← Triangle height depends on chosen base · Area by Decomposition

Triangle height depends on chosen base · 12 practice problems

6.G.A.1

Generated variants — 12

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

Variant 1 easy answer: 68 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 6 cm6\ \text{cm} long and side DC is 11 cm11\ \text{cm} long. The diagonal AC is drawn; its length is 10 cm10\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 4.8 cm4.8\ \text{cm}.

A D B C 6 11 10 4.8
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 6 cm and DC = 11 cm, both perpendicular to BC.
  • The diagonal AC is 10 cm long.
  • The perpendicular from B to AC is 4.8 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 6 cm) and triangle ACD (holding side DC = 11 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 10 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 4.8 cm.
AreaABC=12×10×4.8=24 cm2\text{Area}_{ABC} = \frac{1}{2} \times 10 \times 4.8 = 24\ \text{cm}^2
Triangle ABC covers 24 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 6 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×6×BC=24    BC=8 cm\frac{1}{2} \times 6 \times BC = 24 \;\Rightarrow\; BC = 8\ \text{cm}
The width is 8 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 11 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 8 cm we just found.
AreaACD=12×11×8=44 cm2\text{Area}_{ACD} = \frac{1}{2} \times 11 \times 8 = 44\ \text{cm}^2
Triangle ACD covers 44 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
24+44=68 cm224 + 44 = 68\ \text{cm}^2
The trapezoid covers 68 cm2.
Answer: 68 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 6 and 11, and the distance between them is BC = 8, so (6 + 11) / 2 x 8 = 68 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 6 x 6 + 8 x 8 = 10 x 10, so a diagonal of 10 cm is exactly what a 6 by 8 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 2 easy answer: 69 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 8 cm8\ \text{cm} long and side DC is 15 cm15\ \text{cm} long. The diagonal AC is drawn; its length is 10 cm10\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 4.8 cm4.8\ \text{cm}.

A D B C 8 15 10 4.8
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 8 cm and DC = 15 cm, both perpendicular to BC.
  • The diagonal AC is 10 cm long.
  • The perpendicular from B to AC is 4.8 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 8 cm) and triangle ACD (holding side DC = 15 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 10 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 4.8 cm.
AreaABC=12×10×4.8=24 cm2\text{Area}_{ABC} = \frac{1}{2} \times 10 \times 4.8 = 24\ \text{cm}^2
Triangle ABC covers 24 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 8 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×8×BC=24    BC=6 cm\frac{1}{2} \times 8 \times BC = 24 \;\Rightarrow\; BC = 6\ \text{cm}
The width is 6 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 15 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 6 cm we just found.
AreaACD=12×15×6=45 cm2\text{Area}_{ACD} = \frac{1}{2} \times 15 \times 6 = 45\ \text{cm}^2
Triangle ACD covers 45 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
24+45=69 cm224 + 45 = 69\ \text{cm}^2
The trapezoid covers 69 cm2.
Answer: 69 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 8 and 15, and the distance between them is BC = 6, so (8 + 15) / 2 x 6 = 69 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 8 x 8 + 6 x 6 = 10 x 10, so a diagonal of 10 cm is exactly what a 8 by 6 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 3 easy answer: 156 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 9 cm9\ \text{cm} long and side DC is 17 cm17\ \text{cm} long. The diagonal AC is drawn; its length is 15 cm15\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 7.2 cm7.2\ \text{cm}.

