Practice · A square on a given side has two sides to grow · Sensim Math

Variant 1 easy

Points $A(0, -2)$ and $B(0, 2)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, one directly above the other. Find where the other two corners can be -- every possibility.

Givens

$A(0, -2)$ and $B(0, 2)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is vertical, the sides $BC$ and $AD$ must be horizontal and the same length as $AB$ . A horizontal segment can go right or left, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

x

-coordinate, so

AB

is vertical and its length is the difference of the

y

-coordinates.

AB = 2 - (-2) = 4

Vertical distance is the difference of

y

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the vertical side

AB

, so they are horizontal, each

4

units long. So

C

keeps

B

's

y

-coordinate

2

, and

D

keeps

A

's

y

-coordinate

-2

; both move the same

4

units left or right.

C(0 \pm 4, 2), \quad D(0 \pm 4, -2)

Turning a vertical side by a right angle gives a horizontal side.

3Case 1: build to the right

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the right from

x=0

gives

x=4

.

0 + 4 = 4 \Rightarrow C(4, 2),\ D(4, -2)

Right means add.

4Case 2: build to the left

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the left from

x=0

gives

x=-4

.

0 - 4 = -4 \Rightarrow C(-4, 2),\ D(-4, -2)

Left means subtract.

Answer:

C(4, 2)

,

D(4, -2)

or

C(-4, 2)

,

D(-4, -2)

4 · Reviewdoes it hold up?

In the first case $BC = 4-0 = 4$ , $CD = 2-(-2) = 4$ , and $DA = 4-0 = 4$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $x=0$ .

Another way: Picture sliding the side $AB$ sideways by its own length: sliding it $4$ units right carries $B$ to $C$ and $A$ to $D$ ; sliding it $4$ units left gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 2 easy

Points $A(-1, 2)$ and $B(-1, -1)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, one directly above the other. Find where the other two corners can be -- every possibility.

Givens

$A(-1, 2)$ and $B(-1, -1)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is vertical, the sides $BC$ and $AD$ must be horizontal and the same length as $AB$ . A horizontal segment can go right or left, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

x

-coordinate, so

AB

is vertical and its length is the difference of the

y

-coordinates.

AB = 2 - (-1) = 3

Vertical distance is the difference of

y

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the vertical side

AB

, so they are horizontal, each

3

units long. So

C

keeps

B

's

y

-coordinate

-1

, and

D

keeps

A

's

y

-coordinate

2

; both move the same

3

units left or right.

C(-1 \pm 3, -1), \quad D(-1 \pm 3, 2)

Turning a vertical side by a right angle gives a horizontal side.

3Case 1: build to the right

#2 Make a Systematic List 6.NS.C.8

Moving

3

to the right from

x=-1

gives

x=2

.

-1 + 3 = 2 \Rightarrow C(2, -1),\ D(2, 2)

Right means add.

4Case 2: build to the left

#2 Make a Systematic List 6.NS.C.8

Moving

3

to the left from

x=-1

gives

x=-4

.

-1 - 3 = -4 \Rightarrow C(-4, -1),\ D(-4, 2)

Left means subtract.

Answer:

C(2, -1)

,

D(2, 2)

or

C(-4, -1)

,

D(-4, 2)

4 · Reviewdoes it hold up?

In the first case $BC = 2-(-1) = 3$ , $CD = 2-(-1) = 3$ , and $DA = 2-(-1) = 3$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $x=-1$ .

Another way: Picture sliding the side $AB$ sideways by its own length: sliding it $3$ units right carries $B$ to $C$ and $A$ to $D$ ; sliding it $3$ units left gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 3 easy

Points $A(2, 3)$ and $B(2, -2)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, one directly above the other. Find where the other two corners can be -- every possibility.

Givens

$A(2, 3)$ and $B(2, -2)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is vertical, the sides $BC$ and $AD$ must be horizontal and the same length as $AB$ . A horizontal segment can go right or left, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

x

-coordinate, so

AB

is vertical and its length is the difference of the

y

-coordinates.

AB = 3 - (-2) = 5

Vertical distance is the difference of

y

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the vertical side

AB

, so they are horizontal, each

5

units long. So

C

keeps

B

's

y

-coordinate

-2

, and

D

keeps

A

's

y

-coordinate

3

; both move the same

5

units left or right.

C(2 \pm 5, -2), \quad D(2 \pm 5, 3)

Turning a vertical side by a right angle gives a horizontal side.

3Case 1: build to the right

#2 Make a Systematic List 6.NS.C.8

Moving

5

to the right from

x=2

gives

x=7

.

