Practice · Measure a border path strip by strip · Sensim Math

Variant 1 easy

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $1$ m wide along the top and bottom and $2$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $1$ m wide at top and bottom, $2$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $1$ m wide along the top and bottom.
The path is $2$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>4$ and $y>2$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

2

m on each side across and

1

m at the top and bottom.

\text{width}=x-2-2=x-4,\quad \text{length}=y-1-1=y-2

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

1

m wide. The left and right strips fit between them, so they are

y-2

long and

2

m wide.

A=2\times x+2\times2(y-2)=2x+4y-8

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-4)+2(y-2)=2x+2y-12

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(2x+4y-8)+(2x+2y-12)=4x+6y-20

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

4x+6y-20

4 · Reviewdoes it hold up?

Try $x=18$ , $y=8$ . The field is $14$ by $6$ , so the path's area is $144-84=60$ and $2\times18+4\times8-8=60$ . The field's perimeter is $2\times(14+6)=40$ and $2\times18+2\times8-12=40$ . Then $60+40=100$ , and $4\times18+6\times8-20=100$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-4)(y-2)$ . Expanding gives $xy-(xy-2x-4y+8)=2x+4y-8$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 2 easy

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $3$ m wide along the top and bottom and $2$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $3$ m wide at top and bottom, $2$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $3$ m wide along the top and bottom.
The path is $2$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>4$ and $y>6$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

2

m on each side across and

3

m at the top and bottom.

\text{width}=x-2-2=x-4,\quad \text{length}=y-3-3=y-6

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

3

m wide. The left and right strips fit between them, so they are

y-6

long and

2

m wide.

A=2\times(3x)+2\times2(y-6)=6x+4y-24

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-4)+2(y-6)=2x+2y-20

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(6x+4y-24)+(2x+2y-20)=8x+6y-44

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

8x+6y-44

4 · Reviewdoes it hold up?

Try $x=18$ , $y=12$ . The field is $14$ by $6$ , so the path's area is $216-84=132$ and $6\times18+4\times12-24=132$ . The field's perimeter is $2\times(14+6)=40$ and $2\times18+2\times12-20=40$ . Then $132+40=172$ , and $8\times18+6\times12-44=172$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-4)(y-6)$ . Expanding gives $xy-(xy-6x-4y+24)=6x+4y-24$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 3 easy

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $2$ m wide along the top and bottom and $3$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $2$ m wide at top and bottom, $3$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $2$ m wide along the top and bottom.
The path is $3$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>6$ and $y>4$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

3

m on each side across and

2

m at the top and bottom.

\text{width}=x-3-3=x-6,\quad \text{length}=y-2-2=y-4

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

2

m wide. The left and right strips fit between them, so they are

y-4

long and

3

m wide.

A=2\times(2x)+2\times3(y-4)=4x+6y-24

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-6)+2(y-4)=2x+2y-20

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(4x+6y-24)+(2x+2y-20)=6x+8y-44

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

6x+8y-44

4 · Reviewdoes it hold up?

Try $x=20$ , $y=10$ . The field is $14$ by $6$ , so the path's area is $200-84=116$ and $4\times20+6\times10-24=116$ . The field's perimeter is $2\times(14+6)=40$ and $2\times20+2\times10-20=40$ . Then $116+40=156$ , and $6\times20+8\times10-44=156$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-6)(y-4)$ . Expanding gives $xy-(xy-4x-6y+24)=4x+6y-24$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 4 easy

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $1$ m wide along the top and bottom and $3$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $1$ m wide at top and bottom, $3$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $1$ m wide along the top and bottom.
The path is $3$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>6$ and $y>2$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

3

m on each side across and

1

m at the top and bottom.

\text{width}=x-3-3=x-6,\quad \text{length}=y-1-1=y-2

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

1

m wide. The left and right strips fit between them, so they are

y-2

long and

3

m wide.

