← Same base and height means same area · Area by Decomposition

Same base and height means same area · 12 practice problems

3.MD.C.74.MD.A.3

Generated variants — 12

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

Variant 1 easy answer: 4 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 128 cm2128\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 16 cm16\,\text{cm} and MD is 16 cm16\,\text{cm}; the right side DC of the rectangle is 5 cm5\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

16 cm 16 cm 5 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 16 cm base with the same 5 cm height and overlap in a triangle. The two together cover 128 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 16 cm by 5 cm.
  • The parallelogram has the same 16 cm base and 5 cm height.
  • The shaded region measures 128 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 16 cm wide and 5 cm tall. Multiply width by height.
16×5=80 cm216 \times 5 = 80 \text{ cm}^2
80 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 16 cm base and has the same 5 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
16×5=80 cm216 \times 5 = 80 \text{ cm}^2
The same 80 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
80+80−overlap=128  ⇒  overlap=160−128=32 cm280 + 80 - \text{overlap} = 128 \;\Rightarrow\; \text{overlap} = 160 - 128 = 32 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 16 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
16×GF÷2=32  ⇒  GF=4 cm16 \times \text{GF} \div 2 = 32 \;\Rightarrow\; \text{GF} = 4 \text{ cm}
GF is 4 cm.
Answer: 4 cm
4 · Reviewdoes it hold up?

Put 4 back: the triangle is 16 x 4 / 2, and 160 minus that is 128 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 5 cm height, because G lies inside the figures. 4 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 2 easy answer: 6 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 132 cm2132\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 12 cm12\,\text{cm} and MD is 12 cm12\,\text{cm}; the right side DC of the rectangle is 7 cm7\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

12 cm 12 cm 7 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 12 cm base with the same 7 cm height and overlap in a triangle. The two together cover 132 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 12 cm by 7 cm.
  • The parallelogram has the same 12 cm base and 7 cm height.
  • The shaded region measures 132 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 12 cm wide and 7 cm tall. Multiply width by height.
12×7=84 cm212 \times 7 = 84 \text{ cm}^2
84 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 12 cm base and has the same 7 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
12×7=84 cm212 \times 7 = 84 \text{ cm}^2
The same 84 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
84+84−overlap=132  ⇒  overlap=168−132=36 cm284 + 84 - \text{overlap} = 132 \;\Rightarrow\; \text{overlap} = 168 - 132 = 36 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 12 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
12×GF÷2=36  ⇒  GF=6 cm12 \times \text{GF} \div 2 = 36 \;\Rightarrow\; \text{GF} = 6 \text{ cm}
GF is 6 cm.
Answer: 6 cm
4 · Reviewdoes it hold up?

Put 6 back: the triangle is 12 x 6 / 2, and 168 minus that is 132 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 7 cm height, because G lies inside the figures. 6 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 3 easy answer: 6 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 153 cm2153\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 9 cm9\,\text{cm} and MD is 9 cm9\,\text{cm}; the right side DC of the rectangle is 10 cm10\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

9 cm 9 cm 10 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 9 cm base with the same 10 cm height and overlap in a triangle. The two together cover 153 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 9 cm by 10 cm.
  • The parallelogram has the same 9 cm base and 10 cm height.
  • The shaded region measures 153 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 9 cm wide and 10 cm tall. Multiply width by height.
9×10=90 cm29 \times 10 = 90 \text{ cm}^2
90 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 9 cm base and has the same 10 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
9×10=90 cm29 \times 10 = 90 \text{ cm}^2
The same 90 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
90+90−overlap=153  ⇒  overlap=180−153=27 cm290 + 90 - \text{overlap} = 153 \;\Rightarrow\; \text{overlap} = 180 - 153 = 27 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 9 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
9×GF÷2=27  ⇒  GF=6 cm9 \times \text{GF} \div 2 = 27 \;\Rightarrow\; \text{GF} = 6 \text{ cm}
GF is 6 cm.
Answer: 6 cm
4 · Reviewdoes it hold up?

