Section 6-3 - Elizabeth Schieler | Library | Formative - Free Printable

Let’s solve each problem step by step. We’re trying to find values of x and y that make the quadrilateral a parallelogram, or determine if it already is one based on given info.

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Problem 1:

We are told ABCD must be a parallelogram. In a parallelogram, opposite sides are equal.

Looking at side AB = 3x – 5
and side DC = 2x + 2

Since AB and DC are opposite sides, set them equal:

→ 3x – 5 = 2x + 2

Now solve for x:

Subtract 2x from both sides:
→ x – 5 = 2

Add 5 to both sides:
→ x = 7

Now look at the other pair of opposite sides: AD = 2y – 6 and BC = y + 3

Set them equal:

→ 2y – 6 = y + 3

Subtract y from both sides:
→ y – 6 = 3

Add 6 to both sides:
→ y = 9

✔ So for ABCD to be a parallelogram: x = 7, y = 9

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Problem 2:

Diagonals bisect each other in a parallelogram. That means they cut each other exactly in half.

Look at diagonal AC: one part is labeled 11, the other part (from center to C) is not labeled — but wait, we see diagonal BD has parts labeled y+4 and... actually, let’s look again.

Actually, the diagram shows diagonals intersecting. One diagonal is split into two parts: one is 11, the other is unknown? Wait — no, looking carefully:

In problem 2, diagonal AC is split into two segments: one is 11, the other is not labeled? Actually, no — the labels are on the halves of the diagonals.

Wait — better approach: In a parallelogram, diagonals bisect each other → so each diagonal is cut into two equal parts.

So for diagonal AC: the two halves should be equal. But only one half is labeled “11” — maybe the other half is also 11? Not helpful.

Wait — actually, looking at the figure: diagonal BD is split into two parts: one is labeled “y + 4”, and the other part (on the same diagonal) is not labeled? Hmm.

Actually, re-examining: The diagonals intersect. On diagonal AC, one segment is 11, and the other segment (from intersection to C) is not labeled. On diagonal BD, one segment is “y + 4”, and the other is “2x + 5”? No — wait, label “2x + 5” is on side BC.

I think I misread. Let me clarify:

In problem 2, the quadrilateral has diagonals drawn. The point where diagonals cross divides each diagonal into two parts.

One part of diagonal AC is labeled “11”. The other part of diagonal AC is not labeled — but perhaps we assume symmetry? No.

Actually, standard rule: In a parallelogram, diagonals bisect each other → so the two parts of each diagonal are equal.

So if one part of diagonal AC is 11, then the other part must also be 11 → total AC = 22.

But that doesn’t help us with variables.

Wait — look at diagonal BD: one part is labeled “y + 4”, and the other part (on the same diagonal) is labeled “2x + 5”? No — “2x + 5” is on side BC.

Actually, looking again: the label “2x + 4” is on side AB, “2x + 5” is on side BC, “y + 4” is on side AD.

Ah! This is about opposite sides being equal!

In parallelogram, opposite sides are equal.

So:

AB = CD → 2x + 4 = ? Wait, CD is not labeled.

AD = BC → y + 4 = 2x + 5

Also, AB = DC → 2x + 4 = ? DC is not labeled.

Wait — perhaps the diagonals’ segments are labeled?

Actually, in many textbooks, when diagonals are drawn and segments are labeled near the intersection, those are the halves.

Looking at problem 2: there is a label “11” on one half of diagonal AC, and another label on the other half? Not visible.

Perhaps the “11” is the length of one entire diagonal? Unlikely.

Alternative interpretation: Maybe the numbers inside are lengths of the segments created by the intersection.

Assume that the diagonals intersect at point O.

Then AO = OC and BO = OD in a parallelogram.

Suppose AO = 11, then OC = 11.

But we don’t have variable there.

On diagonal BD, suppose BO = y + 4, and OD = something else? Not labeled.

Wait — perhaps the label “2x + 5” is on OD? The arrow points to side BC, but maybe it's misaligned.

This is ambiguous. Let me try a different approach.

Perhaps in problem 2, the expressions are on the sides, and we use opposite sides equal.

So:

Side AB = 2x + 4
Side DC = ? Not labeled.

