Practice · Half of every kind of piece is half the whole · Sensim Math

Variant 1 easy

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 8 cm is cut into six congruent triangles and six congruent outer pieces. 3 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 8 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
3 triangles and 3 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

3 of the six triangles and 3 of the six outer pieces.

\frac{3}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 3 sixths of every kind of piece takes 3 sixths of the whole circle.

\text{shaded} = \frac{3}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 8 cm.

8 \times 8 \times 3.14 \times \frac{3}{6} = 100.48

100.48 cm2 shaded.

Answer: 100.48 cm²

4 · Reviewdoes it hold up?

100.48 cm2 out of the circle's 200.96 cm2 is 3 sixths, which is what shading 3 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 3 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 2 easy

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 12 cm is cut into six congruent triangles and six congruent outer pieces. 3 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 12 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
3 triangles and 3 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

3 of the six triangles and 3 of the six outer pieces.

\frac{3}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 3 sixths of every kind of piece takes 3 sixths of the whole circle.

\text{shaded} = \frac{3}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 12 cm.

12 \times 12 \times 3.14 \times \frac{3}{6} = 226.08

226.08 cm2 shaded.

Answer: 226.08 cm²

4 · Reviewdoes it hold up?

226.08 cm2 out of the circle's 452.16 cm2 is 3 sixths, which is what shading 3 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 3 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 3 easy

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 16 cm is cut into six congruent triangles and six congruent outer pieces. 3 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 16 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
3 triangles and 3 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

3 of the six triangles and 3 of the six outer pieces.

\frac{3}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 3 sixths of every kind of piece takes 3 sixths of the whole circle.

\text{shaded} = \frac{3}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 16 cm.

16 \times 16 \times 3.14 \times \frac{3}{6} = 401.92

401.92 cm2 shaded.

Answer: 401.92 cm²

4 · Reviewdoes it hold up?

401.92 cm2 out of the circle's 803.84 cm2 is 3 sixths, which is what shading 3 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 3 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 4 easy

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 24 cm is cut into six congruent triangles and six congruent outer pieces. 2 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 24 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
2 triangles and 2 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

2 of the six triangles and 2 of the six outer pieces.

\frac{2}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 2 sixths of every kind of piece takes 2 sixths of the whole circle.

\text{shaded} = \frac{2}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 24 cm.

24 \times 24 \times 3.14 \times \frac{2}{6} = 602.88

602.88 cm2 shaded.

Answer: 602.88 cm²

4 · Reviewdoes it hold up?

602.88 cm2 out of the circle's 1808.64 cm2 is 2 sixths, which is what shading 2 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 2 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 5 medium

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 30 cm is cut into six congruent triangles and six congruent outer pieces. 2 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 30 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
2 triangles and 2 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

2 of the six triangles and 2 of the six outer pieces.

\frac{2}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 2 sixths of every kind of piece takes 2 sixths of the whole circle.

\text{shaded} = \frac{2}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 30 cm.

30 \times 30 \times 3.14 \times \frac{2}{6} = 942

942 cm2 shaded.

Answer: 942 cm²

4 · Reviewdoes it hold up?

942 cm2 out of the circle's 2826 cm2 is 2 sixths, which is what shading 2 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 2 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 6 medium

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 38 cm is cut into six congruent triangles and six congruent outer pieces. 3 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 38 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
3 triangles and 3 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

3 of the six triangles and 3 of the six outer pieces.

\frac{3}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 3 sixths of every kind of piece takes 3 sixths of the whole circle.

\text{shaded} = \frac{3}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 38 cm.

38 \times 38 \times 3.14 \times \frac{3}{6} = 2267.08

2267.08 cm2 shaded.

Answer: 2267.08 cm²

4 · Reviewdoes it hold up?

2267.08 cm2 out of the circle's 4534.16 cm2 is 3 sixths, which is what shading 3 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 3 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 7 medium

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 42 cm is cut into six congruent triangles and six congruent outer pieces. 5 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 42 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
5 triangles and 5 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

5 of the six triangles and 5 of the six outer pieces.

\frac{5}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 5 sixths of every kind of piece takes 5 sixths of the whole circle.

\text{shaded} = \frac{5}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 42 cm.

42 \times 42 \times 3.14 \times \frac{5}{6} = 4615.8

4615.8 cm2 shaded.

Answer: 4615.8 cm²

4 · Reviewdoes it hold up?

4615.8 cm2 out of the circle's 5538.96 cm2 is 5 sixths, which is what shading 5 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 5 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 8 hard

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 42 cm is cut into six congruent triangles and six congruent outer pieces. 3 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 42 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
3 triangles and 3 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

3 of the six triangles and 3 of the six outer pieces.