Put 6 back: the triangle is 9 x 6 / 2, and 180 minus that is 153 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 10 cm height, because G lies inside the figures. 6 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 4 easy answer: 4 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 160 cm2160\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 10 cm10\,\text{cm} and MD is 10 cm10\,\text{cm}; the right side DC of the rectangle is 9 cm9\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

10 cm 10 cm 9 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 10 cm base with the same 9 cm height and overlap in a triangle. The two together cover 160 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 10 cm by 9 cm.
  • The parallelogram has the same 10 cm base and 9 cm height.
  • The shaded region measures 160 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 10 cm wide and 9 cm tall. Multiply width by height.
10×9=90 cm210 \times 9 = 90 \text{ cm}^2
90 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 10 cm base and has the same 9 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
10×9=90 cm210 \times 9 = 90 \text{ cm}^2
The same 90 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
90+90−overlap=160  ⇒  overlap=180−160=20 cm290 + 90 - \text{overlap} = 160 \;\Rightarrow\; \text{overlap} = 180 - 160 = 20 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 10 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
10×GF÷2=20  ⇒  GF=4 cm10 \times \text{GF} \div 2 = 20 \;\Rightarrow\; \text{GF} = 4 \text{ cm}
GF is 4 cm.
Answer: 4 cm
4 · Reviewdoes it hold up?

Put 4 back: the triangle is 10 x 4 / 2, and 180 minus that is 160 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 9 cm height, because G lies inside the figures. 4 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 5 medium answer: 6 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 168 cm2168\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 8 cm8\,\text{cm} and MD is 8 cm8\,\text{cm}; the right side DC of the rectangle is 12 cm12\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

8 cm 8 cm 12 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 8 cm base with the same 12 cm height and overlap in a triangle. The two together cover 168 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 8 cm by 12 cm.
  • The parallelogram has the same 8 cm base and 12 cm height.
  • The shaded region measures 168 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 8 cm wide and 12 cm tall. Multiply width by height.
8×12=96 cm28 \times 12 = 96 \text{ cm}^2
96 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 8 cm base and has the same 12 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
8×12=96 cm28 \times 12 = 96 \text{ cm}^2
The same 96 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
96+96−overlap=168  ⇒  overlap=192−168=24 cm296 + 96 - \text{overlap} = 168 \;\Rightarrow\; \text{overlap} = 192 - 168 = 24 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 8 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
8×GF÷2=24  ⇒  GF=6 cm8 \times \text{GF} \div 2 = 24 \;\Rightarrow\; \text{GF} = 6 \text{ cm}
GF is 6 cm.
Answer: 6 cm
4 · Reviewdoes it hold up?

Put 6 back: the triangle is 8 x 6 / 2, and 192 minus that is 168 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 12 cm height, because G lies inside the figures. 6 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 6 medium answer: 6 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 182 cm2182\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 14 cm14\,\text{cm} and MD is 14 cm14\,\text{cm}; the right side DC of the rectangle is 8 cm8\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

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

A rectangle and a parallelogram sit on the same 14 cm base with the same 8 cm height and overlap in a triangle. The two together cover 182 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 14 cm by 8 cm.
  • The parallelogram has the same 14 cm base and 8 cm height.
  • The shaded region measures 182 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 14 cm wide and 8 cm tall. Multiply width by height.
14×8=112 cm214 \times 8 = 112 \text{ cm}^2
112 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 14 cm base and has the same 8 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
14×8=112 cm214 \times 8 = 112 \text{ cm}^2
The same 112 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
112+112−overlap=182  ⇒  overlap=224−182=42 cm2112 + 112 - \text{overlap} = 182 \;\Rightarrow\; \text{overlap} = 224 - 182 = 42 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 14 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
14×GF÷2=42  ⇒  GF=6 cm14 \times \text{GF} \div 2 = 42 \;\Rightarrow\; \text{GF} = 6 \text{ cm}
GF is 6 cm.
Answer: 6 cm
4 · Reviewdoes it hold up?

Put 6 back: the triangle is 14 x 6 / 2, and 224 minus that is 182 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 8 cm height, because G lies inside the figures. 6 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 7 medium answer: 6 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 195 cm2195\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 15 cm15\,\text{cm} and MD is 15 cm15\,\text{cm}; the right side DC of the rectangle is 8 cm8\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

15 cm 15 cm 8 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 15 cm base with the same 8 cm height and overlap in a triangle. The two together cover 195 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 15 cm by 8 cm.
  • The parallelogram has the same 15 cm base and 8 cm height.
  • The shaded region measures 195 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 15 cm wide and 8 cm tall. Multiply width by height.
15×8=120 cm215 \times 8 = 120 \text{ cm}^2
120 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 15 cm base and has the same 8 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
15×8=120 cm215 \times 8 = 120 \text{ cm}^2
The same 120 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
120+120−overlap=195  ⇒  overlap=240−195=45 cm2120 + 120 - \text{overlap} = 195 \;\Rightarrow\; \text{overlap} = 240 - 195 = 45 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 15 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
15×GF÷2=45  ⇒  GF=6 cm15 \times \text{GF} \div 2 = 45 \;\Rightarrow\; \text{GF} = 6 \text{ cm}
GF is 6 cm.
Answer: 6 cm
4 · Reviewdoes it hold up?

