Geometry, Homework Practice Workbook

Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes

Page 1 Problem 1 Answer

Given,

We need to name a line that contains point E.

Given,

By watching the picture, we can clearly say that the name of the line that contains point E is the line q or EB or BE.

The name of the line that contains point E is the line q or EB or BE.

Page 1 Problem 2 Answer

Given,

We need to name a point contained in line n.

Given,

By watching the picture, we can clearly say that the point contained in line n is A or B.

The point contained in line n is A or B.

Page 1 Problem 3 Answer

Given,

We need to determine another name for line P.

Given,

By watching the picture, we can clearly say that the other name for line p is CDor DC.

The other name for line p is CD or DC.

Page 1 Problem 4 Answer

Given.

We need to name the plane containing lines n and p.

Given,

By watching the picture, we can clearly say that the name of the plane containing n and p is plane G.

The name of the plane containing n and p is plane G.

Page 1 Problem 5 Answer

We need to draw and label a figure on the basis of the given information, the point K lies on RT.

Given information, the point K lies on RT.

The drawn figure –

In the picture, the dark dots represent points, the arrow represents a line, and the page represents a plane.

The drawn figure –

In the picture, the dark dots represent points, the arrow represents a line, and the page represents a plane.

Page 1 Problem 6 Answer

We need to draw and label a figure on the basis of the given information, YP

lies in plane B and contains point C but does not contain point H.

Given information, YP lies in plane B and contains point C, but does not contain point H.

The drawn figure –

In the picture, the dark dots represent points, the arrow represents a line, and the page represents a plane.

The drawn figure –

In the picture, the dark dots represent points, the arrow represents a line, and the page represents a plane.

Page 1 Problem 7 Answer

We need to draw and label a figure on the basis of the given information, line q and f intersect at point Z in-plane U.

Given information, line q and f intersect at point Z in-plane U.

The drawn figure –

In the picture, the dark dots represent points, the arrow represents a line, and the page represents a plane.

According to the given information, the figure is,

Page 1 Problem 8 Answer

Given,

We need to determine how many planes are shown in the figure.

Given,

By watching the picture, we can say that the planes are ABC, FED, EBC, DFA, AFB. So the number of planes in the figure is 5.

After briefly examining the picture, we came to the conclusion that there are 5 planes in the figure.

Page 1 Problem 9 Answer

Given,

We need to determine how many planes contains the points F and E.

Given,

By watching the picture, we can clearly say that the number of planes contains F and E is 2.

The number of planes contains F and E is 2.

Page 1 Problem 10 Answer

Given,

We need to determine four points that are co-planar.

Given,

By watching the picture, we can clearly say that the co-planer points are E, B, C, and D.

The co-planer points are E, B, C, and D.

Page 1 Problem 11 Answer

Given,

We need to determine whether points A, B and C are coplanar or not.

Given,

We know, Co-planer points, mean points situated in the same plane.

So, points A, B, and C are coplanar.

Because they are situated in the same plane.

So, points A, B, and C are coplanar.

Because they are situated in the same plane.

Page 2 Problem 12 Answer

Given,

We need to name a line that contains points T and P.

Given,

By watching the picture, we can clearly say that the line containing points T and P is line g or TP or TN or NP.

The containing points T and P is line g or TP or TN or NP.

Page 2 Problem 13 Answer

Given,

We need to determine a line that intersects the plane containing points Q, N, and P.

Given,

By watching the picture, we can say that the line that intersects the plane containing points Q.N, and P is line J or MT.

The line that intersects the plane containing points Q, N, and P is line J or MT.

Page 2 Problem 14 Answer

Given,

We need to determine the plane that contains  TN and QR.

Given.

By watching the picture, we can clearly say that the plane contains TN and QR is plane S.

The plane contains TN and QR is plane S.

Page 2 Problem 15 Answer

The given relation is, two lines AK and CG intersect at a point M in plane T.

We draw these two lines in a plane where these lines have different slopes i.e. they are not parallel as they intersect at a particular point.

We have the conditions, two lines AK and CG intersect at a point M  in the plane T.

Now, first, we draw a plane T as

Now we draw two lines AK and CG  on this plane, where they are not parallel.

The two lines intersect at a point that is labeled as M.

Finally, we can conclude that the given relation can be drawn and labeled as

Page 2 Problem 16 Answer

The given relation is, a line contains L(-4,-4) and M(2,3). Line q is in the same coordinate plane N but does not intersect the line LM.

The N plane is given as the X-Y plane.

We draw a line that passes through both the given points.

Then we make another line q in the same plane N(x-y) in such a way that it is parallel to LM i.e. they do not intersect.

We have the conditions, one line that connects the two points L(−4,−4) and M(2,3).

First, we draw a line LM through these two points in the N plane (X-Y).

Then we draw a random line in parallel with LM so that they do not intersect with each other.

Finally, we can conclude that the given relation can be drawn and labeled as

Page 2 Problem 17 Answer

We are given a figure,

and we have to find the number of planes in this figure.

From this figure, we can find that the planes are, WPQT, WPNM, TQRS, QPNR, TWMS, and also MNSR that is coinciding in plane A.

We can figure out there are 6 planes in the given figure.

Finally, we can conclude that the given figure has 6 planes.

Page 2 Problem 18 Answer

We have the given figure,

and we have to find three collinear points.

Now, in the whole figure, we can see only one case where three points M, X, and S are in the same line.

Therefore, M, X, S are collinear points.

Finally, we can conclude that the three collinear points are M, X, and S.

Page 2 Problem 19 Answer

We are given the figure,

and we have to check whether N, R, S, and W are coplanar or not.

We know, that for being coplanar these points have to be on the same plane.

Now, we can see that N, R, and S are in the same plane i.e. plane A, but point W is not in the same plane.

Therefore, the points N, R, S, and W are not coplanar.

Finally, we can conclude that the given points N, R, S, and W are not coplanar.

Page 2 Problem 20 Answer

We are given the figure,

Now, we can clearly see, that the strings are connected between two points and also it is straight and one-dimensional, hence the string models Lines.

Finally, we can conclude that the strings model Lines.

Page 2 Problem 21 Answer

We are given the reference of a car antenna,

Now, we can clearly see, that the antenna is a one-dimensional linear structure i.e. referring to a Line and the tip of the antenna is an exact location, hence models a Point.

Finally, we can conclude that the car antenna models a Line and a Point.

Geometry Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes

Page 3 Problem 1 Answer

Given,

We need to determine the length of the object.

Given,

The ruler is marked in inches. The distance between two consecutive numbers is divided into 4 equal parts.

One end of the object starts at the zero mark of the ruler, and the other endpoint is 1/4 marks after 2.

So the length of the object is 2×1/4in.

So the length of the object is 2×1/4in.

Page 3 Problem 2 Answer

Given,

We need to find the length of the segment NQ.

Given,

The length of NQ= The length of NP+ the length of PQ

​=1×1/4in+1in

=(1×1/4+1)in

=2×1/4in

​The length of the segment NQ is 2×1/4in.

Page 3 Problem 3 Answer

Given.

We need to find the length of the segment AC.

Given,

The length of AC= the length of AB + the length of BC

​= 4.9cm+5.2cm

= (4.9+5.2)cm

= 10.1cm

​The length of the segment AC is 10.1cm

Page 3 Problem 4 Answer

Given,

We need to find the value of the segment GH.

Given,

The value of the segment GH= the value of FH- the value of FG

​=15mm−9.7mm

= (15−9.7)mm

= 5.3mm

​The value of the given segment GH is 5.3mm.

Page 3 Problem 5 Answer

Here, the given values are XY=5x, YZ=x, and XY=25.

We need to find the value of x and YZ.

Here, Y is between X and Z.

Given values of XY are 2x and 25.

By equalling both the values, we get,

∴ 5x=25( required linear equation)

⇒ x=25/5

⇒ x=5​

Therefore, the value of YZ=5. (putting the value of x).

The value of x is 5 and the value of YZ is 5.

Page 3 Problem 6 Answer

Here, the given values of XY = 12, YZ = 2x, and XZ = 28.

We need to find the value of x and YZ.

Here, Y is between X and Z.

So the value of XZ = the value of XY + the value of YZ.

∴ 28=12+2x(reaquired linear equation).

By solving the equation, we get,

∴ 28=12+2x

⇒ 28−12=2x

⇒ 16=2x

⇒ x=16/2

⇒ x=8

Therefore, the value of YZ,

= 2×8( putting the value of x)

= 16

​The value of x is 8 and the value of YZ is 16.

Page 3 Problem 7 Answer

Here, the given values of XY = 4x, YZ = 3x, XZ = 42. We need to find the value of x and YZ.

Here, Y is between X and Z.

So the value of XZ = the value of XY + the value of YZ.

∴ 42=4x+3x(reaquired linear equation).

By solving the equation, we get,

∴ 42=4x+3x

⇒ 42=7x

⇒ 42/7

= x

⇒ x = 6

​Therefore, the value of YZ,

= 3x

= 3×6(putting the value of x)

= 18

​The value of x is 6 and the value of YZ is 18.

Page 3 Problem 8 Answer

Here, the given values of XY =2x+1, YZ = 6x and XZ = 81.

We need to find the value of x and YZ.

Here, Y is between X and Z.

So the value of XZ = the value of XY + the value of YZ.

∴ 81=2x+1+6x( reaquired linear equation).

By solving the equation, we get,

∴ 81=2x+1+6x

⇒ 81−1 = 8x

⇒ 80=8x

⇒ 80/8

= x

⇒ x=10

​Therefore, the value of YZ,

= 6x

= 6×10( putting the value of x)

= 60

​The value of x is 10 and the value of YZ is 60.

Page 3 Problem 9 Answer

Given,

We need to determine whether the given pair of segments BE, and CD are congruent.

Segments that have the same measure are called congruent segments.

Given,

Here, BE=3cm and CD =3cm.

Therefore, BE=CD.

So. the given pair of segments BE, CD is congruent.

The given pair of segments BE, CD is congruent.

Page 3 Problem 10 Answer

We are given the figure,

and we have to find whether segments MP and NP are congruent or not.

Now, we can clearly see that the segment MP has a length of 12 yds, and the length of the segment NP is 10 yds.

Hence, as these are different, they are not congruent.

