a. The sum of all angles in a triangle is 180$text{textdegree}$.

$$
angle ITR=angle NGA=180text{textdegree}-90text{textdegree}-28text{textdegree}=62text{textdegree}
$$

Since all three pair of angles are congruent and two pairs of sides are congruent, the third pair of sides also have to be equivalent.

The rigid transformations needed will be a rotation and a translation.

b.
$$
triangle IRT cong triangle NAG
$$

There are multiple correct congruence statements, by reordering the letters in one triangle name and then applying the same reordering to the other triangle name.

a. Rotation and translation

b. $triangle IRT cong triangle NAG$

a. $angle M$ is in between the side of length 15 and the side of length 6, the same is true for $angle U$ and thus $angle M$ is congruent with $angle U$.

$angle N$ is in between the side of length 15 and the side of length 12, the same is true for $angle W$ and thus $angle N$ is congruent with $angle W$.

$angle P$ is in between the side of length 12 and the side of length 6, the same is true for $angle V$ and thus $angle P$ is congruent with $angle V$.

The congruence statement then becomes:

$$
triangle MNPcong triangle UWV
$$

b. A rotation of 180$text{textdegree}$ about $P$ and a translation is sufficient.

a. $triangle MNPcong triangle UWV$

b. A rotation of 180$text{textdegree}$ about $P$ and a translation is sufficient.

Yes, because if you rotate $triangle NLR$ 90$text{textdegree}$ counterclockwise, reflect about $NR$ and then translate it to the right, then you will obtain $triangle BAC$.

$$
triangle NLRcong triangle BAC
$$

$$
triangle NLRcong triangle BAC
$$

a.
$$
SAS: triangle ABC cong triangle FED
$$

You would need a rotation and a translation to map one triangle on the other.

b. Not congruent, because the indicated angle in the first triangle lies between the known sides, but not in the second triangle.

c. Using the vertical angles and alternate interior angles are equivalent:
$$
ASA: triangle MNXcong triangle ZYX
$$

You would need a rotation about $X$ at 180$text{textdegree}$ counterclockwise.

d.
$$
RHS: triangle ACD cong triangle ABD
$$

You would need a reflection about $AD$.

e. Using alternate interior angles are equivalent:
$$
AAS: triangle EFHcong triangle GHF
$$

You would need a rotation and a translation to map one triangle on the other.

f. Not congruent, because the indicated angle in the first triangle lies between the known sides, but not in the second triangle.

a. The two triangles are congruent, because they have two angles of the same measure and the included sides are equal in length.
$$
ASA: triangle ABDcong triangle CBD
$$

b. The two triangles are congruent, because they have two pairs of sides of the same lengths and the included angles have the same measure.

$$
SAS: triangle DBCcong triangle ACB
$$

c. Vertical pairs and alternate interior angles are equivalent, which implies that we have three pairs of equivalent angles in the two triangles. However, the side $CD$ in $triangle CDE$ corresponds with side $AC$ in $triangle ABC$, which are not necessarily equally long (similarly, side $BC$ and side $CE$ are not necessarily equally long) and thus we cannot determine whether the two triangles are congruent.

d. The two triangles are congruent, because they have two angles of the same measure and a pair of non-included corresponding sides of equal length.

$$
AAS: triangle ADBcong triangle BCA
$$

e. Cannot be determined, because we do not know if there is a pair of corresponding sides that are congruent.

f. The two triangles are congruent, because they have three pairs of sides of the same lengths.

$$
SSS: triangle SQRcong triangle HGK
$$

a. Yes, the slope of the segments is given by $dfrac{y_2-y_1}{x_2-x_1}$

$$
Slope AB: dfrac{6-4}{1+3}=dfrac{1}{2}
$$

$$
Slope BC: dfrac{-2-6}{5-1}=dfrac{-8}{4}=-2
$$

$$
Slope CD: dfrac{-4+2}{1-5}=dfrac{1}{2}
$$

$$
Slope DA: dfrac{4+4}{-3-1}=-2
$$

Since parallel lines have the same slope and perpendicular lines have slopes whose product is $-1$, we then know that the shape is a rectangle.

b. The coordinates are $(4,3),(6,-1),(-2-5),(-4,-1)$ (interchange the $x$- and $y$-values and change the new $y$-values of sign).

a. Yes

b. $(4,3),(6,-1),(-2-5),(-4,-1)$

The area of a trapezium is the sum of the bases multiplied by the height and divided by 2:

$$
AREA=dfrac{(5+8)4}{2}=26
$$

Determine the length of the left side using the Pythagorean theorem:

$$
sqrt{1^2+4^2}=sqrt{1+16}=sqrt{17}
$$

The perimeter is the sum of all sides:

$$
PERIMETER=sqrt{17}+5+8+sqrt{20}=13+sqrt{17}+sqrt{20}
$$

Area: 26

Perimeter: $13+sqrt{17}+sqrt{20}$

The length of the third side can be at most the sum of the lengths of the other sides and is at least the difference of the lengths of the other sides:

$$
Maximum: 8+5=13ft
$$

$$
Minimum: 8-5=3ft
$$

Thus the length of the third side is between 3 ft and 13 ft.

a. Yes, because of SSS.

b.
$$
AB=DF
$$

$$
BC=DE
$$

$$
AC=EF
$$

$$
Downarrow SSS
$$

$$
triangle ABCcong triangle FDE
$$

$$
PQ=ST
$$

$$
mangle P = mangle T=120text{textdegree}
$$

$$
mangle QRP = mangle SRTtext{ (vertical angles) }
$$

$$
Downarrow AAS
$$

$$
triangle PQRcong triangle TSR
$$

Use $AAS$

Every pair of corresponding sides and angles are congruent:

$$
UV=XY
$$

$$
VW=YZ
$$

$$
UW=XZ
$$

$$
angle Ucong angle X
$$

$$
angle Vcong angle Y
$$

$$
angle Wcong angle Z
$$

Use $AAS$

a.
$$
AC=CA
$$

$$
mangle B = mangle D
$$

$$
mangle BAC= mangle DCAtext{ (alternate interior angles) }
$$

$$
Downarrow AAS
$$

$$
triangle ABCcong triangle CDA
$$

b. Not congruent, because the triangles have three pair of congruent angles, while a pair of corresponding sides do not have an equal length.

c. Not congruent, because the legs of one right triangle have lengths 2 and 3, while the legs of the other triangle have lengths 3 and 6.

d.
$$
MP=QR=13
$$

$$
KL=PR=12 text{ Using pythagorean theorem}
$$

$$
mangle KLM= mangle QRP=90text{textdegree}
$$

$$
Downarrow SAS
$$

$$
triangle KLMcong triangle RPQ
$$

a.
$$
SE=ME
$$

$$
SY=MY
$$

$$
YE=YE
$$

$$
Downarrow SSS
$$

$$
triangle SYEcong triangle MYE
$$

$$
Downarrow
$$

$$
angle Scong angle M
$$

b. Yes, if you take the bissectrice of the angle in between the two equal sides (which divides the angle into two equal angles), you obtain a line such as $YE$ and then you can prove it in the same way.

