← A square's one side gives every other · Transformations Preserve Measures

A square's one side gives every other · 12 practice problems

4.G.A.26.G.A.1

Generated variants — 12

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

Variant 1 easy answer: 72 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 14 cm14\ \text{cm}. Point HH lies on the top side ADAD with AH=4 cm\overline{AH}=4\ \text{cm}, and the right side DC=6 cm\overline{DC}=6\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=5 cm\overline{EF}=5\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 4 cm 6 cm 14 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 14 cm, and a square HGCD sits in its right-hand corner with side DC = 6 cm. H lies on the top side with AH = 4 cm. We need the trapezoid's area.

Givens
  • BC = 14 cm along the bottom.
  • AH = 4 cm, with H on the top side AD.
  • DC = 6 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 6 cm, so every side of the square is 6 cm. That makes the vertical side HG = 6 cm -- the height of the trapezoid -- and the top piece HD = 6 cm as well.
HD=HG=GC=DC=6 cmHD = HG = GC = DC = 6\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=4+6=10 cmAD = AH + HD = 4 + 6 = 10\ \text{cm}
The top is 10 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 10 cm and the bottom BC = 14 cm, and the height is the square's side, 6 cm. Add the parallel sides, multiply by the height, halve.
(10+14)×6÷2=24×6÷2=72 cm2(10 + 14) \times 6 \div 2 = 24 \times 6 \div 2 = 72\ \text{cm}^2
The trapezoid covers 72 cm2.
Answer: 72 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 14 by 6 rectangle of area 84 cm2 and contains an 10 by 6 one of area 60 cm2. 72 cm2 sits between them, as it must.

Another way: The 5 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 2 easy answer: 52 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 16 cm16\ \text{cm}. Point HH lies on the top side ADAD with AH=6 cm\overline{AH}=6\ \text{cm}, and the right side DC=4 cm\overline{DC}=4\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=8 cm\overline{EF}=8\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 6 cm 4 cm 16 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 16 cm, and a square HGCD sits in its right-hand corner with side DC = 4 cm. H lies on the top side with AH = 6 cm. We need the trapezoid's area.

Givens
  • BC = 16 cm along the bottom.
  • AH = 6 cm, with H on the top side AD.
  • DC = 4 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 4 cm, so every side of the square is 4 cm. That makes the vertical side HG = 4 cm -- the height of the trapezoid -- and the top piece HD = 4 cm as well.
HD=HG=GC=DC=4 cmHD = HG = GC = DC = 4\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=6+4=10 cmAD = AH + HD = 6 + 4 = 10\ \text{cm}
The top is 10 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 10 cm and the bottom BC = 16 cm, and the height is the square's side, 4 cm. Add the parallel sides, multiply by the height, halve.
(10+16)×4÷2=26×4÷2=52 cm2(10 + 16) \times 4 \div 2 = 26 \times 4 \div 2 = 52\ \text{cm}^2
The trapezoid covers 52 cm2.
Answer: 52 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 16 by 4 rectangle of area 64 cm2 and contains an 10 by 4 one of area 40 cm2. 52 cm2 sits between them, as it must.

Another way: The 8 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 3 easy answer: 54 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 18 cm18\ \text{cm}. Point HH lies on the top side ADAD with AH=5 cm\overline{AH}=5\ \text{cm}, and the right side DC=4 cm\overline{DC}=4\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=7 cm\overline{EF}=7\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 5 cm 4 cm 18 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 18 cm, and a square HGCD sits in its right-hand corner with side DC = 4 cm. H lies on the top side with AH = 5 cm. We need the trapezoid's area.