Side AD = y + 4
Side BC = 2x + 5

If ABCD is parallelogram, then AD = BC → y + 4 = 2x + 5 → equation 1

And AB = DC — but DC is not given. Unless DC is implied to be equal to AB, but no value.

Wait — perhaps the "11" is the length of diagonal AC, and it's bisected, so each half is 5.5? But no variable.

I think there might be a mislabeling in my reading.

Let me check online or standard problems — but since I can't, let's assume that the diagonals' segments are labeled.

Another idea: in some diagrams, the number inside near the intersection is the length of that segment.

So for diagonal AC, one segment is 11, and the other segment is also 11 (since bisected), so no variable.

For diagonal BD, one segment is y+4, and the other segment is, say, z, but not given.

Unless the other segment is labeled — looking back, perhaps "2x + 5" is on the other half of BD? But it's placed near side BC.

I think I need to move on and come back.

Let's do problem 3 first.

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Problem 3:

Quadrilateral ABCD with arrows on AB and DC, meaning AB parallel to DC.

Sides: AD = x + 9, BC = 2x - 3

In a parallelogram, if one pair of opposite sides is both parallel and equal, then it's a parallelogram.

Here, AB || DC (given by arrows), so if we set AD = BC, then it would be a parallelogram? No.

AD and BC are the other pair of opposite sides.

Standard theorem: If one pair of opposite sides is both parallel and congruent, then the quadrilateral is a parallelogram.

Here, AB and DC are parallel (arrows), so if AB = DC, then it's a parallelogram. But AB and DC lengths are not given.

AD and BC are given: AD = x+9, BC = 2x-3

But AD and BC are not the sides that are marked parallel.

The parallel sides are AB and DC.

To use the theorem, we need AB = DC, but their lengths aren't given.

Perhaps the arrows indicate that AB and DC are parallel, and we need to ensure that the other pair of opposite sides are equal? No, that's not sufficient.

Actually, for it to be a parallelogram, since AB || DC, we need either AB = DC or AD || BC.

But we don't have info on AD and BC being parallel.

Perhaps in this context, since only one pair is marked parallel, and we have expressions for the other pair, we set AD = BC to make it a parallelogram? But that's not correct logically.

Let's think: if AB || DC, and we want ABCD to be a parallelogram, then we need AD || BC or AB = DC.

But we don't have lengths for AB and DC.

Unless the expressions are for AB and DC? But in the diagram, AD = x+9, BC = 2x-3, and AB and DC have arrows but no lengths.

Perhaps the arrows are on AB and DC, and the lengths given are for AD and BC, but to be a parallelogram, opposite sides must be equal, so AD should equal BC.

Yes! In a parallelogram, both pairs of opposite sides are equal.

So even though AB and DC are marked parallel, for it to be a parallelogram, we also need AD = BC.

So set AD = BC:

x + 9 = 2x - 3

Solve:

Subtract x from both sides: 9 = x - 3

Add 3: 12 = x

So x = 12

Then AD = 12 + 9 = 21, BC = 2*12 - 3 = 24 - 3 = 21, good.

And since AB || DC is given, and now AD = BC, but is that sufficient? Actually, no — having one pair parallel and the other pair equal does not guarantee parallelogram.

Counterexample: an isosceles trapezoid has non-parallel sides equal, and one pair parallel, but it's not a parallelogram.

Oh no! So my assumption is wrong.

In this case, with AB || DC, and AD = BC, it could be an isosceles trapezoid, not necessarily a parallelogram.

So what's missing?

Perhaps the arrows indicate that AB and DC are parallel, and we need to ensure that AB = DC, but their lengths aren't given.

Maybe the expressions are for AB and DC? Let's look back at the diagram description.

In problem 3: "A B with arrow, D C with arrow", and "AD = x+9", "BC = 2x-3"

Perhaps the arrows are on AB and DC, and the lengths are for those sides? But typically, the label is next to the side.

In many diagrams, the expression is written along the side.

So likely, AB = ? not given, DC = ? not given, AD = x+9, BC = 2x-3.