Finally, we can conclude that the two segments MP and NP are not congruent.

Page 3 Problem 11 Answer

We are given the figure,

and we have to find whether segments WX and WZ are congruent or not.

Now, we can clearly see that the segment WX is parallel and equal to YZ i.e. it has a length of 9ft.

Also, it is given that the length of the segment WZ is 5ft.

Hence, they have different lengths, so they are not congruent.

Finally, we can conclude that the two segments WX and WZ are not congruent.

Page 4 Problem 12 Answer

We are given the figure,

We match the line with the measuring scale to find the number of small segments combinedly refer to the length of the line EF.

We have the line EF and a measuring scale (in inches) as,

Now, we can notice that in the scale every inch is segmented into 16 small portions.

The line lies in a way that its starting point is matching with the starting point of the measuring scale.

Now, we can measure that the line is long enough to cover 1 inch and 11 small segments in the scale.

Therefore, the length of the line EF can be written as,

EF=1+11/16 in.

=111/16 in.

Finally, we can conclude that the length of line EF is given as 111/16 in.

Page 4 Problem 13 Answer

We are given the figure,

We can see that the line PS is made of two segments PQ and QS. We add the lengths of these segments to obtain the final length of the line PS.

We have the figure

The length of the segment PQ  is 18.4 cm and the length of the segment QS is 4.7 cm.

Now, the line PS is made of these two consecutive segments, hence, length of PS=length of PQ+length of QS

⇒ length of PS=(18.4+4.7)

⇒ length of PS=23.1 cm.

Finally, we can conclude that the length of the line PS is 23.1 cm.

Page 4 Problem 14 Answer

Given

To find AD.

The value of AD can be calculated as AD=AC+CD.

AD = AC + CD

⇒ AD = 2×3/8 + 1×1/4

⇒ AD =19/8 +5/4

⇒ AD =19 + 2 × 5/8

⇒ AD =29/8 = 3.625in.

Hence the measurement of WX is 35/8in or 3.625in.

Page 4 Problem 15 Answer

Given,

To find WX.

WY=WX+XY

⇒ WX=WY−XY.

​WX = WY − XY

⇒ WX = 100 − 89.6

⇒ WX = 10.4cm.

Hence the measurement of WX is 10.4cm.

Page 4 Problem 16 Answer

Given, JK=6x, KL=3x, JL=27, and K is between JL.

To find x and KL.

Since K is between JL, JK+KL=JL.

Since K is between J and L, therefore​JL=JK+KL

⇒ 27=6x+3x

⇒ 9x=27

⇒ x=27/9

⇒ x=3.

​Hence​KL=3x

= 3×3

= 9.

​Hence we can conclude that the values of x=3, KL =9.

Page 4 Problem 17 Answer

Given,JK=2x,KL=x+2,andJL=5x−10.

K is between J and L.

To find x and KL.

Since K is between J&L, JK+KL=JL.

Since K is between J&L, therefore JL=JK+KL

⇒ 5x−10=(2x)+(x+2)

⇒ 2x=10+2​

⇒ x=12/2

⇒ x=6.

​Hence KL=x+2

= 6+2

= 8.

​Hence we can conclude that the values of x=6,KL=8.

Page 4 Problem 18 Answer

Given

To check whether the line segments TU&SW are congruent or not.

Two line segments are congruent when they are of equal lengths.Length of TU=2ft,

Length of SW=3ft.

Since the two segments do not have equal lengths hence they are not congruent.

Hence we can conclude that since the two segments TU&SW do not have equal lengths hence they are not congruent.

Page 4 Problem 19 Answer

Given

To check whether AD&BC are congruent.

Two line segments are congruent when they are of equal lengths.

The given figure shows two marks on the lengths AD&BC which indicate they are equal.

Since both the line segments are equal they are congruent.

Hence we can conclude that since the two segments AD&BC

have equal lengths hence they are congruent.

Page 4 Problem 20 Answer

Given

To check whether GF&FE are congruent.

Two line segments are congruent when they are of equal lengths.

The similar marks on two different line segments indicate that they are equal.

Hence GF≡HE=6x and GH≡FE=5x.

Thus GF&FE are not equal, hence they are not congruent.

Hence we can conclude that since the two segments GF&FE do not have equal lengths hence they are not congruent.

Page 4 Problem 21 Answer

Given

To find all the congruent pairs of line segments.

The similar marks on two different line segments indicate that they are equal.HenceAF≡AB, EF≡CB, DE≡DC.

All these pairs are congruent. Hence AF≅AB, EF≅CB, DE≅DC.

These are the pairs of congruent segments.

Hence the pairs of congruent segments are AF and AB, EF and CB, DE and DC.

Geometry Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes

Page 7 Problem 1 Answer

The figure is given as

We can clarify that the verte x² denotes an angle.

The angle can be identified as the vertex denotes the middle of the respective angle.

The angle for the following figure can be named as∠2=∠TVW with vertex as V.

The angle for the following figure can be named as∠2=∠TVW with vertex as V.

Page 7 Problem 2 Answer

The figure is given as

We can clarify that the vertex 5 denotes an angle.

The angle can be identified as the vertex denotes the middle of the respective angle.

The angle for the following figure can be named as ∠5=∠STW with vertex as T.

The angle for the following figure can be named as ∠5=∠STW with vertex as T.

Page 7 Problem 3 Answer

The figure is given as

We can clarify that the angle ∠5 is defined in the figure.

The side can be identified as each set of adjacent points creates an edge or side.

The angle for the following figure can be named as ∠5=∠STW.

∠STW is defined by three points S,T,W. Hence the defined sides are ST and TW.

The angle for the following figure can be named as ∠5=∠STW with sides as ST and TW.

Page 7 Problem 4 Answer

The figure is given as

We can clarify that the angle ∠STV is defined in the figure.

The side can be identified as each set of adjacent points creates an edge or side.

In the following figure,∠STV is defined by three points S,T,V.

Hence the defined sides are ST and TV.

The defined side of the following angles are ST and TV.

Page 7 Problem 5 Answer

The figure is given as

We can clarify that the verte x⁴  denotes an angle.

The angle can be identified as the vertex denotes the middle of the respective angle.

The angle for the following figure can be named as ∠4=∠UTS.

The angle for the following figure can be named as∠UTS.

Page 7 Problem 6 Answer

The figure is given as

We can clarify that the angle ∠WTS denotes a defined angle.

The angle can be identified as the vertex denotes the middle of the respective angle.

The angle for the following figure can be named as∠WTS=∠5  with vertex as

The angle for the following figure can be named as∠TSW=∠5 with vertex as T.

Page 7 Problem 7 Answer

Given the figure that is

Here the angle∠2 is located at the point V inside the triangle ΔTWV.So the name of that angle is∠WVT.

Hence, another name of∠2 is∠WVT.

Page 7 Problem 8 Answer

The given figure is,

Here the angle∠OMN is said to be an acute angle because the line OM is not perpendicular to the baseline and it makes the angle in such a way that it does not exceed 90∘.

So, now by using a protractor we can say the measurement of this angle is 40∘.

Therefore, the measurement of the angle is 40∘and so it is an acute angle.

Page 7 Problem 9 Answer

The given figure is,

Here the angle∠QMN is said to be an obtuse angle because the line QM is not perpendicular to the baseline and it makes the angle in such a way that it exceeds 90∘.

So, now by using a protractor we can say the measurement of this angle is 140∘.

Therefore, the measurement of the angle is 140∘and so it is an obtuse angle.

Page 7 Problem 10 Answer

Given the figure is,

Here the angle∠QMO is an obtuse angle because no line making this angle is perpendicular to each other.

Using a protractor the angle is found out as 100∘ .

Therefore, the measurement of the angle is 100∘and so it is an obtuse angle.

Page 7 Problem 11 Answer

Given the figure is,

Here BD bisects the angle∠EBC

so∠EBD is half of∠EBC.

Putting the value of both angles and solving both sides we will get the value of x.

Let’s put the value of x and get∠EBC.

Given that∠EBD=4x−8 and∠EBC=5x+20

From the given figure we can say∠EBD=1/2∠EBC.

So by replacing the value of∠EBD&∠EBC in the above formula we can say 4x−8=1/2(5x+20) or,

2(4x−8)=5x+20 or,

8x−16=5x+20 or,

8x−5x=20+16 or,

3x=36 or,

x=36/3=12.

Now put the value of x in∠EBC=5x+20

we will get ∠EBC=5×12+20=80.

Therefore, the value of∠EBC is 80∘.

Page 8 Problem 12 Answer

Given the figure is,

Here BD⃗bisects the angle∠EBC

so∠EBD is half of∠EBC.

Putting the value of both angles and solving both sides we will get the value ofx.

Let’s put the value of x and get∠EBC.

Given that∠EBD=4x−8 and∠EBC=5x+20

From the given figure we can say∠EBD=1/2∠EBC.

So by replacing the value of∠EBD&∠EBC in the above formula we can say

4x−8=1/2(5x+20)or,

2(4x−8)=5x+20or,

8x−16=5x+20or,

8x−5x=20+16or,

3x=36or,

x=36/3=12.

Now put the value of x in∠EBC=5x+20

we will get ∠EBC=5×12+20 =80.

Therefore, the value of∠EBC is 80∘.

Page 8 Problem 13 Answer

Given that

Here we have to find the vertex of∠8.

As we know The vertex of an angle is the point where two rays or edges begin or meet so∠8 is∠MOP in the triangle Δ MOP and two edges are MO&OP.

So the common point of two edges is O that is the vertex of∠8.

Hence, the vertex of the angle∠8 is O.

Page 8 Problem 14 Answer

Given,

To find the sides of the angle∠6.

In Euclidean geometry, an angle is the figure formed by two rays, referred to as the sides of the angle.

The sides that form the angle∠6 are NM&NR.

The sides that form the angle∠6 are NM&NR.

Page 8 Problem 15 Answer

Given,

To find the sides of the angle∠2.

In Euclidean geometry, an angle is the figure formed by two rays, referred to as the sides of the angle.

The sides that form the angle∠2 are PM&PR.

The sides that form the angle∠2 are PM&PR.

Page 8 Problem 16 Answer

Given,

To find the sides of the angle∠MOP.