a. The area of a trapezium is the sum of the bases multiplied by the height divided by 2:

$$
dfrac{(8-1-1+4)3}{2}=dfrac{30}{2}=15
$$

The area of a rectangle is the product of the length and the width of the rectangle:

$$
8cdot 2=16
$$

Thus the total area is then the sum of the subareas:

$$
15+16=31
$$

b. The area of a triangle is the product of the base and the height divided by 2:

$$
dfrac{5cdot 3}{2}=7.5
$$

$$
dfrac{4cdot 2}{2}=4
$$

The area of a rectangle is the product of the length and the width of the rectangle:

$$
6cdot 5=30
$$

Thus the total area is then the sum of the subareas:

$$
7.5+4+30=41.5
$$

a. Group like terms:

$$
34x-10x=-8+80
$$

Simplify;

$$
24x=72
$$

Divide both sides of the equation by 24:

$$
x=3
$$

b. Group like terms:

$$
4x-4x=10+5
$$

Simplify

$$
0x=15
$$

This is not possible, since zero cannot equal 15 and thus the equation has no solutions.

c. Use distributive property:

$$
3x-15+6x+2=41
$$

Group like terms:

$$
3x+6x=41+15-2
$$

Simplify:

$$
9x=54
$$

Divide both sides of the equation by 9:

$$
x=6
$$

d. Use distributive property:

$$
-2x-8+6=-3
$$

Group like terms:

$$
-2x=-3+8-6
$$

Simplify:

$$
-2x=-1
$$

Divide both sides of the equation by $-2$:

$$
x=dfrac{1}{2}
$$

a. $x=3$

b. No solutions

c. $x=6$

d. $x=dfrac{1}{2}$

The upper number of the diamond contains the product of the left and right number, while the lower number contains the sum of the left and right number.

a. We need to find two numbers whose product is 15 and whose sum is 8, this is true for 5 and 3. Thus 5 and 3 need to be filled in the left and right cells (order is not important).

b. Upper: $3cdot 1=3$ and Lower: $3+1=4$

c. Right: $dfrac{10}{5}=2$ and Lower: $5+2=7$

d. Upper: $dfrac{1}{2}cdot dfrac{1}{3}=dfrac{1}{6}$ and Lower: $dfrac{1}{2}+dfrac{1}{3}=dfrac{3}{6}+dfrac{2}{6}=dfrac{5}{6}$

a. 5 and 3

b. 3 and 4

c. 2 and 7

d. $frac{1}{6}$ and $frac{5}{6}$

a. According to the corresponding angles theorem:

$$
5x+7=9x-63
$$

Group like terms:

$$
5x-9x=-63-7
$$

Simplify:

$$
-4x=-70
$$

Divide both sides of the equation by $-4$:

$$
x=17.5
$$

Determine the angles:

$$
5x+7=5(17.5)+7=94.5text{textdegree}
$$

$$
9x-63=9(17.5)-63=94.5text{textdegree}
$$

b. According to the alternate interior angles theorem:

$$
12x-14=5x+21
$$

Group like terms:

$$
12x-5x=21+14
$$

Simplify:

$$
7x=35
$$

Divide both sides of the equation by $7$:

$$
x=5
$$

Determine the angles:

$$
12x-14=12(5)-14=46text{textdegree}
$$

$$
5x+21=5(5)+21=46text{textdegree}
$$

$$
20x+34=20(5)+34=134text{textdegree}
$$

a. $x=17.5text{textdegree}$, Both angles are 94.5$text{textdegree}$

b. $x=5text{textdegree}$, Angles are 46$text{textdegree}$, 46$text{textdegree}$, and 134$text{textdegree}$

The converse interchanges the statement after if with the statement after then.

a. Yes, this arrow is true because the statement after the arrow is a property of congruent triangles.

b. All pairs of corresponding sides are congruent $Rightarrow$ Triangles congruent

The converse is also true (SSS).

c. Yes, this arrow is true because the statement after the arrow is a property of congruent triangles.

d. All pairs of corresponding angles are congruent $Rightarrow$ Triangles congruent

The converse is not true because we do not know if the two triangles have at least one pair of corresponding sides that are congruent.

e. The area of a polygon is $b cdot h$ $Rightarrow$ the polygon is a rectangle

The converse is not true, because the area of a parallelogram is also $bcdot h$ while a parallelogram is not a rectangle.

The converse interchanges the statement after if with the statement after then.

a. If the ground is wet, then it rains. FALSE (you could have sprayed water on the ground yourself)

b. If a polygon is a rectangle, then the polygon is a square. FALSE A rectangle does not have four equal sides.

c. If a polygon hs four 90$text{textdegree}$ angles, then the polygon is a rectangle. FALSE The polygon could have 5 angles.

d. If a polygon is a triangle, then it has three angles. TRUE

e. If vertical angles are congruent, then two lines intersect. TRUE, this follows from the definition of vertical angles.

a. The sum of all angles in a triangle is 180$text{textdegree}$:

$$
180text{textdegree}-75text{textdegree}-35text{textdegree}=70text{textdegree}
$$

Supplementary angles:

$$
x=180text{textdegree}-70text{textdegree}=110text{textdegree}
$$

b. Supplementary angles:

$$
180text{textdegree}-140text{textdegree}=40text{textdegree}
$$

The sum of all angles in a triangle is 180$text{textdegree}$ and the base angles of an isosceles triangle are equal:

$$
2x=180text{textdegree}-40text{textdegree}=140text{textdegree}
$$

Thus:

$$
x=70text{textdegree}
$$

c. Supplementary angles:

$$
180text{textdegree}-148text{textdegree}=32text{textdegree}
$$

The sum of all angles in a triangle is 180$text{textdegree}$:

$$
x=180text{textdegree}-32text{textdegree}-100text{textdegree}=48text{textdegree}
$$

d. The sum of all angles in a triangle is 180$text{textdegree}$ and the base angles of an isosceles triangle are equal:

$$
180text{textdegree}-36text{textdegree}=144text{textdegree}
$$

Thus:

$$
dfrac{144}{2}text{textdegree}=72text{textdegree}
$$

Supplementary angles:

$$
x=180text{textdegree}-72text{textdegree}=108text{textdegree}
$$

a. 110$text{textdegree}$

b. 70$text{textdegree}$

c. 48$text{textdegree}$

d. 108$text{textdegree}$

a. The sum of the measures of the remote interior angles is equal to the mesuare of the exterior angle.

b.
$$
a+b=x
$$

c. $c$ and $x$ are supplmentary angles:

$$
x=180text{textdegree} -c
$$

The sum of all angles in a triangle is equal to $180text{textdegree}$:

$$
c=180text{textdegree}-a-b
$$

Then we obtain:

$$
x=180text{textdegree} -180text{textdegree}+a+b=a+b
$$

d. The measure of the exterior angle is equal to the sum of the measure of the remote interior angles.

a. The sum of the measures of the remote interior angles is equal to the mesuare of the exterior angle.