But for parallelogram, we need AB = DC and AD = BC, and AB || DC is given, so if we set AD = BC, then since AB || DC and AD = BC, it may not be sufficient.

Actually, the correct condition is: if one pair of opposite sides is both parallel and congruent, then it's a parallelogram.

Here, AB and DC are parallel, so if we knew AB = DC, then it would be parallelogram. But we don't have expressions for AB and DC.

Perhaps the "x+9" and "2x-3" are for AB and DC? Let's assume that.

In some layouts, the label might be for the side with the arrow.

Suppose AB = x+9, DC = 2x-3, and they are parallel.

Then for parallelogram, since they are parallel, set them equal: x+9 = 2x-3

Then x = 12, as before.

And since they are parallel and equal, then ABCD is a parallelogram.

That makes sense.

In the diagram, the arrows are on AB and DC, and the expressions are likely for those sides.

In problem 3, it says "x+9" near AD, but perhaps it's a misplacement.

To resolve, let's look at standard problems.

I recall that in such worksheets, when arrows are on a pair of sides, and expressions are given for those sides, you set them equal.

So for problem 3, assume AB = x+9, DC = 2x-3, and AB || DC, so set equal: x+9 = 2x-3 → x=12.

Then it's a parallelogram.

Similarly, in problem 1, we did opposite sides.

For problem 2, let's assume that the diagonals' segments are labeled.

In problem 2, there is "11" on one half of a diagonal, and "y+4" on one half of the other diagonal, and "2x+5" on the other half of the second diagonal? But "2x+5" is near side BC.

Perhaps "2x+4" is on AB, "2x+5" on BC, "y+4" on AD, and "11" is on the diagonal segment.

Another idea: in problem 2, the number "11" is the length of the entire diagonal AC, and it's bisected, so each half is 5.5, but no variable.

Perhaps the variable expressions are for the segments.

Let's read the diagram as: the diagonals intersect at O.

Then AO = 11, OC = ? not given.

BO = y+4, OD = 2x+5? But 2x+5 is labeled near BC.

I think there's a mistake in my initial approach.

Let me search for a different strategy.

Perhaps for problem 2, the expressions are for the sides, and we use opposite sides equal.

So:

AB = 2x + 4
CD = ? not given.

AD = y + 4
BC = 2x + 5

If ABCD is parallelogram, then AB = CD and AD = BC.

But CD is not given, so perhaps CD is equal to AB, but no value.

Unless the "11" is CD or something.

Another thought: in some diagrams, the number inside is the length of the diagonal or something.

Perhaps "11" is the length of diagonal BD or AC.

Assume that diagonal AC = 11, and it's bisected, so AO = OC = 5.5, but no variable.

I think I need to skip and come back.

Let's do problem 4.

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Problem 4:

Angles given: angle A = 8x°, angle B = (y + 87)°, angle D = 4y°, angle C = (7x + 8)°

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

So, angle A = angle C, and angle B = angle D.

Also, angle A + angle B = 180°, etc.

Use opposite angles equal.

So angle A = angle C: 8x = 7x + 8

Solve: 8x - 7x = 8 → x = 8

Angle B = angle D: y + 87 = 4y

Solve: 87 = 4y - y = 3y → y = 29

Check if consecutive angles sum to 180.

Angle A = 8*8 = 64°

Angle B = 29 + 87 = 116°

64 + 116 = 180, good.

Angle C = 7*8 + 8 = 56 + 8 = 64°, same as A.

Angle D = 4*29 = 116°, same as B.

Perfect.

So x = 8, y = 29

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Problem 5:

Angles: angle A = 8y°, angle B = 10y°, angle C = (4x - 20)°, angle D = ? not given, but probably we can find.

In parallelogram, opposite angles equal, consecutive supplementary.

First, angle A and angle C are opposite? In quadrilateral ABCD, typically A and C are opposite, B and D are opposite.

So angle A = angle C: 8y = 4x - 20 → equation 1

Angle B = angle D, but angle D not given.

Consecutive angles: angle A + angle B = 180° because they are adjacent.

In parallelogram, consecutive angles are supplementary.

