← Folded angles are equal; chain to the unknown · Transformations Preserve Measures

Folded angles are equal; chain to the unknown · 14 practice problems

4.MD.C.7

From the workbook (authentic) — 1

Real practice problems extracted and localized from the source workbook.

Workbook 1 answer: 65 degrees

A sheet of paper shaped like an equilateral triangle is folded as shown, so that the bottom-left corner is folded up onto the triangle. Find the measure of the marked angle.

A B C Q P B' 70° ?
Show solution
1 · Understandwhat's really being asked

An equilateral triangle ABC has its bottom-left corner B folded up. The crease goes from P on the base BC to Q on the left side BA, and corner B lands at B'. The folded edge PB' makes a 70-degree angle with the base. I must find the flap angle at Q, which is angle PQB'.

Givens
  • Triangle ABC is equilateral, so every interior angle is 60 degrees (in particular angle B = 60 degrees)
  • The corner B is folded over crease PQ, so B lands at B' and the fold copies every length and angle exactly
  • The folded edge PB' makes a 70-degree angle with the base toward C (angle B'PC = 70 degrees)
Unknowns
  • The measure of the flap angle at Q, angle PQB'
Constraints
  • Angles in a triangle add to 180 degrees
  • Angles on a straight line add to 180 degrees
  • Folding preserves angle sizes, so a folded angle equals the original angle it came from
2 · Planchoose the strategy

#10 Create a Physical Representation · also uses: #7 Identify Subproblems#1 Draw a Diagram

Folding paper is a hands-on action, so picturing the fold shows that the crease reflects corner B onto B' and keeps angles equal. Then I break the figure into small pieces: the angles meeting at P on the base, and the small triangle BPQ, and chain them to the flap angle at Q.

3 · Execute4 carry out the plan

1Use the equilateral corner

#10 Create a Physical Representation 4.MD.C.7
Because ABC is an equilateral triangle, each corner measures 60 degrees, so the angle at B is 60 degrees.
∠PBQ=60∘\angle PBQ = 60^\circ
All three corners of an equilateral triangle are the same, and three equal corners totaling 180 degrees each measure 60 degrees.

2Find the angle the crease makes with the base at P

#7 Identify Subproblems 4.MD.C.7
At P the three angles on the straight base line BC add to 180 degrees. The folded edge PB' takes up 70 degrees (angle B'PC). Folding reflects the original base part PB onto PB', so the crease PQ splits the leftover angle into two equal halves: angle BPQ = angle B'PQ. The leftover is 180 - 70 = 110 degrees, so each half is 110 / 2 = 55 degrees.
∠BPQ=180∘−70∘2=110∘2=55∘\angle BPQ = \dfrac{180^\circ - 70^\circ}{2} = \dfrac{110^\circ}{2} = 55^\circ
The fold makes a mirror image, so the crease sits exactly in the middle of the angle between the original edge and the folded edge.

3Find the flap angle in triangle BPQ

#7 Identify Subproblems 4.MD.C.7
In triangle BPQ the three angles add to 180 degrees. With angle B = 60 degrees and angle BPQ = 55 degrees, the angle at Q is angle BQP = 180 - 60 - 55 = 65 degrees.
∠BQP=180∘−60∘−55∘=65∘\angle BQP = 180^\circ - 60^\circ - 55^\circ = 65^\circ
Once two angles of a triangle are known, the third is whatever is left over from 180 degrees.

4Carry the angle through the fold to B'

#10 Create a Physical Representation 4.MD.C.7
Folding does not change angle sizes, so the flap angle at Q after folding equals the angle before folding: angle PQB' = angle PQB = 65 degrees.
∠PQB′=∠BQP=65∘\angle PQB' = \angle BQP = 65^\circ
A folded shape lands exactly on top of its original, so the matching angle keeps the same size.
Answer: 65 degrees
4 · Reviewdoes it hold up?

The flap angle 65 degrees is acute, which matches the narrow corner shown at Q. Checking triangle BPQ: 60 + 55 + 65 = 180 degrees, and the folded corner at B' keeps its 60 degrees (180 - 55 - 65 = 60), exactly the equilateral corner that was folded. Everything is consistent.

Another way: Draw the figure to scale (tool 1) and measure the flap angle at Q with a protractor to confirm the 65-degree answer found by angle chasing.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Splitting the straight-line angle at P, using the equilateral 60-degree corner, and chaining through the triangle-sum and the equal folded angle to reach the flap angle at Q.
💡Takeaway. Split the straight line in half at the fold, then use a triangle's three angles adding to 180 degrees -- only Grade 4 angle work needed!