b. $a+b=x$

c. Use that $c$ and $x$ are supplementary angles and thus the sum of all angles in a triangle is equal to $180text{textdegree}$

d. The measure of the exterior angle is equal to the sum of the measure of the remote interior angles.

a. The solution is the intersection of of the two graphs: $(-2,3)$

b. Multiply the first equation by 2:

$$
2y=-2x+2
$$

$$
y=2x+7
$$

Add the two equations:

$$
3y=9
$$

Divide both sides of the equation by 3:

$$
y=3
$$

Determine $x$:

$$
3=-x+1
$$

Add $x$ to both sides of the equation:

$$
3+x=1
$$

Subtract three from both sides of the equation:

$$
x=-2
$$

Thus the solution is $(-2,3)$.

$$
(-2,3)
$$

a. Using the Pythagorean theorem we find that the hypotenuse is:

$$
sqrt{8^2+6^2}=sqrt{100}=10
$$

Then by SAS we know that the two triangles are congruent.

b. Congruent because of AAS.

c. Not congruent, because the pair of congruent angles does not lie in between the two pairs of corresponding congruent sides.

d. Congruent, because of SAS.

The upper number of the diamond contains the product of the left and right number, while the lower number contains the sum of the left and right number.

a. Upper: $(-4)cdot (-7)=28$ and Lower: $-4-7=-11$

b. We need to find two numbers whose product is $-12$ and whose sum is 4, this is true for 6 and $-2$. Thus 6 and $-2$ need to be filled in the left and right cells (order is not important).

c. Right: $dfrac{-8}{1/2}=-16$ and Lower: $dfrac{1}{2}-16=-dfrac{31}{2}$

d. Upper: $dfrac{1}{2}cdot dfrac{1}{5}=dfrac{1}{10}$ and Lower: $dfrac{1}{2}+dfrac{1}{5}=dfrac{5}{10}+dfrac{2}{10}=dfrac{7}{10}$

a. 28, $-11$

b. 6, $-2$

c. $-16$, $-frac{31}{2}$

d. $frac{1}{10}$, $frac{7}{10}$

The converse interchanges the statement after if with the statement after then.

a. The lines that are crossed by a transversal are parallel$Rightarrow x$ and $y$ are supplementary.

b. $x$ and $y$ are supplementary $Rightarrow$ The lines that are crossed by a transversal are parallel

This is true.

c. Assume that same-side interior angles are supplementary but that the lines are not parallel. If the lines are not parallel, then there exists a point where the two lines intersect, let us call this point A. But then we have created a triangle with angles $x$ and $y$ and $angle A$. Since the sum of the angles in a triangle is 180$text{textdegree}$:

$$
x+y+angle A =180text{textdegree}
$$

Since $x$ and $y$ are supplementary:

$$
180text{textdegree}+angle A=180text{textdegree}
$$

Subtract 180$text{textdegree}$ from both sides of the equation:

$$
angle A=0text{textdegree}
$$

But the angle of the triangle cannot be zero, thus we have obtained a contradiction and thus the lines must be parallel.

a. Supplementary

b. Yes

c. Derive $angle A=0text{textdegree}$

a. Because we assumed that the lines were not parallel and thus they have to intersect.

b. The sum of the angles in a triangle is 180$text{textdegree}$:

$$
mangle BED=180text{textdegree}-100text{textdegree}-80text{textdegree}=0text{textdegree}
$$

c. This example does not exist, because an angle of a triangle cannot be zero.

a. We assume that the lines are not parallel

b. $mangle BED=0text{textdegree}$

c. This example doesn’t exist

a. Supplementary angles have measures that sum up to 180$text{textdegree}$:

$$
mangle YBA=180text{textdegree}-x
$$

b. Assume that $overleftrightarrow{BY}$ and $overleftrightarrow{AZ}$ are not parallel.

c.
$$
mangle BAZ=x
$$

$$
mangle YBA=180text{textdegree}-x
$$

$$
overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{ are not parallel}
$$

$$
Downarrow
$$

$$
overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{intersect in a point } C
$$

$$
Downarrow
$$

$$
ABC is a triangle
$$

$$
Downarrow
$$

$$
mangle BAZ+mangle YBA+mangle C=180text{textdegree}
$$

$$
Downarrow
$$

$$
180text{textdegree}+mangle C=180text{textdegree}
$$

$$
Downarrow
$$

$$
mangle C=0text{textdegree}
$$

$$
CONTRADICTIONRightarrow overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{ are parallel}
$$

a. $mangle YBA=180text{textdegree}-x$

b. Assume that $overleftrightarrow{BY}$ and $overleftrightarrow{AZ}$ are not parallel.

c. Assume that the lines are not parallel and derive a contradiction.

d. If the same-side interior angles of two lines with a transversal are supplementary, then the lines are parallel.

a. Corresponding angles are equal

$$
overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{ are not parallel}
$$

$$
Downarrow
$$

$$
overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{intersect in a point } C
$$

$$
Downarrow
$$

$$
ABCtext{ is a triangle}
$$

$$
Downarrow
$$

$$
mangle BAZ+mangle YBA+mangle C=180text{textdegree}
$$

$$
Downarrow
$$

$$
180text{textdegree}+mangle C=180text{textdegree}
$$

$$
Downarrow
$$

$$
mangle C=0text{textdegree}
$$

$$
CONTRADICTIONRightarrow overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{ are parallel}
$$

b. Alternate interior angles are equal

$$
overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{ are not parallel}
$$

$$
Downarrow
$$

$$
overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{intersect in a point } C
$$

$$
Downarrow
$$

$$
ABC is a triangle
$$

$$
Downarrow
$$

$$
mangle BAZ+mangle YBA+mangle C=180text{textdegree}
$$

$$
Downarrow
$$

$$
180text{textdegree}+mangle C=180text{textdegree}
$$

$$
Downarrow
$$

$$
mangle C=0text{textdegree}
$$

$$
CONTRADICTIONRightarrow overleftrightarrow{BY}text{ and }overleftrightarrow{AZ}text{ are parallel}
$$

No, two parallel lines do not have a point of intersection.

For example, $y=1$ and $y=2$ are parallel lines and they cannot have a point of intersection because 1 is never equal to 2.

a. Use cross multiplication:

$$
14(3)=(5)x
$$

Simplify:

$$
42=5x
$$

Divide both sides of the equation by 5:

$$
8.4=x
$$

b. Use cross multiplication:

$$
10(11)=(5)m
$$

Simplify:

$$
110=5m
$$

Divide both sides of the equation by 5:

$$
22=m
$$

c. Use cross multiplication:

$$
(t-2)(8)=(7)12
$$

Simplify:

$$
8t-16=84
$$

Add 16 to both sides of the equation:

$$
8t=100
$$

Divide both sides of the equation by 8:

$$
t=12.5
$$

Use cross multiplication:

$$
3(x+1)=(5)x
$$

Simplify:

$$
3x+3=5x
$$

Subtract $3x$ from both sides of the equation:

$$
3=2x
$$

Divide both sides of the equation by 2:

$$
1.5=x
$$

a. $x=8.4$

b. $m=22$

c. $t=12.5$

d. $x=1.5$

a.
$$
triangle DEFcong triangle LJK
$$

Because $70text{textdegree}+60text{textdegree}+50text{textdegree}=180text{textdegree}$ and $DE=JL$ which are the sides between the angles of 70$text{textdegree}$ and of 50$text{textdegree}$.

b. A reflection, rotation and translation will be needed.

