← A fold moves paper without stretching it · Transformations Preserve Measures

A fold moves paper without stretching it · 12 practice problems

4.MD.A.36.G.A.18.G.B.7

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 120 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 10 cm. Along the bottom BF is 6 cm, and the folded edge FE is 9 cm. We need the area of the original sheet.

Givens

The crease FC is 10 cm.
BF is 6 cm along the bottom edge.
The folded edge FE is 9 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 6 and the sheet's height BC, and its long side is the crease FC = 10.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 6 and the crease FC = 10 is the longest side. The missing side BC satisfies 6 x 6 plus BC x BC equals 10 x 10.

6^2 + BC^2 = 10^2 \Rightarrow 36 + BC^2 = 100 \Rightarrow BC = 8

The sheet is 8 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 9 cm long.

\text{width} = BF + FE = 6 + 9 = 15

The sheet is 15 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 15 cm wide and 8 cm tall, so its area is length times width.

15 \times 8 = 120

The sheet is 120 cm2.

Answer: 120 cm²

4 · Reviewdoes it hold up?

Check the triangle: 6 x 6 + 8 x 8 = 36 + 64 = 100, and 10 x 10 is 100. The crease really is 10 cm.

Another way: Sanity-check the shape: the crease 10 cm has to be longer than either of BF = 6 and BC = 8 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 2 easy answer: 192 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 13 cm. Along the bottom BF is 5 cm, and the folded edge FE is 11 cm. We need the area of the original sheet.

Givens

The crease FC is 13 cm.
BF is 5 cm along the bottom edge.
The folded edge FE is 11 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 5 and the sheet's height BC, and its long side is the crease FC = 13.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 5 and the crease FC = 13 is the longest side. The missing side BC satisfies 5 x 5 plus BC x BC equals 13 x 13.

5^2 + BC^2 = 13^2 \Rightarrow 25 + BC^2 = 169 \Rightarrow BC = 12

The sheet is 12 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 11 cm long.

\text{width} = BF + FE = 5 + 11 = 16

The sheet is 16 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 16 cm wide and 12 cm tall, so its area is length times width.

16 \times 12 = 192

The sheet is 192 cm2.

Answer: 192 cm²

4 · Reviewdoes it hold up?

Check the triangle: 5 x 5 + 12 x 12 = 25 + 144 = 169, and 13 x 13 is 169. The crease really is 13 cm.

Another way: Sanity-check the shape: the crease 13 cm has to be longer than either of BF = 5 and BC = 12 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 3 easy answer: 252 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 15 cm. Along the bottom BF is 9 cm, and the folded edge FE is 12 cm. We need the area of the original sheet.

Givens

The crease FC is 15 cm.
BF is 9 cm along the bottom edge.
The folded edge FE is 12 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 9 and the sheet's height BC, and its long side is the crease FC = 15.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 9 and the crease FC = 15 is the longest side. The missing side BC satisfies 9 x 9 plus BC x BC equals 15 x 15.

9^2 + BC^2 = 15^2 \Rightarrow 81 + BC^2 = 225 \Rightarrow BC = 12

The sheet is 12 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 12 cm long.

\text{width} = BF + FE = 9 + 12 = 21

The sheet is 21 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 21 cm wide and 12 cm tall, so its area is length times width.

21 \times 12 = 252

The sheet is 252 cm2.

Answer: 252 cm²

4 · Reviewdoes it hold up?

Check the triangle: 9 x 9 + 12 x 12 = 81 + 144 = 225, and 15 x 15 is 225. The crease really is 15 cm.

Another way: Sanity-check the shape: the crease 15 cm has to be longer than either of BF = 9 and BC = 12 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 4 easy answer: 330 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 17 cm. Along the bottom BF is 8 cm, and the folded edge FE is 14 cm. We need the area of the original sheet.