Generated variants — 13

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

Variant 1 easy answer: 75 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 55° 50°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 55-degree angle at B and a 50-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 55 degrees and angle DCB = 50 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 55 - 50 = 75 degrees.
∠BDC=180∘−55∘−50∘=75∘\angle BDC = 180^\circ - 55^\circ - 50^\circ = 75^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 55 degrees. Then angle DEB = 180 - 55 - 55 = 70 degrees, and since B, E, C are in a line, angle DEC = 180 - 70 = 110 degrees.
∠DEB=180∘−55∘−55∘=70∘,∠DEC=180∘−70∘=110∘\angle DEB = 180^\circ - 55^\circ - 55^\circ = 70^\circ,\quad \angle DEC = 180^\circ - 70^\circ = 110^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 110 / 2 = 55 degrees. The angle at C is untouched by the fold, so it is still 50 degrees.
∠FEC=110∘2=55∘,∠FCE=50∘\angle FEC = \dfrac{110^\circ}{2} = 55^\circ,\quad \angle FCE = 50^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 50 - 55 = 75 degrees.
∠EFC=180∘−50∘−55∘=75∘\angle EFC = 180^\circ - 50^\circ - 55^\circ = 75^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 75 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 75 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 75 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 2 easy answer: 70 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 50° 60°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 50-degree angle at B and a 60-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 50 degrees and angle DCB = 60 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 50 - 60 = 70 degrees.
∠BDC=180∘−50∘−60∘=70∘\angle BDC = 180^\circ - 50^\circ - 60^\circ = 70^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 50 degrees. Then angle DEB = 180 - 50 - 50 = 80 degrees, and since B, E, C are in a line, angle DEC = 180 - 80 = 100 degrees.
∠DEB=180∘−50∘−50∘=80∘,∠DEC=180∘−80∘=100∘\angle DEB = 180^\circ - 50^\circ - 50^\circ = 80^\circ,\quad \angle DEC = 180^\circ - 80^\circ = 100^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 100 / 2 = 50 degrees. The angle at C is untouched by the fold, so it is still 60 degrees.
∠FEC=100∘2=50∘,∠FCE=60∘\angle FEC = \dfrac{100^\circ}{2} = 50^\circ,\quad \angle FCE = 60^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 60 - 50 = 70 degrees.
∠EFC=180∘−60∘−50∘=70∘\angle EFC = 180^\circ - 60^\circ - 50^\circ = 70^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 70 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 70 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 70 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 3 easy answer: 72 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 42° 66°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 42-degree angle at B and a 66-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 42 degrees and angle DCB = 66 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 42 - 66 = 72 degrees.
∠BDC=180∘−42∘−66∘=72∘\angle BDC = 180^\circ - 42^\circ - 66^\circ = 72^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 42 degrees. Then angle DEB = 180 - 42 - 42 = 96 degrees, and since B, E, C are in a line, angle DEC = 180 - 96 = 84 degrees.
∠DEB=180∘−42∘−42∘=96∘,∠DEC=180∘−96∘=84∘\angle DEB = 180^\circ - 42^\circ - 42^\circ = 96^\circ,\quad \angle DEC = 180^\circ - 96^\circ = 84^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 84 / 2 = 42 degrees. The angle at C is untouched by the fold, so it is still 66 degrees.
∠FEC=84∘2=42∘,∠FCE=66∘\angle FEC = \dfrac{84^\circ}{2} = 42^\circ,\quad \angle FCE = 66^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 66 - 42 = 72 degrees.
∠EFC=180∘−66∘−42∘=72∘\angle EFC = 180^\circ - 66^\circ - 42^\circ = 72^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 72 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 72 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 72 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 4 easy answer: 65 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 45° 70°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 45-degree angle at B and a 70-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 45 degrees and angle DCB = 70 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 45 - 70 = 65 degrees.
∠BDC=180∘−45∘−70∘=65∘\angle BDC = 180^\circ - 45^\circ - 70^\circ = 65^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 45 degrees. Then angle DEB = 180 - 45 - 45 = 90 degrees, and since B, E, C are in a line, angle DEC = 180 - 90 = 90 degrees.
∠DEB=180∘−45∘−45∘=90∘,∠DEC=180∘−90∘=90∘\angle DEB = 180^\circ - 45^\circ - 45^\circ = 90^\circ,\quad \angle DEC = 180^\circ - 90^\circ = 90^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 90 / 2 = 45 degrees. The angle at C is untouched by the fold, so it is still 70 degrees.
∠FEC=90∘2=45∘,∠FCE=70∘\angle FEC = \dfrac{90^\circ}{2} = 45^\circ,\quad \angle FCE = 70^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 70 - 45 = 65 degrees.