Note that the figure is made up out of a rectangle and a triangle.

The area of a rectangle is the product of the length and the width:

$$
4cdot 3=12
$$

The area of a triangle is product of the base and the height divided by 2:

$$
h=sqrt{10^2-6^2}=sqrt{64}=8 text{Pythagorean identity}
$$

$$
dfrac{bcdot h}{2}=dfrac{12cdot 8}{2}=48
$$

The total area is then the sum of the subareas:

$$
A=12+48=60
$$

The perimeter is the sum of the lengths of all sides:

$$
P=10+10+4+3+4+3+4=38
$$

Area $60$

Perimeter 38

The sum of all angles in a triangle is 180$text{textdegree}$:

$$
mangle ADB=180text{textdegree}-11text{textdegree}-90text{textdegree}=79text{textdegree}
$$

$$
mangle BDC=180text{textdegree}-90text{textdegree}-68text{textdegree}=22text{textdegree}
$$

Since longer sides are opposite larger angles, we then know that $overline{AB}$ is longer than $overline{BC}$.

$$
overline{AB}
$$

a.
$$
{ab, ac, ad, ae, bc, bd, be, cd, ce, de}
$$

We did not forget any songs because we worked alphabetically.

b. Yes, every combination is equally likely and this is important because else the songs would not be chosen randomly.

c. We note that the songs $a$, $b$, and $d$ all have the word Mama in its title. Moreover, 3 of the 10 pairs of songs in part (a) contain two letters out of $a$, $b$, and $d$ (that is, the pairs of songs $ab$, $ad$, and $bd$).

$$
dfrac{3}{10}=0.3=30%
$$

d. We note that the songs $a$, $b$, and $d$ all have the word Mama in its title. Moreover, 9 of the 10 pairs of songs in part (a) contain either $a$, $b$, or $d$ (since the only pair of songs not containing one of these letters is $ce$).

$$
dfrac{9}{10}=0.9=90%
$$

e. Because the probability of (d) includes the probability of (c), because if at least one song contains mama, then either both songs contain mama or exactly one of the songs contains mama.

a. ${ab, ac, ad, ae, bc, bd, be, cd, ce, de}$

b. Yes

c. $dfrac{3}{10}=0.3=30%$

d. $dfrac{9}{10}=0.9=90%$

e. Because the probability of (d) includes the probability of (c)

a. Yes, because the second triangle can be obtained by dilating the first triangle by factor 2 (and a reflection, translation and rotation).

b. $D$ has to correspond with $C$; $O$ has to correspond to $T$ and $G$ has to correspond with $A$. Thus (ii) is the correct statement.

a. Determine the side lengths using the distance formula: $sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$$
AB=sqrt{(3-0)^2+(4-0)^2}=sqrt{25}=5
$$

$$
BC=sqrt{(3-3)^2+(4-0)^2}=sqrt{16}=4
$$

$$
AC=sqrt{(3-0)^2+(0-0)^2}=sqrt{9}=3
$$

b. The side lengths of the enlarged triangle is the product of the dilation factor and the side lengths of the original triangle.

$$
A’B’=2cdot 5=10
$$

$$
B’C’=2cdot 4=8
$$

$$
A’C’=2cdot 3=6
$$

c. The perimeter is the sum of all sides:

$$
10+8+6=24
$$

The area of a triangle is the product of the base and the height, divided by 2:

$$
AREA=dfrac{8cdot 6}{2}=24
$$

a. If a rectangle has base $x$ and height $2x$, then the area of the rectangle is $2x^2$. TRUE, because the area of a rectangle is the product of the base and the height.

b. If a rectangle has base $x$ and height $3y$, then the perimeter of the rectangle is $3xy$. FALSE, because the perimeter of a rectangle isthe sum of all sides: $x+x+3y+3y=2x+6y$.

Thus the correct statement is then: If a rectangle has base $x$ and height $3y$, then the perimeter of the rectangle is $2x+6y$.

c. If a rectangle has base $2$ft and height $3$ft, then the area of the rectangle is $864$ square inches. TRUE, because the area of a rectangle is the product of the base and the height.

The area of a triangle is the product of the base and the height, divided by 2:

$$
dfrac{(10cdot h)}{2}=25
$$

Multiply both sides of the equatoin by 2:

$$
10h=50
$$

Divide both sides of the equation by 10:

$$
h=5
$$

Determine the sides of the triangle using the Pythagorean theorem:

$$
sqrt{h^2+6^2}=sqrt{25+36}=sqrt{61}
$$

$$
sqrt{h^2+4^2}=sqrt{25+16}=sqrt{41}
$$

The perimeter is then the sum of all sides:

$$
6+4+sqrt{61}+sqrt{41}=10+sqrt{61}+sqrt{41}
$$

$h=5$ft

Perimeter=$10+sqrt{61}+sqrt{41}$ft

$$
AC=DF
$$

$$
mangle A=mangle D
$$

$$
AB=DE
$$

$$
Downarrow SAS
$$

$$
triangle ABC cong triangle DEF
$$

Use $SAS$

a. The total area is the sum of the subareas and is also the product of the length and the width.

$$
(x+5)(x+3)=x^2+5x+3x+15=x^2+8x+15
$$

b. The total area is the sum of the subareas and is also the product of the length and the width.

$$
(x+5)(2x+3)=2x^2+10x+3x+15=2x^2+13x+15
$$

a. $(x+5)(x+3)=x^2+8x+15$

b. $(x+5)(2x+3)=2x^2+13x+15$

a. The triangle is an isosceles triangle since it has two equal side lengths. The base angles of an isosceles triangles are equal. The sum of all angles in a triangle is 180$text{textdegree}$:

$$
2x+x+x=180text{textdegree}
$$

Combine like terms:

$$
4x=180text{textdegree}
$$

Divide both sides of the equation by 4:

$$
x=45text{textdegree}
$$

b. The triangle is an isosceles triangle since it has two equal side lengths. The base angles of an isosceles triangles are equal. Using supplementary angles we then obtain:

$$
y=180text{textdegree}-71text{textdegree}=109text{textdegree}
$$

a. $x=45text{textdegree}$

b. $y=109text{textdegree}$

a. $overline{C’B’}$ corresponds with $overline{CB}$

$overline{A’B’}$ corresponds with $overline{AB}$

b. $C’B’=2CB$ and $A’B’=AB$, thus the lengths of the sides are multiplied by the dilation factor.

c. Because the point about which you dilated is $A$ and it lies on $overline{AB}$ and $overline{AC}$, but not on $overline{BC}$.

d. No, because you would need to add the length of the original sides itself, which is different for each side.

