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Spot equal sides to build isosceles triangles · 4 practice problems

4.MD.C.74.G.A.2

From the workbook (authentic) — 4

Real practice problems extracted and localized from the source workbook.

Workbook 1 answer: 105 degrees

Quadrilateral ABCDABCD is a square, and triangle CEDCED is an equilateral triangle. Find the measure of ∠a\angle a.

A D B C E F a
Show solution
1 · Understandwhat's really being asked

ABCD is a square and CED is an equilateral triangle built on the outside of side DC, with E pointing right. Segment AE (from the top-left corner A) crosses side DC at F. I must find angle a, the obtuse angle at F between FD (up) and FE.

Givens
  • ABCD is a square, so all sides are equal and every corner angle is 90 degrees
  • Triangle CED is equilateral on side DC, so DC = CE = DE and each of its angles is 60 degrees
  • Side AD of the square equals side DE of the triangle (both equal the square's side)
  • Segment AE meets side DC at point F; angle a is the obtuse angle at F
Unknowns
  • The measure of angle a at point F
Constraints
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees
  • Square corner angles are 90 degrees and equilateral angles are 60 degrees
2 · Planchoose the strategy

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

Spot that side AD (square) equals side DE (equilateral triangle), so triangle ADE is isosceles. Find its base angles, then use triangle DFE to reach the angle at F.

3 · Execute3 carry out the plan

1Find angle ADE, the full angle at D

#7 Identify Subproblems 4.MD.C.7
At corner D the square contributes angle ADC = 90 degrees and the equilateral triangle contributes angle CDE = 60 degrees on the outside. Together angle ADE = 90 + 60 = 150 degrees.
∠ADE=90∘+60∘=150∘\angle ADE = 90^\circ + 60^\circ = 150^\circ
Angles that meet at one point add up, so the square angle plus the triangle angle gives the whole angle at D.

2Use isosceles triangle ADE to find its base angles

#7 Identify Subproblems 4.MD.C.7
Side AD of the square equals side DE of the equilateral triangle, so triangle ADE is isosceles with AD = DE. Its two base angles are equal: angle DAE = angle DEA = (180 - 150) / 2 = 15 degrees.
∠DAE=∠DEA=180∘−150∘2=15∘\angle DAE = \angle DEA = \dfrac{180^\circ - 150^\circ}{2} = 15^\circ
Equal sides give equal base angles, and the three angles still total 180 degrees.

3Find angle a at F using triangle DFE

#7 Identify Subproblems 4.MD.C.7
F lies on side DC, so in triangle DFE the angle at D (angle FDE) is the equilateral angle CDE = 60 degrees, and angle DEF = angle DEA = 15 degrees because F is on segment AE. So angle DFE = 180 - 60 - 15 = 105 degrees. Angle a is exactly this obtuse angle at F.
∠a=∠DFE=180∘−60∘−15∘=105∘\angle a = \angle DFE = 180^\circ - 60^\circ - 15^\circ = 105^\circ
Once two angles of triangle DFE are known the third follows from the 180-degree total.
Answer: 105 degrees
4 · Reviewdoes it hold up?

Angle a = 105 degrees is obtuse, which matches the wide opening that the crossing segment AE makes with the upper part of side DC; its straight-line partner angle CFE = 75 degrees is acute, and 105 + 75 = 180 degrees along side DC, so the result is consistent.

Another way: Place the square on a coordinate grid (tool 1), compute the positions of E and F, and measure angle a directly to confirm the 105-degree result.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems to find unknown angles — Adding the square and triangle angles at D, finding isosceles base angles, and chaining through triangle DFE to angle a.
  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines, or angle size — Using square and equilateral-triangle side and angle properties to spot the isosceles triangle ADE.
💡Takeaway. This only needs Grade 4 angle-adding plus spotting that a square side and a triangle side are equal!
Workbook 2 answer: angle x = 45 degrees, angle y = 60 degrees

Quadrilateral ABCDABCD is a square, and triangle CEDCED is an equilateral triangle. Find the measures of ∠x\angle x and ∠y\angle y.