∠EFC=180∘−70∘−45∘=65∘\angle EFC = 180^\circ - 70^\circ - 45^\circ = 65^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 65 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 65 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 65 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 5 medium answer: 58 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 48° 74°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 48-degree angle at B and a 74-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 48 degrees and angle DCB = 74 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 48 - 74 = 58 degrees.
∠BDC=180∘−48∘−74∘=58∘\angle BDC = 180^\circ - 48^\circ - 74^\circ = 58^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 48 degrees. Then angle DEB = 180 - 48 - 48 = 84 degrees, and since B, E, C are in a line, angle DEC = 180 - 84 = 96 degrees.
∠DEB=180∘−48∘−48∘=84∘,∠DEC=180∘−84∘=96∘\angle DEB = 180^\circ - 48^\circ - 48^\circ = 84^\circ,\quad \angle DEC = 180^\circ - 84^\circ = 96^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 96 / 2 = 48 degrees. The angle at C is untouched by the fold, so it is still 74 degrees.
∠FEC=96∘2=48∘,∠FCE=74∘\angle FEC = \dfrac{96^\circ}{2} = 48^\circ,\quad \angle FCE = 74^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 74 - 48 = 58 degrees.
∠EFC=180∘−74∘−48∘=58∘\angle EFC = 180^\circ - 74^\circ - 48^\circ = 58^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 58 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 58 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 58 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 6 medium answer: 70 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 35° 75°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 35-degree angle at B and a 75-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 35 degrees and angle DCB = 75 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 35 - 75 = 70 degrees.
∠BDC=180∘−35∘−75∘=70∘\angle BDC = 180^\circ - 35^\circ - 75^\circ = 70^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 35 degrees. Then angle DEB = 180 - 35 - 35 = 110 degrees, and since B, E, C are in a line, angle DEC = 180 - 110 = 70 degrees.
∠DEB=180∘−35∘−35∘=110∘,∠DEC=180∘−110∘=70∘\angle DEB = 180^\circ - 35^\circ - 35^\circ = 110^\circ,\quad \angle DEC = 180^\circ - 110^\circ = 70^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 70 / 2 = 35 degrees. The angle at C is untouched by the fold, so it is still 75 degrees.
∠FEC=70∘2=35∘,∠FCE=75∘\angle FEC = \dfrac{70^\circ}{2} = 35^\circ,\quad \angle FCE = 75^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 75 - 35 = 70 degrees.
∠EFC=180∘−75∘−35∘=70∘\angle EFC = 180^\circ - 75^\circ - 35^\circ = 70^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 70 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 70 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 70 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 7 medium answer: 60 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 40° 80°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 40-degree angle at B and a 80-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 40 degrees and angle DCB = 80 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 40 - 80 = 60 degrees.
∠BDC=180∘−40∘−80∘=60∘\angle BDC = 180^\circ - 40^\circ - 80^\circ = 60^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 40 degrees. Then angle DEB = 180 - 40 - 40 = 100 degrees, and since B, E, C are in a line, angle DEC = 180 - 100 = 80 degrees.
∠DEB=180∘−40∘−40∘=100∘,∠DEC=180∘−100∘=80∘\angle DEB = 180^\circ - 40^\circ - 40^\circ = 100^\circ,\quad \angle DEC = 180^\circ - 100^\circ = 80^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 80 / 2 = 40 degrees. The angle at C is untouched by the fold, so it is still 80 degrees.
∠FEC=80∘2=40∘,∠FCE=80∘\angle FEC = \dfrac{80^\circ}{2} = 40^\circ,\quad \angle FCE = 80^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 80 - 40 = 60 degrees.
∠EFC=180∘−80∘−40∘=60∘\angle EFC = 180^\circ - 80^\circ - 40^\circ = 60^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 60 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 60 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 60 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 8 medium answer: 58 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 38° 84°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 38-degree angle at B and a 84-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 38 degrees and angle DCB = 84 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 38 - 84 = 58 degrees.
∠BDC=180∘−38∘−84∘=58∘\angle BDC = 180^\circ - 38^\circ - 84^\circ = 58^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 38 degrees. Then angle DEB = 180 - 38 - 38 = 104 degrees, and since B, E, C are in a line, angle DEC = 180 - 104 = 76 degrees.
∠DEB=180∘−38∘−38∘=104∘,∠DEC=180∘−104∘=76∘\angle DEB = 180^\circ - 38^\circ - 38^\circ = 104^\circ,\quad \angle DEC = 180^\circ - 104^\circ = 76^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 76 / 2 = 38 degrees. The angle at C is untouched by the fold, so it is still 84 degrees.
∠FEC=76∘2=38∘,∠FCE=84∘\angle FEC = \dfrac{76^\circ}{2} = 38^\circ,\quad \angle FCE = 84^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 84 - 38 = 58 degrees.