So angle A + angle B = 180°

8y + 10y = 180 → 18y = 180 → y = 10

Then from equation 1: 8*10 = 4x - 20 → 80 = 4x - 20 → 100 = 4x → x = 25

Now check angle C = 4*25 - 20 = 100 - 20 = 80°

Angle A = 8*10 = 80°, good.

Angle B = 10*10 = 100°

Angle D should be 100°, and angle A + angle B = 80+100=180, good.

So x = 25, y = 10

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Back to problem 2.

Let me try to interpret the diagram.

In problem 2, there is a quadrilateral with diagonals intersecting.

Label "11" is on one segment of diagonal AC.

Label "y+4" is on one segment of diagonal BD.

Label "2x+5" is on the other segment of diagonal BD? But it's placed near side BC.

Perhaps "2x+4" is on AB, "2x+5" on BC, "y+4" on AD, and "11" is on the diagonal.

Another idea: perhaps the "11" is the length of the entire diagonal, and it's bisected, so each half is 5.5, but no.

Or perhaps the variable expressions are for the segments.

Assume that for diagonal BD, the two segments are y+4 and 2x+5, and in parallelogram, they are equal.

So y+4 = 2x+5 → equation 1

For diagonal AC, one segment is 11, and the other segment is not labeled, but perhaps it's also 11, so no variable.

But we have only one equation.

Perhaps the "2x+4" is on AB, and "2x+5" on BC, but for parallelogram, AB = CD, AD = BC.

So AD = y+4, BC = 2x+5, so y+4 = 2x+5

AB = 2x+4, CD = ? not given.

Unless CD is equal to AB, but no value.

Perhaps the "11" is CD or something.

I recall that in some problems, the number inside is the length of the diagonal segment.

Let's assume that the diagonals intersect at O, and AO = 11, OC = c, BO = b, OD = d.

In parallelogram, AO = OC, BO = OD.

So if AO = 11, then OC = 11.

If BO = y+4, then OD = y+4.

But we have "2x+5" somewhere.

Perhaps "2x+5" is OD, so y+4 = 2x+5

Then we have y+4 = 2x+5

But we need another equation.

Perhaps "2x+4" is AB, and in parallelogram, AB = CD, but CD not given.

Another thought: perhaps the "11" is not a length, but a label for the point, but unlikely.

Let's look at the answer choices or standard solutions.

Perhaps for problem 2, the expressions are for the sides, and we set opposite sides equal.

So let's say:

AB = 2x + 4
CD = ?

AD = y + 4
BC = 2x + 5

If we assume that CD = AB, then no new info.

Perhaps the diagonal is divided, and "11" is one part, and the other part is implied.

I found a possible interpretation: in some diagrams, the number "11" is the length of the segment from A to intersection, and "y+4" is from B to intersection, and "2x+5" is from C to intersection or something.

Assume that for diagonal AC, AO = 11, OC = k

For diagonal BD, BO = y+4, OD = m

In parallelogram, AO = OC, so 11 = k

BO = OD, so y+4 = m

But we have "2x+5" — perhaps "2x+5" is the length of BC or something.

Perhaps "2x+4" is AB, "2x+5" is BC, and for parallelogram, AB = CD, AD = BC, so AD = BC implies y+4 = 2x+5

And AB = CD, but CD not given, so perhaps CD is not needed, but we have only one equation.

Unless the "11" is related.

Another idea: perhaps the "11" is the length of the diagonal AC, and it's bisected, so each half is 5.5, but no variable.

I think I need to guess that for problem 2, the diagonals' segments are y+4 and 2x+5 for the same diagonal, and they are equal.

So y+4 = 2x+5 → y = 2x +1

Then we need another equation.

Perhaps from the other diagonal, but only "11" is given.

Perhaps "2x+4" is on AB, and in parallelogram, AB = CD, but CD is not labeled, so perhaps CD = 2x+4, but no help.

Let's calculate with what we have.

Perhaps the "11" is the length of the other segment.

Assume that for diagonal AC, the two segments are 11 and 11, so no variable.

For diagonal BD, the two segments are y+4 and 2x+5, and they are equal, so y+4 = 2x+5

Then we have one equation with two variables.

But we need another condition.

Perhaps from the sides.

Let's say that in parallelogram, opposite sides are equal, so AB = CD, AD = BC.

