← Use a shared base to find composite area · Area by Decomposition

Use a shared base to find composite area · 12 practice problems

3.MD.C.76.G.A.1

Generated variants — 12

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

Variant 1 medium answer: 164 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 8 cm8\,\text{cm} long and its right side DF is 14 cm14\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

14 cm G 8 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 14 cm and the overlap's short side is 8 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 14 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 8 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 14 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 14 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 14 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×14×14=98 cm2\frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2
98 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 8 cm.
12×8×8=32 cm2\frac{1}{2} \times 8 \times 8 = 32 \text{ cm}^2
32 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
98+98−32=164 cm298 + 98 - 32 = 164 \text{ cm}^2
164 cm2 shaded.
Answer: 164 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (98 cm2) and two (196 cm2), closer to two because the overlap is small. 164 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 98 - 32 = 66 cm2 each. That gives 32 + 2 x 66 = 164 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 2 medium answer: 56 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 4 cm4\,\text{cm} long and its right side DF is 8 cm8\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

8 cm G 4 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 8 cm and the overlap's short side is 4 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 8 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 4 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 8 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 8 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 8 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×8×8=32 cm2\frac{1}{2} \times 8 \times 8 = 32 \text{ cm}^2
32 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 4 cm.
12×4×4=8 cm2\frac{1}{2} \times 4 \times 4 = 8 \text{ cm}^2
8 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
32+32−8=56 cm232 + 32 - 8 = 56 \text{ cm}^2
56 cm2 shaded.
Answer: 56 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (32 cm2) and two (64 cm2), closer to two because the overlap is small. 56 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 32 - 8 = 24 cm2 each. That gives 8 + 2 x 24 = 56 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 3 medium answer: 514 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 18 cm18\,\text{cm} long and its right side DF is 26 cm26\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

26 cm G 18 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 26 cm and the overlap's short side is 18 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 26 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 18 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 26 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 26 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 26 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×26×26=338 cm2\frac{1}{2} \times 26 \times 26 = 338 \text{ cm}^2
338 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 18 cm.
12×18×18=162 cm2\frac{1}{2} \times 18 \times 18 = 162 \text{ cm}^2
162 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
338+338−162=514 cm2338 + 338 - 162 = 514 \text{ cm}^2
514 cm2 shaded.
Answer: 514 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (338 cm2) and two (676 cm2), closer to two because the overlap is small. 514 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 338 - 162 = 176 cm2 each. That gives 162 + 2 x 176 = 514 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 4 medium answer: 386 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 14 cm14\,\text{cm} long and its right side DF is 22 cm22\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

22 cm G 14 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 22 cm and the overlap's short side is 14 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 22 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 14 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 22 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 22 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 22 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×22×22=242 cm2\frac{1}{2} \times 22 \times 22 = 242 \text{ cm}^2
242 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 14 cm.
12×14×14=98 cm2\frac{1}{2} \times 14 \times 14 = 98 \text{ cm}^2
98 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
242+242−98=386 cm2242 + 242 - 98 = 386 \text{ cm}^2
386 cm2 shaded.
Answer: 386 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (242 cm2) and two (484 cm2), closer to two because the overlap is small. 386 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 242 - 98 = 144 cm2 each. That gives 98 + 2 x 144 = 386 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 5 medium answer: 328 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 12 cm12\,\text{cm} long and its right side DF is 20 cm20\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

20 cm G 12 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 20 cm and the overlap's short side is 12 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 20 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 12 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 20 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 20 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 20 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×20×20=200 cm2\frac{1}{2} \times 20 \times 20 = 200 \text{ cm}^2
200 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 12 cm.
12×12×12=72 cm2\frac{1}{2} \times 12 \times 12 = 72 \text{ cm}^2
72 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
200+200−72=328 cm2200 + 200 - 72 = 328 \text{ cm}^2
328 cm2 shaded.
Answer: 328 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (200 cm2) and two (400 cm2), closer to two because the overlap is small. 328 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 200 - 72 = 128 cm2 each. That gives 72 + 2 x 128 = 328 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 6 medium answer: 82 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 6 cm6\,\text{cm} long and its right side DF is 10 cm10\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