In Euclidean geometry, an angle is the figure formed by two rays, referred to as the sides of the angle.

The sides that form the angle∠MOP are OM&OP.

Hence we can conclude that the sides that form the angle∠MOP Are OM&OP.

Page 8 Problem 17 Answer

Given,

To find the sides of the angle∠OMN.

In Euclidean geometry, an angle is the figure formed by two rays, referred to as the sides of the angle.

The sides that form the angle∠OMN Are MO&MN.

Hence we can conclude that the sides that form the angle∠OMN Are MO&MN.

Page 8 Problem 18 Answer

Given

To determine∠TZW is acute or obtuse or right-angled.

An angle is acute if it is less than 90∘.

An angle is obtuse if it is more than 90∘.

An angle is right if it is equal to 90∘.

To measure the angle of∠TZW place the protractor along any line.

ZW makes an angle of 110∘with respect to ZT.

Hence the required measure of the angle110∘.

Hence we can conclude that∠TZW is an obtuse angle with a measure of 110∘.

Page 8 Problem 19 Answer

Given

To determine∠UZT is acute or obtuse or right-angled.

An angle is acute if it is less than 90∘.

An angle is obtuse if it is more than 90∘.

An angle is right if it is equal to 90∘.

To measure the angle of∠UZT place the protractor along any line.

ZU makes an angle of 20∘with respect to ZT.

Hence we can conclude that∠UZT is an obtuse angle with a measure of 20∘.

Page 8 Problem 20 Answer

Given

In the figure,CD&CB are opposite rays, CE bisects angle∠DCF and CG bisects angle∠FCB.

m∠FCG=9x+3,m∠GCB=13x−9.

To find∠GCB.

CG bisects angle∠FCB thus∠FCG=∠GCB.

Equate the above values and find the value of x.

Equating∠FCG=∠GCB

We get,

9x+3=13x−9

⇒4x=12

⇒x=12/4

⇒x=3.

Therefore,∠GCB=13x−9

=13×3−9

=39−9

=30.

​Hence we can conclude that the measure of angle∠GCB=30∘.

Page 8 Problem 21 Answer

Given

To measure and classify the angles∠1&∠2.

By measuring with a protractor the angle∠1=90∘and∠2=120∘.

Since∠1 is less than 90∘

it is an acute angle. Since∠2 is more than 90∘it is an obtuse angle.

Hence we can conclude that∠1=90∘and∠2=120∘and they are acute angle and obtuse angle respectively.

Geometry Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes

Page 5 Problem 1 Answer

Given

To find the length of LN.

The absolute value difference between the coordinates of the endpoints will give the length.

Let the coordinate of the endpointL=(l) and the coordinate of N=(n).

The length of​LN=∣l−n∣

=∣3−9∣

=∣−6∣

=6 units.

​Hence we can conclude that the length of LN is 6 units.

Page 5 Problem 2 Answer

Here given a number line

By using this number line we have to measure KN.

Now we will consider K as point x₁ and N as point x₂.

Then we will find the distance between K

And N by using the distance formula that is ∣x₂−x₁∣.

Given that a number line

Now we will consider K as point x₁ and N as point x₂ and to measure the distance we will use the formula of distance that is ∣x₂−x₁∣.

If we give a closer look at this number line we can say x₁ is on −2 and x₂ is on 9.

So, KN= ∣x₂−x₁∣

=∣9−(−2)∣Unit

=∣11∣Unit

=11Unit.

Hence, the length of KN on the number line is 11 units.

Page 5 Problem 3 Answer

Here given a number line

By using this number line we have to measure MN.

Now we will consider M as point x₁ and N as point x₂.

Then we will find the distance between M and N by using the distance formula that is∣x₂−x₁.

Given that a number line

Now we will consider M as point x₁ and N as point x₂ and to measure the distance we will use the formula of distance that is ∣x₂−x₁∣.

If we take a closer look at this number line we can say x₁ is on 6 x₂ is on 9.

So, MN = ∣x₂−x₁∣

= ∣6−9∣Unit

=  ∣−3∣Unit

= 3Unit.

Hence, the length of MN on the number line is 3 units.

Page 5 Problem 4 Answer

Given two points K(2,3) and F(4,4).

Let’s assume (x₁,y₁) and(x₂,y₂) be the coordinates of those points respectively.

Finally, we are going to use the formula∣√(x₂−x₁)²+(y₂−y₁)²∣to measure the distance between those points.

Given that two points are K(2,3) and F(4,4).

Now to calculate the distance between them from the formula that is∣√(x₂−x₁)²+(y₂−y₁)²∣

we will replace the value of x₁,y₁,x₂,y₂.

So KF=∣√(4−2)²+(4−3)²Unit

=∣√22+1/2∣Unit

=∣√5∣Unit

=∣2.23∣Unit

=2.23Unit.

Hence, the distance between the two points is 2.23Unit.

Page 5 Problem 5 Answer

Given two points C(−3,−1) and Q(−2,3).

Let’s assume (x₁,y₁) and (x₂,y₂) be the coordinates of those points respectively.

Finally, we are going to use the formula∣√(x₂−x₁)²+(y₂−y₁)²∣to measure the distance between those points.

Given that two points are C(−3,−1) and Q(−2,3).

Now to calculate the distance between them from the formula that is ∣√(x₂−x₁)²+(y₂−y₁)²∣

we will replace the value of x₁,y₁,x₂,y₂.

So, CQ= ∣√{(−2)−(−3)²+{3−(−1)}²∣Unit

= ∣√12+4/2∣Unit

= ∣√17∣Unit

= ∣4.12∣Unit

= 4.12Unit.

Hence, the distance between the two points is 4.12Unit.

Page 5 Problem 6 Answer

Given two pointsY(2,0) and P(2,6).

Let assume (x₁,y₁) and (x₂,y₂) be the coordinates of those points respectively.

Finally, we are going to use the formula∣√(x₂−x₁)²+(y₂−y₁)²∣to measure the distance between those points.

Given that two points are Y(2,0) and P(2,6).

Now to calculate the distance between them from the formula that is∣√(x₂−x₁)²+(y₂−y₁)²∣

we will replace the value of x₁,y₁,x₂,y₂.

So, YP=∣√(2−2)²+(6−0)²∣Unit

=∣√6/2∣Unit

=∣6∣Unit

=6Unit.

Hence, the distance between the two points is 6 units.

Page 5 Problem 7 Answer

Given two points W(−2,2) and R(5,2).

Let’s assume (x₁,y₁) and (x₂,y₂) be the coordinates of those points respectively.

Finally, we are going to use the formula∣√(x₂−x₁)²+(y₂−y₁)²| to measure the distance between those points.

Given that two points are W(−2,2) and R(5,2).

Now to calculate the distance between them from the formula that is∣√(x₂−x₁)²+(y₂−y₁)²∣

we will replace the value of x₁,y₁,x₂,y₂.

So,WR=∣√{5−(−2)}²+(2−2)²∣Unit

= ∣√72+0/2∣Unit

= ∣√49∣Unit

= ∣7∣Unit

= 7Unit

Hence, the distance between the two points is 7 units.

Page 5 Problem 8 Answer

Given two points A(−7,−3) and B(5,2). Let’s assume (x₁,y₁) and(x₂,y₂) be the coordinates of those points respectively.

Finally, we are going to use the formula∣(x₂−x₁)²+(y₂−y₁)²∣to measure the distance between those points.

Given that two points are A(−7,−3) and B(5,2).

Now to calculate the distance between them from the formula that is∣√(x₂−x₁)²+(y₂−y₁)²∣

we will replace the value of x₁,y₁,x₂,y₂.

So, AB=∣√{5−(−7)}²+{2−(−3)}²∣Unit

= ∣√122+5/2∣Unit

= ∣√144+25∣Unit

= ∣√169∣Unit

= ∣13∣Unit

= 13 Unit.

Hence, the distance between the two points is 13 Units.

Page 5 Problem 9 Answer

Given two points C(−3,1) and Q(2,6).

Let’s assume(x₁,y₁) and (x₂,y₂) be the coordinates of those points respectively.

Finally, we are going to use the formula∣√(x₂−x₁)²+(y₂−y₁)²∣to measure the distance between those points.

Given that two points are C(−3,1) and Q(2,6).

Now to calculate the distance between them from the formula that is∣√(x₂−x₁)²+(y₂−y₁)²∣

we will replace the value of x₁,y₁,x₂,y₂.

So CQ=∣√{2−(−3)}²+(6−1)²∣Unit

= ∣√52+5/2∣Unit

= ∣√25+25∣Unit

= ∣√50∣Unit

= ∣7.07∣Unit

= 7.07Unit.

Hence, the distance between the two points is 7.07 Units.

Page 5 Problem 10 Answer

Given

To find the coordinate of the midpoint of DE.

The midpoint of a line with end coordinates x₁ and x₂ is given by the formula \(\frac{x_1+y_1}{2}\)

The coordinate of the endpoint D is x₁=7.

The coordinate of the endpoint E is y₁=11.

Hence the coordinate of the midpoint of the line DE is​ \(\frac{x_1+y_1}{2}\)=\(\frac{7+11}{2}\)

=\(\frac{8}{2}\)

=9.​

Hence we can conclude that the coordinate of the midpoint of DE is 9.

Page 5 Problem 11 Answer

Given

To find the coordinate of the midpoint of BC.

The midpoint of a line with end coordinates x₁ and y₁ is given by the formula \(\frac{x_1+y_1}{2}\)

The coordinate of the endpoint B is −1.

The coordinate of the endpoint C is 3.

Hence the coordinate of the midpoint of line BC is

⇒ \(\frac{x_1+y_1}{2}\)

= \(\frac{-1+3}{2}\)

= \(\frac{2}{2}\)

=1.

​Hence we can conclude that the coordinate of the midpoint of BC is 1.

Page 5 Problem 12 Answer

Given

To find the coordinate of the midpoint of BD.

The midpoint of a line with end coordinates x₁ and y₁ is given by the formula \(\frac{x_1+y_1}{2}\)

The coordinate of the endpoint B is −1.

The coordinate of the endpoint D is 7.

Hence the coordinate of the midpoint of the line BD is​ \(\frac{x_1+y_1}{2}\)

= \(\frac{-1+7}{2}\)

= \(\frac{6}{2}\)

= 3.​

Hence we can conclude that the coordinate of the midpoint of BD is 3.