∠EFC=180∘−84∘−38∘=58∘\angle EFC = 180^\circ - 84^\circ - 38^\circ = 58^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 58 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 58 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 58 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 9 medium answer: 60 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 33° 87°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 33-degree angle at B and a 87-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 33 degrees and angle DCB = 87 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 33 - 87 = 60 degrees.
∠BDC=180∘−33∘−87∘=60∘\angle BDC = 180^\circ - 33^\circ - 87^\circ = 60^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 33 degrees. Then angle DEB = 180 - 33 - 33 = 114 degrees, and since B, E, C are in a line, angle DEC = 180 - 114 = 66 degrees.
∠DEB=180∘−33∘−33∘=114∘,∠DEC=180∘−114∘=66∘\angle DEB = 180^\circ - 33^\circ - 33^\circ = 114^\circ,\quad \angle DEC = 180^\circ - 114^\circ = 66^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 66 / 2 = 33 degrees. The angle at C is untouched by the fold, so it is still 87 degrees.
∠FEC=66∘2=33∘,∠FCE=87∘\angle FEC = \dfrac{66^\circ}{2} = 33^\circ,\quad \angle FCE = 87^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 87 - 33 = 60 degrees.
∠EFC=180∘−87∘−33∘=60∘\angle EFC = 180^\circ - 87^\circ - 33^\circ = 60^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 60 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 60 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 60 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 10 hard answer: 60 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 30° 90°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 30-degree angle at B and a 90-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 30 degrees and angle DCB = 90 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 30 - 90 = 60 degrees.
∠BDC=180∘−30∘−90∘=60∘\angle BDC = 180^\circ - 30^\circ - 90^\circ = 60^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 30 degrees. Then angle DEB = 180 - 30 - 30 = 120 degrees, and since B, E, C are in a line, angle DEC = 180 - 120 = 60 degrees.
∠DEB=180∘−30∘−30∘=120∘,∠DEC=180∘−120∘=60∘\angle DEB = 180^\circ - 30^\circ - 30^\circ = 120^\circ,\quad \angle DEC = 180^\circ - 120^\circ = 60^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 60 / 2 = 30 degrees. The angle at C is untouched by the fold, so it is still 90 degrees.
∠FEC=60∘2=30∘,∠FCE=90∘\angle FEC = \dfrac{60^\circ}{2} = 30^\circ,\quad \angle FCE = 90^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 90 - 30 = 60 degrees.
∠EFC=180∘−90∘−30∘=60∘\angle EFC = 180^\circ - 90^\circ - 30^\circ = 60^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 60 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 60 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 60 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 11 hard answer: 52 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 36° 92°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 36-degree angle at B and a 92-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 36 degrees and angle DCB = 92 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 36 - 92 = 52 degrees.
∠BDC=180∘−36∘−92∘=52∘\angle BDC = 180^\circ - 36^\circ - 92^\circ = 52^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 36 degrees. Then angle DEB = 180 - 36 - 36 = 108 degrees, and since B, E, C are in a line, angle DEC = 180 - 108 = 72 degrees.
∠DEB=180∘−36∘−36∘=108∘,∠DEC=180∘−108∘=72∘\angle DEB = 180^\circ - 36^\circ - 36^\circ = 108^\circ,\quad \angle DEC = 180^\circ - 108^\circ = 72^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 72 / 2 = 36 degrees. The angle at C is untouched by the fold, so it is still 92 degrees.
∠FEC=72∘2=36∘,∠FCE=92∘\angle FEC = \dfrac{72^\circ}{2} = 36^\circ,\quad \angle FCE = 92^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 92 - 36 = 52 degrees.
∠EFC=180∘−92∘−36∘=52∘\angle EFC = 180^\circ - 92^\circ - 36^\circ = 52^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 52 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 52 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 52 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 12 hard answer: 60 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 25° 95°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 25-degree angle at B and a 95-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 25 degrees and angle DCB = 95 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 25 - 95 = 60 degrees.
∠BDC=180∘−25∘−95∘=60∘\angle BDC = 180^\circ - 25^\circ - 95^\circ = 60^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 25 degrees. Then angle DEB = 180 - 25 - 25 = 130 degrees, and since B, E, C are in a line, angle DEC = 180 - 130 = 50 degrees.
∠DEB=180∘−25∘−25∘=130∘,∠DEC=180∘−130∘=50∘\angle DEB = 180^\circ - 25^\circ - 25^\circ = 130^\circ,\quad \angle DEC = 180^\circ - 130^\circ = 50^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 50 / 2 = 25 degrees. The angle at C is untouched by the fold, so it is still 95 degrees.
∠FEC=50∘2=25∘,∠FCE=95∘\angle FEC = \dfrac{50^\circ}{2} = 25^\circ,\quad \angle FCE = 95^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 95 - 25 = 60 degrees.
∠EFC=180∘−95∘−25∘=60∘\angle EFC = 180^\circ - 95^\circ - 25^\circ = 60^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 60 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 60 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 60 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.
Variant 13 hard answer: 52 degrees