Put 6 back: the triangle is 15 x 6 / 2, and 240 minus that is 195 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 8 cm height, because G lies inside the figures. 6 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 8 medium answer: 4 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 200 cm2200\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 20 cm20\,\text{cm} and MD is 20 cm20\,\text{cm}; the right side DC of the rectangle is 6 cm6\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

20 cm 20 cm 6 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 20 cm base with the same 6 cm height and overlap in a triangle. The two together cover 200 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 20 cm by 6 cm.
  • The parallelogram has the same 20 cm base and 6 cm height.
  • The shaded region measures 200 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 20 cm wide and 6 cm tall. Multiply width by height.
20×6=120 cm220 \times 6 = 120 \text{ cm}^2
120 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 20 cm base and has the same 6 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
20×6=120 cm220 \times 6 = 120 \text{ cm}^2
The same 120 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
120+120−overlap=200  ⇒  overlap=240−200=40 cm2120 + 120 - \text{overlap} = 200 \;\Rightarrow\; \text{overlap} = 240 - 200 = 40 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 20 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
20×GF÷2=40  ⇒  GF=4 cm20 \times \text{GF} \div 2 = 40 \;\Rightarrow\; \text{GF} = 4 \text{ cm}
GF is 4 cm.
Answer: 4 cm
4 · Reviewdoes it hold up?

Put 4 back: the triangle is 20 x 4 / 2, and 240 minus that is 200 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 6 cm height, because G lies inside the figures. 4 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 9 hard answer: 8 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 220 cm2220\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 11 cm11\,\text{cm} and MD is 11 cm11\,\text{cm}; the right side DC of the rectangle is 12 cm12\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

11 cm 11 cm 12 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 11 cm base with the same 12 cm height and overlap in a triangle. The two together cover 220 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 11 cm by 12 cm.
  • The parallelogram has the same 11 cm base and 12 cm height.
  • The shaded region measures 220 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 11 cm wide and 12 cm tall. Multiply width by height.
11×12=132 cm211 \times 12 = 132 \text{ cm}^2
132 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 11 cm base and has the same 12 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
11×12=132 cm211 \times 12 = 132 \text{ cm}^2
The same 132 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
132+132−overlap=220  ⇒  overlap=264−220=44 cm2132 + 132 - \text{overlap} = 220 \;\Rightarrow\; \text{overlap} = 264 - 220 = 44 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 11 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
11×GF÷2=44  ⇒  GF=8 cm11 \times \text{GF} \div 2 = 44 \;\Rightarrow\; \text{GF} = 8 \text{ cm}
GF is 8 cm.
Answer: 8 cm
4 · Reviewdoes it hold up?

Put 8 back: the triangle is 11 x 8 / 2, and 264 minus that is 220 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 12 cm height, because G lies inside the figures. 8 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 10 hard answer: 4 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 288 cm2288\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 24 cm24\,\text{cm} and MD is 24 cm24\,\text{cm}; the right side DC of the rectangle is 7 cm7\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

24 cm 24 cm 7 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 24 cm base with the same 7 cm height and overlap in a triangle. The two together cover 288 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 24 cm by 7 cm.
  • The parallelogram has the same 24 cm base and 7 cm height.
  • The shaded region measures 288 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 24 cm wide and 7 cm tall. Multiply width by height.
24×7=168 cm224 \times 7 = 168 \text{ cm}^2
168 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 24 cm base and has the same 7 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
24×7=168 cm224 \times 7 = 168 \text{ cm}^2
The same 168 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
168+168−overlap=288  ⇒  overlap=336−288=48 cm2168 + 168 - \text{overlap} = 288 \;\Rightarrow\; \text{overlap} = 336 - 288 = 48 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 24 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
24×GF÷2=48  ⇒  GF=4 cm24 \times \text{GF} \div 2 = 48 \;\Rightarrow\; \text{GF} = 4 \text{ cm}
GF is 4 cm.
Answer: 4 cm
4 · Reviewdoes it hold up?