e. 4 times because $AB=5$, the length of the other side is then the product of the dilation factor 4 and the original length:

$$
B”C”=4cdot 4=16
$$

a. $overline{C’B’}$ corresponds with $overline{CB}$

$overline{A’B’}$ corresponds with $overline{AB}$

b. $C’B’=2CB$ and $A’B’=AB$

c. Because the point about which you dilated is $A$ and it lies on $overline{AB}$ and $overline{AC}$, but not on $overline{BC}$.

d. No

e. $B”C”=16$

a. The scale factor is the ratio of the corresponding sides:
$$
dfrac{24}{8}=dfrac{18}{6}=3
$$

b. Yes, the ratios are both equal to $dfrac{4}{3}$ (see part (a)).

c. We note that the base has been multiplied by 12, then we need to multiply the height by the same factor:

$$
x=8cdot 12=96
$$

d. We note that the height has been multiplied by $dfrac{1}{4}$ (to reduce the height of 8 to the height of 2) and thus the scale factor is $dfrac{1}{4}$.

e. If the scale factor is 1, then the figure remains the same and we will thus obtain two congruent figures.

a. 3

b. Yes

c. $x=96$ cm

d. $frac{1}{4}$

e. Congruent figures

a. Because the two triangle have three pairs of congruent angles.

b.
$$
dfrac{4}{10}=dfrac{x}{x+9}
$$

Use cross multiplication:

$$
4x+36=10x
$$

Subtract $4x$ from both sides of the equation:
$$
36=6x
$$

Divide both sides of the equation by 6:

$$
6=x
$$

c.
$$
dfrac{4}{x}=dfrac{10}{x+9}
$$

Use cross multiplication:

$$
4x+36=10x
$$

Subtract $4x$ from both sides of the equation:
$$
36=6x
$$

Divide both sides of the equation by 6:

$$
6=x
$$

d. $AC=x+9=6+9=15$ and $AC’=x=6$

e. $dfrac{4}{10}=dfrac{2}{5}$ (see (b))

f. Parallel lines

a. Three pairs of congruent angles

b. $x=6$

c. $x=6$

d. 15, 6

e. $frac{2}{5}$

f. Parallel lines

No, because the size of the angle was not latered, only the size of the legs and thus the angle remains 60$text{textdegree}$ (Also note that we have an acute angle while 120$text{textdegree}$ is an obtuse angle).

$ABCD$ is the original figure, and the dilated figure then is $A’B’C’D’$.

a. The sides with length 6 and 15 correspond, because they lie between the same corresponding pairs of angles. In the same manner, the sides with length 20 and 8 correspond:

$$
dfrac{15}{6}=dfrac{20}{8}=2.5
$$

b. Use the Pythagorean theorem:

$$
sqrt{15^2+20^2}=sqrt{625}=25
$$

$$
sqrt{6^2+8^2}=sqrt{100}=10
$$

The ratio is then:

$$
dfrac{25}{10}=2.5
$$

Thus we note that the ratio remains unchanged.

c. You will need a dilation, rotation, and translation.

a. The longest side cannot be more than the sum of the other two sides of a triangle.

b. Using the Pythagorean theorem:

$$
5^2+12^2=169=13^2neq 14^2
$$

Thus the hypotenuse should be 13 in stead of 14.

c. Since two sides of the triangle are equal, the base angles are also equal. Summing up all angles we then obtain:

$$
65text{textdegree}+65text{textdegree}+55text{textdegree}=185text{textdegree}neq 180text{textdegree}
$$

Thus this is not possible, because the sum of all angles in a triangle should be equal to 180$text{textdegree}$.

a. The longest side cannot be more than the sum of the other two sides of a triangle.

b. Hypotenuse should be 13 instead of 14

c. The sum of all angles in a triangle should be equal to 180$text{textdegree}$.

The upper number of the diamond contains the product of the left and right number, while the lower number contains the sum of the left and right number.

a. Right: $dfrac{-98}{7}=-14$ and Lower: $7-14=-7$

b. Upper: $dfrac{3}{2}cdot dfrac{3}{2}=dfrac{9}{4}$ and Lower: $dfrac{3}{2}+dfrac{3}{2}=dfrac{6}{2}=3$

c. Left: $dfrac{5/4}{-5/2}=-dfrac{1}{2}$ and Lower: $-dfrac{1}{2}-dfrac{5}{2}=-dfrac{6}{2}=-3$

d. Right: $0+10=10$ and Upper: $(-10)cdot (10)=-100$

a. $-14$ and $-7$

b. $frac{9}{4}$ and 3

c. $-frac{1}{2}$ and $-3$

d. 10 and $-100$

a. Yes, if you let the mp3 player play randomly songs from her playlist.

b. Since the playlist contains 2 country songs out of 5 songs:

$$
dfrac{2}{5}=0.4=40%
$$

c. Since the playlist contains 3 “Mama” songs out of 5 songs:

$$
dfrac{3}{5}=0.6=60%
$$

d. Since the playlist contains 1 Hank Tumbleweed out of 5 songs:

$$
dfrac{1}{5}=0.2=20%
$$

e. Since the playlist contains 4 not R&B songs out of 5 songs:

$$
dfrac{4}{5}=0.8=80%
$$

a. The corresponding angles need to have the same measure, while the all pairs of corresponding sides need to have the same ratio.

b. The two triangles are similar.

c. Perform a dilation with factor $dfrac{1}{2}$ to the larger triangle and then translated onto the smaller triangle.

a. Determine the missing angles using that the sum of all angles in a triangle is 180$text{textdegree}$:

$$
mangle I=180text{textdegree}-68text{textdegree}-87text{textdegree}=25text{textdegree}
$$

$$
mangle E=180text{textdegree}-25text{textdegree}-87text{textdegree}=68text{textdegree}
$$

Thus we note that the corresponding angle measures are all equal and thus the triangle are similar.

b.Two, because the third angle will follow from the fact that the sum of all angles in a triangle is 180$text{textdegree}$. Abreviated we will refer to this as AA.

c. Two triangles are similar if they have two pairs of corresponding sides that are proportional and the included angles are congruent.

d. Let us abbreviate it as $SAS$.

a. Yes

b. No

c. Two triangles are similar if they have two pairs of corresponding sides that are proportional and the included angles are congruent.

d. $SAS$

a. Determine the missing angles using that the sum of all angles in a triangle is 180$text{textdegree}$:

$$
mangle P=180text{textdegree}-32text{textdegree}-52text{textdegree}=96text{textdegree}
$$

Thus we then know that the triangles are similar because of AA.

b. Similar because of SAS, and the common ratio of the sides is
$$
dfrac{64}{20}=dfrac{44.8}{14}=3.2
$$

c. Similar because of AA (top angle is the same angle in both triangles)

d. Not similar because the ratios of the sides are not equal:

$$
dfrac{211.2}{58.5}=3.6neq dfrac{202.4}{44}=4.6neq dfrac{90}{18}=5
$$

a. Similar, because of HH. A dilation about the right vertex is sufficient to transform one triangle to the other.

b. Similar, because of SSS. A dilation about the left vertex and a translation will be sufficient.

c. Not similar because the sides are not proportional:

$$
dfrac{33}{11}=3text{ and } dfrac{20}{8}=2.5text{ and }dfrac{15}{6}=2.5
$$

d. Not similar, because the two triangles cannot have two pairs of congruent angles.