A D B C E P x y
Show solution
1 · Understandwhat's really being asked

ABCD is a square and CED is an equilateral triangle on the outside of side DC, with E pointing right. The square's diagonal AC and the segment BE cross at P. I must find angle x (= angle DEB at E) and angle y (= angle APB at P, opening toward the left).

Givens
  • ABCD is a square, so all sides are equal and every corner angle is 90 degrees
  • Triangle CED is equilateral, so DC = CE = DE and each of its angles is 60 degrees
  • Side BC of the square equals side CE of the triangle
  • AC is the diagonal of the square; it makes a 45-degree angle with each side it meets
  • Segment BE crosses the diagonal AC at P
Unknowns
  • angle x = angle DEB at E
  • angle y = angle APB at P
Constraints
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees
  • A diagonal of a square splits its 90-degree corner into two 45-degree angles
  • Angles on a straight line sum to 180 degrees
2 · Planchoose the strategy

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

First find the equilateral apex angle CED = 60 and the isosceles base angle CEB = 15, subtract to get x. For y, work in triangle BPC: the diagonal makes 45 degrees at C and BE makes 15 degrees at B, so the third angle gives y as a straight-line partner.

3 · Execute3 carry out the plan

1Find the isosceles base angle CEB

#7 Identify Subproblems 4.MD.C.7
At corner C the square angle BCD = 90 degrees and the equilateral angle DCE = 60 degrees add to angle BCE = 150 degrees. Side BC equals side CE, so triangle BCE is isosceles and its base angles are equal: angle CBE = angle CEB = (180 - 150) / 2 = 15 degrees.
∠BCE=90∘+60∘=150∘,∠CEB=180∘−150∘2=15∘\angle BCE = 90^\circ + 60^\circ = 150^\circ,\quad \angle CEB = \dfrac{180^\circ - 150^\circ}{2} = 15^\circ
The square corner and the triangle corner stack at C, and equal sides force equal base angles.

2Find angle x = angle DEB

#7 Identify Subproblems 4.MD.C.7
The whole apex angle of the equilateral triangle is angle DEC = 60 degrees. The segment EB sits inside it, with angle CEB = 15 degrees next to EC. So angle x = angle DEB = angle DEC - angle CEB = 60 - 15 = 45 degrees.
∠x=∠DEC−∠CEB=60∘−15∘=45∘\angle x = \angle DEC - \angle CEB = 60^\circ - 15^\circ = 45^\circ
Take the full 60-degree triangle tip and remove the 15-degree slice to leave the part toward D.

3Find angle y = angle APB at the crossing

#7 Identify Subproblems 4.MD.C.7
Look at triangle BPC, where P is on diagonal AC. At C the diagonal splits the square's 90-degree corner so angle BCP = angle BCA = 45 degrees. At B the segment makes angle CBP = angle CBE = 15 degrees. So angle BPC = 180 - 45 - 15 = 120 degrees. Since A, P, C are in a straight line, angle y = angle APB = 180 - 120 = 60 degrees.
∠BPC=180∘−45∘−15∘=120∘,∠y=180∘−120∘=60∘\angle BPC = 180^\circ - 45^\circ - 15^\circ = 120^\circ,\quad \angle y = 180^\circ - 120^\circ = 60^\circ
Solve the little triangle BPC, then flip across the straight diagonal to land on angle y.
Answer: angle x = 45 degrees, angle y = 60 degrees
4 · Reviewdoes it hold up?

angle x = 45 degrees is less than the 60-degree triangle tip it sits inside, which is consistent. angle y = 60 degrees is the straight-line partner of the 120-degree angle BPC, and 60 + 120 = 180 degrees along the diagonal, so both checks hold.

Another way: Set the square on a coordinate grid (tool 1), compute E, P, and the rays, and measure both angles directly to confirm 45 and 60 degrees.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems to find unknown angles — Stacking the square and triangle angles at C, splitting the 60-degree apex to get x, and chaining triangle BPC plus the straight line to get y.
  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines, or angle size — Using square, diagonal, and equilateral-triangle properties to set up the isosceles triangle BCE and the 45-degree diagonal split.
💡Takeaway. Spot the equal sides to get the 15-degree base angle, then just add and subtract angles - all Grade 4!
Workbook 3 answer: 150 degrees

Quadrilateral ABCDABCD is a square, and triangle ABEABE is an equilateral triangle. Find the measure of ∠DEC\angle DEC.

