Practice · A lens is two sectors with their triangles taken out · Sensim Math

Variant 1 easy

The figure at the right shows parts of circles overlapping inside a square with sides of $8\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 4 cm cross inside a 8 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 8 cm.
Both arcs have radius 4 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 4 cm, and a right triangle with both legs 4 cm.

12.56 - 8 = 4.56

Each piece is 4.56 cm2.

4Double it

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

The two pieces are congruent.

4.56 \times 2 = 9.12

The lens is 9.12 cm2.

Answer: 9.12 cm²

4 · Reviewdoes it hold up?

The lens, 9.12 cm2, is smaller than one quarter circle, 12.56 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 2 easy

The figure at the right shows parts of circles overlapping inside a square with sides of $12\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 6 cm cross inside a 12 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 12 cm.
Both arcs have radius 6 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 6 cm, and a right triangle with both legs 6 cm.

28.26 - 18 = 10.26

Each piece is 10.26 cm2.

4Double it

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

The two pieces are congruent.

10.26 \times 2 = 20.52

The lens is 20.52 cm2.

Answer: 20.52 cm²

4 · Reviewdoes it hold up?

The lens, 20.52 cm2, is smaller than one quarter circle, 28.26 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 3 easy

The figure at the right shows parts of circles overlapping inside a square with sides of $16\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 8 cm cross inside a 16 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 16 cm.
Both arcs have radius 8 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 8 cm, and a right triangle with both legs 8 cm.

50.24 - 32 = 18.24

Each piece is 18.24 cm2.

4Double it

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

The two pieces are congruent.

18.24 \times 2 = 36.48

The lens is 36.48 cm2.

Answer: 36.48 cm²

4 · Reviewdoes it hold up?

The lens, 36.48 cm2, is smaller than one quarter circle, 50.24 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 4 easy

The figure at the right shows parts of circles overlapping inside a square with sides of $20\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 10 cm cross inside a 20 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 20 cm.
Both arcs have radius 10 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 10 cm, and a right triangle with both legs 10 cm.

78.5 - 50 = 28.5

Each piece is 28.5 cm2.

4Double it

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

The two pieces are congruent.

28.5 \times 2 = 57

The lens is 57 cm2.

Answer: 57 cm²

4 · Reviewdoes it hold up?

The lens, 57 cm2, is smaller than one quarter circle, 78.5 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 5 medium

The figure at the right shows parts of circles overlapping inside a square with sides of $24\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 12 cm cross inside a 24 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 24 cm.
Both arcs have radius 12 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 12 cm, and a right triangle with both legs 12 cm.

113.04 - 72 = 41.04

Each piece is 41.04 cm2.

4Double it

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

The two pieces are congruent.

41.04 \times 2 = 82.08

The lens is 82.08 cm2.

Answer: 82.08 cm²

4 · Reviewdoes it hold up?

The lens, 82.08 cm2, is smaller than one quarter circle, 113.04 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 6 medium

The figure at the right shows parts of circles overlapping inside a square with sides of $28\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 14 cm cross inside a 28 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 28 cm.
Both arcs have radius 14 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 14 cm, and a right triangle with both legs 14 cm.

153.86 - 98 = 55.86

Each piece is 55.86 cm2.

4Double it

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

The two pieces are congruent.

55.86 \times 2 = 111.72

The lens is 111.72 cm2.

Answer: 111.72 cm²

4 · Reviewdoes it hold up?

The lens, 111.72 cm2, is smaller than one quarter circle, 153.86 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 7 medium

The figure at the right shows parts of circles overlapping inside a square with sides of $32\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 16 cm cross inside a 32 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 32 cm.
Both arcs have radius 16 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 16 cm, and a right triangle with both legs 16 cm.

200.96 - 128 = 72.96

Each piece is 72.96 cm2.

4Double it

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

The two pieces are congruent.

72.96 \times 2 = 145.92

The lens is 145.92 cm2.

Answer: 145.92 cm²

4 · Reviewdoes it hold up?

The lens, 145.92 cm2, is smaller than one quarter circle, 200.96 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 8 medium

The figure at the right shows parts of circles overlapping inside a square with sides of $36\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 18 cm cross inside a 36 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 36 cm.
Both arcs have radius 18 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 18 cm, and a right triangle with both legs 18 cm.

