Printable math worksheet for finding angles in quadrilaterals.

Let’s solve each problem step by step. Remember: the sum of all interior angles in any quadrilateral is always 360°. We’ll use that fact, plus properties like parallel sides or equal sides (which mean certain angles are equal), to find the missing angles.

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Problem 1:

Quadrilateral ABCD has AB = AD and CB = CD → it’s a kite.
In a kite, one pair of opposite angles are equal — specifically, the angles between the unequal sides. But here, we’re told angle B = 104°, and we need angle D.

Actually, in this diagram, since AB = AD and CB = CD, the kite is symmetric across diagonal AC. That means angles at B and D are NOT necessarily equal — wait, let me think again.

Wait — actually, in a kite with two pairs of adjacent equal sides, the angles between the unequal sides are equal. Here, sides AB=AD and CB=CD, so the “top” and “bottom” vertices are A and C? Actually, looking at the marks: AB and AD have one tick, CB and CD have two ticks → so vertex A connects two equal sides, vertex C connects two equal sides → so angles at B and D should be equal? No — actually, in a kite, the angles between the *different* pairs of equal sides are equal. Let me recall:

Standard kite property: One diagonal is axis of symmetry. The angles on either side of that diagonal are equal. In this case, if AB=AD and CB=CD, then diagonal AC is the axis of symmetry → so angle B = angle D? But angle B is given as 104°, so angle D would also be 104°? But that can’t be right because then total would exceed 360° unless other angles are small.

Wait — no! Actually, in a kite, only one pair of opposite angles are equal — and they are the ones that are NOT between the equal sides. Let me draw mentally: points A, B, C, D. AB=AD, CB=CD → so triangle ABD and CBD? Hmm.

Better approach: Use the fact that sum of angles = 360°. But we don’t know enough yet. Wait — maybe I misread. Looking back: in problem 1, it's labeled with ticks: AB and AD have single tick → equal; BC and DC have double tick → equal. So yes, it’s a kite with symmetry over diagonal AC. Therefore, angles at B and D are equal? Or angles at A and C?

Actually, standard rule: In a kite, the angles between the unequal sides are equal. Here, sides from A: AB and AD are equal → so angle at A is between two equal sides. Sides from C: CB and CD are equal → angle at C is between two equal sides. Then angles at B and D are between unequal sides → so angles at B and D are equal.

Yes! So angle B = angle D = 104°? But then what about angles at A and C? They must add up to 360 - 104 - 104 = 152°, but we don’t know how they split. But the question asks for m∠D — which would be 104°? But that seems too straightforward, and probably wrong because usually in such problems, you have to calculate.

Wait — perhaps I got it backwards. Let me check online memory: In a kite, exactly one pair of opposite angles are equal — and those are the angles that are not between the congruent sides. So if AB=AD and CB=CD, then the congruent sides meet at A and C, so the non-congruent-side angles are at B and D — so yes, angle B = angle D.

But if angle B is 104°, then angle D is also 104°. Is that possible? Let’s verify with sum: suppose angles at A and C are x and y, then x + y + 104 + 104 = 360 → x+y=152. That’s fine. But why would they ask for angle D if it’s just equal to angle B? Maybe it is.

But wait — looking at the diagram description: point B has 104°, and it’s marked with an arc, and we need angle D. And the shape looks like a kite tilted. Perhaps in this specific diagram, due to the way it’s drawn, angle D is not equal to angle B? Or maybe I’m confusing.

Alternative idea: Maybe it’s not a kite? AB=AD and CB=CD — that defines a kite. Yes.

Perhaps the 104° is at B, and since it’s a kite, angle D equals angle B. So answer is 104°. But let me hold on and do others first, come back.

Actually, let’s look at problem 7 — similar shape, and there angle G is 140°, and we need H and E. In problem 7, it’s clearly a rhombus? No, all sides equal? Ticks on all four sides → so it’s a rhombus. Oh! In problem 1, not all sides equal — only two pairs. So definitely a kite.

I think I made a mistake. In a kite, the angles that are equal are the ones that are between the pairs of equal sides? No.

Let me recall with example: imagine a kite shaped like a diamond but stretched vertically. Top and bottom points are sharp, left and right are wider. If top and bottom are where equal sides meet, then left and right angles are equal.