Givens

The crease FC is 17 cm.
BF is 8 cm along the bottom edge.
The folded edge FE is 14 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 8 and the sheet's height BC, and its long side is the crease FC = 17.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 8 and the crease FC = 17 is the longest side. The missing side BC satisfies 8 x 8 plus BC x BC equals 17 x 17.

8^2 + BC^2 = 17^2 \Rightarrow 64 + BC^2 = 289 \Rightarrow BC = 15

The sheet is 15 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 14 cm long.

\text{width} = BF + FE = 8 + 14 = 22

The sheet is 22 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 22 cm wide and 15 cm tall, so its area is length times width.

22 \times 15 = 330

The sheet is 330 cm2.

Answer: 330 cm²

4 · Reviewdoes it hold up?

Check the triangle: 8 x 8 + 15 x 15 = 64 + 225 = 289, and 17 x 17 is 289. The crease really is 17 cm.

Another way: Sanity-check the shape: the crease 17 cm has to be longer than either of BF = 8 and BC = 15 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 5 medium answer: 432 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 20 cm. Along the bottom BF is 12 cm, and the folded edge FE is 15 cm. We need the area of the original sheet.

Givens

The crease FC is 20 cm.
BF is 12 cm along the bottom edge.
The folded edge FE is 15 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 12 and the sheet's height BC, and its long side is the crease FC = 20.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 12 and the crease FC = 20 is the longest side. The missing side BC satisfies 12 x 12 plus BC x BC equals 20 x 20.

12^2 + BC^2 = 20^2 \Rightarrow 144 + BC^2 = 400 \Rightarrow BC = 16

The sheet is 16 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 15 cm long.

\text{width} = BF + FE = 12 + 15 = 27

The sheet is 27 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 27 cm wide and 16 cm tall, so its area is length times width.

27 \times 16 = 432

The sheet is 432 cm2.

Answer: 432 cm²

4 · Reviewdoes it hold up?

Check the triangle: 12 x 12 + 16 x 16 = 144 + 256 = 400, and 20 x 20 is 400. The crease really is 20 cm.

Another way: Sanity-check the shape: the crease 20 cm has to be longer than either of BF = 12 and BC = 16 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 6 medium answer: 660 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 25 cm. Along the bottom BF is 15 cm, and the folded edge FE is 18 cm. We need the area of the original sheet.

Givens

The crease FC is 25 cm.
BF is 15 cm along the bottom edge.
The folded edge FE is 18 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 15 and the sheet's height BC, and its long side is the crease FC = 25.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 15 and the crease FC = 25 is the longest side. The missing side BC satisfies 15 x 15 plus BC x BC equals 25 x 25.

15^2 + BC^2 = 25^2 \Rightarrow 225 + BC^2 = 625 \Rightarrow BC = 20

The sheet is 20 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 18 cm long.

\text{width} = BF + FE = 15 + 18 = 33

The sheet is 33 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 33 cm wide and 20 cm tall, so its area is length times width.

33 \times 20 = 660

The sheet is 660 cm2.

Answer: 660 cm²

4 · Reviewdoes it hold up?

Check the triangle: 15 x 15 + 20 x 20 = 225 + 400 = 625, and 25 x 25 is 625. The crease really is 25 cm.

Another way: Sanity-check the shape: the crease 25 cm has to be longer than either of BF = 15 and BC = 20 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 7 medium answer: 648 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 25 cm. Along the bottom BF is 7 cm, and the folded edge FE is 20 cm. We need the area of the original sheet.

Givens

The crease FC is 25 cm.
BF is 7 cm along the bottom edge.
The folded edge FE is 20 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 7 and the sheet's height BC, and its long side is the crease FC = 25.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 7 and the crease FC = 25 is the longest side. The missing side BC satisfies 7 x 7 plus BC x BC equals 25 x 25.

7^2 + BC^2 = 25^2 \Rightarrow 49 + BC^2 = 625 \Rightarrow BC = 24

The sheet is 24 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 20 cm long.

\text{width} = BF + FE = 7 + 20 = 27

The sheet is 27 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 27 cm wide and 24 cm tall, so its area is length times width.