A B C D E
Show solution
1 · Understandwhat's really being asked

ABCD is a square and ABE is an equilateral triangle built on the left side AB with vertex E pointing inward. Segments ED and EC join E to the two right-hand corners D and C. I must find angle DEC at E.

Givens
  • ABCD is a square, so all sides are equal and every corner angle is 90 degrees
  • Triangle ABE is equilateral on side AB, so AB = AE = BE and each of its angles is 60 degrees
  • Side AD of the square equals side AE of the triangle (both equal the square's side); likewise BC = BE
  • Angle DEC is the angle at E between ED and EC
Unknowns
  • The measure of angle DEC at E
Constraints
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees
  • The four angles around point E add up to 360 degrees
2 · Planchoose the strategy

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

Triangle ADE is isosceles (AD = AE), so I can find angle AED; by symmetry angle BEC equals it. Then subtract both of those and the 60-degree triangle apex from the full 360 degrees around E to get angle DEC.

3 · Execute3 carry out the plan

1Find the apex angle of triangle ADE at A

#7 Identify Subproblems 4.MD.C.7
At corner A the square angle DAB = 90 degrees. The equilateral triangle takes angle EAB = 60 degrees of it, so the leftover angle DAE = 90 - 60 = 30 degrees.
∠DAE=90∘−60∘=30∘\angle DAE = 90^\circ - 60^\circ = 30^\circ
The triangle's 60-degree corner eats part of the square's 90-degree corner, leaving 30 degrees for angle DAE.

2Use isosceles triangle ADE to find angle AED

#7 Identify Subproblems 4.MD.C.7
Side AD of the square equals side AE of the triangle, so triangle ADE is isosceles with AD = AE. Its two base angles are equal: angle ADE = angle AED = (180 - 30) / 2 = 75 degrees. By the left-right symmetry of the figure, triangle BCE gives angle BEC = 75 degrees as well.
∠AED=180∘−30∘2=75∘=∠BEC\angle AED = \dfrac{180^\circ - 30^\circ}{2} = 75^\circ = \angle BEC
Equal sides give equal base angles, and the same setup on the bottom gives the matching 75 degrees.

3Find angle DEC using the full turn around E

#7 Identify Subproblems 4.MD.C.7
The four angles around point E are angle AEB (the equilateral apex = 60 degrees), angle AED = 75 degrees, angle BEC = 75 degrees, and angle DEC. They make a full turn: 60 + 75 + 75 + angle DEC = 360, so angle DEC = 360 - 210 = 150 degrees.
∠DEC=360∘−(60∘+75∘+75∘)=150∘\angle DEC = 360^\circ - (60^\circ + 75^\circ + 75^\circ) = 150^\circ
Everything around a point adds to a full 360-degree turn, so the leftover slice is angle DEC.
Answer: 150 degrees
4 · Reviewdoes it hold up?

angle DEC = 150 degrees is obtuse, which fits the wide opening that ED and EC make as they fan out to the two far corners; the four angles 60 + 75 + 75 + 150 = 360 degrees complete one full turn around E, so the result is consistent.

Another way: Place the square on a coordinate grid (tool 1), put E at the equilateral apex, and measure angle DEC directly to confirm 150 degrees.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems to find unknown angles — Subtracting to get angle DAE, finding isosceles base angles, and using the 360-degree total around E to get angle DEC.
  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines, or angle size — Using square and equilateral-triangle side and angle properties to spot the isosceles triangles ADE and BCE.
💡Takeaway. Find two equal sides to get the 75-degree base angles, then fill the 360-degree turn around E - all Grade 4!
Workbook 4 answer: 85 degrees

In the figure, square ABCDABCD and triangle CEFCEF all have sides of equal length (so triangle CEFCEF is equilateral with each side equal to a side of the square). Find the measure of ∠g\angle g.

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

ABCD is a square and CEF is an equilateral triangle sharing corner C, with side CF making a 40-degree angle with the square's side CD. Segment CE and segment BF cross at G. I must find angle g = angle EGF at G.