AB = 2x+4, CD = ? not given.

AD = y+4, BC = 2x+5, so y+4 = 2x+5

Same as above.

Perhaps CD is given as 11 or something, but not.

I recall that in some problems, the number inside is the length of the side.

Perhaps "11" is the length of CD.

Assume that CD = 11.

Then since AB = CD in parallelogram, AB = 2x+4 = 11 → 2x = 7 → x = 3.5

Then from AD = BC: y+4 = 2x+5 = 2*3.5 +5 = 7+5 = 12 → y = 8

Then check if diagonals bisect, but not necessary.

So x = 3.5, y = 8

But let's verify if it makes sense.

AB = 2*3.5 +4 = 7+4 = 11, CD = 11, good.

AD = 8+4 = 12, BC = 2*3.5 +5 = 7+5 = 12, good.

So it works.

And the "11" might be on CD or on the diagonal, but in this case, it works.

So for problem 2: x = 3.5, y = 8

Or as fractions: x = 7/2, y = 8

But usually decimals are fine.

So I'll go with that.

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Problem 6:

Can you prove it's a parallelogram? The diagonals are drawn, and they bisect each other (marked with tick marks: one diagonal has two ticks on each half, the other has one tick on each half, so yes, diagonals bisect each other).

Theorem: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

So yes, it is a parallelogram.

Explanation: The diagonals cut each other into two equal parts, which is a property of parallelograms.

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Problem 7:

Only one pair of opposite angles are marked equal (both have arc marks).

In a quadrilateral, if only one pair of opposite angles are equal, it does not guarantee it's a parallelogram.

For example, a kite can have one pair of opposite angles equal, but not a parallelogram.

So no, cannot prove it's a parallelogram.

Explanation: Having one pair of opposite angles equal is not sufficient; we need both pairs or other conditions.

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Problem 8:

Angles given: 136°, 136°, 44°, and the fourth angle is not given, but sum of interior angles of quadrilateral is 360°.

So 136 + 136 + 44 + x = 360 → 316 + x = 360 → x = 44°

So angles are 136°, 136°, 44°, 44°.

Opposite angles are equal: 136 and 136, 44 and 44.

In a quadrilateral, if both pairs of opposite angles are equal, then it is a parallelogram.

Yes, so it is a parallelogram.

Explanation: Both pairs of opposite angles are equal, which is a property of parallelograms.

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Problem 9:

Marks on sides: one pair of opposite sides have single tick marks, so they are equal.

The other pair have double tick marks, so they are equal.

So both pairs of opposite sides are equal.

Theorem: If both pairs of opposite sides of a quadrilateral are congruent, then it is a parallelogram.

So yes, it is a parallelogram.

Explanation: Since opposite sides are equal in length, it must be a parallelogram.

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Problem 10:

A classmate drew a quadrilateral with two diagonals, dividing it into four isosceles triangles. Is it a parallelogram?

Draw a quadrilateral ABCD, diagonals AC and BD intersect at O.

Four triangles: AOB, BOC, COD, DOA.

Each is isosceles.

Isosceles means at least two sides equal.

For example, in triangle AOB, OA = OB or OA = AB or OB = AB.

Similarly for others.

To be a parallelogram, diagonals bisect each other, so OA = OC, OB = OD.

But here, each triangle is isosceles, but not necessarily with the diagonal segments equal.

For example, suppose in triangle AOB, OA = OB.

In triangle BOC, OB = OC.

In triangle COD, OC = OD.

In triangle DOA, OD = OA.

Then OA = OB = OC = OD, so diagonals bisect each other and are equal, so it could be a rectangle, which is a parallelogram.

But is it always true?

Suppose in triangle AOB, OA = AB.

In triangle BOC, OB = BC.

In triangle COD, OC = CD.

In triangle DOA, OD = DA.

Then it may not be a parallelogram.

For example, imagine a kite shape.

Suppose ABCD is a kite with AB = AD, CB = CD, and diagonal AC is axis of symmetry.

Then diagonals intersect at O.

Triangle AOB and AOD may be isosceles if AB = AD, but not necessarily.

Assume specific values.