Page 5 Problem 13 Answer

Given

To find the coordinate of the midpoint of AD.

The midpoint of a line with end coordinates x₁ and y₁ is given by the formula \(\frac{x_1+y_1}{2}\)

The coordinate of the endpoint A is −4.

The coordinate of the endpoint D is 7.

Hence the coordinate of the midpoint of the line AD is \(\frac{x_1+y_1}{2}\)

= \(\frac{-4+7}{2}\)

= \(\frac{3}{2}\)

=1.5

Hence we can conclude that the coordinate of the midpoint of AD is 1.5.

Page 5 Problem 14 Answer

Given endpoints T(3,1) and U(5,3).

To find the coordinates of the midpoint of a segment with the given endpoints.

The midpoint of a line with end coordinates(x₁,y₁)and(x₂,y₂) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

The coordinates of the endpoint T is (3,1).

The coordinates of the endpoint U is (5,3).

Hence the coordinates of the midpoint of the line TU are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

= \(\frac{3+5}{2}\), \(\frac{1+3}{2}\)

= \(\frac{8}{2}\), \(\frac{4}{2}\)

=(4,2).​

Hence we can conclude that the coordinates of the midpoint of TU is (4,2).

Page 5 Problem 15 Answer

Given endpoints J(−4,2) and F(5,−2).

To find the coordinates of the midpoint of a segment with the given endpoints.

The midpoint of a line with end coordinates(x₁,y₁) and (x₂,y₂) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

The coordinates of the endpoint J are (−4,2).

The coordinates of the endpoint F is (5,−2).

Hence the coordinates of the midpoint of the line JF are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

= \(\frac{-4+5}{2}\), \(\frac{2-2}{2}\)

= \(\frac{1}{2},0\)

​Hence we can conclude that the coordinates of the midpoint of JF are \(\frac{1}{2},0\)

Page 5 Problem 16 Answer

Given endpoint N (2,0) and the midpoint of the line NQ is P(5,2).

To find the coordinates of the missing endpoint.

The midpoint of a line with end coordinates(x₁,y₁) and (x₂,y₂) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

Let the coordinates of Q be (x,y).

The coordinates of the endpoint N are (2,0).

The coordinates of the midpoint P is (5,2).

Hence the coordinates of the midpoint of the line NQ

are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) =(5,2)

⇒ \(\frac{2+x}{2}\), \(\frac{0+y}{2}\) =(5,2)

⇒ 2+x=10, y=4

⇒ x=8, y=4.​

Thus, the coordinates of Q are (8,4).

Hence we can conclude that the coordinates of the missing endpoint are(8,4).

Page 5 Problem 17 Answer

Given endpoint N(5,4) and the midpoint of the line NQ is P(6,3).

To find the coordinates of the missing endpoint.The midpoint of a line with end coordinates(x₁,y₁)

And (x₂,y₂) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

Let the coordinates of Q be (x,y).

The coordinates of the endpoint N are (5,4).

The coordinates of the midpoint P are (6,3).

Hence the coordinates of the midpoint of the line NQ

are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) =(6,3)

⇒ \(\frac{5+x}{2}\), \(\frac{4+y}{2}\) =(6,3)

⇒ 5+x=12,4+y=6

⇒ x=7,y=2.

​Thus, the coordinates of Q are (7,2).

Hence we can conclude that the coordinates of the missing endpoint are(7,2).

Page 5 Problem 18 Answer

Given endpoint Q(3,9) and the midpoint of the line NQ is P(−1,5).

To find the coordinates of the missing endpoint.

The midpoint of a line with end coordinates (x₁,y₁) and (x₂,y₂) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

Let the coordinates of N be(x,y).

The coordinates of the endpoint Q are (3,9).

The coordinates of the midpoint P are (−1,5).

Hence the coordinates of the midpoint of the line NQ are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)=(−1,5)

⇒ \(\frac{x+3}{2}\), \(\frac{y+9}{2}\) =(−1,5)

⇒ x+3=−2,y+9=10

⇒ x=−5,y=1.

​Thus, the coordinates of N are (−5,1).

Hence we can conclude that the coordinates of the missing endpoint are (−5,1).

Page 6 Problem 19 Answer

Given

To find the length of VW.

The absolute value difference between the coordinates of the endpoints will give the length.

Let the coordinate of the endpoint V=(v) and the coordinate of W=(w).

The length of ​VW=∣v−w∣

= ∣1−5∣

= ∣−4∣

= 4 units.

​Hence we can conclude that the length of VW is 4 units.

Page 6 Problem 20 Answer

Here given a number line By using this number line we have to measure TV.

Now we will consider T as point x₁ and V as point x₂.

Then we will find the distance between T and V by using the distance formula that is ∣x₂−x₁∣.

Given that a number line

Now we will consider T as point x₁ and V as point x₂ and to measure the distance we will use the formula of distance that is ∣x₂−x₁∣.

If we give a closer look at this number line we can say T is on −4 and V is on 1.

So, TV= ∣x₂−x₁∣

= ∣−4−1∣Unit

= ∣−5∣Unit

= 5Unit.

Hence the length of TV on the number line is 5 units.

Page 6 Problem 21 Answer

Here given a number line  By using this number line we have to measure ST

Now we will consider S as point x₁ and T as point x₂.

Then we will find the distance between S And T by using the distance formula that is ∣x₂−x₁∣.

Given that a number line

Now we will consider S as point x₁ and T as point x₂ and to measure the distance we will use the formula of distance that is ∣x₂−x₁∣.

If we give a closer look at this number line we can say x₁ is on−7 and x₂ is on−4.

So, ST= ∣x₂−x₁∣

= ∣−7−(−4)∣Unit

= ∣−3∣Unit

= 3Unit.

Hence the length of ST on the number line is 3 Units.

Page 6 Problem 22 Answer

Here given a number line By using this number line we have to measure SV.

Now we will consider S as point x₁ and V as point x₂.

Then we will find the distance between S and V

by using the distance formula that is ∣x₂−x₁∣.

Given that a number line

Now we will consider S as point x₁ and V as point x₂ and to measure the distance we will use the formula of distance that is ∣x₂−x₁∣.

If we give a closer look at this number line we can say S is on −7 and V is on 1.

So SV=∣x2−x1∣

= ∣−7−1∣Unit

= ∣−8∣Unit

= 8 Unit.

Hence the length of SV on the number line is 8Unit.

Page 6 Problem 23 Answer

Given that

By using this graph we have to measure the distance between two given points.

Now find the coordinates of those points and find out the distance between them using∣√(x₂−x₁)²+(y₂−y₁)²∣

where(x₁,y₁) and (x₂,y₂) are the coordinates of S and E respectively.

Given that

By using this graph we have to measure the distance between two given points.

From the graph the coordinates of S and E respectively (4,4) and (−3,−4).

Now we are going to use the formula ∣√(x₂−x₁)²+(y₂−y₁)²∣ to calculate the distance between each pair of points.

Putting the value of x₁,x₂,y₁,y₂

The distance of SE

GF= ∣√(−3−4)²+(−4−4)²∣Unit

= ∣√(−7)²+(−8)²∣Unit

= ∣√49+64∣Unit

= ∣√113∣Unit

= 10.630Unit

≈10.6Unit.

Hence, the distance between each pair of points is 10.6 Units.

Page 6 Problem 24 Answer

The given points are.

We need to find the distance between the two points.

The distance can be calculated as d=2√(x2−x1)²+(y2−y1)².

Using the formula d=2√(x2−x1)²+(y2−y1)²

we can say that,

d= 2 √(5+7)²+(9−0)²

= 2√(12)²+(9)²

= √144+81

= √225

= 15

The total distance is found to be 15 units.

Page 6 Problem 25 Answer

The given points are U(1,3),B(4,6). We need to find the distance between the two points.

The distance can be calculated as d = 2/√(x2−x1)²+(y2−y1)².

Using the formula d = 2/√(x2−x1)²+(y2−y1)²

we can say that,

d=2/√(4−1)²+(6−3)²

=2√32+32

=√9+9

=√18

=3/√2

The total distance is found to be 3/√2 units.

Page 6 Problem 26 Answer

The given points are V(−2,5),M(0,−4). We need to find the distance between the two points.

The distance can be calculated as d = 2/√(x2−x1)²+(y2−y1)².

Using the formula we can say that,

d = 2/√(x2−x1)²+(y2−y1)²

=2/√(0+2)²+(−4−5)²

=2/√(2)²+(−9)²

=2/√4+81

=√85

The total distance is found to be √85 units.

Page 6 Problem 27 Answer

The given points are C(−2,−1),K(8,3). We need to find the distance between the two points.

The distance can be calculated as d=2√(x2−x1)²+(y2−y1)².

Using the formula we can say that,

d=2√(x2−x1)²+(y2−y1)² or,d=2√(8+2)²+(3+1)²=2

√(10)²+(4)²=√100+16

=√116

=2/√29

The total distance is found to be 2/√29 units.

Page 6 Problem 28 Answer

The given line is RT.

We must find the endpoints of the line.

The midpoint can be found using the formula m= \(\frac{x_1+x_2}{2}\)

We can see that the line has endpoints as−2 and 4.

Thereby using the formula m=\(\frac{x_1+x_2}{2}\),

m= \(\frac{-2+4}{2}\)

m=\(\frac{2}{2}\)

m=1.

The midpoint is found to be (1,0).

Page 6 Problem 29 Answer

The given line is QR.

We must find the endpoints of the line.

The midpoint can be found using the formula m= \(\frac{x_1+x_2}{2}\)

We can see that the line has endpoints as−6and−2,

Thereby using the formula m=\(\frac{x_1+x_2}{2}\)

m= \(\frac{-6-2}{2}\)

= \(\frac{-8}{2}\)

=−4

The midpoint is found to be (−4,0).

Page 6 Problem 30 Answer

The given line is ST.

We must find the endpoints of the line.

The midpoint can be found using the formula m= x1+x2/2.

We can see that the line has endpoints 1 and 4.

Thereby using the formula m= \(\frac{x_1+x_2}{2}\)

m= \(\frac{1+4}{2}\)

= \(\frac{5}{2}\)

The midpoint is found to be \(\frac{5}{2},0\)

Page 6 Problem 31 Answer

The given line is PR.

