← Triangles that share a height are in the ratio of their bases · Proportion and Proportional Division

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

6.G.A.17.RP.A.2

Generated variants — 12

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

Variant 1 easy answer: 2 : 7

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.

A B C E D 5 cm 2 cm 9 cm
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.
AE=5+2=7,DE=2\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.
BE×7=EC×2\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.
BE=2,EC=7\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=2:72 : 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 answer: 1 : 2

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.

A B C E D 2 cm 2 cm 9 cm
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.
AE=2+2=4,DE=2\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.
BE×4=EC×2\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.
BE=3,EC=6\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=1:23 : 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 answer: 2 : 5

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.

A B C E D 6 cm 4 cm 14 cm
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.
AE=6+4=10,DE=4\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.
BE×10=EC×4\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.
BE=4,EC=10\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=2:54 : 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 answer: 1 : 3

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.

A B C E D 6 cm 3 cm 16 cm
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.
AE=6+3=9,DE=3\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.
BE×9=EC×3\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.
BE=4,EC=12\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=1:34 : 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 answer: 5 : 11

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.

A B C E D 12 cm 10 cm 16 cm
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.
AE=12+10=22,DE=10\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.
BE×22=EC×10\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.
BE=5,EC=11\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=5:115 : 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 answer: 1 : 2

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.

A B C E D 10 cm 10 cm 18 cm
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.
AE=10+10=20,DE=10\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.
BE×20=EC×10\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.
BE=6,EC=12\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=1:26 : 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 answer: 8 : 11

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.

A B C E D 3 cm 8 cm 19 cm
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.
AE=3+8=11,DE=8\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.
BE×11=EC×8\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.
BE=8,EC=11\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=8:118 : 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 answer: 1 : 4

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.

A B C E D 6 cm 2 cm 25 cm
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.
AE=6+2=8,DE=2\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.
BE×8=EC×2\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.
BE=5,EC=20\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=1:45 : 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 answer: 5 : 9

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.

A B C E D 8 cm 10 cm 28 cm
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.
AE=8+10=18,DE=10\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.
BE×18=EC×10\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.
BE=10,EC=18\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=5:910 : 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 answer: 1 : 2

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.

A B C E D 7 cm 7 cm 33 cm
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.
AE=7+7=14,DE=7\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.
BE×14=EC×7\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.
BE=11,EC=22\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=1:211 : 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 answer: 1 : 2

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.

A B C E D 4 cm 4 cm 33 cm
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.
AE=4+4=8,DE=4\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.
BE×8=EC×4\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.
BE=11,EC=22\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=1:211 : 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 answer: 1 : 2

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.

A B C E D 5 cm 5 cm 36 cm
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.
AE=5+5=10,DE=5\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.
BE×10=EC×5\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.
BE=12,EC=24\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=1:212 : 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.