A=2\times x+2\times3(y-2)=2x+6y-12

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-6)+2(y-2)=2x+2y-16

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(2x+6y-12)+(2x+2y-16)=4x+8y-28

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

4x+8y-28

4 · Reviewdoes it hold up?

Try $x=20$ , $y=8$ . The field is $14$ by $6$ , so the path's area is $160-84=76$ and $2\times20+6\times8-12=76$ . The field's perimeter is $2\times(14+6)=40$ and $2\times20+2\times8-16=40$ . Then $76+40=116$ , and $4\times20+8\times8-28=116$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-6)(y-2)$ . Expanding gives $xy-(xy-2x-6y+12)=2x+6y-12$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 5 hard

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $1$ m wide along the top and bottom and $4$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $1$ m wide at top and bottom, $4$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $1$ m wide along the top and bottom.
The path is $4$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>8$ and $y>2$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

4

m on each side across and

1

m at the top and bottom.

\text{width}=x-4-4=x-8,\quad \text{length}=y-1-1=y-2

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

1

m wide. The left and right strips fit between them, so they are

y-2

long and

4

m wide.

A=2\times x+2\times4(y-2)=2x+8y-16

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-8)+2(y-2)=2x+2y-20

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(2x+8y-16)+(2x+2y-20)=4x+10y-36

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

4x+10y-36

4 · Reviewdoes it hold up?

Try $x=22$ , $y=8$ . The field is $14$ by $6$ , so the path's area is $176-84=92$ and $2\times22+8\times8-16=92$ . The field's perimeter is $2\times(14+6)=40$ and $2\times22+2\times8-20=40$ . Then $92+40=132$ , and $4\times22+10\times8-36=132$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-8)(y-2)$ . Expanding gives $xy-(xy-2x-8y+16)=2x+8y-16$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 6 medium

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $4$ m wide along the top and bottom and $2$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $4$ m wide at top and bottom, $2$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $4$ m wide along the top and bottom.
The path is $2$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>4$ and $y>8$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

2

m on each side across and

4

m at the top and bottom.

\text{width}=x-2-2=x-4,\quad \text{length}=y-4-4=y-8

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

4

m wide. The left and right strips fit between them, so they are

y-8

long and

2

m wide.

A=2\times(4x)+2\times2(y-8)=8x+4y-32

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-4)+2(y-8)=2x+2y-24

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(8x+4y-32)+(2x+2y-24)=10x+6y-56

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

10x+6y-56

4 · Reviewdoes it hold up?

Try $x=18$ , $y=14$ . The field is $14$ by $6$ , so the path's area is $252-84=168$ and $8\times18+4\times14-32=168$ . The field's perimeter is $2\times(14+6)=40$ and $2\times18+2\times14-24=40$ . Then $168+40=208$ , and $10\times18+6\times14-56=208$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-4)(y-8)$ . Expanding gives $xy-(xy-8x-4y+32)=8x+4y-32$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 7 medium

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $4$ m wide along the top and bottom and $3$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $4$ m wide at top and bottom, $3$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $4$ m wide along the top and bottom.
The path is $3$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>6$ and $y>8$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

3

m on each side across and

4

m at the top and bottom.

\text{width}=x-3-3=x-6,\quad \text{length}=y-4-4=y-8

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

4

m wide. The left and right strips fit between them, so they are

y-8

long and

3

m wide.

A=2\times(4x)+2\times3(y-8)=8x+6y-48

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-6)+2(y-8)=2x+2y-28

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(8x+6y-48)+(2x+2y-28)=10x+8y-76

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

10x+8y-76

4 · Reviewdoes it hold up?

