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← 4-2 · Adjacent angles of a parallelogram sum to 180 · Angle Facts in a Figure

Adjacent angles of a parallelogram sum to 180 · 10 practice problems

4.G.A.2

Generated variants — 10

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

Variant 1 answer: 100 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 40∘40^\circ, and the angle marked near vertex D is 20∘20^\circ. The angle to find, ⓐ, is the one marked at vertex A.

40° 20° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 40 deg and the angle near D (angle MDC, between DM and side DC) is 20 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 40 deg.
  • Angle MDC at D is 20 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 40 deg, so the base angle at D inside the triangle is also 40 deg.
∠ADM=∠AMD=40∘\angle ADM = \angle AMD = 40^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−40∘−40∘=100∘a = 180^\circ - 40^\circ - 40^\circ = 100^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 100 degrees

Review

With 40 deg base angles, the apex at A is the leftover 180 - 2(40) = 100 deg. Check: 40 + 40 + 100 = 180 deg. The extra 20 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 40 + 20 = 60 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 40 + 20 = 60 deg, so angle DAB = 120 deg; combined with the isosceles triangle this cross-checks angle DAM = 100 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 40 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 40 deg corners leave a 100 deg angle at A: just the triangle's leftover of 180 deg!
Variant 2 answer: 110 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 35∘35^\circ, and the angle marked near vertex D is 30∘30^\circ. The angle to find, ⓐ, is the one marked at vertex A.

35° 30° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 35 deg and the angle near D (angle MDC, between DM and side DC) is 30 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 35 deg.
  • Angle MDC at D is 30 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 35 deg, so the base angle at D inside the triangle is also 35 deg.
∠ADM=∠AMD=35∘\angle ADM = \angle AMD = 35^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−35∘−35∘=110∘a = 180^\circ - 35^\circ - 35^\circ = 110^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 110 degrees

Review

With 35 deg base angles, the apex at A is the leftover 180 - 2(35) = 110 deg. Check: 35 + 35 + 110 = 180 deg. The extra 30 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 35 + 30 = 65 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 35 + 30 = 65 deg, so angle DAB = 115 deg; combined with the isosceles triangle this cross-checks angle DAM = 110 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 35 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 35 deg corners leave a 110 deg angle at A: just the triangle's leftover of 180 deg!
Variant 3 answer: 70 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 55∘55^\circ, and the angle marked near vertex D is 10∘10^\circ. The angle to find, ⓐ, is the one marked at vertex A.

55° 10° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 55 deg and the angle near D (angle MDC, between DM and side DC) is 10 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 55 deg.
  • Angle MDC at D is 10 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 55 deg, so the base angle at D inside the triangle is also 55 deg.
∠ADM=∠AMD=55∘\angle ADM = \angle AMD = 55^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−55∘−55∘=70∘a = 180^\circ - 55^\circ - 55^\circ = 70^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 70 degrees

Review

With 55 deg base angles, the apex at A is the leftover 180 - 2(55) = 70 deg. Check: 55 + 55 + 70 = 180 deg. The extra 10 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 55 + 10 = 65 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 55 + 10 = 65 deg, so angle DAB = 115 deg; combined with the isosceles triangle this cross-checks angle DAM = 70 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 55 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 55 deg corners leave a 70 deg angle at A: just the triangle's leftover of 180 deg!
Variant 4 answer: 130 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 25∘25^\circ, and the angle marked near vertex D is 40∘40^\circ. The angle to find, ⓐ, is the one marked at vertex A.

25° 40° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 25 deg and the angle near D (angle MDC, between DM and side DC) is 40 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 25 deg.
  • Angle MDC at D is 40 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 25 deg, so the base angle at D inside the triangle is also 25 deg.
∠ADM=∠AMD=25∘\angle ADM = \angle AMD = 25^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−25∘−25∘=130∘a = 180^\circ - 25^\circ - 25^\circ = 130^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 130 degrees

Review

With 25 deg base angles, the apex at A is the leftover 180 - 2(25) = 130 deg. Check: 25 + 25 + 130 = 180 deg. The extra 40 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 25 + 40 = 65 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 25 + 40 = 65 deg, so angle DAB = 115 deg; combined with the isosceles triangle this cross-checks angle DAM = 130 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 25 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 25 deg corners leave a 130 deg angle at A: just the triangle's leftover of 180 deg!
Variant 5 answer: 60 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 60∘60^\circ, and the angle marked near vertex D is 15∘15^\circ. The angle to find, ⓐ, is the one marked at vertex A.