We must find the endpoints of the line.

The midpoint can be found using the formula m = \(\frac{x_1+x_2}{2}\)

We can see that the line has endpoints as −9 and −2.

Thereby using the formula m = \(\frac{x_1+x_2}{2}\),m = \(\frac{-9-2}{2}\)

= \(\frac{-11}{2}\)

The midpoint is found to be \(\frac{-11}{2},0\)

Page 6 Problem 32 Answer

The given points are K(−9,3) and H(5,7).

We must find the endpoints of the line.

The midpoint can be found using the formula m= \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

Using the formula,

m = \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) or,m=(−9+5)

(2,3+\(\frac{7}{2}\)) or,m= \(\frac{-4}{2}\), \(\frac{10}{2}\)or,m=(−2,5)

The midpoint is found to be (−2,5).

Page 6 Problem 33 Answer

The given points are W(−12,−7);T(−8,−4).We must find the endpoints of the line.

The midpoint can be found using the formula m=\(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)

Using the formula,

m= \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) or,m=(−12−8/2,−7−4/2)

or,m= \(\frac{-20}{2}\)\(\frac{-11}{2}\)or,m=(−10,\(\frac{-11}{2}\))

The midpoint is found to be (−10,−11/2).

Page 6 Problem 34 Answer

The given points are R(−1,3),S(3,3),T(5,−1),U(−2,−1) and.

We must find the length of each side.

The perimeter can be found by summing up all the lengths of the sides.

The perimeter The length found is given below,

RS= √(3+1)²+(3−3)²

=√4²

=4

ST=√(5−3)²+(−1−3)²

=√2²+(−4)²

=√20

TU=√(5+2)²+(1−1)²

=√7²

=7

RU=√(−2+1)²+(−1−3)²

=√1+4²

=√17

Perimeter of quadrilateral RSTU=4+7+√20+√17

=19.59

The perimeter is found to be 19.59 sq. units.

Geometry Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes

Page 9 Problem 1 Answer

The given diagram is,

In this figure, You can use the corner of a piece of paper to see that ∠EKF and ∠GKH are each larger than a right angle.

Therefore, ∠EKF and ∠GKH are two obtuse vertical angles.

Page 9 Problem 2 Answer

The given diagram is,

In the figure, the angles “∠EKF & ∠FKG”, “∠FKG & ∠GKH”, “∠GKH & ∠HKE”, and “∠HKE & ∠EKF” are adjacent to each other after the intersection of the two lines and The sum of angles of a linear pair is always equal to 180°.

Therefore, the linear pair are “∠EKF & ∠FKG”, “∠FKG & ∠GKH”, “∠GKH & ∠HKE”, and “∠HKE & ∠EKF”.

Page 9 Problem 3 Answer

The given diagram is,

In the figure, the angles are  “∠EKH & ∠HKJ”, and “∠HKJ & ∠JKG” are less than 90° and a common vertex and side KH & KJ, but have no common interior points.

Therefore, two acute adjacent angles are “∠EKH & ∠HKJ”, and “∠HKJ & ∠JKG”.

Page 9 Problem 4 Answer

The given diagram is,

In the figure, the sum of the angles ∠FKG & ∠GKH form a straight angle.Again, the sum of the angles ∠FKG & ∠FKE forms a straight angle.

Therefore, the supplementary angles of ∠FKG are ∠GKH and ∠FKE.

Page 9 Problem 5 Answer

Let the measures of an angle is x°. Then the other measure of angle which is complement to x°, is(x−24)°.  [∵ one angle measures 24 degrees more than the other]So, (x)°+(x+24)°=90°

Or,  2x° +24=90°

Or,  2x°=90°−24°=66°

Or, x°=66°/2=33°

Therefore, the measure of an angle is 33° and its complement if one angle measures 24 degrees more than the other is (90−33)°=57°.

Page 9 Problem 6 Answer

Let the measures of an angle is x°.

Then the other measure of angle which is a supplement to x°, is(x−36)°.[∵the supplement of an angle is 36 less than the measure of the angle.]

So, (x°)+(x−36)°=180°

Or, 2x°−36°=180°

Or, 2x°=180°+36°=216°

Or, x°=216/2°=108°

Therefore, the measure of an angle is108° and its supplement of an angle is 36 less than the measure of the angle is (108°−36°)=72°.

Page 9 Problem 7 Answer

The given diagram is,

In the figure, TR⃗⊥TS⃗,

Then the angle between TR⃗&TS⃗=m∠RTS=90°.

So, by the problem, 8x+18=90°

Or, 8x=90°−18°=72°

Or, x=72/8=9.

Therefore, the value of x is 9.

Page 9 Problem 8 Answer

The given diagram is,

In the figure, it is clear that the angle ∠WZU forms an angle 90°, that means right angle.

Yes, ∠WZU  is a right angle.

Page 9 Problem 9 Answer

The given diagram is,

In the figure, it is clear that∠YZU &∠UZV  forms a straight angle when they are put together.

That means∠YZV=180°.

Yes, ∠YUZ & ∠UZV are supplementary.

Page 10 Problem 10 Answer

The given diagram is,

In the picture, ∠GFH & ∠CFE angles that are both vertical and obtuse, which means both angles are less than a right angle but less than a straight angle and both angles are lying on opposite sides of two intersecting lines.

Therefore, two obtuse vertical angles are ∠GFH & ∠CFE.

Page 10 Problem 11 Answer

The given diagram is,

In the figure, there are only two angles with vertex B that is ∠ABC & ∠GBC And we can see that, ∠ABC +∠GBC=180°, So these are the linear pair with vertex B.

Therefore, ∠ABC & ∠GBC is a linear pair with vertex B.

Page 10 Problem 12 Answer

The given diagram is,

In the figure, the angle ∠FED=50° and ∠FGH=50°.Then, ∠CGF=90°−50°=40°[∵∠CGH=90°].

∠FED is not an adjacent point to∠FGC because there is no common vertex, But complementary to ∠FGC Because ∠FGC+∠FED=50°+40°=90°.

Therefore, ∠FED is not adjacent to, but complementary to ∠FGC.

Page 10 Problem 13 Answer

We need to find an angle adjacent and supplementary to∠DCB.

This can be found by using the definition of adjacent angles and supplementary angles

A figure is given

We know that the angles on a straight line add up to180o.

∠DCB and∠DCF lie on the straight line BCF.

Hence, ∠DCB+∠DCF=180o

∠DCB and∠DCF also lie beside each other.

Hence, ∠DCF is the angle adjacent and supplementary to∠DCB.

∠DCF is an angle adjacent and supplementary to∠DCB.

Page 10 Problem 14 Answer

We are given two angles that are complementary.

It is also given that the measure of one angle is 21 more than twice the measure of the other angle.

We need to find the measures of these angles.

Consider the required angles as variables and construct a mathematical equation with the definition of complementary angles.

Let,x and y be the values of the two angles.

It is given that the measure of one angle is 21 more than twice the measure of the other angle.

Hence, we get,

y=21+2x……………………(1)

We know that complementary angles add up to 90o.

It is given that x and y are complementary to each other.

Therefore,

x+y=90o

⇒ x+(21+2x)=90o

[From (1), y=21+x]

⇒ x+21+2x=90o

⇒ 21+3x=90o

⇒ 3x=69

⇒ x=69/3

⇒ x=23

From (1), y=21+2x

⇒ y=21+2(23)

⇒ y=67

The measurements of the required angles are 34.5o and 55.5o.

Page 10 Problem 15 Answer

It is given that the supplement of an angle has a measure78 less than the measure of the angle.

We need to find the measure of the angle.

This can be found by considering the required angles as variables and constructing a mathematical equation with the definition of supplementary angles.

Let,x and y be the values of two angles.

It is given that the supplement of an angle has a measure78 less than the measure of the angle.

Hence, y=x−78………….(1)

The two angles are a supplement to each other.

Therefore,

x+y=180o

⇒ x+(x−78)=180

[From (1), y=x−78]

⇒ x+x−78=180

⇒ 2x−78=180

⇒ 2x=258

⇒ x=258/2

⇒ x=129

From (1), y=x−78

⇒ y=129−78

⇒ y=51

The measures of the required angles are 129o and 51o.

Page 10 Problem 16 Answer

It is also given that m∠BGC=16x−4 and m∠CGD=2x+13.

We need to find the value of x, so that∠BGD is a right angle.

This can be found by adding up the given angle expressions to 90o.

A figure is given

From the given figure, we can say that ∠BGC+∠CGD=∠BGD

m∠BGC=16x−4,m∠CGD=2x+13

It is said that∠BGD is a right angle which is equal to 90o.

Hence, m∠BGC+m∠CGD=90o

⇒(16x−4)+(2x+13)=90

⇒16x−4+2x+13=90

⇒18x+9=90

⇒18x=81

⇒ x=81/18

⇒ x=4.5

The value of x so that∠BGD is a right angle is 2.4.

Page 10 Problem 17 Answer

A figure is given,

We need to explain the correctness of the given statement.

This can be explained on the basis of the statement that two angles are said to be complementary when they add up to 90o.

We see in the figure that ∠NQP is a right angle, i.e, ∠NQP=90o…………………(1)

We see that ∠NQP is made up of two angles, namely ∠NQO and ∠OQP, i.e, ∠NQP=∠NQO+∠OQP………………..(2)

From (1) and (2), we get,∠NQO+∠OQP=90o

Hence, these two angles are complimentary as they add up to give 90o.

The given statement can be assumed from the figure given.

Page 10 Problem 18 Answer

A figure is given,

We need to explain the correctness of the given statement.

This can be explained with the help of the statement that vertical angles refer to each of the pairs of opposite angles made by two intersecting lines.

We see that the lines NR and MP intersect each other at point Q.

The angles ∠MQN,∠MQR lie on the line NR.

Hence, ∠MQN+∠MQR=180o and they are a pair of opposite angles.

Therefore, these two angles are vertical angles

∠MQN and∠MQR are vertical angles.

Page 10 Problem 19 Answer

It is given that Darren sketched a map of the cross streets nearest to his home for his friend Miguel.

We need to describe two different angle relationships between the streets.