A D B C 9 17 15 7.2
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 9 cm and DC = 17 cm, both perpendicular to BC.
  • The diagonal AC is 15 cm long.
  • The perpendicular from B to AC is 7.2 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 9 cm) and triangle ACD (holding side DC = 17 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 15 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 7.2 cm.
AreaABC=12×15×7.2=54 cm2\text{Area}_{ABC} = \frac{1}{2} \times 15 \times 7.2 = 54\ \text{cm}^2
Triangle ABC covers 54 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 9 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×9×BC=54    BC=12 cm\frac{1}{2} \times 9 \times BC = 54 \;\Rightarrow\; BC = 12\ \text{cm}
The width is 12 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 17 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 12 cm we just found.
AreaACD=12×17×12=102 cm2\text{Area}_{ACD} = \frac{1}{2} \times 17 \times 12 = 102\ \text{cm}^2
Triangle ACD covers 102 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
54+102=156 cm254 + 102 = 156\ \text{cm}^2
The trapezoid covers 156 cm2.
Answer: 156 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 9 and 17, and the distance between them is BC = 12, so (9 + 17) / 2 x 12 = 156 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 9 x 9 + 12 x 12 = 15 x 15, so a diagonal of 15 cm is exactly what a 9 by 12 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 4 easy answer: 264 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 12 cm12\ \text{cm} long and side DC is 21 cm21\ \text{cm} long. The diagonal AC is drawn; its length is 20 cm20\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 9.6 cm9.6\ \text{cm}.

A D B C 12 21 20 9.6
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 12 cm and DC = 21 cm, both perpendicular to BC.
  • The diagonal AC is 20 cm long.
  • The perpendicular from B to AC is 9.6 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 12 cm) and triangle ACD (holding side DC = 21 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 20 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 9.6 cm.
AreaABC=12×20×9.6=96 cm2\text{Area}_{ABC} = \frac{1}{2} \times 20 \times 9.6 = 96\ \text{cm}^2
Triangle ABC covers 96 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 12 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×12×BC=96    BC=16 cm\frac{1}{2} \times 12 \times BC = 96 \;\Rightarrow\; BC = 16\ \text{cm}
The width is 16 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 21 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 16 cm we just found.
AreaACD=12×21×16=168 cm2\text{Area}_{ACD} = \frac{1}{2} \times 21 \times 16 = 168\ \text{cm}^2
Triangle ACD covers 168 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
96+168=264 cm296 + 168 = 264\ \text{cm}^2
The trapezoid covers 264 cm2.
Answer: 264 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 12 and 21, and the distance between them is BC = 16, so (12 + 21) / 2 x 16 = 264 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 12 x 12 + 16 x 16 = 20 x 20, so a diagonal of 20 cm is exactly what a 12 by 16 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 5 medium answer: 240 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 7 cm7\ \text{cm} long and side DC is 13 cm13\ \text{cm} long. The diagonal AC is drawn; its length is 25 cm25\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 6.72 cm6.72\ \text{cm}.

A D B C 7 13 25 6.72
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 7 cm and DC = 13 cm, both perpendicular to BC.
  • The diagonal AC is 25 cm long.
  • The perpendicular from B to AC is 6.72 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 7 cm) and triangle ACD (holding side DC = 13 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 25 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 6.72 cm.
AreaABC=12×25×6.72=84 cm2\text{Area}_{ABC} = \frac{1}{2} \times 25 \times 6.72 = 84\ \text{cm}^2
Triangle ABC covers 84 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 7 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×7×BC=84    BC=24 cm\frac{1}{2} \times 7 \times BC = 84 \;\Rightarrow\; BC = 24\ \text{cm}
The width is 24 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 13 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 24 cm we just found.
AreaACD=12×13×24=156 cm2\text{Area}_{ACD} = \frac{1}{2} \times 13 \times 24 = 156\ \text{cm}^2
Triangle ACD covers 156 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
84+156=240 cm284 + 156 = 240\ \text{cm}^2
The trapezoid covers 240 cm2.
Answer: 240 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 7 and 13, and the distance between them is BC = 24, so (7 + 13) / 2 x 24 = 240 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 7 x 7 + 24 x 24 = 25 x 25, so a diagonal of 25 cm is exactly what a 7 by 24 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 6 medium answer: 345 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 20 cm20\ \text{cm} long and side DC is 26 cm26\ \text{cm} long. The diagonal AC is drawn; its length is 25 cm25\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 12 cm12\ \text{cm}.