\frac{3}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 3 sixths of every kind of piece takes 3 sixths of the whole circle.

\text{shaded} = \frac{3}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 42 cm.

42 \times 42 \times 3.14 \times \frac{3}{6} = 2769.48

2769.48 cm2 shaded.

Answer: 2769.48 cm²

4 · Reviewdoes it hold up?

2769.48 cm2 out of the circle's 5538.96 cm2 is 3 sixths, which is what shading 3 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 3 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 9 medium

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 42 cm is cut into six congruent triangles and six congruent outer pieces. 4 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 42 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
4 triangles and 4 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

4 of the six triangles and 4 of the six outer pieces.

\frac{4}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 4 sixths of every kind of piece takes 4 sixths of the whole circle.

\text{shaded} = \frac{4}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 42 cm.

42 \times 42 \times 3.14 \times \frac{4}{6} = 3692.64

3692.64 cm2 shaded.

Answer: 3692.64 cm²

4 · Reviewdoes it hold up?

3692.64 cm2 out of the circle's 5538.96 cm2 is 4 sixths, which is what shading 4 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 4 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 10 hard

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 54 cm is cut into six congruent triangles and six congruent outer pieces. 4 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 54 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
4 triangles and 4 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

4 of the six triangles and 4 of the six outer pieces.

\frac{4}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 4 sixths of every kind of piece takes 4 sixths of the whole circle.

\text{shaded} = \frac{4}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 54 cm.

54 \times 54 \times 3.14 \times \frac{4}{6} = 6104.16

6104.16 cm2 shaded.

Answer: 6104.16 cm²

4 · Reviewdoes it hold up?

6104.16 cm2 out of the circle's 9156.24 cm2 is 4 sixths, which is what shading 4 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 4 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 11 hard

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 54 cm is cut into six congruent triangles and six congruent outer pieces. 1 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 54 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
1 triangles and 1 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

1 of the six triangles and 1 of the six outer pieces.

\frac{1}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 1 sixths of every kind of piece takes 1 sixths of the whole circle.

\text{shaded} = \frac{1}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 54 cm.

54 \times 54 \times 3.14 \times \frac{1}{6} = 1526.04

1526.04 cm2 shaded.

Answer: 1526.04 cm²

4 · Reviewdoes it hold up?

1526.04 cm2 out of the circle's 9156.24 cm2 is 1 sixths, which is what shading 1 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 1 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Variant 12 hard

The figure at the right shows congruent equilateral triangles drawn inside a circle, with some of them shaded. What is the total area of the shaded parts, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A circle of radius 60 cm is cut into six congruent triangles and six congruent outer pieces. 4 of each kind are shaded, and we want their total area.

Givens

The circle's radius is 60 cm.
Six congruent equilateral triangles make a regular hexagon inside it.
4 triangles and 4 outer pieces are shaded.

Unknowns

The total area of the shaded parts.

Constraints

The six triangles are congruent, and so are the six outer pieces.

2 · Planchoose the strategy

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

The two kinds of shaded piece are both awkward to measure, and neither needs to be. The same number of each kind is shaded, so sliding them together makes exactly that share of the whole circle.

3 · Execute4 carry out the plan

1Sort the circle into two kinds of piece

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

Six congruent triangles inside the hexagon, and six congruent pieces between the hexagon and the circle.

6 + 6 = 12

Two kinds, six of each.

2Count how many of each are shaded

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

4 of the six triangles and 4 of the six outer pieces.

\frac{4}{6} \text{ of each kind}

The same fraction of both.

3Slide them together

#16 Count the Complement 7.G.B.4

Taking 4 sixths of every kind of piece takes 4 sixths of the whole circle.

\text{shaded} = \frac{4}{6} \times \text{circle}

No awkward shape left to measure.

4Measure that share of the circle

#7 Identify Subproblems 7.G.B.4

A circle of radius 60 cm.

60 \times 60 \times 3.14 \times \frac{4}{6} = 7536

7536 cm2 shaded.

Answer: 7536 cm²

4 · Reviewdoes it hold up?

7536 cm2 out of the circle's 11304 cm2 is 4 sixths, which is what shading 4 of every six pieces has to give.

Another way: Measuring one triangle and one outer piece separately and multiplying by 4 each reaches the same total, but the outer piece needs the circle's area anyway -- so the sliding saves the harder half.

Standardsmin grade 7

7.G.B.4 Know the formulas for area and circumference of a circle — Finding the circle's area and taking a share of it.
6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Sorting the circle into congruent pieces of two kinds.

💡Takeaway. If the shading takes the same share of every kind of piece, it takes that share of the whole thing.

Half of every kind of piece is half the whole · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?