10 cm G 6 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 10 cm and the overlap's short side is 6 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 10 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 6 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 10 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 10 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 10 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×10×10=50 cm2\frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2
50 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 6 cm.
12×6×6=18 cm2\frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2
18 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
50+50−18=82 cm250 + 50 - 18 = 82 \text{ cm}^2
82 cm2 shaded.
Answer: 82 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (50 cm2) and two (100 cm2), closer to two because the overlap is small. 82 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 50 - 18 = 32 cm2 each. That gives 18 + 2 x 32 = 82 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 7 medium answer: 206 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 10 cm10\,\text{cm} long and its right side DF is 16 cm16\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

16 cm G 10 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 16 cm and the overlap's short side is 10 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 16 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 10 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 16 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 16 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 16 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×16×16=128 cm2\frac{1}{2} \times 16 \times 16 = 128 \text{ cm}^2
128 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 10 cm.
12×10×10=50 cm2\frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2
50 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
128+128−50=206 cm2128 + 128 - 50 = 206 \text{ cm}^2
206 cm2 shaded.
Answer: 206 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (128 cm2) and two (256 cm2), closer to two because the overlap is small. 206 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 128 - 50 = 78 cm2 each. That gives 50 + 2 x 78 = 206 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 8 medium answer: 98 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 2 cm2\,\text{cm} long and its right side DF is 10 cm10\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

10 cm G 2 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 10 cm and the overlap's short side is 2 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 10 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 2 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 10 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 10 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 10 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×10×10=50 cm2\frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2
50 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 2 cm.
12×2×2=2 cm2\frac{1}{2} \times 2 \times 2 = 2 \text{ cm}^2
2 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
50+50−2=98 cm250 + 50 - 2 = 98 \text{ cm}^2
98 cm2 shaded.
Answer: 98 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (50 cm2) and two (100 cm2), closer to two because the overlap is small. 98 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 50 - 2 = 48 cm2 each. That gives 2 + 2 x 48 = 98 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 9 medium answer: 34 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 2 cm2\,\text{cm} long and its right side DF is 6 cm6\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

6 cm G 2 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 6 cm and the overlap's short side is 2 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 6 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 2 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 6 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 6 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 6 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×6×6=18 cm2\frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2
18 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 2 cm.
12×2×2=2 cm2\frac{1}{2} \times 2 \times 2 = 2 \text{ cm}^2
2 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
18+18−2=34 cm218 + 18 - 2 = 34 \text{ cm}^2
34 cm2 shaded.
Answer: 34 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (18 cm2) and two (36 cm2), closer to two because the overlap is small. 34 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 18 - 2 = 16 cm2 each. That gives 2 + 2 x 16 = 34 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 10 medium answer: 306 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 6 cm6\,\text{cm} long and its right side DF is 18 cm18\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

18 cm G 6 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 18 cm and the overlap's short side is 6 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 18 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 6 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 18 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 18 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 18 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×18×18=162 cm2\frac{1}{2} \times 18 \times 18 = 162 \text{ cm}^2
162 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 6 cm.
12×6×6=18 cm2\frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2
18 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
162+162−18=306 cm2162 + 162 - 18 = 306 \text{ cm}^2
306 cm2 shaded.
Answer: 306 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (162 cm2) and two (324 cm2), closer to two because the overlap is small. 306 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 162 - 18 = 144 cm2 each. That gives 18 + 2 x 144 = 306 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 11 medium answer: 136 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 4 cm4\,\text{cm} long and its right side DF is 12 cm12\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