A D B C 20 26 25 12
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 20 cm and DC = 26 cm, both perpendicular to BC.
  • The diagonal AC is 25 cm long.
  • The perpendicular from B to AC is 12 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 20 cm) and triangle ACD (holding side DC = 26 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 25 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 12 cm.
AreaABC=12×25×12=150 cm2\text{Area}_{ABC} = \frac{1}{2} \times 25 \times 12 = 150\ \text{cm}^2
Triangle ABC covers 150 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 20 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×20×BC=150    BC=15 cm\frac{1}{2} \times 20 \times BC = 150 \;\Rightarrow\; BC = 15\ \text{cm}
The width is 15 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 26 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 15 cm we just found.
AreaACD=12×26×15=195 cm2\text{Area}_{ACD} = \frac{1}{2} \times 26 \times 15 = 195\ \text{cm}^2
Triangle ACD covers 195 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
150+195=345 cm2150 + 195 = 345\ \text{cm}^2
The trapezoid covers 345 cm2.
Answer: 345 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 20 and 26, and the distance between them is BC = 15, so (20 + 26) / 2 x 15 = 345 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 20 x 20 + 15 x 15 = 25 x 25, so a diagonal of 25 cm is exactly what a 20 by 15 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 7 medium answer: 440 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 15 cm15\ \text{cm} long and side DC is 29 cm29\ \text{cm} long. The diagonal AC is drawn; its length is 25 cm25\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 12 cm12\ \text{cm}.

A D B C 15 29 25 12
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 15 cm and DC = 29 cm, both perpendicular to BC.
  • The diagonal AC is 25 cm long.
  • The perpendicular from B to AC is 12 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 15 cm) and triangle ACD (holding side DC = 29 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 25 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 12 cm.
AreaABC=12×25×12=150 cm2\text{Area}_{ABC} = \frac{1}{2} \times 25 \times 12 = 150\ \text{cm}^2
Triangle ABC covers 150 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 15 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×15×BC=150    BC=20 cm\frac{1}{2} \times 15 \times BC = 150 \;\Rightarrow\; BC = 20\ \text{cm}
The width is 20 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 29 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 20 cm we just found.
AreaACD=12×29×20=290 cm2\text{Area}_{ACD} = \frac{1}{2} \times 29 \times 20 = 290\ \text{cm}^2
Triangle ACD covers 290 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
150+290=440 cm2150 + 290 = 440\ \text{cm}^2
The trapezoid covers 440 cm2.
Answer: 440 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 15 and 29, and the distance between them is BC = 20, so (15 + 29) / 2 x 20 = 440 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 15 x 15 + 20 x 20 = 25 x 25, so a diagonal of 25 cm is exactly what a 15 by 20 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 8 medium answer: 516 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 18 cm18\ \text{cm} long and side DC is 25 cm25\ \text{cm} long. The diagonal AC is drawn; its length is 30 cm30\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 14.4 cm14.4\ \text{cm}.

A D B C 18 25 30 14.4
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 18 cm and DC = 25 cm, both perpendicular to BC.
  • The diagonal AC is 30 cm long.
  • The perpendicular from B to AC is 14.4 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 18 cm) and triangle ACD (holding side DC = 25 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 30 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 14.4 cm.
AreaABC=12×30×14.4=216 cm2\text{Area}_{ABC} = \frac{1}{2} \times 30 \times 14.4 = 216\ \text{cm}^2
Triangle ABC covers 216 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 18 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×18×BC=216    BC=24 cm\frac{1}{2} \times 18 \times BC = 216 \;\Rightarrow\; BC = 24\ \text{cm}
The width is 24 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 25 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 24 cm we just found.
AreaACD=12×25×24=300 cm2\text{Area}_{ACD} = \frac{1}{2} \times 25 \times 24 = 300\ \text{cm}^2
Triangle ACD covers 300 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
216+300=516 cm2216 + 300 = 516\ \text{cm}^2
The trapezoid covers 516 cm2.
Answer: 516 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 18 and 25, and the distance between them is BC = 24, so (18 + 25) / 2 x 24 = 516 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 18 x 18 + 24 x 24 = 30 x 30, so a diagonal of 30 cm is exactly what a 18 by 24 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 9 hard answer: 700 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 21 cm21\ \text{cm} long and side DC is 29 cm29\ \text{cm} long. The diagonal AC is drawn; its length is 35 cm35\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 16.8 cm16.8\ \text{cm}.