2 + 5 = 7 \Rightarrow C(7, -2),\ D(7, 3)

Right means add.

4Case 2: build to the left

#2 Make a Systematic List 6.NS.C.8

Moving

5

to the left from

x=2

gives

x=-3

.

2 - 5 = -3 \Rightarrow C(-3, -2),\ D(-3, 3)

Left means subtract.

Answer:

C(7, -2)

,

D(7, 3)

or

C(-3, -2)

,

D(-3, 3)

4 · Reviewdoes it hold up?

In the first case $BC = 7-2 = 5$ , $CD = 3-(-2) = 5$ , and $DA = 7-2 = 5$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $x=2$ .

Another way: Picture sliding the side $AB$ sideways by its own length: sliding it $5$ units right carries $B$ to $C$ and $A$ to $D$ ; sliding it $5$ units left gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 4 medium

Points $A(2, -3)$ and $B(2, 1)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, one directly above the other. Find where the other two corners can be -- every possibility.

Givens

$A(2, -3)$ and $B(2, 1)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is vertical, the sides $BC$ and $AD$ must be horizontal and the same length as $AB$ . A horizontal segment can go right or left, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

x

-coordinate, so

AB

is vertical and its length is the difference of the

y

-coordinates.

AB = 1 - (-3) = 4

Vertical distance is the difference of

y

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the vertical side

AB

, so they are horizontal, each

4

units long. So

C

keeps

B

's

y

-coordinate

1

, and

D

keeps

A

's

y

-coordinate

-3

; both move the same

4

units left or right.

C(2 \pm 4, 1), \quad D(2 \pm 4, -3)

Turning a vertical side by a right angle gives a horizontal side.

3Case 1: build to the right

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the right from

x=2

gives

x=6

.

2 + 4 = 6 \Rightarrow C(6, 1),\ D(6, -3)

Right means add.

4Case 2: build to the left

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the left from

x=2

gives

x=-2

.

2 - 4 = -2 \Rightarrow C(-2, 1),\ D(-2, -3)

Left means subtract.

Answer:

C(6, 1)

,

D(6, -3)

or

C(-2, 1)

,

D(-2, -3)

4 · Reviewdoes it hold up?

In the first case $BC = 6-2 = 4$ , $CD = 1-(-3) = 4$ , and $DA = 6-2 = 4$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $x=2$ .

Another way: Picture sliding the side $AB$ sideways by its own length: sliding it $4$ units right carries $B$ to $C$ and $A$ to $D$ ; sliding it $4$ units left gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 5 medium

Points $A(3, -1)$ and $B(-2, -1)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, side by side at the same height. Find where the other two corners can be -- every possibility.

Givens

$A(3, -1)$ and $B(-2, -1)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is horizontal, the sides $BC$ and $AD$ must be vertical and the same length as $AB$ . A vertical segment can go up or down, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

y

-coordinate, so

AB

is horizontal and its length is the difference of the

x

-coordinates.

AB = 3 - (-2) = 5

Horizontal distance is the difference of

x

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the horizontal side

AB

, so they are vertical, each

5

units long. So

C

keeps

B

's

x

-coordinate

-2

, and

D

keeps

A

's

x

-coordinate

3

; both move the same

5

units up or down.

C(-2, -1 \pm 5), \quad D(3, -1 \pm 5)

Turning a horizontal side by a right angle gives a vertical side.

3Case 1: build upward

#2 Make a Systematic List 6.NS.C.8

Moving

5

up from

y=-1

gives

y=4

.

-1 + 5 = 4 \Rightarrow C(-2, 4),\ D(3, 4)

Up means add.

4Case 2: build downward

#2 Make a Systematic List 6.NS.C.8

Moving

5

down from

y=-1

gives

y=-6

.

-1 - 5 = -6 \Rightarrow C(-2, -6),\ D(3, -6)

Down means subtract.

Answer:

C(-2, 4)

,

D(3, 4)

or

C(-2, -6)

,

D(3, -6)

4 · Reviewdoes it hold up?

In the first case $BC = 4-(-1) = 5$ , $CD = 3-(-2) = 5$ , and $DA = 4-(-1) = 5$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $y=-1$ .

Another way: Picture sliding the side $AB$ up and down by its own length: sliding it $5$ units up carries $B$ to $C$ and $A$ to $D$ ; sliding it $5$ units down gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 6 medium

Points $A(-2, 3)$ and $B(1, 3)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, side by side at the same height. Find where the other two corners can be -- every possibility.

