Practice · Add the shared part back and the two wholes must be equal · Sensim Math

Variant 1 easy

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 4 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 4 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 4 cm.

4 \times 4 \times 3.14 \div 4 = 12.56

12.56 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 4 cm along BC, and its area is now known.

4 \times \text{AB} = 12.56

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

12.56 \div 4 = 3.14

AB is 3.14 cm.

Answer: 3.14 cm

4 · Reviewdoes it hold up?

AB is 3.14 cm, less than the 4 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 2 easy

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 6 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 6 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 6 cm.

6 \times 6 \times 3.14 \div 4 = 28.26

28.26 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 6 cm along BC, and its area is now known.

6 \times \text{AB} = 28.26

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

28.26 \div 6 = 4.71

AB is 4.71 cm.

Answer: 4.71 cm

4 · Reviewdoes it hold up?

AB is 4.71 cm, less than the 6 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 3 easy

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 8 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 8 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 8 cm.

8 \times 8 \times 3.14 \div 4 = 50.24

50.24 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 8 cm along BC, and its area is now known.

8 \times \text{AB} = 50.24

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

50.24 \div 8 = 6.28

AB is 6.28 cm.

Answer: 6.28 cm

4 · Reviewdoes it hold up?

AB is 6.28 cm, less than the 8 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 4 easy

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 10 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 10 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 10 cm.

10 \times 10 \times 3.14 \div 4 = 78.5

78.5 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 10 cm along BC, and its area is now known.

10 \times \text{AB} = 78.5

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

78.5 \div 10 = 7.85

AB is 7.85 cm.

Answer: 7.85 cm

4 · Reviewdoes it hold up?

AB is 7.85 cm, less than the 10 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 5 medium

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 12 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 12 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 12 cm.

12 \times 12 \times 3.14 \div 4 = 113.04

113.04 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 12 cm along BC, and its area is now known.

12 \times \text{AB} = 113.04

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

113.04 \div 12 = 9.42

AB is 9.42 cm.

Answer: 9.42 cm

4 · Reviewdoes it hold up?

AB is 9.42 cm, less than the 12 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 6 medium

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 14 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 14 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 14 cm.

14 \times 14 \times 3.14 \div 4 = 153.86

153.86 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 14 cm along BC, and its area is now known.

14 \times \text{AB} = 153.86

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

153.86 \div 14 = 10.99

AB is 10.99 cm.

Answer: 10.99 cm

4 · Reviewdoes it hold up?

AB is 10.99 cm, less than the 14 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 7 medium

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 16 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 16 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 16 cm.

16 \times 16 \times 3.14 \div 4 = 200.96

200.96 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 16 cm along BC, and its area is now known.

16 \times \text{AB} = 200.96

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

200.96 \div 16 = 12.56

AB is 12.56 cm.

Answer: 12.56 cm

4 · Reviewdoes it hold up?

AB is 12.56 cm, less than the 16 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 8 medium

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 20 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 20 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 20 cm.

20 \times 20 \times 3.14 \div 4 = 314

314 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 20 cm along BC, and its area is now known.

20 \times \text{AB} = 314

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

314 \div 20 = 15.7

AB is 15.7 cm.

Answer: 15.7 cm

4 · Reviewdoes it hold up?

AB is 15.7 cm, less than the 20 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 9 hard

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 24 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 24 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 24 cm.

24 \times 24 \times 3.14 \div 4 = 452.16

452.16 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 24 cm along BC, and its area is now known.

24 \times \text{AB} = 452.16

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

452.16 \div 24 = 18.84

AB is 18.84 cm.

Answer: 18.84 cm

4 · Reviewdoes it hold up?

AB is 18.84 cm, less than the 24 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 10 hard

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 30 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 30 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 30 cm.

30 \times 30 \times 3.14 \div 4 = 706.5

706.5 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 30 cm along BC, and its area is now known.

30 \times \text{AB} = 706.5

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

706.5 \div 30 = 23.55

AB is 23.55 cm.

Answer: 23.55 cm

4 · Reviewdoes it hold up?

AB is 23.55 cm, less than the 30 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 11 hard

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 40 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 40 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 40 cm.

40 \times 40 \times 3.14 \div 4 = 1256

1256 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 40 cm along BC, and its area is now known.

40 \times \text{AB} = 1256

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

1256 \div 40 = 31.4

AB is 31.4 cm.

Answer: 31.4 cm

4 · Reviewdoes it hold up?

AB is 31.4 cm, less than the 40 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Variant 12 hard

The figure at the right shows a rectangle overlapping part of a circle. If the two shaded parts have equal areas, how long is segment AB, in $\text{cm}$ ?

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1 · Understandwhat's really being asked

A rectangle ABCD overlaps a quarter circle of radius 50 cm centred at C. The two parts that stick out of each other have equal areas, and we want AB.

Givens

BC is 50 cm and is also the quarter circle's radius.
The rectangle's part outside the quarter circle and the quarter circle's part outside the rectangle have equal areas.

Unknowns

The length of AB.

Constraints

Neither shaded part can be measured directly: both have a curved edge.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#7 Identify Subproblems

Both shaded pieces are what is left after taking away the same overlap. Add that overlap back to each and they become the whole rectangle and the whole quarter circle -- which must therefore be equal.

3 · Execute4 carry out the plan

1Put the overlap back

#16 Count the Complement 6.G.A.1

Each shaded piece is its own shape minus the shared middle. Adding the middle to both keeps them equal.

\text{rectangle} = \text{quarter circle}

Two equal wholes, not two awkward pieces.

2Measure the quarter circle

#7 Identify Subproblems 7.G.B.4

A quarter of a circle of radius 50 cm.

50 \times 50 \times 3.14 \div 4 = 1962.5

1962.5 cm2.

3Use it as the rectangle's area

#13 Convert to Algebra 6.EE.B.7

The rectangle is 50 cm along BC, and its area is now known.

50 \times \text{AB} = 1962.5

One equation, one unknown.

4Divide for AB

#13 Convert to Algebra 6.EE.B.7

The other side of a rectangle from its area.

1962.5 \div 50 = 39.25

AB is 39.25 cm.

Answer: 39.25 cm

4 · Reviewdoes it hold up?

AB is 39.25 cm, less than the 50 cm radius, so the rectangle really does sit lower than the top of the arc -- which is what makes both shaded parts exist.

Another way: Trying to measure either shaded part on its own needs the area under the arc, which is not something this unit gives a formula for; putting the overlap back is what avoids it.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the quarter circle's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Reasoning about the two regions and their shared part.
6.EE.B.7 Solve one-step equations x+p=q and px=q — Solving for AB from the rectangle's area.

💡Takeaway. If two leftovers are equal after losing the same piece, the wholes they came from were equal too.

Add the shared part back and the two wholes must be equal · 12 practice problems

Generated variants — 12

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