254.34 - 162 = 92.34

Each piece is 92.34 cm2.

4Double it

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

The two pieces are congruent.

92.34 \times 2 = 184.68

The lens is 184.68 cm2.

Answer: 184.68 cm²

4 · Reviewdoes it hold up?

The lens, 184.68 cm2, is smaller than one quarter circle, 254.34 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 9 hard

The figure at the right shows parts of circles overlapping inside a square with sides of $40\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 20 cm cross inside a 40 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 40 cm.
Both arcs have radius 20 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 20 cm, and a right triangle with both legs 20 cm.

314 - 200 = 114

Each piece is 114 cm2.

4Double it

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

The two pieces are congruent.

114 \times 2 = 228

The lens is 228 cm2.

Answer: 228 cm²

4 · Reviewdoes it hold up?

The lens, 228 cm2, is smaller than one quarter circle, 314 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 10 hard

The figure at the right shows parts of circles overlapping inside a square with sides of $44\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 22 cm cross inside a 44 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 44 cm.
Both arcs have radius 22 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 22 cm, and a right triangle with both legs 22 cm.

379.94 - 242 = 137.94

Each piece is 137.94 cm2.

4Double it

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

The two pieces are congruent.

137.94 \times 2 = 275.88

The lens is 275.88 cm2.

Answer: 275.88 cm²

4 · Reviewdoes it hold up?

The lens, 275.88 cm2, is smaller than one quarter circle, 379.94 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 11 hard

The figure at the right shows parts of circles overlapping inside a square with sides of $48\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 24 cm cross inside a 48 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 48 cm.
Both arcs have radius 24 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 24 cm, and a right triangle with both legs 24 cm.

452.16 - 288 = 164.16

Each piece is 164.16 cm2.

4Double it

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

The two pieces are congruent.

164.16 \times 2 = 328.32

The lens is 328.32 cm2.

Answer: 328.32 cm²

4 · Reviewdoes it hold up?

The lens, 328.32 cm2, is smaller than one quarter circle, 452.16 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

Variant 12 hard

The figure at the right shows parts of circles overlapping inside a square with sides of $52\ \text{cm}$ . What is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Two arcs of radius 26 cm cross inside a 52 cm square, and the lens between them is shaded. We want its area.

Givens

The square's side is 52 cm.
Both arcs have radius 26 cm.
They cross at two points.

Unknowns

The area of the lens between the two arcs.

Constraints

The lens is bounded by two curves and no straight edge.

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#16 Count the Complement

A lens has no formula, so cut it into pieces that do. Joining the two crossings splits it into two identical halves, each a sector with a right triangle removed.

3 · Execute4 carry out the plan

1Join the two crossings

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

That chord cuts the lens into two matching pieces, each bounded by one arc and the chord.

\text{lens} = 2 \times \text{piece}

Two identical halves.

2Name one piece

#7 Identify Subproblems 7.G.B.4

The two radii to the crossings meet at a right angle, so the piece is a quarter circle with that triangle taken out.

\text{piece} = \text{quarter} - \text{triangle}

Now both parts have formulas.

3Measure both

#7 Identify Subproblems 7.G.B.4

A quarter circle of radius 26 cm, and a right triangle with both legs 26 cm.

530.66 - 338 = 192.66

Each piece is 192.66 cm2.

4Double it

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

The two pieces are congruent.

192.66 \times 2 = 385.32

The lens is 385.32 cm2.

Answer: 385.32 cm²

4 · Reviewdoes it hold up?

The lens, 385.32 cm2, is smaller than one quarter circle, 530.66 cm2, and bigger than nothing -- which is where a sliver between two arcs has to sit.

Another way: Taking the union of the two quarter circles and subtracting it from their total gives the same lens, since the overlap is exactly what the sum counts twice.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the sector's area.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Cutting the lens into pieces that can be measured.

💡Takeaway. When a shape has no formula, cut it into ones that do. A chord across a lens does exactly that.

A lens is two sectors with their triangles taken out · 12 practice problems

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