Givens
  • BC = 18 cm along the bottom.
  • AH = 5 cm, with H on the top side AD.
  • DC = 4 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 4 cm, so every side of the square is 4 cm. That makes the vertical side HG = 4 cm -- the height of the trapezoid -- and the top piece HD = 4 cm as well.
HD=HG=GC=DC=4 cmHD = HG = GC = DC = 4\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=5+4=9 cmAD = AH + HD = 5 + 4 = 9\ \text{cm}
The top is 9 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 9 cm and the bottom BC = 18 cm, and the height is the square's side, 4 cm. Add the parallel sides, multiply by the height, halve.
(9+18)×4÷2=27×4÷2=54 cm2(9 + 18) \times 4 \div 2 = 27 \times 4 \div 2 = 54\ \text{cm}^2
The trapezoid covers 54 cm2.
Answer: 54 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 18 by 4 rectangle of area 72 cm2 and contains an 9 by 4 one of area 36 cm2. 54 cm2 sits between them, as it must.

Another way: The 7 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 4 easy answer: 102 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 20 cm20\ \text{cm}. Point HH lies on the top side ADAD with AH=8 cm\overline{AH}=8\ \text{cm}, and the right side DC=6 cm\overline{DC}=6\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=9 cm\overline{EF}=9\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 8 cm 6 cm 20 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 20 cm, and a square HGCD sits in its right-hand corner with side DC = 6 cm. H lies on the top side with AH = 8 cm. We need the trapezoid's area.

Givens
  • BC = 20 cm along the bottom.
  • AH = 8 cm, with H on the top side AD.
  • DC = 6 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 6 cm, so every side of the square is 6 cm. That makes the vertical side HG = 6 cm -- the height of the trapezoid -- and the top piece HD = 6 cm as well.
HD=HG=GC=DC=6 cmHD = HG = GC = DC = 6\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=8+6=14 cmAD = AH + HD = 8 + 6 = 14\ \text{cm}
The top is 14 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 14 cm and the bottom BC = 20 cm, and the height is the square's side, 6 cm. Add the parallel sides, multiply by the height, halve.
(14+20)×6÷2=34×6÷2=102 cm2(14 + 20) \times 6 \div 2 = 34 \times 6 \div 2 = 102\ \text{cm}^2
The trapezoid covers 102 cm2.
Answer: 102 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 20 by 6 rectangle of area 120 cm2 and contains an 14 by 6 one of area 84 cm2. 102 cm2 sits between them, as it must.

Another way: The 9 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 5 medium answer: 148 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 22 cm22\ \text{cm}. Point HH lies on the top side ADAD with AH=7 cm\overline{AH}=7\ \text{cm}, and the right side DC=8 cm\overline{DC}=8\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=10 cm\overline{EF}=10\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 7 cm 8 cm 22 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 22 cm, and a square HGCD sits in its right-hand corner with side DC = 8 cm. H lies on the top side with AH = 7 cm. We need the trapezoid's area.

Givens
  • BC = 22 cm along the bottom.
  • AH = 7 cm, with H on the top side AD.
  • DC = 8 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 8 cm, so every side of the square is 8 cm. That makes the vertical side HG = 8 cm -- the height of the trapezoid -- and the top piece HD = 8 cm as well.
HD=HG=GC=DC=8 cmHD = HG = GC = DC = 8\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=7+8=15 cmAD = AH + HD = 7 + 8 = 15\ \text{cm}
The top is 15 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 15 cm and the bottom BC = 22 cm, and the height is the square's side, 8 cm. Add the parallel sides, multiply by the height, halve.
(15+22)×8÷2=37×8÷2=148 cm2(15 + 22) \times 8 \div 2 = 37 \times 8 \div 2 = 148\ \text{cm}^2
The trapezoid covers 148 cm2.
Answer: 148 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 22 by 8 rectangle of area 176 cm2 and contains an 15 by 8 one of area 120 cm2. 148 cm2 sits between them, as it must.

Another way: The 10 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 6 medium answer: 168 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 24 cm24\ \text{cm}. Point HH lies on the top side ADAD with AH=10 cm\overline{AH}=10\ \text{cm}, and the right side DC=8 cm\overline{DC}=8\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=11 cm\overline{EF}=11\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 10 cm 8 cm 24 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 24 cm, and a square HGCD sits in its right-hand corner with side DC = 8 cm. H lies on the top side with AH = 10 cm. We need the trapezoid's area.