60° 15° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 60 deg and the angle near D (angle MDC, between DM and side DC) is 15 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 60 deg.
  • Angle MDC at D is 15 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 60 deg, so the base angle at D inside the triangle is also 60 deg.
∠ADM=∠AMD=60∘\angle ADM = \angle AMD = 60^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−60∘−60∘=60∘a = 180^\circ - 60^\circ - 60^\circ = 60^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 60 degrees

Review

With 60 deg base angles, the apex at A is the leftover 180 - 2(60) = 60 deg. Check: 60 + 60 + 60 = 180 deg. The extra 15 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 60 + 15 = 75 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 60 + 15 = 75 deg, so angle DAB = 105 deg; combined with the isosceles triangle this cross-checks angle DAM = 60 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 60 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 60 deg corners leave a 60 deg angle at A: just the triangle's leftover of 180 deg!
Variant 6 answer: 50 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 65∘65^\circ, and the angle marked near vertex D is 20∘20^\circ. The angle to find, ⓐ, is the one marked at vertex A.

65° 20° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 65 deg and the angle near D (angle MDC, between DM and side DC) is 20 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 65 deg.
  • Angle MDC at D is 20 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 65 deg, so the base angle at D inside the triangle is also 65 deg.
∠ADM=∠AMD=65∘\angle ADM = \angle AMD = 65^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−65∘−65∘=50∘a = 180^\circ - 65^\circ - 65^\circ = 50^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 50 degrees

Review

With 65 deg base angles, the apex at A is the leftover 180 - 2(65) = 50 deg. Check: 65 + 65 + 50 = 180 deg. The extra 20 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 65 + 20 = 85 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 65 + 20 = 85 deg, so angle DAB = 95 deg; combined with the isosceles triangle this cross-checks angle DAM = 50 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 65 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 65 deg corners leave a 50 deg angle at A: just the triangle's leftover of 180 deg!
Variant 7 answer: 120 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 30∘30^\circ, and the angle marked near vertex D is 25∘25^\circ. The angle to find, ⓐ, is the one marked at vertex A.

30° 25° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 30 deg and the angle near D (angle MDC, between DM and side DC) is 25 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 30 deg.
  • Angle MDC at D is 25 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 30 deg, so the base angle at D inside the triangle is also 30 deg.
∠ADM=∠AMD=30∘\angle ADM = \angle AMD = 30^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−30∘−30∘=120∘a = 180^\circ - 30^\circ - 30^\circ = 120^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 120 degrees

Review

With 30 deg base angles, the apex at A is the leftover 180 - 2(30) = 120 deg. Check: 30 + 30 + 120 = 180 deg. The extra 25 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 30 + 25 = 55 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 30 + 25 = 55 deg, so angle DAB = 125 deg; combined with the isosceles triangle this cross-checks angle DAM = 120 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 30 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 30 deg corners leave a 120 deg angle at A: just the triangle's leftover of 180 deg!
Variant 8 answer: 140 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 20∘20^\circ, and the angle marked near vertex D is 35∘35^\circ. The angle to find, ⓐ, is the one marked at vertex A.

20° 35° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 20 deg and the angle near D (angle MDC, between DM and side DC) is 35 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 20 deg.
  • Angle MDC at D is 35 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 20 deg, so the base angle at D inside the triangle is also 20 deg.
∠ADM=∠AMD=20∘\angle ADM = \angle AMD = 20^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−20∘−20∘=140∘a = 180^\circ - 20^\circ - 20^\circ = 140^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 140 degrees

Review

With 20 deg base angles, the apex at A is the leftover 180 - 2(20) = 140 deg. Check: 20 + 20 + 140 = 180 deg. The extra 35 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 20 + 35 = 55 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 20 + 35 = 55 deg, so angle DAB = 125 deg; combined with the isosceles triangle this cross-checks angle DAM = 140 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 20 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 20 deg corners leave a 140 deg angle at A: just the triangle's leftover of 180 deg!
Variant 9 answer: 90 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 45∘45^\circ, and the angle marked near vertex D is 20∘20^\circ. The angle to find, ⓐ, is the one marked at vertex A.