12 cm G 4 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 12 cm and the overlap's short side is 4 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 12 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 4 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 12 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 12 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 12 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×12×12=72 cm2\frac{1}{2} \times 12 \times 12 = 72 \text{ cm}^2
72 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 4 cm.
12×4×4=8 cm2\frac{1}{2} \times 4 \times 4 = 8 \text{ cm}^2
8 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
72+72−8=136 cm272 + 72 - 8 = 136 \text{ cm}^2
136 cm2 shaded.
Answer: 136 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (72 cm2) and two (144 cm2), closer to two because the overlap is small. 136 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 72 - 8 = 64 cm2 each. That gives 8 + 2 x 64 = 136 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.
Variant 12 medium answer: 448 cm²

In the figure at right, triangle ABC and triangle DEF have the same shape and size, and triangle AGD and triangle GEC also have the same shape and size. What is the area of the shaded region, in cm2\text{cm}^2?

Figure description: Two right triangles overlap. Along the bottom, the points B, E, C, F lie in that order on one straight line. The larger left right triangle is ABC, and the angle at vertex A measures 45∘45^\circ. The right-hand right triangle is DEF; its top side AD is 16 cm16\,\text{cm} long and its right side DF is 24 cm24\,\text{cm} long. The point where the two triangles cross is labeled G, and right angles are formed at C and F where the slanted sides meet the base.

24 cm G 16 A
Show solution
1 · Understandwhat's really being asked

Two congruent right triangles with a 45 degree angle overlap along one line. One side is 24 cm and the overlap's short side is 16 cm. We need the area the two of them cover together.

Givens
  • The triangles are congruent, with DF = 24 cm.
  • The angle at A is 45 degrees and the angles at C and F are right angles.
  • AD = 16 cm.
Unknowns
  • The area of the shaded region.
Constraints
  • The shaded region counts the overlap once, not twice.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#15 Organize Information in More Ways

The 45 degrees is doing more work than it looks: it makes each triangle isosceles, so one length gives both legs. After that it is two areas added and one subtracted.

3 · Execute4 carry out the plan

1Find each leg length

#1 Draw a Diagram 6.G.A.1
Triangle ABC is a right triangle with a 45 degree angle at A. The third angle must also be 45 degrees, so it is right isosceles and its two legs are equal -- both 24 cm.
angle B=180∘−90∘−45∘=45∘\text{angle B} = 180^\circ - 90^\circ - 45^\circ = 45^\circ
Both legs are 24 cm.

2Area of one whole triangle

#7 Identify Subproblems 3.MD.C.7
Each right triangle has two perpendicular legs of 24 cm (one along the base, one vertical). The area of a triangle is half of base times height.
12×24×24=288 cm2\frac{1}{2} \times 24 \times 24 = 288 \text{ cm}^2
288 cm2 each.

3Area of the overlap

#7 Identify Subproblems 3.MD.C.7
The overlap of the two triangles is exactly triangle GEC. It is right isosceles too, with both short sides 16 cm.
12×16×16=128 cm2\frac{1}{2} \times 16 \times 16 = 128 \text{ cm}^2
128 cm2 belongs to both.

4Add the two triangles and subtract the shared overlap

#15 Organize Information in More Ways 6.G.A.1
The shaded figure is the two triangles together. Adding both areas counts the overlap twice, so subtract it once.
288+288−128=448 cm2288 + 288 - 128 = 448 \text{ cm}^2
448 cm2 shaded.
Answer: 448 cm²
4 · Reviewdoes it hold up?

The answer has to sit between one triangle (288 cm2) and two (576 cm2), closer to two because the overlap is small. 448 cm2 does.

Another way: Count the pieces instead: the shaded region is the overlap once plus the two leftover parts, 288 - 128 = 160 cm2 each. That gives 128 + 2 x 160 = 448 cm2.

Standardsmin grade 6
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The area of each right triangle from its two legs.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Reading the 45 degrees as isosceles, and combining the pieces.
💡Takeaway. When two shapes overlap, adding them counts the shared part twice -- so take it off once.