The total area is the sum of the subareas and is also the product of the length and the width.

a.
$$
(2y)(x+4)=2xy+8y
$$

b.
$$
(x+8)(x+5)=x^2+8x+5x+40=x^2+13x+40
$$

c.
$$
(x-4)(2z-3y+5)=2xz-3xy+5x-8z+12y-20
$$

d.
$$
(x+1)(x+6)=x^2+6x+x+6=x^2+7x+6
$$

a. $2xy+8y$

b. $x^2+13x+40$

c. $2xz-3xy+5x-8z+12y-20$

d. $x^2+7x+6$

The converse interchanges the statement after if with the statement after then.

a. If a triangle is isosceles, then the base angles of the triangle are congruent. TRUE (property isosceles triangles)

b. If the sum of the angles in a polygon is 180$text{textdegree}$, then the polygon is a triangle. TRUE (triangles are the only polygons whose angles sum up to 180$text{textdegree}$)

c. If my mom is happy, then I have cleaned my room. FALSE (Your mom could be happy for another reason)

Determine the height of the trapezium using the Pythagoran theorem:

$$
h=sqrt{13^2-5^2}=sqrt{169-25}=sqrt{144}=12
$$

The area of a trapezium is the sum of its bases, multiplied by the height divided by 2:

$$
dfrac{(12+18+5)12}{2}=210
$$

$$
210
$$

Make sure that every vertex corresponds to the vertex of the other polygon (that has the same angle or lies between the same pairs of corresponding sides).

a.
$$
ABCDsim EVOL
$$

b.
$$
RIGHTsim RONGW
$$

c.
$$
triangle DOGsim triangle ACT
$$

a.
$$
ABCDsim EVOL
$$

b.
$$
RIGHTsim RONGW
$$

c.
$$
triangle DOGsim triangle ACT
$$

b. The coordinates of every point have been multiplied by the dilation factor.

c. The perimeter is the sum of all sides:

$$
P_1=3+8+3+sqrt{4^2+3^2}+sqrt{4^2+3^2}=3+8+3+5+5=24
$$

$$
P_2=6+16+6+sqrt{8^2+6^2}+sqrt{8^2+6^2}=6+16+6+10+10=48
$$

The area is the sum of the area of the triangle (roof) and the area of the rectangle:

$$
A_1=3cdot 8+dfrac{8cdot 3}{2}=24+12=36
$$

$$
A_2=6cdot 16+dfrac{16cdot 6}{2}=96+48=144
$$

a. Determine the missing angle using that the sum of all angles in a triangle is 180$text{textdegree}$:

$$
180text{textdegree}-60text{textdegree}-40text{textdegree}=80text{textdegree}
$$

Thus we then know that the triangles are NOT similar because they have two pairs of angles that are not congruent.

b. Determine the missing angle using that the sum of all angles in a triangle is 180$text{textdegree}$:

$$
180text{textdegree}-45text{textdegree}-50text{textdegree}=85text{textdegree}
$$

Similar because of AA.

c. Similar because the ratios of the sides are equal and thus SSS:

$$
dfrac{12}{8}=dfrac{14}{7}=dfrac{12}{6}=2
$$

Since $A$ corresponds with $D$, $B$ corresponds with $E$ and $C$ corresponds with $F$:

$$
dfrac{AC}{BC}=dfrac{DF}{EF}
$$

$$
dfrac{AB}{BC}=dfrac{DE}{EF}
$$

$$
dfrac{AC}{BC}=dfrac{DF}{EF}
$$

$$
dfrac{AB}{BC}=dfrac{DE}{EF}
$$

a. A rotation, reflection and translation will be needed (AA).

b. A rotation and a translation will be sufficient.

c. A dilation of factor 2 and a rotation and a translation will be suffiicient.

d. A dilation of factor $dfrac{1}{2}$, a rotation, a reflection and a translation will all be needed.

Replace $y$ with $-dfrac{2}{5}x-2$ in the first equation:

$$
-dfrac{2}{5}x-2=-dfrac{2}{5}x+1
$$

Group like terms:

$$
-dfrac{2}{5}x+dfrac{2}{5}x=1+2
$$

Simplify:

$$
0x=3
$$

This last equation has no solution because 0 never equals 3. You can also see that the system has no solutions, because the lines are parallel (have the same slope).

a. Since one of the angles of the triangle is 90$text{textdegree}$, then the sum of the other two angles need to be 90$text{textdegree}$ too. Since 3 is less than 4, the angle opposite the side of length 3 needs to be smaller than the angle opposite the side of length 4. This then means that $x$ must be less than 45$text{textdegree}$.

b. Since one of the angles of the triangle is 90$text{textdegree}$, then the sum of the other two angles need to be 90$text{textdegree}$ too. Since both sides have length 18, the angle opposite the side of length 18 needs to be equal to the angle opposite the other side of length 18. This then means that $x$ must be equal than 45$text{textdegree}$.

c. Since one of the angles of the triangle is 90$text{textdegree}$, then the sum of the other two angles need to be 90$text{textdegree}$ too. Since 2 is less than 8, the angle opposite the side of length 2 needs to be smaller than the angle opposite the side of length 8. This then means that $x$ must be more than 45$text{textdegree}$.

Group like terms:

$$
(3x^2-4x^2)+(x-x)+(-2-(-5))
$$

Combine like terms:

$$
-1x^2+0x+3
$$

Simplify:

$$
-x^2+3
$$

$$
-x^2+3
$$

a. Similar because of SAS, and the common ratio of the sides is
$$
dfrac{16}{12}=dfrac{12}{9}=dfrac{4}{3}
$$

b. Not similar, because two of the corresponding sides have doubled in length (that is, the longest sides and the shortest sides), but the third one has not.

c. Similar because of AA (right angle is the same angle in both triangles)

a. Use the Pythagorean theorem:

$$
JK=sqrt{12^2+9^2}=sqrt{144+81}=sqrt{225}=15
$$

b. It is not possible to determine $overline{VT}$ because we do not know the length of the corresponding side $overline{SN}$.

c. Using that the lengths of corresponding sides in the triangles have the same propertions:

$$
dfrac{x+11}{11}=dfrac{2.5+9}{9}
$$

Multiply both sides of the equation by 11:

$$
x+11=dfrac{126.5}{9}
$$

Subtract 11 from both sides of the equation:

$$
x=dfrac{126.5}{9}-11approx 3
$$

a. 15

b. Not possible

c. $dfrac{126.5}{9}-11$

a. Yes (moreover they are even congruent)
$$
angle Q=angle X
$$

$$
dfrac{DQ}{XZ}=1
$$

$$
dfrac{PQ}{XY}=1
$$

$$
Downarrow SAS
$$

$$
triangle QDPsim triangle XZY
$$

b. Yes, because the ratios are 1.