This can be found by using the angle definitions and the fact that the angles on a straight line add up to 180o.

A figure is given

We see that Olive and Main add up to give 90o.

We see that Bacon and Main are perpendicular to each other.

The two different angle relationships between the streets are

Olive and Main are complementary and

Bacon and Main are perpendicular to each other.

Geometry Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes

Page 13 Problem 1 Answer

The given shape is

We must determine the figure and which category it belongs to.

If all the sides are straight line then the figure is stated to be polyhedron.

The given figure is

Clearly, we can observe that the figure is a cuboid which is nothing but a three-dimensional projection of a rectangle.

As a result, the shape is a polyhedron in nature.

The following bases, edges, and vertices are as follows:

1 Vertex-

2 Edges-

The shape is proved as polyhedron and it is identified as cuboid with :

1 Vertex- U, T, S, R, Y, X, W, V.

2 Edges-UT, TS, SR, RU, SX, XY, YR, YV, VW, WX, WT, VU.

Page 13 Problem 2 Answer

The given shape is

We must determine the figure and which category it belongs to.

If all the sides are straight line then the figure is stated to be polyhedrons.

The given figure is

Clearly, we can observe that the figure is a pentagonal pyramid which is nothing but a three-dimensional projection of a pentagon extended to a pyramid.

As a result, the shape is a polyhedron in nature.

The following bases, edges, and vertices are as follows:

1 Vertices-

2 Edges-

3 Bases-

The shape is proved as polyhedron and it is identified as a pentagonal pyramid with :

1 Vertices-A, B, C, D, E, F.

2 Edges-FA, FB, FC, FD, FE.

3 Bases-AB, BC, CD, DE, EA.

Page 13 Problem 3 Answer

The given figure is

We must find the slant height and radius in order to find the surface area.

After we find the slant height, we can calculate the surface area.

Given, l=6;r=3

Therefore,s=πr(l+r)

⇒ s=3.14×3×9

⇒ s=84.78 inches

The surface area is found to be 84.78≈84.8inches

Page 13 Problem 4 Answer

The given figure is

We must find the height and radius in order to find the surface area.

After we find the height and radius, we can calculate the surface area.

Given,r=3;h=8

Therefore,s=2πr(h+r)

⇒ s=2×3.14×3×11

⇒ s=207.24

The surface area is found to be 207.24≈and 207.2cm.

Page 13 Problem 5 Answer

The given figure is

We must find the slant height and radius in order to find the surface area.

After we find the slant height, we can calculate the surface area.

Given, b = 4; h = 5

∴ s = b(b+4h)​

⇒ s=4(4+20)

⇒ s=4×24

⇒ s=96m

The surface area is found to be 96m.

Page 13 Problem 6 Answer

Given

To find the volume of the solid.

The given solid is a cuboid and its volume is calculated by the formula Length(l)×Width(w)×Height(h).

Length(l) = 8yd.

Width(w) = 5yd.

Height(h) = 6yd.

The volume of the cuboid is Length(l) × Width(w) × Height(h) = 8 × 5 × 6= 240yd .

Hence we can conclude that the volume of the cuboid is 240y d³.

Page 13 Problem 7 Answer

Given

To find the volume of the solid.

The given diagram is of a cylinder and its volume is calculated by the formula V=Π×r²×h.

(where are the radius and height of the cylinder respectively)

Radius(r) = 2cm.

Height(h) = 10 cm.

The volume of the cylinder is Π × r × h = Π × 2 × 10 2 2

= 3.14 × 40

= 125.6cm.

Hence we can conclude that the volume of the cylinder is 125.6 cm².

Page 14 Problem 8 Answer

Given

To determine whether the solid is a polyhedron.

Then identify the solid.

This solid is a trapezoid.

A polyhedron is a 3D shape that has flat faces, straight edges, and sharp vertices or corners.

The trapezoid has flat faces, straight edges, and sharp vertices; hence it is a polyhedron.

There are 8 vertices of the trapezoid and they are H,I,J,K,L,M,N,O.

There are 6 faces of the trapezoid and they are KLOH,​JMNI,​JKHI,​MLON,​ONIH,​JKLM.

There are 12 edges of the trapezoid and they are, JK, KH, HI, OH, NI, KL, JM, LO, LM, MN, NO.

Hence we can conclude that the given solid is a polyhedron. The polyhedron is a trapezoid.

Page 14 Problem 9 Answer

Given

To determine whether the solid is a polyhedron.

Then identify the solid.

This solid is a cylinder.

A polyhedron is a 3D shape that has flat faces, straight edges, and sharp vertices or corners.

A cylinder has a curved surface hence it is not a polyhedron.

Hence we can finally conclude that the given solid is not a polyhedron. The given solid is a cylinder.

Page 14 Problem 10 Answer

Here given a cone that is

From the given solid we have to calculate the surface area and volume of this cone.

To calculate the Surface Area we will use the formula of surface area that is Surface Area=πrl+πr² and to calculate the Volume the formula is Volume=1/3πr²h wherer=radius,l=Slant Height,h=height.

Given that from the above cone of the diagram, we can say radius(r)=9in.

And height(h)=12in.

So, we have to calculate the slant height of the cone by using the Pythagoras theorem

Slant Height(l)=√r²+h²

=√9²+122in.

=√81+144in.

=√225in.

=15in.

Now to calculate the surface area of the cone we will replace the value of the radius, height, and slant height in the formula that is Surface Area=πrl+πr².

So, Surface Area=(π×9×15+π×92)in².

=216 πin²

=678.58 in²

And the volume of the cone will be Volume=1/3πr²h. Replacing the value of radius and height we will get

Volume=(1/3×π×92×12)in³.

=1017.876in³.

Hence, the surface area of the cone is 678.58 in 2 and the volume of the cone is 1017.876 in 3.

Page 14 Problem 11 Answer

Given that,

Here the given solid is a cube and we have to calculate the Surface Area and Volume of the cuboid solid.

To calculate the Surface Area and Volume at first, we will calculate the perimeter of any side of the cube here the formula is Perimeter=2×(length+breadth) and the area of any side of the cube is Base = length × breadth Now the formula of Surface Area is Surface Area=Perimeter×height+2×Base and the formula of volume that is Volume=Base×Height.

Here given a cuboid-shaped solid

Now if we take a closer look at the diagram we can say that length=17m,breadth=15m,height=16m.

So if we replace the value of length and breadth in the formula that is Perimeter=2×(length+breadth) to calculate the perimeter of one side of the cube we will get

Perimeter=2×(17+15)m

=64m

And to calculate the area of one side the formula isBase=length×breadth

Base=(17×15)m²

=255m²

Now the perimeter base and height will be replaced by their value in the formula of Surface Area=Perimeter×height+2×Base

Surface Area=(64×16+2×255)m²

=1534m²

Replace base and height by their value in the formula Volume=Base×Height to calculate the volume of the cuboid and we will get

Volume=(255×16)m³

=4080m³.

Hence, the surface area of the cubic solid is 1534m^2, and the volume of the cubic solid is 4080m³.

Page 14 Problem 12 Answer

Given that,

Here the given solid is a cylinder and we have to calculate the Surface Area and volume of the solid.

To calculate the surface Area

we will use2πrh+2πr² and forVolume

we will useπr²h where r is the radius and h is the height of the solid cylinder.

Here given a cylindrical solid

If we give a closer look at the solid we can say the radius (r) is 5cm and height(h) is 7cm.

Now if we replace the value of r and h to calculate the surface area of the solid in the formula surface area(S)=2πrh+2πr²

we will get

S=(2×π×5×7+2×π×52)cm²

=376.99cm²

≈377.0cm²

And if we replace the value of r and h to calculate the volume of the solid in the formula Volume(V)=πr²h

we will get

V=(π×52×7)cm³

=549.77cm³

≈549.8cm³.

Hence, the surface area and volume of the cylindrical solid are 377.0 cm² and 549.8 cm³ respectively.

Page 14 Problem 13 Answer

Given that, boxes to hold a stack of 8.5inch by 11inch papers and volume of the box to be 500 Cubic inches.

Now we have to find out the height of the boxes.

The box is in cuboid shape so the box’s volume will be length × breadth × height.

Given that, boxes to hold a stack of 8.5inch by11 inch papers.

so the length of the box is 8.5 inches and the breadth of the box is 11 inches.

Now given that the volume of the box is 500 Cubic inches.

So to calculate the height of the cuboid, we will replace the above information in the formula Volume of the Cuboid=length×breadth×height

500=8.5×11×height or,height=8.5×11 /500inch

=5.34759358inch

≈5.35inch.

Hence, the height of the box should be 5.35 inches.

Geometry Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes

Page 11 Problem 1 Answer

The given diagram is,

The polygon has 3 sides and a polygon with 3 sides is a triangle.

None of the lines containing the sides will have points in the interior of the polygon.

So, the polygon is convex.

All sides and all angles are equal in the polygon, so it is regular.

The polygon is a triangle, convex and regular.

Page 11 Problem 2 Answer

The given diagram is,

The polygon has 5 sides and a polygon with 5 sides is a pentagon.

Some of the lines containing the sides will have points in the interior of the polygon.

So, the polygon is concave.

Some angles and sides are not equal to each other in the polygon, hence it is irregular.

Therefore, the polygon is a pentagon. And it is concave and irregular.

Page 11 Problem 3 Answer

The given diagram is,

To find the perimeter of the figure, we have to add all lengths of the sides.

So, the perimeter of the figure =(18+20+20+40)yd.=98 yd.

Therefore, the perimeter of the figure is 98 yd.

Page 11 Problem 4 Answer

The given diagram is,

To find the perimeter of the figure, we have to add all lengths of the sides.

So, the perimeter of the figure =(4+6+3+5+2)m.=20m.

Therefore, the perimeter of the figure is 20 m.

Page 11 Problem 5 Answer

The given diagram is,

To find the circumference of the figure, we have to use the formula 2Πr where r is the radius of the figure.

So the circumference of the figure =2×Π×11 m.

= 22×Π m.

Therefore, the circumference of the figure is 22×Π≈69.115.

Page 11 Problem 6 Answer

The given diagram is,

The area of the circle =Π×r²sq.cm.

= Π×72 sq. cm.

= 49Π sq. cm.

The area of the figure is=49Π sq. cm.

