Practice · Triangles that share a height are in the ratio of their bases · Sensim Math

Variant 1 easy

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 9 cm base, a perpendicular segment AE meets the base at E, with D on it 5 cm from A and 2 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 9 cm.
AD is 5 cm and DE is 2 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 5 plus 2, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 5 + 2 = 7,\quad \text{DE} = 2

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 7 = \text{EC} \times 2

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 9, that equation gives both.

\text{BE} = 2,\quad \text{EC} = 7

2 cm and 7 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

2 : 7 = 2 : 7

2 to 7.

Answer: 2 : 7

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 7 and DEC is 7 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 2 easy

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 9 cm base, a perpendicular segment AE meets the base at E, with D on it 2 cm from A and 2 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 9 cm.
AD is 2 cm and DE is 2 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 2 plus 2, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 2 + 2 = 4,\quad \text{DE} = 2

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 4 = \text{EC} \times 2

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 9, that equation gives both.

\text{BE} = 3,\quad \text{EC} = 6

3 cm and 6 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

3 : 6 = 1 : 2

1 to 2.

Answer: 1 : 2

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 6 and DEC is 6 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 3 easy

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 14 cm base, a perpendicular segment AE meets the base at E, with D on it 6 cm from A and 4 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 14 cm.
AD is 6 cm and DE is 4 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 6 plus 4, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 6 + 4 = 10,\quad \text{DE} = 4

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 10 = \text{EC} \times 4

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 14, that equation gives both.

\text{BE} = 4,\quad \text{EC} = 10

4 cm and 10 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

4 : 10 = 2 : 5

2 to 5.

Answer: 2 : 5

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 20 and DEC is 20 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 4 medium

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 16 cm base, a perpendicular segment AE meets the base at E, with D on it 6 cm from A and 3 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 16 cm.
AD is 6 cm and DE is 3 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 6 plus 3, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 6 + 3 = 9,\quad \text{DE} = 3

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 9 = \text{EC} \times 3

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 16, that equation gives both.

\text{BE} = 4,\quad \text{EC} = 12

4 cm and 12 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

4 : 12 = 1 : 3

1 to 3.

Answer: 1 : 3

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 18 and DEC is 18 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 5 easy

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 16 cm base, a perpendicular segment AE meets the base at E, with D on it 12 cm from A and 10 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 16 cm.
AD is 12 cm and DE is 10 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 12 plus 10, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 12 + 10 = 22,\quad \text{DE} = 10

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 22 = \text{EC} \times 10

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 16, that equation gives both.

\text{BE} = 5,\quad \text{EC} = 11

5 cm and 11 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

5 : 11 = 5 : 11

5 to 11.

Answer: 5 : 11

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 55 and DEC is 55 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 6 medium

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 18 cm base, a perpendicular segment AE meets the base at E, with D on it 10 cm from A and 10 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 18 cm.
AD is 10 cm and DE is 10 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 10 plus 10, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 10 + 10 = 20,\quad \text{DE} = 10

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 20 = \text{EC} \times 10

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 18, that equation gives both.

\text{BE} = 6,\quad \text{EC} = 12

6 cm and 12 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

6 : 12 = 1 : 2

1 to 2.

Answer: 1 : 2

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 60 and DEC is 60 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 7 medium

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 19 cm base, a perpendicular segment AE meets the base at E, with D on it 3 cm from A and 8 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 19 cm.
AD is 3 cm and DE is 8 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 3 plus 8, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 3 + 8 = 11,\quad \text{DE} = 8

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 11 = \text{EC} \times 8

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 19, that equation gives both.

\text{BE} = 8,\quad \text{EC} = 11

8 cm and 11 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

8 : 11 = 8 : 11

8 to 11.

Answer: 8 : 11

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 44 and DEC is 44 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 8 medium

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 25 cm base, a perpendicular segment AE meets the base at E, with D on it 6 cm from A and 2 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 25 cm.
AD is 6 cm and DE is 2 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 6 plus 2, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 6 + 2 = 8,\quad \text{DE} = 2

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 8 = \text{EC} \times 2

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 25, that equation gives both.

\text{BE} = 5,\quad \text{EC} = 20

5 cm and 20 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

5 : 20 = 1 : 4

1 to 4.