27 \times 24 = 648

The sheet is 648 cm2.

Answer: 648 cm²

4 · Reviewdoes it hold up?

Check the triangle: 7 x 7 + 24 x 24 = 49 + 576 = 625, and 25 x 25 is 625. The crease really is 25 cm.

Another way: Sanity-check the shape: the crease 25 cm has to be longer than either of BF = 7 and BC = 24 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 8 medium answer: 696 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 26 cm. Along the bottom BF is 10 cm, and the folded edge FE is 19 cm. We need the area of the original sheet.

Givens

The crease FC is 26 cm.
BF is 10 cm along the bottom edge.
The folded edge FE is 19 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 10 and the sheet's height BC, and its long side is the crease FC = 26.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 10 and the crease FC = 26 is the longest side. The missing side BC satisfies 10 x 10 plus BC x BC equals 26 x 26.

10^2 + BC^2 = 26^2 \Rightarrow 100 + BC^2 = 676 \Rightarrow BC = 24

The sheet is 24 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 19 cm long.

\text{width} = BF + FE = 10 + 19 = 29

The sheet is 29 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 29 cm wide and 24 cm tall, so its area is length times width.

29 \times 24 = 696

The sheet is 696 cm2.

Answer: 696 cm²

4 · Reviewdoes it hold up?

Check the triangle: 10 x 10 + 24 x 24 = 100 + 576 = 676, and 26 x 26 is 676. The crease really is 26 cm.

Another way: Sanity-check the shape: the crease 26 cm has to be longer than either of BF = 10 and BC = 24 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 9 hard answer: 756 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 29 cm. Along the bottom BF is 20 cm, and the folded edge FE is 16 cm. We need the area of the original sheet.

Givens

The crease FC is 29 cm.
BF is 20 cm along the bottom edge.
The folded edge FE is 16 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 20 and the sheet's height BC, and its long side is the crease FC = 29.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 20 and the crease FC = 29 is the longest side. The missing side BC satisfies 20 x 20 plus BC x BC equals 29 x 29.

20^2 + BC^2 = 29^2 \Rightarrow 400 + BC^2 = 841 \Rightarrow BC = 21

The sheet is 21 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 16 cm long.

\text{width} = BF + FE = 20 + 16 = 36

The sheet is 36 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 36 cm wide and 21 cm tall, so its area is length times width.

36 \times 21 = 756

The sheet is 756 cm2.

Answer: 756 cm²

4 · Reviewdoes it hold up?

Check the triangle: 20 x 20 + 21 x 21 = 400 + 441 = 841, and 29 x 29 is 841. The crease really is 29 cm.

Another way: Sanity-check the shape: the crease 29 cm has to be longer than either of BF = 20 and BC = 21 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 10 hard answer: 960 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 30 cm. Along the bottom BF is 18 cm, and the folded edge FE is 22 cm. We need the area of the original sheet.

Givens

The crease FC is 30 cm.
BF is 18 cm along the bottom edge.
The folded edge FE is 22 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 18 and the sheet's height BC, and its long side is the crease FC = 30.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 18 and the crease FC = 30 is the longest side. The missing side BC satisfies 18 x 18 plus BC x BC equals 30 x 30.

18^2 + BC^2 = 30^2 \Rightarrow 324 + BC^2 = 900 \Rightarrow BC = 24

The sheet is 24 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 22 cm long.

\text{width} = BF + FE = 18 + 22 = 40

The sheet is 40 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 40 cm wide and 24 cm tall, so its area is length times width.

40 \times 24 = 960

The sheet is 960 cm2.

Answer: 960 cm²

4 · Reviewdoes it hold up?

Check the triangle: 18 x 18 + 24 x 24 = 324 + 576 = 900, and 30 x 30 is 900. The crease really is 30 cm.

Another way: Sanity-check the shape: the crease 30 cm has to be longer than either of BF = 18 and BC = 24 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 11 hard answer: 1600 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 40 cm. Along the bottom BF is 24 cm, and the folded edge FE is 26 cm. We need the area of the original sheet.