≈153.93 sq. cm.

Page 11 Problem 7 Answer

The given diagram is,

In the figure, the height of the triangle is 14 ft.

and the base of the triangle is 16 ft.

The area of the triangle is =1/2×(height of the triangle)×(base of the triangle) square unit.

=1/2×14ft. × 16ft.

=7×16ft. 2

=112ft. 2

Therefore, the area of a triangle is 112 ft.2

Page 11 Problem 8 Answer

Four vertices are given, which are Q(−3,2),R(1,2),S(1,−4),T(−3,−4).

We need to graph each figure with the given vertices and identify the figure, and finally find the perimeter and area of the figure.

This can be found by plotting the points on a graph paper and joining them to find the area and perimeter of the figure formed.

The points have been plotted on the graph and then the points have been joined.

The graph looks like as follows

Here, the x-axis denotes the values of the x coordinate and the y-axis denotes the values of the y-coordinate.

The figure formed is a rectangle.

Length of the rectangle=y2−y1

= 2−(−4)

= 2+4

= 6

Breadth of the rectangle=x2−x1

=1−(−3)

=1+3

= 4

Hence, the perimeter of the rectangle will be,2×(length+width)

= 2×(6+4)

= 2×10

= 20units

Hence, the area of the rectangle will be, length × width

= 6×4

= 24unit²

The figure formed by joining the points is a rectangle. Its perimeter is 20 units and its area is 24 unit².

Page 11 Problem 9 Answer

We are given 3 different points.

We need to find the figure formed by joining the points given and finally find the area and perimeter of the figure.

This can be formed by plotting the given points on a graph paper and finding the area and perimeter of the figure formed by joining the points on the graph paper.

The given points are plotted on a graph paper and then joined to get the figure.

The graph is as follows.

The x-axis denotes the values of the x-coordinates and the y-axis denotes the values of the y-coordinate.

The figure formed is a triangle.

Length of the base of the triangle=4−(−4)=8units

Length of the height of the triangle=1−(−2)=3units

Length of the two sides of the triangle=√42+32,√42+32

(Applying Pythagoras theorem)=5units,5units

Hence, the area of the triangle=1/2×base×height

=1/2×8×3

=12unit²

Perimeter of the triangle=5+5+8=18units

The figure formed is a triangle.

The area of the triangle is 12 units 2 and its perimeter is 18 units.

Page 11 Problem 10 Answer

We are given 4 different points.

We need to find the figure formed by joining the points given and finally find the area and perimeter of the figure.

This can be formed by plotting the given points on a graph paper and finding the area and perimeter of the figure formed by joining the points on the graph paper.

The given points are plotted on a graph paper and then joined to get the figure.

The graph is as follows.

The x-axis denotes the values of the x-coordinates and the y-axis denotes the values of the y-coordinates.

The figure formed is a parallelogram.

The figure formed is a parallelogram.

Page 12 Problem 11 Answer

A figure is given. We need to find that whether the given polygon is concave or convex or regular or irregular.

This can be found by considering the definitions of convex, concave, regular, irregular polygons and comparing that with the figure given.

The given polygon has an angle greater than180o. Hence, it is a concave polygon.

In the given polygon, all the sides or angles are not equal to each other. Hence, it is an irregular polygon.

The given polygon is an irregular, concave polygon.

Page 12 Problem 12 Answer

A figure is given. We need to find the area and the perimeter of the given figure.

This can be found by using the formula to calculate the area and circumference of a circle.

The radius of the circle(r)=7mi

Circumference of the circle=2Πr

= 2×3.14×7

= 43.96mi

= 44mi (rounded to nearest tenth)

Area of the circle=Πr²

= 3.14×72

=153.86mi²

=153.9mi²(rounded to nearest tenth)

The perimeter of the circle is 44mi and the area of the circle is 153.9 mi².

Page 12 Problem 13 Answer

A figure is given. We need to find the perimeter and area of the given figure.

This can be found by using the formula to calculate the area and perimeter of a triangle

The length of two sides of the triangle=8.1mm

Length of the base of the triangle=8mm

Height of the triangle=7mm

Perimeter of the triangle=(8.1+8.1+8)

= 24.2mm

=20mm (rounded to nearest tenth)

Area of the triangle=1/2×8×7

= 28mm²

= 30mm² (rounded to nearest tenth)

The perimeter of the triangle is 20mm and the area of the triangle is 30 mm2.

Page 12 Problem 14 Answer

We are given 3 different points.

We need to find the figure formed by joining the points given and finally find the area and perimeter of the figure.

This can be formed by plotting the given points on a graph paper and finding the area and perimeter of the figure formed by joining the points on the graph paper.

The given points are plotted on a graph paper and then joined to get the figure.

The graph is as follows.

The base of the triangle=8−0=8units

Height of the triangle=0−(−2)=2units

The left side of the triangle =√22+32

= √4+9

= √13

=3.6units​

The right side of the triangle=√52+22

=√25+4

=√29

=5.38units​

The perimeter of the triangle =8+3.6+5.38

=16.98units​

Area of the triangle=1/2×base×height

=1/2×8×2

=8unit2

The figure formed is a triangle.

The perimeter of the triangle is 16.98 units and the area of the triangle is 8 unit 2.

Page 12 Problem 15 Answer

The given rectangle is

We have to find find the relation between their perimeters if their dimensions are doubled.

We can find the relation between them using the ratio.

Initial PerimeterP=2(17+4) feets

Final Perimeterp=2(34+8) feets

∴P/p=2(17+4)/2(34+8)

⇒P/p

=2(17+4)/4(17+4)

⇒P/p

=1/2

⇒p=2P

We can clearly observe that if the dimensions are doubled then the perimeter is also doubled.

Page 12 Problem 16 Answer

The given rectangle is

We have to find find the relation between their areas if their dimensions are doubled.

We can find the relation between them using the ratio.

Initial area A=17×4 sq.feets

Final area a=34×8 sq.feets

∴A/a=17×4/34×8

⇒A/a=17×4/2(17×4)

⇒A/a=1/2

⇒a=2A

We can clearly observe that if the dimensions are doubled then the area is also doubled.

Page 12 Problem 17 Answer

The given figure is

As jasmine wants to sew around the pillow, hence we must find the perimeter.

The perimeter can be calculated using C=2πr.

The perimeter is found to be

C=2πr

⇒C=2π(5)

⇒C=2×3.14×5

⇒C=31.4 inches

The amount of fringe Jasmine needs to use is 31.4 inches.

Page 12 Problem 18 Answer

The new radius of the circle(r) is 2×5 in.=10 in.

Then the new area =Π×r² sq. in.

=Π×(10)²sq. in.

= 100×Π sq.in.

Therefore, the new area of the pillow is 100Π sq. in. ≈ 314.159 sq. in.

Geometry Homework Practice Workbook 1st Edition Chapter 2 Inductive Reasoning and Conjecture

Page 29 Problem 1 Answer

We are given a figure, in which m∠5=22 and one angle measures 90.

We are required to find the measure of m∠6.

Here, we will use the fact that total sum of angles that lie on a line is 180.

As all the angles lie on the same line, by angles on a straight line property we have,

90+m∠5+m∠6=180

90+22+m∠6=180

m∠6=180−112

m∠6=68

​In the given figure, by angles on a straight line property, we have m∠6=68 when m∠5=22.

Page 29 Problem 2 Answer

Here in the question, we have been given a figure in which m∠1=38∘.

We have to find the measure of each angle given in the figure.

In the figure, we can see that we only have two angles ∠1,∠2.

And by the vertical opposite angle m∠1=m∠2.Implies that m∠2=38∘.

Hence from the given figure and m∠1=38∘, by the vertically opposite angles, we get m∠2 as 38∘.

Page 29 Problem 3 Answer

In the question, we have been given a figure and also we have:

m∠13=4x+11

m∠14=3x+1

​We have been asked to find the measure of each numbered angle and name the theorems that justify our work.

Using the linear pair theorem, we will find the result.

Here in the given figure, by the linear pair, we get

∠13+∠14=180∘

4x+11+3x+1=180∘

7x=180−12

x=168/7

x=24∘

Now substituting the value of x=24∘ in the angles, we get

m∠13=4×24+11

=107∘m∠14=3×24+1

=73∘

So, by linear pair theorem, we get the measure of each angle.

Hence we get the measure of each angle by the linear pair in the given figuer as:

m∠13=107∘

m∠14=73∘​

Page 29 Problem 4 Answer

We have been given that in the given figure, ∠9,∠10 are complementary angles, ∠7 congruent to ∠9, m∠8=41.

We have to find the measure of each angle.

Using the linear pair theorem, we will get the measure.

By the linear pair, we get

∠7+∠8+∠9+∠10=180

∠7+41+90=180

∠7=49

Now given that ∠7≅∠9.

Thus we get 49∘=∠9.

Also, we have given that ∠9,∠10 are complementary angles.

Thus we get

∠9+∠10=90∘

49+∠10=90∘

∠10=41∘

So, we get the measure of all the angles.

Hence we get the measure of all the angles in the figure as: ∠7=49∘

by linear pair ∠10=41∘

by complement ∠9=49∘by congruence

Page 29 Problem 5 Answer

In the question, we have been given a figure and also given that ∠QPS≅∠TPR and also an incomplete table.

We have to complete the table with the proof of ∠QPR≅∠TPS.

Using the given information and figure we will complete it.

Here the complete table is:

Hence with the help of the given figure, we had completed the table and proved that ∠QPR≅∠TPS

Page 30 Problem 6 Answer

Here, we have given a figure in which a straight line is divided by a line into two angles, that are ∠1,∠2.Further, we have given

m∠1=x+10

m∠2=3x+18.

Thus we just have to find the value of m∠1,m∠2.

As we have given

m∠1=x+10……(1)

m∠2=3x+18……(2)

So we also know according to the property of straight line,m∠1+m∠2=180∘

i.e.x+10+3x+18=180∘

4x+28=180∘

4x=180−28

x=152/4

We get

x=38∘

Now putting x in eq. 1 and 2, we will have

m∠1=x+10

=38+10

=48∘and

m∠2=3x+18

=3⋅38+18

=114+18

=132∘

Hence by solving

m∠1=x+10

m∠2=3x+18​

by using the given Fig.1 we get m∠1=48∘

m∠2=132∘.