Put 4 back: the triangle is 24 x 4 / 2, and 336 minus that is 288 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 7 cm height, because G lies inside the figures. 4 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 11 hard answer: 8 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 324 cm2324\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 18 cm18\,\text{cm} and MD is 18 cm18\,\text{cm}; the right side DC of the rectangle is 11 cm11\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

18 cm 18 cm 11 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 18 cm base with the same 11 cm height and overlap in a triangle. The two together cover 324 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 18 cm by 11 cm.
  • The parallelogram has the same 18 cm base and 11 cm height.
  • The shaded region measures 324 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 18 cm wide and 11 cm tall. Multiply width by height.
18×11=198 cm218 \times 11 = 198 \text{ cm}^2
198 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 18 cm base and has the same 11 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
18×11=198 cm218 \times 11 = 198 \text{ cm}^2
The same 198 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
198+198−overlap=324  ⇒  overlap=396−324=72 cm2198 + 198 - \text{overlap} = 324 \;\Rightarrow\; \text{overlap} = 396 - 324 = 72 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 18 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
18×GF÷2=72  ⇒  GF=8 cm18 \times \text{GF} \div 2 = 72 \;\Rightarrow\; \text{GF} = 8 \text{ cm}
GF is 8 cm.
Answer: 8 cm
4 · Reviewdoes it hold up?

Put 8 back: the triangle is 18 x 8 / 2, and 396 minus that is 324 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 11 cm height, because G lies inside the figures. 8 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.
Variant 12 hard answer: 6 cm

The figure at right shows rectangle MBCD and parallelogram ABCF placed so that they overlap. When the area of the shaded region is 330 cm2330\,\text{cm}^2, what is the length of segment GF, in cm\text{cm}?

Figure description: The parallelogram ABCF on the left and the rectangle MBCD on the right overlap each other. Along the top edge, FM is 22 cm22\,\text{cm} and MD is 22 cm22\,\text{cm}; the right side DC of the rectangle is 9 cm9\,\text{cm}. The point where the two figures cross is labeled G. The shaded region is the part where the two figures overlap.

22 cm 22 cm 9 cm G F
Show solution
1 · Understandwhat's really being asked

A rectangle and a parallelogram sit on the same 22 cm base with the same 9 cm height and overlap in a triangle. The two together cover 330 cm2, and we need the height GF of that triangle.

Givens
  • The rectangle is 22 cm by 9 cm.
  • The parallelogram has the same 22 cm base and 9 cm height.
  • The shaded region measures 330 cm2.
Unknowns
  • The length of segment GF.
Constraints
  • The overlap belongs to both figures, so adding the two areas counts it twice.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#11 Work Backwards

Work out both areas first -- they turn out to be equal, which is the point. Then the shaded total tells you the overlap, and the overlap is a triangle whose base you already know.

3 · Execute4 carry out the plan

1Area of the rectangle

#7 Identify Subproblems 4.MD.A.3
The rectangle MBCD is 22 cm wide and 9 cm tall. Multiply width by height.
22×9=198 cm222 \times 9 = 198 \text{ cm}^2
198 cm2.

2Area of the parallelogram

#7 Identify Subproblems 3.MD.C.7
The parallelogram ABCF sits on the same 22 cm base and has the same 9 cm height as the rectangle. Base times height gives its area -- the slant makes no difference.
22×9=198 cm222 \times 9 = 198 \text{ cm}^2
The same 198 cm2.

3Find the overlap area by working backwards

#11 Work Backwards 4.MD.A.3
Adding the two figures counts the overlap triangle twice, but the shaded region counts it once. So the two areas added minus the overlap equals the shaded region.
198+198−overlap=330  ⇒  overlap=396−330=66 cm2198 + 198 - \text{overlap} = 330 \;\Rightarrow\; \text{overlap} = 396 - 330 = 66 \text{ cm}^2
The overlap is what the shaded total is short by.

4Use the triangle area to find GF

#7 Identify Subproblems 3.MD.C.7
The overlap is triangle GBC with base BC = 22 cm. Its height is the segment GF, dropped straight down from G to that base. Triangle area is base times height divided by 2.
22×GF÷2=66  ⇒  GF=6 cm22 \times \text{GF} \div 2 = 66 \;\Rightarrow\; \text{GF} = 6 \text{ cm}
GF is 6 cm.
Answer: 6 cm
4 · Reviewdoes it hold up?

Put 6 back: the triangle is 22 x 6 / 2, and 396 minus that is 330 cm2 -- the shaded area given.

Another way: A quick check on size: GF must be less than the 9 cm height, because G lies inside the figures. 6 cm fits.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — The parallelogram and triangle areas from base and height.
  • 4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems — The rectangle's area and unpicking the shaded total.
💡Takeaway. A shape leaning over covers the same area as one standing straight, as long as the base and the height stay the same.