c. If two triangles are congruent, then they are also similar. This follows directly from the transformations, since a congruence only uses rotations, reflections and translations, while a similarity uses rotations, reflections, translations and dilations.

d. Triangles are congruent $Rightarrow$ Triangles are similar

a. Yes

b. Yes

c. If two triangles are congruent, then they are also similar.

c. Triangles are congruent $Rightarrow$ Triangles are similar

a. Yes
$$
angle R=angle C
$$

$$
dfrac{RN}{CT}=2
$$

$$
dfrac{RU}{CA}=2
$$

$$
Downarrow SAS
$$

$$
triangle RUNsim triangle CAT
$$

b. They are both correct, because Leesa and Charles have the same letters that correspond (and to our solution from (a)).

c. No because the triangles use a dilation of factor $2$ to transform one triangle into another and congruence does not use dilations.

d.
$$
AT=dfrac{12.3}{2}=6.15
$$

$$
angle U=angle A=103text{textdegree}
$$

The sum of all angles in a triangle is 180$text{textdegree}$:

$$
angle N=angle T=180text{textdegree}-103text{textdegree}-47text{textdegree}=30text{textdegree}
$$

a. Yes

b. Both are correct

c. No

d. 6.15, 103$text{textdegree}$, 30$text{textdegree}$

a. Similar (SSS), because the ratios of the sides are equal:

$$
dfrac{6}{5}=dfrac{4.8}{4}=dfrac{3.6}{3}=1.2
$$

b. Similar because of AA

c. Not able to determine, since only one pair of side lengths and one pair of angle measures have been given.

Using vertical angles:
$$
angle y =angle x = 48text{textdegree}
$$

Using reflection properties:

$$
angle z =angle x = 48text{textdegree}
$$

$$
angle y =angle z = 48text{textdegree}
$$

$$
angle B =angle E = 90text{textdegree}
$$

$$
angle C=angle Ctext{ (common angle)}
$$

$$
Downarrow AA
$$

$$
triangle ABC sim triangle DEC
$$

An exponential function is of the form: $f(x)=ab^x$ with $a$ and $b$ constants. Evaluate the general function at the known points:

$$
32=acdot b^0=a
$$

$$
4=acdot b^3=32cdot b^3
$$

Thus we know $a=32$, divide both sides of the last equation by 32:

$$
dfrac{1}{8}=b^3
$$

Take the 3th root of both sides of the equation:

$$
dfrac{1}{2}=b
$$

Thus the equation then becomes:

$$
f(x)=32left( dfrac{1}{2}right)^x
$$

$$
f(x)=32left( dfrac{1}{2}right)^x
$$

The length of the third side is at most the sum of the lengths of the two other sides and is at least the difference of the lengths of the two other sides:

$$
8”+13”=21”
$$

$$
13”-8”=5”
$$

Thus the length of the third side is between 5” and 21”.

a. The playlist contains 2 country songs out of 5 songs:

$$
dfrac{2}{5}=0.4=40%
$$

b. No, because the songs are played without repeating and thus there is less chance of the second song being a country song if we know that the first song was already a country song.

c. There will be 1 out of the remaining 2 songs that will be by Sapphire:

$$
dfrac{1}{2}=0.5=50%
$$

d. Since writing a letter is 1 out of 3 possibilities and taking bus 41 is 1 out of 4 possibilities:

$$
dfrac{1}{3}cdot dfrac{1}{4}=dfrac{1}{12}approx 8.3%
$$

Since writing a letter is 1 out of 3 possibilities and taking bus 55 is 1 out of 4 possibilities:

$$
dfrac{1}{3}cdot dfrac{1}{4}=dfrac{1}{12}approx 8.3%
$$

a. $frac{2}{5}=0.4=40%$

b. No

c. $frac{1}{2}=0.5=50%$

d. $frac{1}{12}=8.3%$, $frac{1}{12}=8.3%$

a. By AA (since the triangles have angle $A$ in common and both contain a right angle)
$$
triangle AEIsim triangle AGH
$$

b.
$$
dfrac{x}{930+12}=dfrac{4}{12}
$$

Simplify:

$$
dfrac{x}{942}=dfrac{4}{12}
$$

Multiply both sides of the equation by 942:

$$
x=dfrac{4cdot 942}{12}=314
$$

Finally, adding the 2 ft to the height as well, we obtain that the total height of 314 ft + 2 ft = 316 ft.

a.
$$
angle B=angle E=9text{textdegree}
$$

$$
angle A=angle D=74text{textdegree}
$$

$$
Downarrow AA
$$

$$
triangle ABCsim triangle DEF
$$

b. Yes, because of AAS (two pair of angles are congruent and a pair of sides is congruent)

The converse interchanges the statement after if with the statement after then.

a. If the cat runs away frightened, then it has knocked over the lamp. FALSE (the cat could be frightened because of a different reason)

b. If the probability of getting a 3 is $dfrac{1}{6}$, then a six-sided dice is rolled. FALSE (it could also be drawing a number out of a bowl containing the numbers 1 to 6 once each).

c. If a triangle has a right angle, then the triangle has a 90$text{textdegree}$ angle. TRUE A right angle is a 90$text{textdegree}$ angle.

William is correct, because the right angle is a 90$text{textdegree}$ angle and thus has to be the largest angle in the triangle (since the sum of all angles is 180$text{textdegree}$), but then we know that the largest side is opposite the right angle and this side is called the hypotenuse.

The total area is the sum of the subareas and is also the product of the length and the width.

a.
$$
(x+3)(2x+1)=2x^2+6x+x+3=x^2+7x+3
$$

b.
$$
2x(x+5)=2x^2+10x
$$

c.
$$
x(2x+y)=2x^2+xy
$$

d.
$$
(2x+5)(x+y+2)=2x^2+2xy+4x+5x+5y+10=2x^2+2xy+9x+5y+10
$$

a. $x^2+7x+3$

b. $2x^2+10x$

c. $2x^2+xy$

d. $2x^2+2xy+9x+5y+10$

a. The area of a parallelogram is the product of the base and the height:

$$
16cdot 9 = 144 text{cm}^2
$$

The perimeter is the sum of the lengths of all sides:

$$
P=16+10+16+10=52text{ cm}
$$

b. The area of a trapezoid is the sum of the bases multiplied by the height divided by 2:

$$
A=dfrac{(25+44.67)20}{2}=696.7 text{m}^2
$$

The perimeter is the sum of the lengths of all sides:

$$
P=25+24+44.67+21=114.67text{ m}
$$

c. Note that the figure is a rectangle with two smaller rectangle cut out at the left and the right. The area of a rectangle is the product of the length and the width:

$$
Total: 9cdot 12=108text{ cm}^2
$$

$$
Left: 7cdot 3=21text{ cm}^2
$$

$$
Right: 5cdot 3=15text{ cm}^2
$$

The area of the shape is then the area of the total rectangle decreased by the two rectangles that have been cutout:

$$
A=108-21-15=72text{ cm}^2
$$

The perimeter is the sum of the lengths of all sides:

$$
P=12+2+3+5+3+2+9+7+3+2=48text{ cm}
$$

d. Note that the figure is a trapezium with a rectangle cut out at the top. The area of a trapezoid is the sum of the bases multiplied by the height divided by 2:

$$
dfrac{(23+11)8}{2}=136text{ ft}^2
$$

The area of a rectangle is the product of the length and the width:

$$
2cdot 3=6text{ ft}^2
$$

The area of the shape is then the area of the trapezium decreased by tje area of the rectangle:

$$
A=136′-6’=130text{ ft}^2
$$

The perimeter is the sum of the lengths of all sides:

$$
P=10’+4’+2’+3’+2’+4’+10’+23’=58′
$$

a. Area 114 cm$^2$ and Perimeter 52 cm

b. Area 696.7 m$^2$ and Perimeter 114.67 m

c. Area 72 cm$^2$ and Perimeter 48 cm

d. Area 130 ft$^2$ and Perimeter 58′

Similar triangles have propertional sides:

a.
$$
dfrac{n}{2}=dfrac{112}{7}
$$

Multiply both sides of the equation by 2:

$$
n=dfrac{112cdot 2}{7}=32
$$

b.
$$
dfrac{m}{7}=dfrac{49}{23}
$$

Multiply both sides of the equation by 7:

$$
m=dfrac{49cdot 7}{23}=dfrac{343}{23}approx 15
$$

a. $n=32$

b. $m=dfrac{343}{23}approx 15$

a. Yes, the third pair of angles also has to be congruent, because the sum of all angles in a triangle is equal to 180$text{textdegree}$.

b.Yes, Eliana is correct.

c. They are parallel because $angle Q=angle U$.

d. The scale factor is $dfrac{UV}{QR}$ (or you can use any pair of corresponding sides)

e. Yes, because the diagram has only been rotated and nothing has been changed about the lengths of the sides of the triangles.

a. Yes

b. Yes

c. Parallel

d. $frac{UV}{QR}$

e. Yes

a. Yes, Use cross multiplication:

$$
4(5+x)=5(6)
$$

Use distributive property:

$$
20+4x=30
$$

Subtract 20 from both sides of the equation:

$$
4x=10
$$

Divide both sides of the equation by 4:

$$
x=2.5
$$

b. Janelle use the propertions of the two sides instead of the sides of the triangles.

c. Kamraan should add that $overline{DE}$ and $overline{BC}$ are parallel. He wants to proof that the proportions of both sides are equal:

$$
dfrac{a}{b}=dfrac{c}{d}
$$

d. By similar triangles we know:

$$
dfrac{a+b}{a}=dfrac{c+d}{c}
$$

e. Rewrite the expression:

$$
dfrac{a}{a}+dfrac{b}{a}=dfrac{c}{c}+dfrac{d}{c}
$$

Simplify:

$$
1+dfrac{b}{a}=1+dfrac{d}{c}
$$

Subtract 1 from both sides of the equation:

$$
dfrac{b}{a}=dfrac{d}{c}
$$

a. Yes

b. Proportions of the two sides

c. $frac{a}{b}=frac{c}{d}$

d. $frac{a+b}{a}=frac{c+d}{c}$

e. $frac{b}{a}=frac{d}{c}$

a. If $dfrac{a}{b}=dfrac{c}{d}$, then $triangle MSNsim triangle MVT$.

b.
$$
dfrac{a}{b}=dfrac{c}{d}
$$

c.
$$
angle M=angle Mtext{ (same angle)}
$$

$$
dfrac{a+b}{a}=dfrac{a}{a}+dfrac{b}{a}=1+dfrac{d}{c}=dfrac{c}{c}+dfrac{d}{c}=dfrac{d+c}{c}
$$

$$
Downarrow SAS
$$

$$
triangle MSNsim triangle MVT
$$

d. Since $triangle MSNsim triangle MVT$, we know that the corresponding angles $angle V$ and $angle S$ are equal and thus the lines $overleftrightarrow{VT}$ and $overleftrightarrow{SN}$are parallel.

a. If $dfrac{a}{b}=dfrac{c}{d}$, then $triangle MSNsim triangle MVT$.

b. $dfrac{a}{b}=dfrac{c}{d}$

c. Yes (Use SAS)

d. $angle V$ and $angle S$ are equal

a. The ratio of each pair is 1.5:
$$
dfrac{12}{8}=dfrac{9}{6}=dfrac{6}{4}=1.5
$$

b. Multiply each length by the scale factor:

$$
D’E’=12cdot dfrac{2}{3}=8
$$

$$
D’F’=6cdot dfrac{2}{3}=4
$$

$$
E’F’=9cdot dfrac{2}{3}=6
$$

c. The triangle are congruent because the two triangles have three pairs of congruent sides.

d. It remains to rotate and translate $triangle D’E’F’$ to make it overlap $triangle ABC$ and thus we then know that the two triangles are similar (since only dilations, rotations and translations were used).

e. Yes, but then you need to dilate the larger triangle over factor $dfrac{1}{k}$.

a. Similar because the ratios of the sides are equal (SSS):

$$
dfrac{15}{5}=dfrac{12}{4}=dfrac{6}{2}=3
$$

However, the triangles are not congruent since the ratios are not equal to 1.

b. Determine the missing angles using that the sum of all angles in a triangle is 180$text{textdegree}$:

$$
180text{textdegree}-48text{textdegree}-20text{textdegree}=112text{textdegree}
$$

Thus we then know that the three pairs of angles are equivalent and one of the pair of sides is also equivalent, thus the triangles are congruent by ASA and are also similar (since congruent triangles are also similar triangle).

c. Determine the missing sides using the Pythagorean theorem:

$$
12^2+5^2=169=13^2
$$

Thus we then know that one pair of angles are equivalent and three pair of sides are also equivalent, thus the triangles are congruent by SAS and are also similar (since congruent triangles are also similar triangle).

d. Similar because of AA. Not congruent because the known lengths of the sides are not of corresponding sides.

First rotate one of the triangles about 180$text{textdegree}$ and then place that triangle over the other (which will result in a perfect overlap). Since only rigid transformations (rotation, reflection, translation) were used, the triangles are then congruent.

First rotate one of the triangles about 180$text{textdegree}$ and then place that triangle over the other.

Corresponding sides of similar polygons have the same proportions.

a.
$$
dfrac{x}{12}=dfrac{10}{6}
$$

Multiply both sides of the equation by 12:

$$
x=dfrac{10cdot 12}{6}=20
$$

b.
$$
dfrac{w}{7}=dfrac{39}{3}
$$

Multiply both sides of the equation by 7:

$$
w=dfrac{39cdot 7}{3}=91
$$

a. $x=20$

b. $w=91$

The total area is the sum of the subareas and is also the product of the length and the width.

a.
$$
(2x+1)(2x+1)=4x^2+2x+2x+1=4x^2+4x+1
$$

b.
$$
(2x)(4x)=8x^2
$$

c.
$$
2(3x+5)=6x+10
$$

d.
$$
y(2x+y+3)=2xy+y^2+3y
$$

a. $4x^2+4x+1$

b. $8x^2$

c. $6x+10$

d. $2xy+y^2+3y$