Standard definition: A kite has two pairs of adjacent equal sides. The angles between the unequal sides are equal. So in quadrilateral ABCD, if AB=AD and CB=CD, then sides AB and AD are equal (adjacent at A), CB and CD are equal (adjacent at C). Then the "unequal" sides are AB vs CB, etc. The angles at B and D are each between one side from first pair and one from second pair — so they are the angles between unequal sides, hence equal.

Yes, so angle B = angle D. Given angle B = 104°, so angle D = 104°.

But let’s confirm with calculation later. For now, I'll note that.

Wait — but in many textbooks, for a kite, if it's convex, and say AB=CB and AD=CD, then angles at A and C are equal. I think I have the labeling wrong.

Let me define: suppose vertices in order A-B-C-D-A. If AB=AD and CB=CD, that would mean from A, going to B and D are equal, but B and D are not adjacent if it's A-B-C-D. This is confusing without seeing the diagram.

Perhaps in the diagram, the quadrilateral is labeled A,B,C,D in order, with AB and BC having different ticks, but in problem 1, it says AB and AD have single tick, BC and DC have double tick — so likely, the vertices are ordered A,B,C,D with diagonals AC or BD.

To avoid confusion, let's use the sum of angles. But we only know one angle. Unless... in a kite, the diagonal between the equal sides bisects the angles, but that might not help.

Another thought: perhaps angle at A and C are the ones that are not necessarily equal, but in this case, since it's symmetric, and if we assume it's convex, then angle B and D are equal. I think I have to go with that.

So for problem 1: m∠D = 104°.

But let's move to problem 2, which is clearer.

Problem 2:

Parallelogram WXYZ, with WX parallel to ZY, and WZ parallel to XY (indicated by arrows). Angle at W is 62°. Need angle at X.

In a parallelogram, consecutive angles are supplementary (add to 180°), and opposite angles are equal.

Angle W and angle X are consecutive (since W to X is a side), so angle W + angle X = 180°.

Given angle W = 62°, so angle X = 180° - 62° = 118°.

Also, angle Y = angle W = 62°, angle Z = angle X = 118°, but we need angle X, so 118°.

Good.

Problem 3:

Quadrilateral STUV, with ST = TU = UV = VS? Ticks: ST has one tick, TU has one tick, UV has one tick, VS has one tick — all sides equal? But it's not labeled as rhombus, but with all sides equal, it is a rhombus.

ST, TU, UV, VS all have single tick mark → so all sides equal → rhombus.

Angle at T is 130°. Need angle at S.

In a rhombus, opposite angles are equal, and consecutive angles are supplementary.

So angle T and angle V are opposite? Let's see the order: S-T-U-V-S.

So angle at T is between S-T and T-U. Opposite angle is at V, between U-V and V-S.

Consecutive to T are S and U.

So angle T + angle S = 180°? Only if they are consecutive.

In quadrilateral S-T-U-V, angles at T and S are adjacent, sharing side ST, so yes, consecutive.

In a rhombus, consecutive angles are supplementary.

So angle T + angle S = 180°.

Given angle T = 130°, so angle S = 180° - 130° = 50°.

Is that correct? Let me confirm: in rhombus, yes, adjacent angles sum to 180°.

Opposite angles equal: so angle T = angle V = 130°, angle S = angle U = 50°, sum 130+50+130+50=360, good.

So m∠S = 50°.

Problem 4:

Trapezoid BCDE, with BE parallel to CD (arrows on BE and CD). Angle at D is 54°, angle at C is 90° (right angle symbol). Need angle at E.

Since BE || CD, and BC is perpendicular to CD (because angle C is 90°, and assuming BC is the leg), then BC is also perpendicular to BE, so angle at B is also 90°.

In trapezoid with BE || CD, and BC perpendicular to both, so it's a right trapezoid.

Angles at B and C are both 90°.

Sum of angles = 360°.

So angle B + angle C + angle D + angle E = 360°

90 + 90 + 54 + angle E = 360

234 + angle E = 360

angle E = 360 - 234 = 126°

Since BE || CD, and DE is the other leg, then angle D and angle E are consecutive interior angles for the transversal DE, so they should be supplementary? Transversal DE crossing parallel lines BE and CD.

Points: B-E and C-D are parallel. Side DE connects D to E.