Try $x=20$ , $y=14$ . The field is $14$ by $6$ , so the path's area is $280-84=196$ and $8\times20+6\times14-48=196$ . The field's perimeter is $2\times(14+6)=40$ and $2\times20+2\times14-28=40$ . Then $196+40=236$ , and $10\times20+8\times14-76=236$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-6)(y-8)$ . Expanding gives $xy-(xy-8x-6y+48)=8x+6y-48$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 8 medium

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $2$ m wide along the top and bottom and $4$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $2$ m wide at top and bottom, $4$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $2$ m wide along the top and bottom.
The path is $4$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>8$ and $y>4$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

4

m on each side across and

2

m at the top and bottom.

\text{width}=x-4-4=x-8,\quad \text{length}=y-2-2=y-4

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

2

m wide. The left and right strips fit between them, so they are

y-4

long and

4

m wide.

A=2\times(2x)+2\times4(y-4)=4x+8y-32

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-8)+2(y-4)=2x+2y-24

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(4x+8y-32)+(2x+2y-24)=6x+10y-56

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

6x+10y-56

4 · Reviewdoes it hold up?

Try $x=22$ , $y=10$ . The field is $14$ by $6$ , so the path's area is $220-84=136$ and $4\times22+8\times10-32=136$ . The field's perimeter is $2\times(14+6)=40$ and $2\times22+2\times10-24=40$ . Then $136+40=176$ , and $6\times22+10\times10-56=176$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-8)(y-4)$ . Expanding gives $xy-(xy-4x-8y+32)=4x+8y-32$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 9 medium

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $3$ m wide along the top and bottom and $4$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $3$ m wide at top and bottom, $4$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $3$ m wide along the top and bottom.
The path is $4$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>8$ and $y>6$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

4

m on each side across and

3

m at the top and bottom.

\text{width}=x-4-4=x-8,\quad \text{length}=y-3-3=y-6

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

3

m wide. The left and right strips fit between them, so they are

y-6

long and

4

m wide.

A=2\times(3x)+2\times4(y-6)=6x+8y-48

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-8)+2(y-6)=2x+2y-28

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(6x+8y-48)+(2x+2y-28)=8x+10y-76

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

8x+10y-76

4 · Reviewdoes it hold up?

Try $x=22$ , $y=12$ . The field is $14$ by $6$ , so the path's area is $264-84=180$ and $6\times22+8\times12-48=180$ . The field's perimeter is $2\times(14+6)=40$ and $2\times22+2\times12-28=40$ . Then $180+40=220$ , and $8\times22+10\times12-76=220$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-8)(y-6)$ . Expanding gives $xy-(xy-6x-8y+48)=6x+8y-48$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 10 hard

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $4$ m wide along the top and bottom and $5$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $4$ m wide at top and bottom, $5$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $4$ m wide along the top and bottom.
The path is $5$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>10$ and $y>8$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

5

m on each side across and

4

m at the top and bottom.

\text{width}=x-5-5=x-10,\quad \text{length}=y-4-4=y-8

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

4

m wide. The left and right strips fit between them, so they are

y-8

long and

5

m wide.

A=2\times(4x)+2\times5(y-8)=8x+10y-80

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-10)+2(y-8)=2x+2y-36

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(8x+10y-80)+(2x+2y-36)=10x+12y-116

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

10x+12y-116

4 · Reviewdoes it hold up?

Try $x=24$ , $y=14$ . The field is $14$ by $6$ , so the path's area is $336-84=252$ and $8\times24+10\times14-80=252$ . The field's perimeter is $2\times(14+6)=40$ and $2\times24+2\times14-36=40$ . Then $252+40=292$ , and $10\times24+12\times14-116=292$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-10)(y-8)$ . Expanding gives $xy-(xy-8x-10y+80)=8x+10y-80$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 11 hard

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $2$ m wide along the top and bottom and $5$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $2$ m wide at top and bottom, $5$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $2$ m wide along the top and bottom.
The path is $5$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>10$ and $y>4$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

5

m on each side across and

2

m at the top and bottom.

\text{width}=x-5-5=x-10,\quad \text{length}=y-2-2=y-4

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

2

m wide. The left and right strips fit between them, so they are

y-4

long and

5

m wide.