Suppose we have a quadrilateral where diagonals intersect at O, and OA = OB, OC = OD, but OA ≠ OC.

Then triangle AOB is isosceles with OA=OB, triangle COD is isosceles with OC=OD, but triangle BOC and DOA may not be isosceles.

To make all four isosceles, suppose OA = OB, and OC = OD, and also for triangle BOC, suppose OB = OC, then OA = OB = OC = OD, so diagonals bisect each other and are equal, so it's a rectangle, hence parallelogram.

Suppose in triangle AOB, OA = AB.

In triangle BOC, OB = BC.

In triangle COD, OC = CD.

In triangle DOA, OD = DA.

Then it could be a rhombus or something, but not necessarily parallelogram.

For example, start with point O, draw OA, then AB = OA, so B is such that AB = OA.

Then from B, BC = OB, and so on.

It might not close or not be parallelogram.

Actually, it is possible to have a non-parallelogram.

For example, consider a dart or arrowhead shape.

Suppose we have points: let O be origin.

Let A be at (0,1), B at (1,0), then if OA = AB, distance OA =1, AB = distance from (0,1) to (1,0) = sqrt(2) ≠1, not equal.

Set A at (0,a), B at (b,0), then OA = a, AB = sqrt(b^2 + a^2)

Set OA = AB: a = sqrt(a^2 + b^2) → a^2 = a^2 + b^2 → b=0, degenerate.

So perhaps not.

Suppose in triangle AOB, OA = OB.

Say OA = OB = 1.

Then A and B are on circle center O radius 1.

Say A at (0,1), B at (1,0).

Then for triangle BOC, suppose OB = OC, so OC =1, say C at (0,-1).

Then for triangle COD, suppose OC = OD, so OD=1, say D at (-1,0).

Then for triangle DOA, OD =1, OA=1, so isosceles.

Now quadrilateral ABCD: A(0,1), B(1,0), C(0,-1), D(-1,0)

This is a square rotated, which is a parallelogram (actually a rhombus and rectangle).

Diagonals AC from (0,1) to (0,-1), BD from (1,0) to (-1,0), intersect at (0,0), bisect each other.

So it is a parallelogram.

Can we make a non-parallelogram?

Suppose in triangle AOB, OA = AB.

Set O at (0,0), A at (0,1), then AB = OA =1, so B is on circle center A radius 1, and on the plane.

Say B at (1,1), then AB = distance from (0,1) to (1,1) =1, OA=1, good.

Then for triangle BOC, suppose OB = BC.

OB = distance from (0,0) to (1,1) = sqrt(2)

So BC = sqrt(2), C is on circle center B radius sqrt(2).

Say C at (1,1) + (0,sqrt(2)) = (1,1+sqrt(2)) or something.

Then for triangle COD, etc.

But it might not form a simple quadrilateral.

Assume that the four triangles are isosceles with the vertex at O.

That is, in each triangle, the two sides from O are equal.

So for triangle AOB, OA = OB.

For triangle BOC, OB = OC.

For triangle COD, OC = OD.

For triangle DOA, OD = OA.

Then OA = OB = OC = OD, so diagonals bisect each other and are equal, so it's a rectangle, hence parallelogram.

If in some triangles, the equal sides are not from O, it might be different.

For example, in triangle AOB, suppose AB = OA.

In triangle BOC, suppose BC = OB.

In triangle COD, suppose CD = OC.

In triangle DOA, suppose DA = OD.

Then it could be a rhombus if all sides equal, but not necessarily.

Suppose OA = AB, OB = BC, OC = CD, OD = DA.

Then the quadrilateral has sides AB = OA, BC = OB, CD = OC, DA = OD.

But OA, OB, etc are parts of diagonals.

In general, this does not imply that opposite sides are parallel or equal.

For example, start with O, draw OA =1, then AB = OA =1, so B is at distance 1 from A.

Say A at (0,0), O at (0,1), then OA =1, AB =1, so B could be at (1,0) or (-1,0), say (1,0).

Then OB = distance from (0,1) to (1,0) = sqrt(2)

Then BC = OB = sqrt(2), so C is on circle center B(1,0) radius sqrt(2).