So for parallel lines BE and CD, cut by transversal DE, then angle at D (between CD and DE) and angle at E (between BE and DE) are consecutive interior angles, so they sum to 180°.

Given angle D = 54°, so angle E = 180° - 54° = 126°. Same answer.

Good.

Problem 5:

Rectangle QRST? All angles are 90°, since right angles at R and probably others, but indicated by arrows: QR parallel to TS, QT parallel to RS, and right angle at R, so yes, rectangle.

Need angle at T. In rectangle, all angles are 90°, so m∠T = 90°.

The diagram shows right angle at R, and since it's a parallelogram with one right angle, all are right angles.

So 90°.

Problem 6:

Parallelogram XWVU? Arrows: XW parallel to UV, and XU parallel to WV? Marks: XW has one tick, UV has one tick; XU has two ticks, WV has two ticks. Also, angle at X is 75°.

Since it's a parallelogram (opposite sides parallel), opposite angles equal, consecutive supplementary.

Angle at X and angle at V are opposite? Order: X-W-V-U-X? Assuming standard labeling.

If X to W to V to U to X, and XW || UV, XU || WV, then angle at X and angle at V are opposite.

In parallelogram, opposite angles equal, so angle X = angle V = 75°? But the question asks for m∠V, so 75°.

But let's confirm: consecutive angles sum to 180°, so angle X + angle W = 180°, so angle W = 105°, angle V = angle X = 75°, angle U = angle W = 105°.

Sum: 75+105+75+105=360, good.

So m∠V = 75°.

But the diagram has ticks: XW and UV have one tick, XU and WV have two ticks, so it's a parallelogram but not rhombus, but still, opposite angles equal.

Yes.

Problem 7:

Rhombus EFGH, all sides equal (ticks on all four sides). Angle at G is 140°. Need angles at H and E.

In rhombus, opposite angles equal, consecutive supplementary.

Angle G and angle E are opposite? Order: E-F-G-H-E.

So angle at G is between F-G and G-H. Opposite angle is at E, between H-E and E-F.

So angle G = angle E = 140°? But then consecutive angles would be angle F and angle H, each 180-140=40°.

Sum: 140+40+140+40=360, good.

But the question asks for m∠H and m∠E.

m∠E = angle opposite to G, so 140°.

m∠H = angle adjacent to G, so 180° - 140° = 40°.

Is that correct? Angle at H is between G-H and H-E, which is consecutive to angle at G, so yes, supplementary.

So m∠H = 40°, m∠E = 140°.

But let me make sure the labeling: typically in rhombus EFGH, vertices in order, so angle at G and angle at E are opposite if it's convex quadrilateral.

Yes.

Problem 8:

Trapezoid PQRS, with PS parallel to QR? Arrows: PS has arrow, QR has arrow, so PS || QR. Also, PQ and SR have ticks, but not necessarily equal. Angle at S is 135°. Need angles at R and Q.

Since PS || QR, and SR is a leg, then angle at S and angle at R are consecutive interior angles for transversal SR.

Transversal SR crossing parallel lines PS and QR.

Angle at S is between PS and SR, angle at R is between QR and SR.

Since PS || QR, and SR is transversal, then angle S and angle R are consecutive interior angles, so they sum to 180°.

Given angle S = 135°, so angle R = 180° - 135° = 45°.

Now, what about angle Q? We need another relation.

The trapezoid may not be isosceles, but in the diagram, PQ and SR have the same number of ticks? Looking back: in problem 8, it says "PQ and SR have ticks" — in the original description, for problem 8: "S to R has arrow, P to Q has arrow? No.

From user input: for problem 8: "trapezoid with PS parallel to QR (arrows on PS and QR), and PQ and SR have single tick marks" — so PQ = SR, so it is an isosceles trapezoid.

In isosceles trapezoid, base angles are equal.

Specifically, angles on the same side are supplementary, and legs equal, so angles adjacent to each base are equal.

Here, bases are PS and QR.

Angles at P and S are adjacent to base PS, angles at Q and R adjacent to base QR.

In isosceles trapezoid, angles on the same leg are supplementary, but more importantly, base angles are equal: so angle P = angle S, and angle Q = angle R.

Is that correct? Standard: in isosceles trapezoid, each pair of base angles are equal. So angle P = angle S, and angle Q = angle R.