Givens
  • BC = 24 cm along the bottom.
  • AH = 10 cm, with H on the top side AD.
  • DC = 8 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 8 cm, so every side of the square is 8 cm. That makes the vertical side HG = 8 cm -- the height of the trapezoid -- and the top piece HD = 8 cm as well.
HD=HG=GC=DC=8 cmHD = HG = GC = DC = 8\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=10+8=18 cmAD = AH + HD = 10 + 8 = 18\ \text{cm}
The top is 18 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 18 cm and the bottom BC = 24 cm, and the height is the square's side, 8 cm. Add the parallel sides, multiply by the height, halve.
(18+24)×8÷2=42×8÷2=168 cm2(18 + 24) \times 8 \div 2 = 42 \times 8 \div 2 = 168\ \text{cm}^2
The trapezoid covers 168 cm2.
Answer: 168 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 24 by 8 rectangle of area 192 cm2 and contains an 18 by 8 one of area 144 cm2. 168 cm2 sits between them, as it must.

Another way: The 11 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 7 medium answer: 120 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 25 cm25\ \text{cm}. Point HH lies on the top side ADAD with AH=9 cm\overline{AH}=9\ \text{cm}, and the right side DC=6 cm\overline{DC}=6\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=11 cm\overline{EF}=11\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 9 cm 6 cm 25 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 25 cm, and a square HGCD sits in its right-hand corner with side DC = 6 cm. H lies on the top side with AH = 9 cm. We need the trapezoid's area.

Givens
  • BC = 25 cm along the bottom.
  • AH = 9 cm, with H on the top side AD.
  • DC = 6 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 6 cm, so every side of the square is 6 cm. That makes the vertical side HG = 6 cm -- the height of the trapezoid -- and the top piece HD = 6 cm as well.
HD=HG=GC=DC=6 cmHD = HG = GC = DC = 6\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=9+6=15 cmAD = AH + HD = 9 + 6 = 15\ \text{cm}
The top is 15 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 15 cm and the bottom BC = 25 cm, and the height is the square's side, 6 cm. Add the parallel sides, multiply by the height, halve.
(15+25)×6÷2=40×6÷2=120 cm2(15 + 25) \times 6 \div 2 = 40 \times 6 \div 2 = 120\ \text{cm}^2
The trapezoid covers 120 cm2.
Answer: 120 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 25 by 6 rectangle of area 150 cm2 and contains an 15 by 6 one of area 90 cm2. 120 cm2 sits between them, as it must.

Another way: The 11 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 8 medium answer: 82 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 26 cm26\ \text{cm}. Point HH lies on the top side ADAD with AH=11 cm\overline{AH}=11\ \text{cm}, and the right side DC=4 cm\overline{DC}=4\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=12 cm\overline{EF}=12\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 11 cm 4 cm 26 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 26 cm, and a square HGCD sits in its right-hand corner with side DC = 4 cm. H lies on the top side with AH = 11 cm. We need the trapezoid's area.

Givens
  • BC = 26 cm along the bottom.
  • AH = 11 cm, with H on the top side AD.
  • DC = 4 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 4 cm, so every side of the square is 4 cm. That makes the vertical side HG = 4 cm -- the height of the trapezoid -- and the top piece HD = 4 cm as well.
HD=HG=GC=DC=4 cmHD = HG = GC = DC = 4\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=11+4=15 cmAD = AH + HD = 11 + 4 = 15\ \text{cm}
The top is 15 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 15 cm and the bottom BC = 26 cm, and the height is the square's side, 4 cm. Add the parallel sides, multiply by the height, halve.
(15+26)×4÷2=41×4÷2=82 cm2(15 + 26) \times 4 \div 2 = 41 \times 4 \div 2 = 82\ \text{cm}^2
The trapezoid covers 82 cm2.
Answer: 82 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 26 by 4 rectangle of area 104 cm2 and contains an 15 by 4 one of area 60 cm2. 82 cm2 sits between them, as it must.