Givens
  • ABCD is a square, so all sides are equal and every corner angle is 90 degrees
  • Triangle CEF is equilateral with each side equal to a side of the square, so CE = CF = BC and each triangle angle is 60 degrees
  • Side CF makes a 40-degree angle with side CD: angle FCD = 40 degrees
  • Segment CE meets segment BF at G; angle g is angle EGF
Unknowns
  • The measure of angle g = angle EGF at G
Constraints
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees
  • Vertical (opposite) angles at a crossing are equal
  • Square corner angles are 90 degrees and equilateral angles are 60 degrees
2 · Planchoose the strategy

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

Find angle ECB at C and angle FBC at B (using the equal sides BC = CF to make triangle BCF isosceles). Those are the angles of triangle BCG at C and B, so its third angle BGC follows, and angle g is its vertical angle.

3 · Execute3 carry out the plan

1Find angle ECB at corner C

#7 Identify Subproblems 4.MD.C.7
The equilateral apex angle ECF = 60 degrees, and side CF makes angle FCD = 40 degrees with CD, so the part inside the corner is angle DCE = 60 - 40 = 20 degrees. The square corner is angle DCB = 90 degrees, and E lies inside it, so angle ECB = 90 - 20 = 70 degrees.
∠DCE=60∘−40∘=20∘,∠ECB=90∘−20∘=70∘\angle DCE = 60^\circ - 40^\circ = 20^\circ,\quad \angle ECB = 90^\circ - 20^\circ = 70^\circ
Peel the 40-degree slice off the 60-degree triangle tip to find how far CE sits from CD, then take it away from the 90-degree corner.

2Use isosceles triangle BCF to find angle FBC

#7 Identify Subproblems 4.MD.C.7
Side BC of the square equals side CF of the equilateral triangle, so triangle BCF is isosceles with BC = CF. Its top angle is angle BCF = angle BCD + angle DCF = 90 + 40 = 130 degrees. The two equal base angles are angle FBC = angle BFC = (180 - 130) / 2 = 25 degrees.
∠BCF=90∘+40∘=130∘,∠FBC=180∘−130∘2=25∘\angle BCF = 90^\circ + 40^\circ = 130^\circ,\quad \angle FBC = \dfrac{180^\circ - 130^\circ}{2} = 25^\circ
Spotting that a square side equals a triangle side makes triangle BCF isosceles, so its base angles are equal.

3Find angle g from triangle BCG

#7 Identify Subproblems 4.MD.C.7
G is the crossing of CE and BF, so in triangle BCG the angle at C is angle GCB = angle ECB = 70 degrees and the angle at B is angle GBC = angle FBC = 25 degrees. The third angle is angle BGC = 180 - 70 - 25 = 85 degrees. Angle g = angle EGF is the vertical (opposite) angle to angle BGC, so angle g = 85 degrees.
∠BGC=180∘−70∘−25∘=85∘,∠g=∠EGF=85∘\angle BGC = 180^\circ - 70^\circ - 25^\circ = 85^\circ,\quad \angle g = \angle EGF = 85^\circ
Solve the little triangle BCG, then use that opposite angles at a crossing are equal to land on angle g.
Answer: 85 degrees
4 · Reviewdoes it hold up?

angle g = 85 degrees is just under a right angle, which fits the nearly-square crossing seen where CE and BF meet. Its triangle partner angle BGC = 85 degrees is the vertical angle, and the triangle BCG angles 70 + 25 + 85 = 180 degrees, so the work is consistent.

Another way: Place the square on a coordinate grid (tool 1), build E and F from the 40-degree direction, find G as the intersection of CE and BF, and measure angle EGF directly to confirm 85 degrees.

Standardsmin grade 4
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems to find unknown angles — Combining and subtracting angles at C, finding the isosceles base angles of triangle BCF, and chaining through triangle BCG to angle g.
  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines, or angle size — Using square and equilateral-triangle side and angle properties to spot the isosceles triangle BCF.
💡Takeaway. Once you spot the equal sides (square side = triangle side), the rest is just adding and subtracting angles - all Grade 4!