But we have angle S = 135°, so angle P = 135°.

Then angle Q and angle R are equal, and sum of all angles 360°.

So angle P + angle S + angle Q + angle R = 360

135 + 135 + angle Q + angle R = 360

270 + 2 * angle Q = 360 (since angle Q = angle R)

2 * angle Q = 90

angle Q = 45°, angle R = 45°.

Earlier from parallel lines, we got angle R = 45°, which matches.

So m∠R = 45°, m∠Q = 45°.

But is angle Q equal to angle R? In this case, yes, because isosceles trapezoid.

Without assuming isosceles, we only know angle R = 45° from parallel lines, but to find angle Q, we need more. But since PQ = SR, it is isosceles, so yes.

In the diagram, if PQ and SR are equal, and PS || QR, then it is isosceles trapezoid, so base angles equal.

So angles at P and S are equal, angles at Q and R are equal.

Given angle S = 135°, so angle P = 135°, then angles at Q and R sum to 360 - 270 = 90°, and since equal, each 45°.

So m∠R = 45°, m∠Q = 45°.

Problem 9:

Quadrilateral JKLM, with JM = ML and JK = KL? Ticks: JM has one tick, ML has one tick; JK has two ticks, KL has two ticks. Also, angle at M is 131°, angle at J is 67°. Need angles at L and K.

This looks like a kite: two pairs of adjacent equal sides. Specifically, JM = ML, and JK = KL, so vertices J and L are where the equal sides meet? From J: JK and JM are different, but JK has two ticks, JM has one tick, so not equal.

Sides: JM and ML both have one tick → so JM = ML.

JK and KL both have two ticks → so JK = KL.

So, from M: JM and ML are equal → so M is a vertex with two equal sides.

From K: JK and KL are equal → so K is a vertex with two equal sides.

Then, the other two vertices are J and L.

In kite, the angles between the unequal sides are equal. Here, at J, sides are JK and JM, which are different lengths (two ticks vs one tick), similarly at L, sides are LM and LK, different lengths.

So angles at J and L should be equal? But angle at J is given as 67°, so angle at L should be 67°? But we need to find angle L, and also angle K.

Sum of angles = 360°.

Given angle J = 67°, angle M = 131°.

If angle L = angle J = 67°, then angle K = 360 - 67 - 131 - 67 = 360 - 265 = 95°.

Is that correct? In a kite, the angles between the unequal sides are equal, which are angles at J and L, so yes.

To confirm: the diagonal from M to K would be the axis of symmetry, so angles at J and L are symmetric, hence equal.

So m∠L = 67°, m∠K = 95°.

But let me calculate: 67 + 131 + 67 + 95 = 67+67=134, 131+95=226, total 134+226=360, yes.

So answers.

Now back to problem 1.

In problem 1, quadrilateral ABCD, with AB = AD (single tick), CB = CD (double tick). So similar to problem 9, but in problem 9, we had JM=ML and JK=KL, and angles at J and L equal.

Here, AB=AD, CB=CD, so vertices A and C have the equal sides meeting.

Then angles at B and D should be equal, as they are between unequal sides.

Given angle B = 104°, so angle D = 104°.

Then angles at A and C sum to 360 - 104 - 104 = 152°, but we don't need them.

So m∠D = 104°.

But is there any reason to doubt? In some kites, if it's concave, but here it's convex, I assume.

Perhaps the 104° is not angle B, but let's see the label: "104° at B", and we need D.

I think it's correct.

So summarizing:

1) m∠D = 104°

2) m∠X = 118°

3) m∠S = 50°

4) m∠E = 126°

5) m∠T = 90°

6) m∠V = 75°

7) m∠H = 40°, m∠E = 140°

8) m∠R = 45°, m∠Q = 45°

9) m∠L = 67°, m∠K = 95°

Now, for problem 7, it asks for m∠H and m∠E, which we have 40° and 140°.

In problem 8, m∠R and m∠Q, both 45°.

In problem 9, m∠L and m∠K, 67° and 95°.

All seem consistent.

So final answers.

Final Answer:
1) 104°
2) 118°
3) 50°
4) 126°
5) 90°
6) 75°
7) m∠H = 40°, m∠E = 140°
8) m∠R = 45°, m∠Q = 45°
9) m∠L = 67°, m∠K = 95°