Another way: The 12 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 9 hard answer: 129 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 28 cm28\ \text{cm}. Point HH lies on the top side ADAD with AH=9 cm\overline{AH}=9\ \text{cm}, and the right side DC=6 cm\overline{DC}=6\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=13 cm\overline{EF}=13\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 9 cm 6 cm 28 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 28 cm, and a square HGCD sits in its right-hand corner with side DC = 6 cm. H lies on the top side with AH = 9 cm. We need the trapezoid's area.

Givens
  • BC = 28 cm along the bottom.
  • AH = 9 cm, with H on the top side AD.
  • DC = 6 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 6 cm, so every side of the square is 6 cm. That makes the vertical side HG = 6 cm -- the height of the trapezoid -- and the top piece HD = 6 cm as well.
HD=HG=GC=DC=6 cmHD = HG = GC = DC = 6\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=9+6=15 cmAD = AH + HD = 9 + 6 = 15\ \text{cm}
The top is 15 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 15 cm and the bottom BC = 28 cm, and the height is the square's side, 6 cm. Add the parallel sides, multiply by the height, halve.
(15+28)×6÷2=43×6÷2=129 cm2(15 + 28) \times 6 \div 2 = 43 \times 6 \div 2 = 129\ \text{cm}^2
The trapezoid covers 129 cm2.
Answer: 129 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 28 by 6 rectangle of area 168 cm2 and contains an 15 by 6 one of area 90 cm2. 129 cm2 sits between them, as it must.

Another way: The 13 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 10 hard answer: 260 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 30 cm30\ \text{cm}. Point HH lies on the top side ADAD with AH=12 cm\overline{AH}=12\ \text{cm}, and the right side DC=10 cm\overline{DC}=10\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=14 cm\overline{EF}=14\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 12 cm 10 cm 30 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 30 cm, and a square HGCD sits in its right-hand corner with side DC = 10 cm. H lies on the top side with AH = 12 cm. We need the trapezoid's area.

Givens
  • BC = 30 cm along the bottom.
  • AH = 12 cm, with H on the top side AD.
  • DC = 10 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 10 cm, so every side of the square is 10 cm. That makes the vertical side HG = 10 cm -- the height of the trapezoid -- and the top piece HD = 10 cm as well.
HD=HG=GC=DC=10 cmHD = HG = GC = DC = 10\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=12+10=22 cmAD = AH + HD = 12 + 10 = 22\ \text{cm}
The top is 22 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 22 cm and the bottom BC = 30 cm, and the height is the square's side, 10 cm. Add the parallel sides, multiply by the height, halve.
(22+30)×10÷2=52×10÷2=260 cm2(22 + 30) \times 10 \div 2 = 52 \times 10 \div 2 = 260\ \text{cm}^2
The trapezoid covers 260 cm2.
Answer: 260 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 30 by 10 rectangle of area 300 cm2 and contains an 22 by 10 one of area 220 cm2. 260 cm2 sits between them, as it must.

Another way: The 14 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 11 hard answer: 220 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 32 cm32\ \text{cm}. Point HH lies on the top side ADAD with AH=15 cm\overline{AH}=15\ \text{cm}, and the right side DC=8 cm\overline{DC}=8\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=17 cm\overline{EF}=17\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 15 cm 8 cm 32 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 32 cm, and a square HGCD sits in its right-hand corner with side DC = 8 cm. H lies on the top side with AH = 15 cm. We need the trapezoid's area.

Givens
  • BC = 32 cm along the bottom.
  • AH = 15 cm, with H on the top side AD.
  • DC = 8 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 8 cm, so every side of the square is 8 cm. That makes the vertical side HG = 8 cm -- the height of the trapezoid -- and the top piece HD = 8 cm as well.
HD=HG=GC=DC=8 cmHD = HG = GC = DC = 8\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=15+8=23 cmAD = AH + HD = 15 + 8 = 23\ \text{cm}
The top is 23 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 23 cm and the bottom BC = 32 cm, and the height is the square's side, 8 cm. Add the parallel sides, multiply by the height, halve.
(23+32)×8÷2=55×8÷2=220 cm2(23 + 32) \times 8 \div 2 = 55 \times 8 \div 2 = 220\ \text{cm}^2
The trapezoid covers 220 cm2.
Answer: 220 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 32 by 8 rectangle of area 256 cm2 and contains an 23 by 8 one of area 184 cm2. 220 cm2 sits between them, as it must.