A=2\times(2x)+2\times5(y-4)=4x+10y-40

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-10)+2(y-4)=2x+2y-28

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(4x+10y-40)+(2x+2y-28)=6x+12y-68

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

6x+12y-68

4 · Reviewdoes it hold up?

Try $x=24$ , $y=10$ . The field is $14$ by $6$ , so the path's area is $240-84=156$ and $4\times24+10\times10-40=156$ . The field's perimeter is $2\times(14+6)=40$ and $2\times24+2\times10-28=40$ . Then $156+40=196$ , and $6\times24+12\times10-68=196$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-10)(y-4)$ . Expanding gives $xy-(xy-4x-10y+40)=4x+10y-40$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Variant 12 hard

A rectangular plot of land is $x$ m wide and $y$ m long. A path is built along its edges, on the inside of the plot, and the rest of the plot becomes a field. As shown, the path is $3$ m wide along the top and bottom and $5$ m wide along the left and right sides. Let the area of the path be $A\ \text{m}^2$ and the perimeter of the field be $B$ m. Write $A+B$ in terms of $x$ and $y$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A path runs around the inside edge of an $x$ by $y$ rectangle: $3$ m wide at top and bottom, $5$ m wide at the sides. Find the path's area $A$ and the inner field's perimeter $B$ , then add them.

Givens

The plot is $x$ m wide and $y$ m long.
The path is $3$ m wide along the top and bottom.
The path is $5$ m wide along the left and right.

Unknowns

$A+B$ as an expression in $x$ and $y$ , where $A$ is the path's area and $B$ is the field's perimeter.

Constraints

The path is inside the plot, so the field is smaller than the plot on every side.
$x>10$ and $y>6$ so the field exists.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

First find the field's side lengths from the picture. Then cut the path into four strips whose areas are easy linear expressions, and get the field's perimeter from its sides.

3 · Execute4 carry out the plan

1Field dimensions

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

The field loses

5

m on each side across and

3

m at the top and bottom.

\text{width}=x-5-5=x-10,\quad \text{length}=y-3-3=y-6

Subtract both borders, not just one.

2Path area, strip by strip

#7 Identify Subproblems 7.G.B.6

The top and bottom strips run the full width

x

and are

3

m wide. The left and right strips fit between them, so they are

y-6

long and

5

m wide.

A=2\times(3x)+2\times5(y-6)=6x+10y-60

Cutting the border into rectangles keeps every piece linear.

3Field perimeter

#13 Convert to Algebra 7.EE.A.1

Use the field's width and length.

B=2(x-10)+2(y-6)=2x+2y-32

Distribute the

2

and combine the constants.

4Add

#13 Convert to Algebra 7.EE.A.1

Combine like terms.

A+B=(6x+10y-60)+(2x+2y-32)=8x+12y-92

x

-terms with

x

-terms,

y

-terms with

y

-terms, numbers with numbers.

Answer:

8x+12y-92

4 · Reviewdoes it hold up?

Try $x=24$ , $y=12$ . The field is $14$ by $6$ , so the path's area is $288-84=204$ and $6\times24+10\times12-60=204$ . The field's perimeter is $2\times(14+6)=40$ and $2\times24+2\times12-32=40$ . Then $204+40=244$ , and $8\times24+12\times12-92=244$ .

Another way: Find the path area as whole plot minus field: $A=xy-(x-10)(y-6)$ . Expanding gives $xy-(xy-6x-10y+60)=6x+10y-60$ , the same as the strip method.

Standardsmin grade 7

7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Writing side lengths, perimeter and the total as linear expressions and combining like terms.
7.G.B.6 Solve real-world problems involving area, surface area, and volume — Finding the area of a border region by splitting it into rectangles.

💡Takeaway. Break a frame-shaped region into simple strips and the algebra stays easy.

Measure a border path strip by strip · 12 practice problems

Generated variants — 12

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Givens

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Constraints

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Where next?