A triangular sheet of paper is folded as shown so that side BEBE and side DEDE have the same length. Find the measure of angle EFCEFC.

B E C D F 28° 100°
Show solution
1 · Understandwhat's really being asked

Triangle BDC has a 28-degree angle at B and a 100-degree angle at C. E sits on the base with BE = DE, and folding the apex down puts the crease through F on side DC. I want angle EFC.

Givens
  • Angle DBC = 28 degrees and angle DCB = 100 degrees.
  • BE = DE, so triangle BED is isosceles.
  • B, E and C lie on one straight line.
Unknowns
  • The measure of angle EFC.
Constraints
  • A fold never changes the size of an angle.
  • Angles in a triangle sum to 180 degrees.
  • Angles on a straight line sum to 180 degrees.
2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #1 Draw a Diagram#17 Visualize Spatial Relationships

Work through the small triangles one at a time. The equal sides fix the angles at E, the straight base carries them to the other side of E, and the fold moves an angle without resizing it.

3 · Execute4 carry out the plan

1Find the apex angle of the whole triangle

#7 Identify Subproblems 4.MD.C.7
In triangle BDC the angles add to 180 degrees, so the apex angle at D is 180 - 28 - 100 = 52 degrees.
∠BDC=180∘−28∘−100∘=52∘\angle BDC = 180^\circ - 28^\circ - 100^\circ = 52^\circ
Worth having in hand: the fold is about to move this angle.

2Use BE = DE to find angle DEC

#7 Identify Subproblems 4.MD.C.7
Since BE = DE, triangle BED is isosceles, so its base angles are equal: angle BDE = angle DBE = 28 degrees. Then angle DEB = 180 - 28 - 28 = 124 degrees, and since B, E, C are in a line, angle DEC = 180 - 124 = 56 degrees.
∠DEB=180∘−28∘−28∘=124∘,∠DEC=180∘−124∘=56∘\angle DEB = 180^\circ - 28^\circ - 28^\circ = 124^\circ,\quad \angle DEC = 180^\circ - 124^\circ = 56^\circ
Equal sides give equal base angles, and the straight base does the rest.

3Use the fold to bring the angle down to F

#17 Visualize Spatial Relationships 4.MD.C.7
Folding along EF lays side ED down onto the base, so the crease splits angle DEC into two equal halves: angle FEC = 56 / 2 = 28 degrees. The angle at C is untouched by the fold, so it is still 100 degrees.
∠FEC=56∘2=28∘,∠FCE=100∘\angle FEC = \dfrac{56^\circ}{2} = 28^\circ,\quad \angle FCE = 100^\circ
The crease is the mirror line, so it sits exactly halfway.

4Find angle EFC in triangle EFC

#7 Identify Subproblems 4.MD.C.7
In triangle EFC the three angles add to 180 degrees: angle EFC = 180 - 100 - 28 = 52 degrees.
∠EFC=180∘−100∘−28∘=52∘\angle EFC = 180^\circ - 100^\circ - 28^\circ = 52^\circ
The last unknown angle of a triangle is whatever 180 has left over.
Answer: 52 degrees
4 · Reviewdoes it hold up?

Angle EFC came out 52 degrees, the same as the apex angle BDC from the first step -- which is the point of the problem: the fold carried that angle down to F without changing it.

Another way: Draw the triangle to scale, fold it along EF, and measure angle EFC with a protractor. It reads 52 degrees, because folding paper is what the angle chasing describes.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding and subtracting angle measures through the isosceles triangle, the straight base and the folded triangle EFC.
💡Takeaway. Folding paper moves an angle but never resizes it, so an angle you found at the top can be the answer at the bottom.