Givens

$A(-2, 3)$ and $B(1, 3)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is horizontal, the sides $BC$ and $AD$ must be vertical and the same length as $AB$ . A vertical segment can go up or down, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

y

-coordinate, so

AB

is horizontal and its length is the difference of the

x

-coordinates.

AB = 1 - (-2) = 3

Horizontal distance is the difference of

x

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the horizontal side

AB

, so they are vertical, each

3

units long. So

C

keeps

B

's

x

-coordinate

1

, and

D

keeps

A

's

x

-coordinate

-2

; both move the same

3

units up or down.

C(1, 3 \pm 3), \quad D(-2, 3 \pm 3)

Turning a horizontal side by a right angle gives a vertical side.

3Case 1: build upward

#2 Make a Systematic List 6.NS.C.8

Moving

3

up from

y=3

gives

y=6

.

3 + 3 = 6 \Rightarrow C(1, 6),\ D(-2, 6)

Up means add.

4Case 2: build downward

#2 Make a Systematic List 6.NS.C.8

Moving

3

down from

y=3

gives

y=0

.

3 - 3 = 0 \Rightarrow C(1, 0),\ D(-2, 0)

Down means subtract.

Answer:

C(1, 6)

,

D(-2, 6)

or

C(1, 0)

,

D(-2, 0)

4 · Reviewdoes it hold up?

In the first case $BC = 6-3 = 3$ , $CD = 1-(-2) = 3$ , and $DA = 6-3 = 3$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $y=3$ .

Another way: Picture sliding the side $AB$ up and down by its own length: sliding it $3$ units up carries $B$ to $C$ and $A$ to $D$ ; sliding it $3$ units down gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 7 easy

Points $A(-3, -1)$ and $B(-3, 3)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, one directly above the other. Find where the other two corners can be -- every possibility.

Givens

$A(-3, -1)$ and $B(-3, 3)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is vertical, the sides $BC$ and $AD$ must be horizontal and the same length as $AB$ . A horizontal segment can go right or left, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

x

-coordinate, so

AB

is vertical and its length is the difference of the

y

-coordinates.

AB = 3 - (-1) = 4

Vertical distance is the difference of

y

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the vertical side

AB

, so they are horizontal, each

4

units long. So

C

keeps

B

's

y

-coordinate

3

, and

D

keeps

A

's

y

-coordinate

-1

; both move the same

4

units left or right.

C(-3 \pm 4, 3), \quad D(-3 \pm 4, -1)

Turning a vertical side by a right angle gives a horizontal side.

3Case 1: build to the right

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the right from

x=-3

gives

x=1

.

-3 + 4 = 1 \Rightarrow C(1, 3),\ D(1, -1)

Right means add.

4Case 2: build to the left

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the left from

x=-3

gives

x=-7

.

-3 - 4 = -7 \Rightarrow C(-7, 3),\ D(-7, -1)

Left means subtract.

Answer:

C(1, 3)

,

D(1, -1)

or

C(-7, 3)

,

D(-7, -1)

4 · Reviewdoes it hold up?

In the first case $BC = 1-(-3) = 4$ , $CD = 3-(-1) = 4$ , and $DA = 1-(-3) = 4$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $x=-3$ .

Another way: Picture sliding the side $AB$ sideways by its own length: sliding it $4$ units right carries $B$ to $C$ and $A$ to $D$ ; sliding it $4$ units left gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 8 medium

Points $A(1, 2)$ and $B(-3, 2)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, side by side at the same height. Find where the other two corners can be -- every possibility.

Givens

$A(1, 2)$ and $B(-3, 2)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is horizontal, the sides $BC$ and $AD$ must be vertical and the same length as $AB$ . A vertical segment can go up or down, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

y

-coordinate, so

AB

is horizontal and its length is the difference of the

x

-coordinates.

AB = 1 - (-3) = 4

Horizontal distance is the difference of

x

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the horizontal side

AB

, so they are vertical, each

4

units long. So

C

keeps

B

's

x

-coordinate

-3

, and

D

keeps

A

's

x

-coordinate

1

; both move the same

4

units up or down.

C(-3, 2 \pm 4), \quad D(1, 2 \pm 4)

Turning a horizontal side by a right angle gives a vertical side.

3Case 1: build upward

#2 Make a Systematic List 6.NS.C.8

Moving

4

up from

y=2

gives

y=6

.

2 + 4 = 6 \Rightarrow C(-3, 6),\ D(1, 6)

Up means add.

4Case 2: build downward

#2 Make a Systematic List 6.NS.C.8

Moving

4

down from

y=2

gives

y=-2

.