Page 30 Problem 7 Answer

Here, we have given a figure in which we have given three angles, that are ∠3,∠4,∠5 .Further, we can say we have given

m∠3=90∘

m∠4+m∠5=90∘.

Thus we just have to find the value of m∠4,m∠5.

As we have given ​m∠4=2x−5……(1)

m∠5=4x−13……(2)​

And from the given figure we also have m∠4+m∠5=90∘

Thus we will have

2x−5+4x−13=90∘

6x−18=90

6x=108

x=18∘

Now we will just put the value of x in eq. 1 and 2 respectively,

we will have

m∠4=2x−5

=2⋅18−5

=36−5

=31∘and

m∠5=4x−13

=4⋅18−13

=59∘

Hence by solving

m∠4=2x−5

m∠5=4x−13

​by using the given Fig.1.

we get m∠4=31∘

m∠5=59∘

Page 30 Problem 8 Answer

Here, we have given

And we have to find the unknown value of the angles m∠6 and m∠7 by using the given figure.

As we know that a theorem states that the two opposite angles are equal when two straight lines intersect, forming four angles.

So now we can write m∠6=m∠7

That implies

7x−24=5x+14

7x−5x=14+24

2x=38

x=19∘

Thus putting x in eq. 1 and 2, we will have

m∠6=7x−24

=7⋅19−24

=133−24

=109∘and

m∠7=5x+14

=5⋅19+14

=95+14

=109∘

Hence by solving m∠6=7x−24

m∠7=5x+14​ by using the given Fig.1.we get m∠6=m∠7=109∘.

Page 30 Problem 9 Answer

Here we have given straight lines referred to as the road names.

So as the figure says, Barton rode and Olive tree lane is making 90∘angle where Tryon street and Olive tree lane is making 57∘angle.

So as we have been told, we just need to find the acute angle made by Tryon Street with Barton Road.

As we know that

The angle between Barton Road and Olive tree lane+The angle between Barton Road and Tyron St.+

Tyron St. and Olive tree lane=180∘

⇒90∘+The angle between Barton Road and Tyron St.+57∘

⇒The angle between Barton Road and Tyron St.=180∘−90∘−57∘

=33∘

So the measure of the acute angle Tryon Street forms with Barton Road will be

= 90∘+The angle between Barton Road and Tyron St.

=90∘+33∘

=123∘

Hence by using the given Fig.1, we have measured the acute angle Tryon Street forms with Barton Road is 123∘.

Geometry Homework Practice Workbook 1st Edition Chapter 2 Inductive Reasoning and Conjecture

Page 27 Problem 1 Answer

We have been given QA=QA in the question.

Here, we are asked to find the property of the given comparison.

Here, we have seen that both side parameter is same along with equal sign.

Therefore, we can say that it follows the reflexive property of equality which defines that a number or variable is equal to itself.

Hence, according to the given comparison QA=QA we have found that it follows the reflexive property of equality.

Page 27 Problem 2 Answer

We have been given AB≅BC, BC≅CE,AB≅CE in the question.

Here, we are asked to find the property among these comparisons.

Now, we have seen that all three sides are congruent to each other.

Finally, we will conclude that the given statement follows the transitive property of congruency which tells about similar shape and size.

Hence, using the given comparison AB≅BC,BC≅CE,AB≅CE we got that it followed the transitive property of congruency.

Page 27 Problem 3 Answer

We have been given PR=PQ+QR in the question.

We are asked to determine the property of the given comparison.

Here, we have seen that there are three points P,Q,R and Q is the midpoint.

Now, after evaluating the given comparison we can say that it follows the segment addition postulate which tells about the point on the line.

Hence, according to the given statement PR=PQ+QR we have found that it follows the segment addition postulate.

Page 27 Problem 4 Answer

We have been given AB+BC=EF+FG

AB+BC=AC

EF+FG=AC

​Here, we are asked to find the property of the given statement in the question.

Now, we have seen that there is a comparison between the sum of two points with other one or two points.

Now, we can say that it follows substitution and transitive property which tells about replacing the values and similarity.

Hence, according to the given statements

AB+BC=EF+FG

AB+BC=AC

EF+FG=AC

We have got that this statement follows substitution and transitive property.

Page 27 Problem 5 Answer

We have been given two statements SU≅LR

TU≅LN  in the question.

Here, we will prove the congruency of the given statement ST≅LR

Finally, using the given points and mathematical concepts we will conclude the final result.

We have SU≅LR

TU≅LN

Here, we have seen that T, and N are the midpoints of the lines SU,LR

Now, use each mathematical concept to prove ST≅LR

Using the given information

Hence, using the given statement

SU≅LR

TU≅LN

We prove that ST≅NR

Page 27 Problem 6 Answer

We are given a partially filled proof for the congruence of CD≅AB.

And it is given that, AB≅CD.

We are required to complete the proof.

Here, we will use properties of congruence to fill this.

We will compare the reasons and the statements to each other and then fill the given proof as,

We can fill the given proof as

Page 27 Problem 7 Answer

We are given that,AB≅DE

B is the midpoint of AC

E is the midpoint of DF

We are required to prove that, BC≅EF.

Here, we will use properties of congruence and equality to complete the given proof.

We will compare the reasons and the statements to each other and then fill the given proof as,

The given proof can be completed as,

Page 28 Problem 8 Answer

We are given a figure, in which the distance from Grays on to Apex is the same as the distance from Redding to Pine Bluff.

That is, GA≅RP.

We are required to prove that the distance from Grays on to Redding is equal to the distance from Apex to Pine Bluff. GR≅AP.

Here, we will use properties of equality to prove this.

The given statement can be proved as,

The proof of the distance from Grayson to Redding is equal to the distance from Apex to Pine Bluff is given by,

 

Geometry Homework Practice Workbook 1st Edition Chapter 2 Inductive Reasoning and Conjecture

Page 25 Problem 1 Answer

In this question, we have been given the value: 80=m∠A

We need to state the property that justifies the statement.

By using Inductive Reasoning and Conjecture, we will calculate the result.

As we see this statement uses the symmetric property.

The symmetric property of equality states that regardless of which side of an equal sign they are on, both sides are equal.

Hence, the given statement 80=m∠A, then 80=m∠A uses the symmetric property.

It states that regardless of which side of an equal sign they are on, both sides are equal.

Page 25 Problem 2 Answer

In this question, we have been given the value:

RS=TU

TU=YP

RS=YP

We need to state the property that justifies the statement.

By using Inductive Reasoning and Conjecture, we will calculate the result.

As we see this statement uses the transitive property of equality.

The transitive property of equality states that if a is related to b and b is related to c, then a is related to c.

Hence, the given statement RS=TU and TU=YP, then RS=YP uses the transitive property of equality. It states that if a is related to b and b is related to c, then a is related to c.

Page 25 Problem 3 Answer

In this question, we have been given the value:

7x=28

x=4

We need to state the property that justifies the statement.

By using Inductive Reasoning and Conjecture, we will calculate the result.

As we see this statement uses the division property of equality.

It states that when both sides of an equation are divided by the same non-zero number, the two sides remain equal, according to the division property of equality.

Hence, the given statement 7x=28,  then x=4 uses the division property of equality.

It states that when both sides of an equation are divided by the same non-zero number, the two sides remain equal, according to the division property of equality.

Page 25 Problem 4 Answer

In this question, we have been given the value:

VR+TY=EN+TY

VR=EN​

We need to state the property that justifies the statement.

By using Inductive Reasoning and Conjecture, we will calculate the result.

As we see this statement uses the subtraction property of equality.

The subtraction property of equality states that if one side of an equation is subtracted, the other side must likewise be subtracted to maintain the equation the same.

Hence, the given statement VR+TY=EN+TY,  then VR=EN uses the subtraction property of equality.

It states that if one side of an equation is subtracted, the other side must likewise be subtracted to maintain the equation the same.

Page 25 Problem 5 Answer

We are given m∠1=30 & m∠1=m∠2

​We have to justify the statement that if the above two equations are true then m∠2=30

We will be using some algebraic properties to justify the required statement.

We are given m∠1=30&

m∠1=m∠2​

By using transitive property we can say that m∠2=30

Thus, the transitive property is shown by the statement if ​m∠1=30&

m∠1=m∠2 then m∠2=30

Page 25 Problem 6 Answer

We are given a table as shown

Also, we are given an equation 8x−5=2x+1

We have to complete the table and prove that x=1

We will be using some algebraic properties to complete the table and prove the required result

We are given a table and an equation8x−5=2x+1

After completing the table we get,

Here, the highlighted portion in the table represent the blanks given in the table of question

Thus, we have completed the table and proved x=1 for the given equation 8x−5=2x+1

Page 25 Problem 7 Answer

We are given a figure and PQˉ≅QSˉ&QSˉ=STˉ

We have to prove PQ=ST

We will be using some algebraic properties and congruent lines concept to prove the same.

We are given a figure

Two-column proof to verify the conjecture is

Hence, we have written a two-column proof to verify the conjecture “if PQˉ≅QSˉ&QSˉ≅STˉ then

PQ=ST”, for the given figure

The two-column proof is

Page 26 Problem 8 Answer

In this question, we have been given the value:

m∠ABC+m∠CBD=90

m∠ABC=3x−5

m∠CBD=x+1/2

We need to write a two-column proof to verify each conjecture.

By using Inductive Reasoning and Conjecture, we will calculate the result.

We analyze the diagram

Since the ∠ABC and m∠CBD are complementary angles.

We substitute the values

m∠ABC+m∠CBD=90

3x−5+x+1/2=90

6x−10+x+1=90×2

7x−9=180

Further solving, we get

7x=189

x=189/7

x=27

​Hence, the value of x=27.

Since the ∠ABC and ∠CBD are complementary angles.

We substitute the values and analyze the diagram

Page 26 Problem 9 Answer

In this question, we have been given the value: I=prt

We need to solve the formula for r and justify each step.

By using Inductive Reasoning and Conjecture, we will calculate the result.

We divide the whole formula by pt to solve for r

I/p×t=p×r×t/p×t

I/p×t=r

​Hence, the formula for r is r=I/p×t. Since we divide the whole formula by pt and using the formula of simple interest.