← The axis bisects the segment joining a point to its image · Transformations Preserve Measures

The axis bisects the segment joining a point to its image · 12 practice problems

3.MD.C.74.G.A.34.MD.C.5

Generated variants — 12

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

Variant 1 medium answer: 81 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 18 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 18 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 18 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 18 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 18 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 18 cm.
(18060)÷2=60,BB=18 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 18\ \text{cm}
BB' is 18 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 18 divided by 2.
18÷2=9 cm18 \div 2 = 9\ \text{cm}
B stands 9 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 18 cm. Taking AC = 18 cm as the base -- it lies along the axis -- the matching height is the 9 cm just found.
18×9÷2=81 (cm2)18 \times 9 \div 2 = 81\ (\text{cm}^2)
The triangle covers 81 cm2.
Answer: 81 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 9 by 18 rectangle, whose area is 162 cm2. Half of that is 81 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 18, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 2 medium answer: 36 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 12 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 12 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 12 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 12 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 12 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 12 cm.
(18060)÷2=60,BB=12 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 12\ \text{cm}
BB' is 12 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 12 divided by 2.
12÷2=6 cm12 \div 2 = 6\ \text{cm}
B stands 6 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 12 cm. Taking AC = 12 cm as the base -- it lies along the axis -- the matching height is the 6 cm just found.
12×6÷2=36 (cm2)12 \times 6 \div 2 = 36\ (\text{cm}^2)
The triangle covers 36 cm2.
Answer: 36 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 6 by 12 rectangle, whose area is 72 cm2. Half of that is 36 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 12, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 3 medium answer: 196 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 28 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 28 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 28 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 28 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 28 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 28 cm.
(18060)÷2=60,BB=28 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 28\ \text{cm}
BB' is 28 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 28 divided by 2.
28÷2=14 cm28 \div 2 = 14\ \text{cm}
B stands 14 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 28 cm. Taking AC = 28 cm as the base -- it lies along the axis -- the matching height is the 14 cm just found.
28×14÷2=196 (cm2)28 \times 14 \div 2 = 196\ (\text{cm}^2)
The triangle covers 196 cm2.
Answer: 196 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 14 by 28 rectangle, whose area is 392 cm2. Half of that is 196 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 28, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 4 medium answer: 64 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 16 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 16 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 16 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 16 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 16 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 16 cm.
(18060)÷2=60,BB=16 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 16\ \text{cm}
BB' is 16 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 16 divided by 2.
16÷2=8 cm16 \div 2 = 8\ \text{cm}
B stands 8 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 16 cm. Taking AC = 16 cm as the base -- it lies along the axis -- the matching height is the 8 cm just found.
16×8÷2=64 (cm2)16 \times 8 \div 2 = 64\ (\text{cm}^2)
The triangle covers 64 cm2.
Answer: 64 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 8 by 16 rectangle, whose area is 128 cm2. Half of that is 64 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 16, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 5 medium answer: 144 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 24 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 24 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 24 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 24 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 24 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 24 cm.
(18060)÷2=60,BB=24 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 24\ \text{cm}
BB' is 24 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 24 divided by 2.
24÷2=12 cm24 \div 2 = 12\ \text{cm}
B stands 12 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 24 cm. Taking AC = 24 cm as the base -- it lies along the axis -- the matching height is the 12 cm just found.
24×12÷2=144 (cm2)24 \times 12 \div 2 = 144\ (\text{cm}^2)
The triangle covers 144 cm2.
Answer: 144 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 12 by 24 rectangle, whose area is 288 cm2. Half of that is 144 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 24, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 6 medium answer: 169 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 26 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 26 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 26 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 26 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 26 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 26 cm.
(18060)÷2=60,BB=26 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 26\ \text{cm}
BB' is 26 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 26 divided by 2.
26÷2=13 cm26 \div 2 = 13\ \text{cm}
B stands 13 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 26 cm. Taking AC = 26 cm as the base -- it lies along the axis -- the matching height is the 13 cm just found.
26×13÷2=169 (cm2)26 \times 13 \div 2 = 169\ (\text{cm}^2)
The triangle covers 169 cm2.
Answer: 169 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 13 by 26 rectangle, whose area is 338 cm2. Half of that is 169 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 26, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 7 medium answer: 49 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 14 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 14 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 14 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 14 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 14 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 14 cm.
(18060)÷2=60,BB=14 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 14\ \text{cm}
BB' is 14 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 14 divided by 2.
14÷2=7 cm14 \div 2 = 7\ \text{cm}
B stands 7 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 14 cm. Taking AC = 14 cm as the base -- it lies along the axis -- the matching height is the 7 cm just found.
14×7÷2=49 (cm2)14 \times 7 \div 2 = 49\ (\text{cm}^2)
The triangle covers 49 cm2.
Answer: 49 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 7 by 14 rectangle, whose area is 98 cm2. Half of that is 49 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 14, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 8 medium answer: 16 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 8 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 8 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 8 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 8 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 8 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 8 cm.
(18060)÷2=60,BB=8 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 8\ \text{cm}
BB' is 8 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 8 divided by 2.
8÷2=4 cm8 \div 2 = 4\ \text{cm}
B stands 4 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 8 cm. Taking AC = 8 cm as the base -- it lies along the axis -- the matching height is the 4 cm just found.
8×4÷2=16 (cm2)8 \times 4 \div 2 = 16\ (\text{cm}^2)
The triangle covers 16 cm2.
Answer: 16 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 4 by 8 rectangle, whose area is 32 cm2. Half of that is 16 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 8, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 9 medium answer: 100 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 20 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 20 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 20 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 20 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 20 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 20 cm.
(18060)÷2=60,BB=20 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 20\ \text{cm}
BB' is 20 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 20 divided by 2.
20÷2=10 cm20 \div 2 = 10\ \text{cm}
B stands 10 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 20 cm. Taking AC = 20 cm as the base -- it lies along the axis -- the matching height is the 10 cm just found.
20×10÷2=100 (cm2)20 \times 10 \div 2 = 100\ (\text{cm}^2)
The triangle covers 100 cm2.
Answer: 100 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 10 by 20 rectangle, whose area is 200 cm2. Half of that is 100 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 20, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 10 medium answer: 121 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 22 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 22 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 22 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 22 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 22 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 22 cm.
(18060)÷2=60,BB=22 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 22\ \text{cm}
BB' is 22 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 22 divided by 2.
22÷2=11 cm22 \div 2 = 11\ \text{cm}
B stands 11 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 22 cm. Taking AC = 22 cm as the base -- it lies along the axis -- the matching height is the 11 cm just found.
22×11÷2=121 (cm2)22 \times 11 \div 2 = 121\ (\text{cm}^2)
The triangle covers 121 cm2.
Answer: 121 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 11 by 22 rectangle, whose area is 242 cm2. Half of that is 121 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 22, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 11 medium answer: 25 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 10 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 10 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 10 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 10 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 10 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 10 cm.
(18060)÷2=60,BB=10 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 10\ \text{cm}
BB' is 10 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 10 divided by 2.
10÷2=5 cm10 \div 2 = 5\ \text{cm}
B stands 5 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 10 cm. Taking AC = 10 cm as the base -- it lies along the axis -- the matching height is the 5 cm just found.
10×5÷2=25 (cm2)10 \times 5 \div 2 = 25\ (\text{cm}^2)
The triangle covers 25 cm2.
Answer: 25 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 5 by 10 rectangle, whose area is 50 cm2. Half of that is 25 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 10, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.
Variant 12 medium answer: 9 cm²

