Given: The sides of the triangle.To find: Distance to be paddled across the lake.
Let ,a,b, c be sides of the triangle.
⇒ a = 80 + 80 = 160 (Say)
⇒ b = 250 (Say)
⇒ c = 150 + 150 = 300 (Say)
Distance to be paddled across the lake = The perimeter of the triangle × 3.5
= (a + b + c) × 3.5
= (160 + 250 + 300) × 3.5
= 710 × 3.5
= 2485 ft
The distance to be paddled across the lake is 2485ft.
Savvas Learning Co Geometry Student Edition Chapter 5 Page 289 Exercise 22 Problem 23
Given: The following triangular face of the Rock and Roll Hall of Fame in Cleveland, Ohio, that is isosceles is provided with congruent red segments and the length of the base is 229 feet 6 Inches or 229 ft 6 in.
Converting the given measurement of the base into inches 229 ft 6 in
= (229 × 12) + 6 in
= 2748 + 6 in
= 2754 in
The red segments divide the legs into four congruent parts and the white segment divides each leg into two congruent parts. The white segment is a midsegment of the triangular face of the building, so its length is half the length of the base.
So, the length of the white segment
= \(\frac{2754}{2}\)
= 1377 in
Converting to feet and inches
Length of the white segment
= \(\frac{1377}{12}\) ft
= 114. 75 ft
= 114 ft (0.75 × 12) in
= 114 ft 9 in
The length of the white segment is 114 ft 9in.
Midsegments Of Triangles Solutions Chapter 5 Exercise 5.1 Savvas Geometry Page 289 Exercise 23 Problem 24
Given: A figure.
To find – m ∠ QPR
S is the mid-point of QP
T is the mid-point of QR
Using the mid-point theorem, we get ST ∥ PR
QP is a transversal and ST ∥ PR
Therefore,∠QPR = ∠QST (Corresponding angles)
∠QPR = 40° (Since,∠QST = 40°)
Using the mid-point theorem,m ∠QPR = 40°.
Savvas Learning Co Geometry Student Edition Chapter 5 Page 289 Exercise 24 Problem 25
Given: The coordinates of the vertices of a triangle EFG.
To find – The coordinates of H & J.
Since, H is the mid-point of EG
Therefore, coordinates of
H = \(\left(\frac{1+3}{2}, \frac{2-2}{2}\right)\)
H = (2,0)
Since, J is the mid-point of FG
Therefore, coordinates of
J = \(\left(\frac{5+3}{2}, \frac{6-2}{2}\right)\)
J = (4,2)
The coordinates of H = (2,0) and of J = (4,2).
Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 289 Exercise 24 Problem 26
Given: H, J are the mid-points of the sides EG, FG respectively.
To prove – HJ ∥ EF
Since, H,J are the mid-points of the sides EG, FG respectively.
Therefore, using the mid-point theorem we get, HJ ∥ EF
Using the mid-point theorem we get, HJ ∥ EF
Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 289 Exercise 24 Problem 27
Given: H,J are the mid-points of the sides EG, FG respectively.
To prove – HJ = \(\frac{1}{2}\) EF
Since, H,J are the mid-points of the sides EG,FG, respectively.
Therefore, using the mid-point theorem we get
⇒ HJ = \(\frac{1}{2}\) EF
Using the mid-point theorem we get,HJ = \(\frac{1}{2}\) EF
Savvas Learning Co Geometry Student Edition Chapter 5 Page 289 Exercise 25 Problem 28
Given: X,Y are the mid-points of the sides UV, UW respectively.m∠UXY = 60
To find – m∠V
Since, X, Y are the mid-points of the sides UV, UW respectively.
Therefore, using the mid-point theorem we get, XY ∥ VW
Hence,∠V = ∠UXY (Corresponding angles)
⇒ m∠V = 60 (Since m∠UXY = 60)
According to the question,m∠V = 60°.
Chapter 5 Exercise 5.1 Midsegments Of Triangles Savvas Learning Co Geometry Explanation Page 289 Exercise 26 Problem 29
Given: X,Y are the mid-points of the sides UV,UW respectively.
To find – m∠UYX
Since ,X,Y are the mid-points of the sides UV, UW respectively.
Therefore, using the mid-point theorem we get, XY ∥ VW
Hence,∠UYX = ∠W (Corresponding angles)
⇒ m∠UYX = 45 (Since m∠W = 45)
According to the given condition, m∠UYX = 45°