A D B C 21 29 35 16.8
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 21 cm and DC = 29 cm, both perpendicular to BC.
  • The diagonal AC is 35 cm long.
  • The perpendicular from B to AC is 16.8 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 21 cm) and triangle ACD (holding side DC = 29 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 35 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 16.8 cm.
AreaABC=12×35×16.8=294 cm2\text{Area}_{ABC} = \frac{1}{2} \times 35 \times 16.8 = 294\ \text{cm}^2
Triangle ABC covers 294 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 21 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×21×BC=294    BC=28 cm\frac{1}{2} \times 21 \times BC = 294 \;\Rightarrow\; BC = 28\ \text{cm}
The width is 28 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 29 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 28 cm we just found.
AreaACD=12×29×28=406 cm2\text{Area}_{ACD} = \frac{1}{2} \times 29 \times 28 = 406\ \text{cm}^2
Triangle ACD covers 406 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
294+406=700 cm2294 + 406 = 700\ \text{cm}^2
The trapezoid covers 700 cm2.
Answer: 700 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 21 and 29, and the distance between them is BC = 28, so (21 + 29) / 2 x 28 = 700 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 21 x 21 + 28 x 28 = 35 x 35, so a diagonal of 35 cm is exactly what a 21 by 28 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 10 hard answer: 880 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 24 cm24\ \text{cm} long and side DC is 31 cm31\ \text{cm} long. The diagonal AC is drawn; its length is 40 cm40\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 19.2 cm19.2\ \text{cm}.

A D B C 24 31 40 19.2
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 24 cm and DC = 31 cm, both perpendicular to BC.
  • The diagonal AC is 40 cm long.
  • The perpendicular from B to AC is 19.2 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 24 cm) and triangle ACD (holding side DC = 31 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 40 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 19.2 cm.
AreaABC=12×40×19.2=384 cm2\text{Area}_{ABC} = \frac{1}{2} \times 40 \times 19.2 = 384\ \text{cm}^2
Triangle ABC covers 384 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 24 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×24×BC=384    BC=32 cm\frac{1}{2} \times 24 \times BC = 384 \;\Rightarrow\; BC = 32\ \text{cm}
The width is 32 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 31 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 32 cm we just found.
AreaACD=12×31×32=496 cm2\text{Area}_{ACD} = \frac{1}{2} \times 31 \times 32 = 496\ \text{cm}^2
Triangle ACD covers 496 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
384+496=880 cm2384 + 496 = 880\ \text{cm}^2
The trapezoid covers 880 cm2.
Answer: 880 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 24 and 31, and the distance between them is BC = 32, so (24 + 31) / 2 x 32 = 880 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 24 x 24 + 32 x 32 = 40 x 40, so a diagonal of 40 cm is exactly what a 24 by 32 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 11 hard answer: 1080 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 27 cm27\ \text{cm} long and side DC is 33 cm33\ \text{cm} long. The diagonal AC is drawn; its length is 45 cm45\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 21.6 cm21.6\ \text{cm}.

A D B C 27 33 45 21.6
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 27 cm and DC = 33 cm, both perpendicular to BC.
  • The diagonal AC is 45 cm long.
  • The perpendicular from B to AC is 21.6 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 27 cm) and triangle ACD (holding side DC = 33 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 45 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 21.6 cm.
AreaABC=12×45×21.6=486 cm2\text{Area}_{ABC} = \frac{1}{2} \times 45 \times 21.6 = 486\ \text{cm}^2
Triangle ABC covers 486 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 27 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×27×BC=486    BC=36 cm\frac{1}{2} \times 27 \times BC = 486 \;\Rightarrow\; BC = 36\ \text{cm}
The width is 36 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 33 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 36 cm we just found.
AreaACD=12×33×36=594 cm2\text{Area}_{ACD} = \frac{1}{2} \times 33 \times 36 = 594\ \text{cm}^2
Triangle ACD covers 594 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
486+594=1080 cm2486 + 594 = 1080\ \text{cm}^2
The trapezoid covers 1080 cm2.
Answer: 1080 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 27 and 33, and the distance between them is BC = 36, so (27 + 33) / 2 x 36 = 1080 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 27 x 27 + 36 x 36 = 45 x 45, so a diagonal of 45 cm is exactly what a 27 by 36 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.
Variant 12 hard answer: 1420 cm²

What is the area of trapezoid ABCD on the right, in cm2\text{cm}^2?