In the figure, line \ell is used as the axis of symmetry to draw a line-symmetric figure. If triangle ABC is an isosceles triangle, find the area of triangle ABC in cm2\text{cm}^2.

A B B' C 6 cm 30 l
Show solution
1 · Understandwhat's really being asked

An isosceles triangle ABC has its apex A on a vertical axis of symmetry, with AB = 6 cm making 30 degrees with the axis. We need its area.

Givens
  • AB = 6 cm.
  • The angle between the axis and AB is 30 degrees.
  • A and C both lie on the axis, and the triangle is isosceles.
Unknowns
  • The area of triangle ABC.
Constraints
  • The axis is a line of symmetry, so it reflects B to a matching point on the other side.
2 · Planchoose the strategy

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

Reflect B across the axis. That builds a triangle whose angles are all known, and the axis cuts the segment joining B to its image exactly in half -- which is the height the area formula needs.

3 · Execute4 carry out the plan

1Reflect B across the axis

#1 Draw a Diagram 4.G.A.3
Because the figure is line-symmetric about the axis, reflect B to its corresponding point B'. By symmetry AB' = AB = 6 cm, and the angle at A doubles to 30 + 30.
BAB=30+30=60\angle BAB' = 30^{\circ} + 30^{\circ} = 60^{\circ}
A 60 degree angle between two equal sides.

2Recognize the equilateral triangle

#7 Identify Subproblems 4.MD.C.5
Triangle ABB' has two equal legs (AB = AB' = 6 cm) and 60 degrees between them. The two base angles must be equal and the three angles add to 180, so each base angle is 60 degrees too -- all three angles are 60, so the triangle is equilateral and BB' = 6 cm.
(18060)÷2=60,BB=6 cm(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}, \quad BB' = 6\ \text{cm}
BB' is 6 cm, the same as the legs.

3Find the height of triangle ABC

#1 Draw a Diagram 4.G.A.3
The axis joins B to its mirror image B', so it cuts BB' exactly in half. The distance from B to the axis is therefore 6 divided by 2.
6÷2=3 cm6 \div 2 = 3\ \text{cm}
B stands 3 cm from the axis.

4Compute the area

#7 Identify Subproblems 3.MD.C.7
Triangle ABC is isosceles with AB = AC = 6 cm. Taking AC = 6 cm as the base -- it lies along the axis -- the matching height is the 3 cm just found.
6×3÷2=9 (cm2)6 \times 3 \div 2 = 9\ (\text{cm}^2)
The triangle covers 9 cm2.
Answer: 9 cm²
4 · Reviewdoes it hold up?

Sanity check the shape: triangle ABC sits inside a 3 by 6 rectangle, whose area is 18 cm2. Half of that is 9 cm2, which is what a triangle on the same base and height should be.

Another way: Triangle ABB' is equilateral with side 6, and ABC is exactly half of a rhombus made from it -- another way of seeing why the height is half of BB'.

Standardsmin grade 4
  • 3.MD.C.7 Relate area to the operations of multiplication and addition — Turning a base and a height into an area.
  • 4.G.A.3 Recognize a line of symmetry for a two-dimensional figure — Reflecting B across the axis and using that the axis bisects BB'.
  • 4.MD.C.5 Recognize angles as geometric shapes formed by two rays sharing an endpoint — Doubling the angle at A and splitting 180 among three of them.
💡Takeaway. The axis always cuts the line from a point to its mirror image exactly in half, so half that line is the distance to the axis.