Givens

The crease FC is 40 cm.
BF is 24 cm along the bottom edge.
The folded edge FE is 26 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

Draw the rectangle and the crease FC. Corner B of the rectangle is a square corner, so triangle BFC has a right angle at B. Its two short sides are BF = 24 and the sheet's height BC, and its long side is the crease FC = 40.

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 24 and the crease FC = 40 is the longest side. The missing side BC satisfies 24 x 24 plus BC x BC equals 40 x 40.

24^2 + BC^2 = 40^2 \Rightarrow 576 + BC^2 = 1600 \Rightarrow BC = 32

The sheet is 32 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 26 cm long.

\text{width} = BF + FE = 24 + 26 = 50

The sheet is 50 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 50 cm wide and 32 cm tall, so its area is length times width.

50 \times 32 = 1600

The sheet is 1600 cm2.

Answer: 1600 cm²

4 · Reviewdoes it hold up?

Check the triangle: 24 x 24 + 32 x 32 = 576 + 1024 = 1600, and 40 x 40 is 1600. The crease really is 40 cm.

Another way: Sanity-check the shape: the crease 40 cm has to be longer than either of BF = 24 and BC = 32 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.

Variant 12 hard answer: 1560 cm²

A rectangular sheet of paper is folded as shown in the figure. Find the area of the original sheet of paper in $\text{cm}^2$ .

Show solution

1 · Understandwhat's really being asked

A rectangular sheet is folded so its lower-right corner turns up along a crease FC of 41 cm. Along the bottom BF is 9 cm, and the folded edge FE is 30 cm. We need the area of the original sheet.

Givens

The crease FC is 41 cm.
BF is 9 cm along the bottom edge.
The folded edge FE is 30 cm.

Unknowns

The area of the original rectangle.

Constraints

Folding moves paper without stretching it, so a folded edge keeps its length.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #10 Create a Physical Representation#7 Identify Subproblems

Draw the sheet and the crease. The corner at B is square, which turns the crease into a hypotenuse and gives the height. Then the fold's own rule -- no stretching -- gives the width.

3 · Execute4 carry out the plan

1Draw and label the fold

#1 Draw a Diagram 6.G.A.1

The crease is the hypotenuse of a right triangle.

2Find the height with the right triangle

#7 Identify Subproblems 8.G.B.7

In right triangle BFC, BF = 9 and the crease FC = 41 is the longest side. The missing side BC satisfies 9 x 9 plus BC x BC equals 41 x 41.

9^2 + BC^2 = 41^2 \Rightarrow 81 + BC^2 = 1681 \Rightarrow BC = 40

The sheet is 40 cm tall.

3Find the width using the fold

#10 Create a Physical Representation 6.G.A.1

Folding the lower-right corner up lays the bottom edge from F to the old corner onto the segment FE. A fold never stretches the paper, so that stretch of the bottom edge is exactly FE = 30 cm long.

\text{width} = BF + FE = 9 + 30 = 39

The sheet is 39 cm wide.

4Multiply to get the area

#7 Identify Subproblems 4.MD.A.3

The rectangle is 39 cm wide and 40 cm tall, so its area is length times width.

39 \times 40 = 1560

The sheet is 1560 cm2.

Answer: 1560 cm²

4 · Reviewdoes it hold up?

Check the triangle: 9 x 9 + 40 x 40 = 81 + 1600 = 1681, and 41 x 41 is 1681. The crease really is 41 cm.

Another way: Sanity-check the shape: the crease 41 cm has to be longer than either of BF = 9 and BC = 40 but shorter than the two added together -- and it is.

Standardsmin grade 8

4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area from its width and height.
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the fold as a triangle laid over the sheet.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths — Getting the sheet's height from the crease and BF.

💡Takeaway. Paper does not stretch when you fold it, so a folded edge is the same length before and after.