Figure description: In trapezoid ABCD, A is the top-left vertex, D the top-right, B the bottom-left, and C the bottom-right. The angles at B and at C are right angles, so side AB and side DC are parallel to each other and both perpendicular to side BC. Side AB is 30 cm30\ \text{cm} long and side DC is 41 cm41\ \text{cm} long. The diagonal AC is drawn; its length is 50 cm50\ \text{cm}, and the perpendicular segment dropped from vertex B to the diagonal AC (the height to that diagonal) is 24 cm24\ \text{cm}.

A D B C 30 41 50 24
Show solution
1 · Understandwhat's really being asked

Trapezoid ABCD has right angles at B and C, so AB and DC are the two parallel sides and BC is the width. We know AB, DC, the diagonal AC and the height from B to that diagonal, and we need the area.

Givens
  • AB = 30 cm and DC = 41 cm, both perpendicular to BC.
  • The diagonal AC is 50 cm long.
  • The perpendicular from B to AC is 24 cm.
Unknowns
  • The area of trapezoid ABCD.
Constraints
  • The width BC is not given directly; it has to come out of the triangle we already know the area of.
2 · Planchoose the strategy

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

The diagonal already splits the trapezoid into two triangles. One of them can be measured straight away, and its area -- read a second time off a different base -- gives the width the other triangle needs.

3 · Execute5 carry out the plan

1Split the trapezoid with the diagonal

#7 Identify Subproblems 6.G.A.1
Diagonal AC cuts trapezoid ABCD into triangle ABC (holding side AB = 30 cm) and triangle ACD (holding side DC = 41 cm). The two areas add back to the whole.
AreaABCD=AreaABC+AreaACD\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{ACD}
Two triangles instead of one trapezoid.

2Area of triangle ABC using base AC

#1 Draw a Diagram 6.G.A.1
Take the diagonal AC = 50 cm as the base. Its matching height is the perpendicular dropped from B onto AC, which is 24 cm.
AreaABC=12×50×24=600 cm2\text{Area}_{ABC} = \frac{1}{2} \times 50 \times 24 = 600\ \text{cm}^2
Triangle ABC covers 600 cm2.

3Use the same triangle to find the width BC

#1 Draw a Diagram 6.G.A.1
That triangle has one area, whichever side we call the base. Take AB = 30 cm as the base instead; because AB is perpendicular to BC, the matching height is BC itself.
12×30×BC=600    BC=40 cm\frac{1}{2} \times 30 \times BC = 600 \;\Rightarrow\; BC = 40\ \text{cm}
The width is 40 cm -- got for free from an area we already had.

4Area of triangle ACD using base DC

#1 Draw a Diagram 6.G.A.1
For triangle ACD take DC = 41 cm as the base. DC is perpendicular to BC, so the matching height is the width BC = 40 cm we just found.
AreaACD=12×41×40=820 cm2\text{Area}_{ACD} = \frac{1}{2} \times 41 \times 40 = 820\ \text{cm}^2
Triangle ACD covers 820 cm2.

5Add the two triangle areas

#7 Identify Subproblems 6.G.A.1
The trapezoid's area is the sum of the two triangles.
600+820=1420 cm2600 + 820 = 1420\ \text{cm}^2
The trapezoid covers 1420 cm2.
Answer: 1420 cm²
4 · Reviewdoes it hold up?

The trapezoid formula agrees: the two parallel sides are 30 and 41, and the distance between them is BC = 40, so (30 + 41) / 2 x 40 = 1420 cm2.

Another way: Because the angle at B is a right angle, the diagonal also checks out on its own: 30 x 30 + 40 x 40 = 50 x 50, so a diagonal of 50 cm is exactly what a 30 by 40 corner produces.

Standardsmin grade 6
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing — Splitting the trapezoid into triangles and reading one triangle's area from two different bases.
💡Takeaway. A triangle has only one area, so measuring it off an easy base can tell you a length you were missing on a harder one.