← Congruent pieces turn an area into a count · Area by Decomposition

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

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

Generated variants — 12

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

Variant 1 medium answer: 311153\frac{11}{15} cm²

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 1115 cm211\frac{1}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
1115÷12=141511\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.
1415×4=31115\frac{14}{15} \times 4 = 3\frac{11}{15}
The shaded part is 3 and 11/15 cm2.
Answer: 311153\frac{11}{15} cm²
4 · Reviewdoes it hold up?

4 of 12 is 13\frac{1}{3} of the star, and 13×1115=31115\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 answer: 25\frac{2}{5} cm²

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 445 cm24\frac{4}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
445÷12=254\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.
25×1=25\frac{2}{5} \times 1 = \frac{2}{5}
The shaded part is 2/5 cm2.
Answer: 25\frac{2}{5} cm²
4 · Reviewdoes it hold up?

1 of 12 is 112\frac{1}{12} of the star, and 112×445=25\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 answer: 5165\frac{1}{6} cm²

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 1225 cm212\frac{2}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
1225÷12=113012\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.
1130×5=5161\frac{1}{30} \times 5 = 5\frac{1}{6}
The shaded part is 5 and 1/6 cm2.
Answer: 5165\frac{1}{6} cm²
4 · Reviewdoes it hold up?

5 of 12 is 512\frac{5}{12} of the star, and 512×1225=516\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 answer: 4124\frac{1}{2} cm²

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 1045 cm210\frac{4}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
1045÷12=91010\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.
910×5=412\frac{9}{10} \times 5 = 4\frac{1}{2}
The shaded part is 4 and 1/2 cm2.
Answer: 4124\frac{1}{2} cm²
4 · Reviewdoes it hold up?

5 of 12 is 512\frac{5}{12} of the star, and 512×1045=412\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 answer: 11101\frac{1}{10} cm²

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 635 cm26\frac{3}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
635÷12=11206\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.
1120×2=1110\frac{11}{20} \times 2 = 1\frac{1}{10}
The shaded part is 1 and 1/10 cm2.
Answer: 11101\frac{1}{10} cm²
4 · Reviewdoes it hold up?

2 of 12 is 16\frac{1}{6} of the star, and 16×635=1110\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 answer: 1451\frac{4}{5} cm²

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 715 cm27\frac{1}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
715÷12=357\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.
35×3=145\frac{3}{5} \times 3 = 1\frac{4}{5}
The shaded part is 1 and 4/5 cm2.
Answer: 1451\frac{4}{5} cm²
4 · Reviewdoes it hold up?

3 of 12 is 14\frac{1}{4} of the star, and 14×715=145\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 answer: 910\frac{9}{10} cm²

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 525 cm25\frac{2}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
525÷12=9205\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.
920×2=910\frac{9}{20} \times 2 = \frac{9}{10}
The shaded part is 9/10 cm2.
Answer: 910\frac{9}{10} cm²
4 · Reviewdoes it hold up?

2 of 12 is 16\frac{1}{6} of the star, and 16×525=910\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 answer: 21102\frac{1}{10} cm²

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 825 cm28\frac{2}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
825÷12=7108\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.
710×3=2110\frac{7}{10} \times 3 = 2\frac{1}{10}
The shaded part is 2 and 1/10 cm2.
Answer: 21102\frac{1}{10} cm²
4 · Reviewdoes it hold up?

3 of 12 is 14\frac{1}{4} of the star, and 14×825=2110\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 answer: 310\frac{3}{10} cm²

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 335 cm23\frac{3}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
335÷12=3103\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.
310×1=310\frac{3}{10} \times 1 = \frac{3}{10}
The shaded part is 3/10 cm2.
Answer: 310\frac{3}{10} cm²
4 · Reviewdoes it hold up?

1 of 12 is 112\frac{1}{12} of the star, and 112×335=310\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 answer: 3153\frac{1}{5} cm²

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 935 cm29\frac{3}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
935÷12=459\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.
45×4=315\frac{4}{5} \times 4 = 3\frac{1}{5}
The shaded part is 3 and 1/5 cm2.
Answer: 3153\frac{1}{5} cm²
4 · Reviewdoes it hold up?

4 of 12 is 13\frac{1}{3} of the star, and 13×935=315\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 answer: 23102\frac{3}{10} cm²

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 1345 cm213\frac{4}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
1345÷12=132013\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.
1320×2=23101\frac{3}{20} \times 2 = 2\frac{3}{10}
The shaded part is 2 and 3/10 cm2.
Answer: 23102\frac{3}{10} cm²
4 · Reviewdoes it hold up?

2 of 12 is 16\frac{1}{6} of the star, and 16×1345=2310\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 answer: 311203\frac{11}{20} cm²

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 1415 cm214\frac{1}{5}\ \text{cm}^2, what is the area of the shaded part, in cm2\text{cm}^2?

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=126 + 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.
1415÷12=1116014\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.
11160×3=311201\frac{11}{60} \times 3 = 3\frac{11}{20}
The shaded part is 3 and 11/20 cm2.
Answer: 311203\frac{11}{20} cm²
4 · Reviewdoes it hold up?

3 of 12 is 14\frac{1}{4} of the star, and 14×1415=31120\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.