Answer: 1 : 4

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 20 and DEC is 20 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 9 hard

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 28 cm base, a perpendicular segment AE meets the base at E, with D on it 8 cm from A and 10 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 28 cm.
AD is 8 cm and DE is 10 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 8 plus 10, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 8 + 10 = 18,\quad \text{DE} = 10

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 18 = \text{EC} \times 10

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 28, that equation gives both.

\text{BE} = 10,\quad \text{EC} = 18

10 cm and 18 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

10 : 18 = 5 : 9

5 to 9.

Answer: 5 : 9

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 90 and DEC is 90 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 10 hard

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 33 cm base, a perpendicular segment AE meets the base at E, with D on it 7 cm from A and 7 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 33 cm.
AD is 7 cm and DE is 7 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 7 plus 7, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 7 + 7 = 14,\quad \text{DE} = 7

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 14 = \text{EC} \times 7

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 33, that equation gives both.

\text{BE} = 11,\quad \text{EC} = 22

11 cm and 22 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

11 : 22 = 1 : 2

1 to 2.

Answer: 1 : 2

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 77 and DEC is 77 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 11 hard

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 33 cm base, a perpendicular segment AE meets the base at E, with D on it 4 cm from A and 4 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 33 cm.
AD is 4 cm and DE is 4 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 4 plus 4, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 4 + 4 = 8,\quad \text{DE} = 4

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 8 = \text{EC} \times 4

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 33, that equation gives both.

\text{BE} = 11,\quad \text{EC} = 22

11 cm and 22 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

11 : 22 = 1 : 2

1 to 2.

Answer: 1 : 2

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 44 and DEC is 44 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Variant 12 hard

In the figure at the right, triangle ABE and triangle DEC have equal areas. Write the ratio of the area of triangle ABE to the area of triangle AEC as a ratio of whole numbers in simplest form.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

In a triangle on a 36 cm base, a perpendicular segment AE meets the base at E, with D on it 5 cm from A and 5 cm above the base. Triangles ABE and DEC have equal areas, and we want the ratio ABE to AEC.

Givens

BC is 36 cm.
AD is 5 cm and DE is 5 cm.
AE meets BC at a right angle.
Triangles ABE and DEC have equal areas.

Unknowns

The ratio of triangle ABE to triangle AEC, in simplest form.

Constraints

E lies on BC, so BE and EC add to the whole base.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#7 Identify Subproblems

The right angle hands over two heights at once: AE for the triangles standing on the whole segment, DE for the one on its lower part. Turn the equal-area fact into an equation in the two base pieces, and the final ratio then needs no areas at all.

3 · Execute4 carry out the plan

1Read both heights off the right angle

#1 Draw a Diagram 6.G.A.1

AE is 5 plus 5, and both AE and DE are perpendicular to the base, so both are heights onto it.

\text{AE} = 5 + 5 = 10,\quad \text{DE} = 5

Two heights, no measuring.

2Write the equal-area condition

#13 Convert to Algebra 6.G.A.1

Each triangle is half its base times its height, and the halves cancel.

\text{BE} \times 10 = \text{EC} \times 5

One equation in the two pieces of the base.

3Split the base

#7 Identify Subproblems 7.RP.A.2

With BE and EC adding to 36, that equation gives both.

\text{BE} = 12,\quad \text{EC} = 24

12 cm and 24 cm.

4Compare ABE with AEC

#7 Identify Subproblems 7.RP.A.2

Both stand on the base with the same height AE, so their areas are in the ratio of their bases.

12 : 24 = 1 : 2

1 to 2.

Answer: 1 : 2

4 · Reviewdoes it hold up?

Checking the equal areas: ABE is 60 and DEC is 60 square centimetres -- the same, as the problem says.

Another way: Working out both areas and dividing gives the same ratio, but the heights cancel, which is why the bases alone answer it.

Standardsmin grade 7

6.G.A.1 Find area of triangles, quadrilaterals, polygons by composing/decomposing — Using the triangle area rule with the height the right angle gives.
7.RP.A.2 Recognize and represent proportional relationships between quantities — Reading the area ratio off the base ratio.

💡Takeaway. Two triangles with the same height compare exactly as their bases do. You never have to work out either area.

Triangles that share a height are in the ratio of their bases · 12 practice problems

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