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Congruent pieces turn an area into a count · 12 practice problems

6.G.A.15.NF.B.4

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Generated variants — 12

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

Variant 1 medium

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $11\frac{1}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 11 and 1/5 square centimetres. Four of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 11 and 1/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Four of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

11 and 1/5 cm2 shared equally between 12 tiles.

11\frac{1}{5} \div 12 = \frac{14}{15}

Each tile is 14/15 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 4 of them.

\frac{14}{15} \times 4 = 3\frac{11}{15}

The shaded part is 3 and 11/15 cm2.

Answer:

3\frac{11}{15}

cm²

4 · Reviewdoes it hold up?

4 of 12 is $\frac{1}{3}$ of the star, and $\frac{1}{3} \times 11\frac{1}{5} = 3\frac{11}{15}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 2 easy

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $4\frac{4}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 4 and 4/5 square centimetres. One of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 4 and 4/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
One of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

4 and 4/5 cm2 shared equally between 12 tiles.

4\frac{4}{5} \div 12 = \frac{2}{5}

Each tile is 2/5 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 1 of them.

\frac{2}{5} \times 1 = \frac{2}{5}

The shaded part is 2/5 cm2.

Answer:

\frac{2}{5}

cm²

4 · Reviewdoes it hold up?

1 of 12 is $\frac{1}{12}$ of the star, and $\frac{1}{12} \times 4\frac{4}{5} = \frac{2}{5}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 3 hard

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $12\frac{2}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 12 and 2/5 square centimetres. Five of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 12 and 2/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Five of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

12 and 2/5 cm2 shared equally between 12 tiles.

12\frac{2}{5} \div 12 = 1\frac{1}{30}

Each tile is 1 and 1/30 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 5 of them.

1\frac{1}{30} \times 5 = 5\frac{1}{6}

The shaded part is 5 and 1/6 cm2.

Answer:

5\frac{1}{6}

cm²

4 · Reviewdoes it hold up?

5 of 12 is $\frac{5}{12}$ of the star, and $\frac{5}{12} \times 12\frac{2}{5} = 5\frac{1}{6}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 4 medium

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $10\frac{4}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 10 and 4/5 square centimetres. Five of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 10 and 4/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Five of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

10 and 4/5 cm2 shared equally between 12 tiles.

10\frac{4}{5} \div 12 = \frac{9}{10}

Each tile is 9/10 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 5 of them.

\frac{9}{10} \times 5 = 4\frac{1}{2}

The shaded part is 4 and 1/2 cm2.

Answer:

4\frac{1}{2}

cm²

4 · Reviewdoes it hold up?

5 of 12 is $\frac{5}{12}$ of the star, and $\frac{5}{12} \times 10\frac{4}{5} = 4\frac{1}{2}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 5 medium

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $6\frac{3}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 6 and 3/5 square centimetres. Two of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 6 and 3/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Two of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

6 and 3/5 cm2 shared equally between 12 tiles.

6\frac{3}{5} \div 12 = \frac{11}{20}

Each tile is 11/20 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 2 of them.

\frac{11}{20} \times 2 = 1\frac{1}{10}

The shaded part is 1 and 1/10 cm2.

Answer:

1\frac{1}{10}

cm²

4 · Reviewdoes it hold up?

2 of 12 is $\frac{1}{6}$ of the star, and $\frac{1}{6} \times 6\frac{3}{5} = 1\frac{1}{10}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 6 easy

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $7\frac{1}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 7 and 1/5 square centimetres. Three of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 7 and 1/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Three of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

7 and 1/5 cm2 shared equally between 12 tiles.

7\frac{1}{5} \div 12 = \frac{3}{5}

Each tile is 3/5 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 3 of them.

\frac{3}{5} \times 3 = 1\frac{4}{5}

The shaded part is 1 and 4/5 cm2.

Answer:

1\frac{4}{5}

cm²

4 · Reviewdoes it hold up?

3 of 12 is $\frac{1}{4}$ of the star, and $\frac{1}{4} \times 7\frac{1}{5} = 1\frac{4}{5}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 7 easy

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $5\frac{2}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 5 and 2/5 square centimetres. Two of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 5 and 2/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Two of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

5 and 2/5 cm2 shared equally between 12 tiles.

5\frac{2}{5} \div 12 = \frac{9}{20}

Each tile is 9/20 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 2 of them.

\frac{9}{20} \times 2 = \frac{9}{10}

The shaded part is 9/10 cm2.

Answer:

\frac{9}{10}

cm²

4 · Reviewdoes it hold up?

2 of 12 is $\frac{1}{6}$ of the star, and $\frac{1}{6} \times 5\frac{2}{5} = \frac{9}{10}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 8 medium

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $8\frac{2}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 8 and 2/5 square centimetres. Three of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 8 and 2/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Three of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

8 and 2/5 cm2 shared equally between 12 tiles.