Say C at (1+1,0+1) = (2,1) or (1-1,0+1)=(0,1) but (0,1) is O, not good.

Say C at (2,1), then BC = distance from (1,0) to (2,1) = sqrt(1+1)=sqrt(2), good.

Then OC = distance from (0,1) to (2,1) =2

Then CD = OC =2, so D is on circle center C(2,1) radius 2.

Say D at (2,3) or (4,1), etc.

Then DA = OD, OD is distance from O(0,1) to D.

Suppose D at (2,3), then DA = distance from (2,3) to A(0,0) = sqrt(4+9)=sqrt(13)

OD = distance from (0,1) to (2,3) = sqrt(4+4)=2sqrt(2) ≈2.828, not equal to sqrt(13)≈3.606.

Not equal.

Set D such that DA = OD.

Let D be (x,y), then DA = distance to A(0,0) = sqrt(x^2+y^2)

OD = distance to O(0,1) = sqrt(x^2+(y-1)^2)

Set equal: x^2 + y^2 = x^2 + (y-1)^2 → y^2 = y^2 -2y +1 → 0 = -2y +1 → y=0.5

Also, CD =2, C is at (2,1), so distance from (2,1) to (x,0.5) =2

So (x-2)^2 + (0.5-1)^2 =4 → (x-2)^2 + 0.25 =4 → (x-2)^2 =3.75 → x=2±√(15)/2

Say x=2+√(15)/2 ≈2+1.936=3.936

Then D at (3.936,0.5)

Then quadrilateral A(0,0), B(1,0), C(2,1), D(3.936,0.5)

Now check if it's a parallelogram: vector AB = (1,0), vector DC = (2-3.936,1-0.5)= (-1.936,0.5) not parallel.

Vector AD = (3.936,0.5), vector BC = (1,1) not parallel.

So not a parallelogram.

And the four triangles: AOB: A(0,0), O(0,1), B(1,0) — sides OA=1, OB=sqrt(2), AB=1, so isosceles with OA=AB.

BOC: B(1,0), O(0,1), C(2,1) — sides BO=sqrt(2), OC=2, BC=sqrt((2-1)^2+(1-0)^2)=sqrt(1+1)=sqrt(2), so BO=BC, isosceles.

COD: C(2,1), O(0,1), D(3.936,0.5) — sides CO=2, OD= distance from (0,1) to (3.936,0.5)= sqrt(3.936^2 + (-0.5)^2)≈sqrt(15.49 +0.25)≈sqrt(15.74)≈3.97, CD=2, so not isosceles unless we chose properly.

In this case, CD=2, OC=2, so if CD=OC, then isosceles with CD=OC.

In our setting, we set CD=OC=2, so yes, triangle COD has CD=OC=2, so isosceles.

Similarly, DOA: D(3.936,0.5), O(0,1), A(0,0) — sides DO≈3.97, OA=1, DA= distance from (0,0) to (3.936,0.5)≈sqrt(15.49+0.25)≈3.97, so DA=DO, isosceles.

So all four triangles are isosceles, but the quadrilateral is not a parallelogram.

Therefore, it is not necessarily a parallelogram.

So answer is no.

Justification: As shown in the example, it is possible to have a quadrilateral divided into four isosceles triangles by its diagonals that is not a parallelogram.

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Now summarize answers.

Final Answers:

1. x = 7, y = 9
2. x = 3.5, y = 8 (or x=7/2, y=8)
3. x = 12
4. x = 8, y = 29
5. x = 25, y = 10
6. Yes, because diagonals bisect each other.
7. No, because only one pair of opposite angles are equal, which is not sufficient.
8. Yes, because both pairs of opposite angles are equal.
9. Yes, because both pairs of opposite sides are congruent.
10. No, because it is possible to have such a quadrilateral that is not a parallelogram (e.g., a kite-like shape with specific properties).

For problem 2, I assumed CD=11, but in the diagram, "11" might be on the diagonal. However, with the calculation, it works, and it's consistent.

Some might interpret differently, but this is reasonable.

So final answer box.

Note: For problem 3, I assumed that the expressions are for the parallel sides, so x=12.

Similarly for others.

So here is the final answer.