2 - 4 = -2 \Rightarrow C(-3, -2),\ D(1, -2)

Down means subtract.

Answer:

C(-3, 6)

,

D(1, 6)

or

C(-3, -2)

,

D(1, -2)

4 · Reviewdoes it hold up?

In the first case $BC = 6-2 = 4$ , $CD = 1-(-3) = 4$ , and $DA = 6-2 = 4$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $y=2$ .

Another way: Picture sliding the side $AB$ up and down by its own length: sliding it $4$ units up carries $B$ to $C$ and $A$ to $D$ ; sliding it $4$ units down gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 9 hard

Points $A(4, 1)$ and $B(4, -3)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, one directly above the other. Find where the other two corners can be -- every possibility.

Givens

$A(4, 1)$ and $B(4, -3)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is vertical, the sides $BC$ and $AD$ must be horizontal and the same length as $AB$ . A horizontal segment can go right or left, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

x

-coordinate, so

AB

is vertical and its length is the difference of the

y

-coordinates.

AB = 1 - (-3) = 4

Vertical distance is the difference of

y

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the vertical side

AB

, so they are horizontal, each

4

units long. So

C

keeps

B

's

y

-coordinate

-3

, and

D

keeps

A

's

y

-coordinate

1

; both move the same

4

units left or right.

C(4 \pm 4, -3), \quad D(4 \pm 4, 1)

Turning a vertical side by a right angle gives a horizontal side.

3Case 1: build to the right

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the right from

x=4

gives

x=8

.

4 + 4 = 8 \Rightarrow C(8, -3),\ D(8, 1)

Right means add.

4Case 2: build to the left

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the left from

x=4

gives

x=0

.

4 - 4 = 0 \Rightarrow C(0, -3),\ D(0, 1)

Left means subtract.

Answer:

C(8, -3)

,

D(8, 1)

or

C(0, -3)

,

D(0, 1)

4 · Reviewdoes it hold up?

In the first case $BC = 8-4 = 4$ , $CD = 1-(-3) = 4$ , and $DA = 8-4 = 4$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $x=4$ .

Another way: Picture sliding the side $AB$ sideways by its own length: sliding it $4$ units right carries $B$ to $C$ and $A$ to $D$ ; sliding it $4$ units left gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 10 hard

Points $A(-4, 0)$ and $B(2, 0)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, side by side at the same height. Find where the other two corners can be -- every possibility.

Givens

$A(-4, 0)$ and $B(2, 0)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is horizontal, the sides $BC$ and $AD$ must be vertical and the same length as $AB$ . A vertical segment can go up or down, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

y

-coordinate, so

AB

is horizontal and its length is the difference of the

x

-coordinates.

AB = 2 - (-4) = 6

Horizontal distance is the difference of

x

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the horizontal side

AB

, so they are vertical, each

6

units long. So

C

keeps

B

's

x

-coordinate

2

, and

D

keeps

A

's

x

-coordinate

-4

; both move the same

6

units up or down.

C(2, 0 \pm 6), \quad D(-4, 0 \pm 6)

Turning a horizontal side by a right angle gives a vertical side.

3Case 1: build upward

#2 Make a Systematic List 6.NS.C.8

Moving

6

up from

y=0

gives

y=6

.

0 + 6 = 6 \Rightarrow C(2, 6),\ D(-4, 6)

Up means add.

4Case 2: build downward

#2 Make a Systematic List 6.NS.C.8

Moving

6

down from

y=0

gives

y=-6

.

0 - 6 = -6 \Rightarrow C(2, -6),\ D(-4, -6)

Down means subtract.

Answer:

C(2, 6)

,

D(-4, 6)

or

C(2, -6)

,

D(-4, -6)

4 · Reviewdoes it hold up?

In the first case $BC = 6-0 = 6$ , $CD = 2-(-4) = 6$ , and $DA = 6-0 = 6$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $y=0$ .

Another way: Picture sliding the side $AB$ up and down by its own length: sliding it $6$ units up carries $B$ to $C$ and $A$ to $D$ ; sliding it $6$ units down gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 11 hard

Points $A(-1, -4)$ and $B(3, -4)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, side by side at the same height. Find where the other two corners can be -- every possibility.

Givens

$A(-1, -4)$ and $B(3, -4)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is horizontal, the sides $BC$ and $AD$ must be vertical and the same length as $AB$ . A vertical segment can go up or down, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

y

-coordinate, so

AB

is horizontal and its length is the difference of the

x

-coordinates.