45° 20° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 45 deg and the angle near D (angle MDC, between DM and side DC) is 20 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 45 deg.
  • Angle MDC at D is 20 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 45 deg, so the base angle at D inside the triangle is also 45 deg.
∠ADM=∠AMD=45∘\angle ADM = \angle AMD = 45^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−45∘−45∘=90∘a = 180^\circ - 45^\circ - 45^\circ = 90^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 90 degrees

Review

With 45 deg base angles, the apex at A is the leftover 180 - 2(45) = 90 deg. Check: 45 + 45 + 90 = 180 deg. The extra 20 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 45 + 20 = 65 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 45 + 20 = 65 deg, so angle DAB = 115 deg; combined with the isosceles triangle this cross-checks angle DAM = 90 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 45 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 45 deg corners leave a 90 deg angle at A: just the triangle's leftover of 180 deg!
Variant 10 answer: 80 degrees

Quadrilateral ABCD is a parallelogram. Segment AM and segment AD have the same length. Find the measure of angle ⓐ.

Figure description: Parallelogram ABCD lies tilted to one side (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below the bottom side BC, and segments are drawn from vertex A to point M and to vertex D so that segment AM equals segment AD in length. The angle marked at point M is 50∘50^\circ, and the angle marked near vertex D is 15∘15^\circ. The angle to find, ⓐ, is the one marked at vertex A.

50° 15° a A D C B M
Show solution

Understand

ABCD is a parallelogram (A top-left, D top-right, B bottom-left, C bottom-right). A point M lies below side BC, and segments AM and AD are drawn with AM = AD. The angle at M (angle AMD) is 50 deg and the angle near D (angle MDC, between DM and side DC) is 15 deg. I need angle a at vertex A, which is angle DAM.

Givens
  • ABCD is a parallelogram.
  • Segment AM equals segment AD (triangle AMD is isosceles with apex A).
  • Angle AMD at M is 50 deg.
  • Angle MDC at D is 15 deg.
  • M is below side BC.
Unknowns
  • The measure of angle a = angle DAM at vertex A.
Constraints
  • In an isosceles triangle the two base angles (opposite the equal sides) are equal.
  • The three angles of a triangle add to 180 deg.

Plan

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

Focus on triangle AMD. Since AM = AD, it is isosceles, so the base angles at M and at D inside this triangle are equal. The given angle at M forces the matching base angle at D, and then the apex angle a at A is whatever is left to make the triangle's angles total 180 deg.

Execute

#1 Draw a Diagram 4.G.A.2
Because AM = AD, triangle AMD is isosceles with apex A, so its two base angles, angle AMD (at M) and angle ADM (at D), are equal.
∠ADM=∠AMD\angle ADM = \angle AMD
Two equal sides always sit opposite two equal angles, so the corners at M and D in this triangle match.
#7 Identify Subproblems 4.MD.C.6
The base angle at M is given as 50 deg, so the base angle at D inside the triangle is also 50 deg.
∠ADM=∠AMD=50∘\angle ADM = \angle AMD = 50^\circ
Once one base angle of an isosceles triangle is known, the other base angle is the same number.
#7 Identify Subproblems 4.MD.C.7
The three angles of triangle AMD add to 180 deg, so the apex angle a at A is 180 deg minus the two base angles.
a=180∘−50∘−50∘=80∘a = 180^\circ - 50^\circ - 50^\circ = 80^\circ
After taking away the two equal base angles, the leftover of 180 deg is the top angle at A.
Answer: 80 degrees

Review

With 50 deg base angles, the apex at A is the leftover 180 - 2(50) = 80 deg. Check: 50 + 50 + 80 = 180 deg. The extra 15 deg at D (angle MDC) is consistent: angle ADC at the parallelogram corner is 50 + 15 = 65 deg, a valid parallelogram angle.

Use the parallelogram relations (tool 7): angle ADC = angle ADM + angle MDC = 50 + 15 = 65 deg, so angle DAB = 115 deg; combined with the isosceles triangle this cross-checks angle DAM = 80 deg.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing triangle AMD as isosceles from AM = AD.
  • 4.MD.C.6 Measure angles in whole-number degrees using a protractor — Transferring the 50 deg base angle to the equal base angle at D.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Using the 180 deg triangle total to find the apex angle a.
💡 Equal sides mean equal base angles, so two 50 deg corners leave a 80 deg angle at A: just the triangle's leftover of 180 deg!