Another way: The 17 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.
Variant 12 hard answer: 300 cm²

In the figure, quadrilateral ABCDABCD and quadrilateral EFCGEFCG are congruent. Find the area, in cm2\text{cm}^2, of quadrilateral ABCDABCD.

The figure consists of a lower trapezoid ABCDABCD and a congruent quadrilateral EFCGEFCG tilted above and to its right; the two quadrilaterals share vertex CC. In trapezoid ABCDABCD, the bottom side BC\overline{BC} has length 36 cm36\ \text{cm}. Point HH lies on the top side ADAD with AH=14 cm\overline{AH}=14\ \text{cm}, and the right side DC=10 cm\overline{DC}=10\ \text{cm}. In the upper quadrilateral EFCGEFCG, side EF=15 cm\overline{EF}=15\ \text{cm}. Quadrilateral HGCDHGCD is a square.

A H D B C G 14 cm 10 cm 36 cm
Show solution
1 · Understandwhat's really being asked

A trapezoid ABCD has a bottom of 36 cm, and a square HGCD sits in its right-hand corner with side DC = 10 cm. H lies on the top side with AH = 14 cm. We need the trapezoid's area.

Givens
  • BC = 36 cm along the bottom.
  • AH = 14 cm, with H on the top side AD.
  • DC = 10 cm, and HGCD is a square.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The trapezoid's height is not given directly -- it has to come out of the square.
2 · Planchoose the strategy

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

The square is the key: one side of a square gives all four. That fills in both the rest of the top side and the height, and then the trapezoid formula finishes it.

3 · Execute3 carry out the plan

1Use the square to find the height and the missing top piece

#1 Draw a Diagram 4.G.A.2
HGCD is a square and one of its sides is DC = 10 cm, so every side of the square is 10 cm. That makes the vertical side HG = 10 cm -- the height of the trapezoid -- and the top piece HD = 10 cm as well.
HD=HG=GC=DC=10 cmHD = HG = GC = DC = 10\ \text{cm}
One length, four places to use it.

2Find the full top side AD

#7 Identify Subproblems 6.G.A.1
Point H sits on the top side AD, splitting it into AH and HD. Add the two pieces to get the whole top side.
AD=AH+HD=14+10=24 cmAD = AH + HD = 14 + 10 = 24\ \text{cm}
The top is 24 cm.

3Apply the trapezoid area formula

#7 Identify Subproblems 6.G.A.1
The two parallel sides are the top AD = 24 cm and the bottom BC = 36 cm, and the height is the square's side, 10 cm. Add the parallel sides, multiply by the height, halve.
(24+36)×10÷2=60×10÷2=300 cm2(24 + 36) \times 10 \div 2 = 60 \times 10 \div 2 = 300\ \text{cm}^2
The trapezoid covers 300 cm2.
Answer: 300 cm²
4 · Reviewdoes it hold up?

The trapezoid sits inside a 36 by 10 rectangle of area 360 cm2 and contains an 24 by 10 one of area 240 cm2. 300 cm2 sits between them, as it must.

Another way: The 15 cm side EF of the congruent upper quadrilateral is not needed at all. Congruence guarantees it matches a side of ABCD, but no step of the area calculation asks for it -- a given can be there without being required.

Standardsmin grade 6
  • 4.G.A.2 Classify two-dimensional figures based on parallel or perpendicular lines — Using that all four sides of a square are equal.
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Assembling the top side and applying the trapezoid formula.
💡Takeaway. Spotting a square in a figure is worth four measurements, not one.