8\frac{2}{5} \div 12 = \frac{7}{10}

Each tile is 7/10 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 3 of them.

\frac{7}{10} \times 3 = 2\frac{1}{10}

The shaded part is 2 and 1/10 cm2.

Answer:

2\frac{1}{10}

cm²

4 · Reviewdoes it hold up?

3 of 12 is $\frac{1}{4}$ of the star, and $\frac{1}{4} \times 8\frac{2}{5} = 2\frac{1}{10}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 9 hard

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $3\frac{3}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 3 and 3/5 square centimetres. One of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 3 and 3/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
One of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

3 and 3/5 cm2 shared equally between 12 tiles.

3\frac{3}{5} \div 12 = \frac{3}{10}

Each tile is 3/10 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 1 of them.

\frac{3}{10} \times 1 = \frac{3}{10}

The shaded part is 3/10 cm2.

Answer:

\frac{3}{10}

cm²

4 · Reviewdoes it hold up?

1 of 12 is $\frac{1}{12}$ of the star, and $\frac{1}{12} \times 3\frac{3}{5} = \frac{3}{10}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 10 easy

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $9\frac{3}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 9 and 3/5 square centimetres. Four of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 9 and 3/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Four of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

9 and 3/5 cm2 shared equally between 12 tiles.

9\frac{3}{5} \div 12 = \frac{4}{5}

Each tile is 4/5 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 4 of them.

\frac{4}{5} \times 4 = 3\frac{1}{5}

The shaded part is 3 and 1/5 cm2.

Answer:

3\frac{1}{5}

cm²

4 · Reviewdoes it hold up?

4 of 12 is $\frac{1}{3}$ of the star, and $\frac{1}{3} \times 9\frac{3}{5} = 3\frac{1}{5}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 11 hard

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $13\frac{4}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 13 and 4/5 square centimetres. Two of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 13 and 4/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Two of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

13 and 4/5 cm2 shared equally between 12 tiles.

13\frac{4}{5} \div 12 = 1\frac{3}{20}

Each tile is 1 and 3/20 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 2 of them.

1\frac{3}{20} \times 2 = 2\frac{3}{10}

The shaded part is 2 and 3/10 cm2.

Answer:

2\frac{3}{10}

cm²

4 · Reviewdoes it hold up?

2 of 12 is $\frac{1}{6}$ of the star, and $\frac{1}{6} \times 13\frac{4}{5} = 2\frac{3}{10}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Variant 12 hard

Two congruent equilateral triangles are overlapped so that the shared part is a regular hexagon, making the star shown at the right. If the whole star covers $14\frac{1}{5}\ \text{cm}^2$ , what is the area of the shaded part, in $\text{cm}^2$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

A six-pointed star covers 14 and 1/5 square centimetres. Three of its six points are shaded, and we want how much area that is.

Givens

The whole star covers 14 and 1/5 cm2.
The dotted lines cut the star into 12 congruent small triangles.
Three of the six points are shaded.

Unknowns

The area of the shaded part.

Constraints

The two big triangles are congruent and the middle is a regular hexagon, so all 12 small triangles are the same size.

2 · Planchoose the strategy

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

The dotted lines have already done the hard part: they cut the star into pieces that are all the same size. That turns an area question into a counting question -- how many of the 12 are shaded.

3 · Execute3 carry out the plan

1Check the pieces really are equal

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

The hexagon is regular and the triangles are equilateral, so all 12 small triangles are congruent: 6 in the hexagon and 6 in the points.

6 + 6 = 12

Twelve identical tiles make the star.

2Find one piece

#7 Identify Subproblems 5.NF.B.4

14 and 1/5 cm2 shared equally between 12 tiles.

14\frac{1}{5} \div 12 = 1\frac{11}{60}

Each tile is 1 and 11/60 cm2.

3Count the shaded tiles

#1 Draw a Diagram 5.NF.B.4

Each shaded point is exactly one tile, and there are 3 of them.

1\frac{11}{60} \times 3 = 3\frac{11}{20}

The shaded part is 3 and 11/20 cm2.

Answer:

3\frac{11}{20}

cm²

4 · Reviewdoes it hold up?

3 of 12 is $\frac{1}{4}$ of the star, and $\frac{1}{4} \times 14\frac{1}{5} = 3\frac{11}{20}$ -- the same answer by a different route.

Another way: Working out one small triangle's side and using the equilateral area formula would also work, but it needs a square root and the counting does not.

Standardsmin grade 6

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Decomposing the star into congruent triangles.
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Dividing a fractional area into equal pieces and scaling back up.

💡Takeaway. When a picture is cut into pieces that are all the same, measuring turns into counting.

Where next?

← Back to the problem: Congruent pieces turn an area into a count