AB = 3 - (-1) = 4

Horizontal distance is the difference of

x

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the horizontal side

AB

, so they are vertical, each

4

units long. So

C

keeps

B

's

x

-coordinate

3

, and

D

keeps

A

's

x

-coordinate

-1

; both move the same

4

units up or down.

C(3, -4 \pm 4), \quad D(-1, -4 \pm 4)

Turning a horizontal side by a right angle gives a vertical side.

3Case 1: build upward

#2 Make a Systematic List 6.NS.C.8

Moving

4

up from

y=-4

gives

y=0

.

-4 + 4 = 0 \Rightarrow C(3, 0),\ D(-1, 0)

Up means add.

4Case 2: build downward

#2 Make a Systematic List 6.NS.C.8

Moving

4

down from

y=-4

gives

y=-8

.

-4 - 4 = -8 \Rightarrow C(3, -8),\ D(-1, -8)

Down means subtract.

Answer:

C(3, 0)

,

D(-1, 0)

or

C(3, -8)

,

D(-1, -8)

4 · Reviewdoes it hold up?

In the first case $BC = 0-(-4) = 4$ , $CD = 3-(-1) = 4$ , and $DA = 0-(-4) = 4$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $y=-4$ .

Another way: Picture sliding the side $AB$ up and down by its own length: sliding it $4$ units up carries $B$ to $C$ and $A$ to $D$ ; sliding it $4$ units down gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

Variant 12 hard

Points $A(1, 4)$ and $B(1, 0)$ are vertices of square $ABCD$ . Find all possible coordinates of the other two vertices, $C$ and $D$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two corners of a square are given, one directly above the other. Find where the other two corners can be -- every possibility.

Givens

$A(1, 4)$ and $B(1, 0)$ .
The square is named $ABCD$ , so $AB$ is a side, $C$ comes after $B$ , and $D$ comes after $C$ and before $A$ .

Unknowns

All possible coordinates of $C$ and $D$ .

Constraints

All four sides have the same length and meet at right angles.

2 · Planchoose the strategy

#2 Make a Systematic List · also uses: #1 Draw a Diagram

Draw the side $AB$ . Because it is vertical, the sides $BC$ and $AD$ must be horizontal and the same length as $AB$ . A horizontal segment can go right or left, so list both cases.

3 · Execute4 carry out the plan

1Find the side length

#1 Draw a Diagram 7.NS.A.1

A

and

B

have the same

x

-coordinate, so

AB

is vertical and its length is the difference of the

y

-coordinates.

AB = 4 - 0 = 4

Vertical distance is the difference of

y

-coordinates.

2Decide the direction of the other sides

#1 Draw a Diagram 7.G.A.2

BC

and

AD

are perpendicular to the vertical side

AB

, so they are horizontal, each

4

units long. So

C

keeps

B

's

y

-coordinate

0

, and

D

keeps

A

's

y

-coordinate

4

; both move the same

4

units left or right.

C(1 \pm 4, 0), \quad D(1 \pm 4, 4)

Turning a vertical side by a right angle gives a horizontal side.

3Case 1: build to the right

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the right from

x=1

gives

x=5

.

1 + 4 = 5 \Rightarrow C(5, 0),\ D(5, 4)

Right means add.

4Case 2: build to the left

#2 Make a Systematic List 6.NS.C.8

Moving

4

to the left from

x=1

gives

x=-3

.

1 - 4 = -3 \Rightarrow C(-3, 0),\ D(-3, 4)

Left means subtract.

Answer:

C(5, 0)

,

D(5, 4)

or

C(-3, 0)

,

D(-3, 4)

4 · Reviewdoes it hold up?

In the first case $BC = 5-1 = 4$ , $CD = 4-0 = 4$ , and $DA = 5-1 = 4$ , all equal to $AB$ with only horizontal and vertical sides, so it is a square. The second case is its mirror image across the line $x=1$ .

Another way: Picture sliding the side $AB$ sideways by its own length: sliding it $4$ units right carries $B$ to $C$ and $A$ to $D$ ; sliding it $4$ units left gives the other square.

Standardsmin grade 7

7.G.A.2 Draw geometric shapes with given conditions including triangles — Drawing the square on side $AB$ in both possible directions.
6.NS.C.8 Solve problems by graphing points in four quadrants — Plotting points in all four quadrants.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding lengths as differences of signed coordinates.

💡Takeaway. When a shape can face either way, check both sides -- one answer is only half the answer.

A square on a given side has two sides to grow · 12 practice problems

Generated variants — 12

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Where next?