NCERT Solutions for Class 8 Maths Chapter 4 Quadrilaterals (PDF) – 2027

Because only (i), (ii) and (iii) are closed figures made of exactly four line segments, joined end to end, with no segment crossing another and no free ends.

Figures (iv) and (v) are closed, but their boundaries are not made only of straight line segments: (iv) has three straight sides and one curved side, and (v) has two straight sides and two curved sides. A curve is not a side, so neither figure is built from four line segments.

Yes. The chapter finds two more, and proves that all three pick out exactly the same figures.

The other strip must also be 8 cm, and the two strips must be joined at their midpoints — that is, 4 cm from each end.

8 cm — the same as the first, because the diagonals of a rectangle are always equal.

Deduction 1. In rectangle ABCD, compare ∆ADC and ∆DAB:

It is the midpoint of both diagonals — the diagonals bisect each other.

Deduction 2. Let the diagonals of rectangle ABCD meet at O, and compare ∆AOB and ∆COD.

So O is the midpoint of AC and also the midpoint of BD.

Any angle at all. This is the surprising part of the Carpenter’s Problem: the angle is not forced.

Deduction 3 takes two equal diagonals that bisect each other with an arbitrary angle x between them, and shows every angle of the resulting quadrilateral works out to 90°. So the carpenter may join the strips at any angle she likes and still get a rectangle — different angles simply give rectangles of different shapes.

No. Two of the three facts do not belong to the triangles being compared.

• AO = CO — usable. AO is a side of ∆AOD and CO is a side of ∆COB.
• ∠AOB = ∠COD — not usable. ∠AOB is not an angle of ∆AOD, and ∠COD is not an angle of ∆COB. The angles we need are ∠AOD and ∠COB (which are indeed equal, being vertically opposite).
• AD = CB — usable, but it is a third side, not the second side next to the angle.

A correct set is:

Every angle in the figure is determined. Start with the four angles at the crossing point O.

Now use the four isosceles triangles. Because the diagonals are equal and bisect each other, OA = OB = OC = OD.

So at every vertex of ABCD the two half-angles are 30° and 60°, giving 30° + 60° = 90°. All four angles of ABCD are right angles, so ABCD is a rectangle.

a = 60°.

So ∆AOB has all three angles equal to 60° — it is equilateral, which also tells us AB = OA = OB.

It is a rectangle.

At each vertex the diagonal splits the angle into a 30° part and a 60° part, and 30° + 60° = 90°. So all four angles of ABCD are 90°, which is exactly the condition for a rectangle.

The opposite sides are equal: AB = CD and AD = CB.

Each congruence follows from SAS: OA = OC, OB = OD (the diagonals bisect each other), and the angle at O between them is the same in both triangles because vertically opposite angles are equal.

Yes — it stays a rectangle for every angle. That is exactly what Deduction 3 generalises.

Let one of the angles between the diagonals be x. Then the four angles at O are x, x, 180 – x, 180 – x.

a = 90 – ^x⁄₂ degrees.

AB = CD and AD = CB — the opposite sides are equal, whatever the angle x is.

Putting this together with Q3: if two segments are equal and bisect each other, the quadrilateral formed by their endpoints has all angles 90° and equal opposite sides. So it is a rectangle — and the Carpenter’s Problem is completely solved.

No, we would not be wrong. The shorter definition is enough.

Condition (a) is not an extra requirement at all — it follows from condition (b). Deduction 4 shows that once all four angles are 90°, the opposite sides are forced to be equal; there is no way to build a quadrilateral with four right angles and unequal opposite sides.

No such quadrilateral exists — the construction always closes up into a rectangle, and you can see why while drawing it.

Try it: draw AB of any length. At B turn 90° and draw BC of any length. At C turn 90° and draw CD. To make the angle at D equal 90° as well, D must lie directly above A — and that forces

∠1 = ∠2, and that is exactly the missing piece — the same argument as in Deduction 2.

Let ∠3 = ∠CBD, the other part of the angle at B.

Now compare ∆BAD and ∆DCB:

So a quadrilateral with all angles 90° automatically has equal opposite sides — it is a rectangle.

Yes, it is wrong. In a congruence statement the letters must be written in corresponding order.

The correct statement is ∆BAD ≅ ∆DCB, which asserts:

Writing ∆BAD ≅ ∆CDB instead would match B ↔ C, A ↔ D and D ↔ B. It would then claim AD = DB (a side equal to the diagonal) and ∠ABD = ∠DCB = 90° (a part of the corner at B equal to a whole right angle). Both are false.

Yes, and it can be proved from the right angles alone — no measurement needed.

Take AB as a transversal cutting AD and BC.

This gives Property 3: the opposite sides of a rectangle are parallel to each other.

Yes — repeat the argument with a different transversal.

So both pairs of opposite sides of a rectangle are parallel. This is also why every rectangle is a parallelogram — it satisfies the parallelogram’s definition and, on top of that, has all its angles 90°.

No — all four are rectangles, including (iv).

Every one of the four figures carries a right-angle mark at all four corners, and that is the only thing the definition asks for. Being tilted on the page (as (i), (ii) and (iii) are) changes nothing — angles do not depend on how a figure is turned.

Square: a quadrilateral in which all the angles are equal to 90°, and all the sides are of equal length.

The strips must be equal in length, joined at their midpoints, and set at right angles to each other.

The first two conditions were already needed for a rectangle. Making the angle exactly 90° is the one extra demand, and Deduction 5 shows it is the right one.

Yes — set the angle between the diagonals to 90°. That single change is all that is needed.

Recall from Deduction 3 that the four half-angles at a vertex are a = 90 – ^x⁄₂ and b = ^x⁄₂, where x is the angle between the diagonals. The sides come out equal exactly when the four triangles round O are all congruent, and that happens when the four angles at O are all the same:

At x = 90° we also get a = b = 45°, so each diagonal cuts each corner into two 45° halves.

Take ∆BOA and ∆BOC. They give the angles at once.

This is Deduction 5: the diagonals of a square bisect each other at right angles.

Steps.

1. Draw AC = 8 cm.
2. Construct the perpendicular bisector of AC with a compass — arcs of equal radius from A and from C on both sides, joined. Let it cut AC at O, so OA = OC = 4 cm.
3. On this perpendicular mark B and D with OB = OD = 4 cm, one on each side of AC.
4. Join AB, BC, CD and DA.

ABCD is the required square.

Both apply word for word, because neither deduction ever used anything except the right angles and the equal opposite sides — which a square has.

So a square has equal diagonals that bisect each other — and, from Deduction 5, they also cross at 90°.

All four are 45°.

So the diagonal cuts each 90° corner into two 45° halves. This is Property 5: the diagonals of a square bisect the angles of the square.

(i) Rectangle ABCD, ∠CAB = 30°. The diagonals of a rectangle are equal and bisect each other, so OA = OB = OC = OD.

(ii) Rectangle PQRS, ∠QOR = 110°. Again OP = OQ = OR = OS.

The construction is the same in every case; only the angle at O changes.

1. Draw AB = 8 cm and mark its midpoint O (so OA = OB = 4 cm).
2. At O draw a ray making the required angle with OB.
3. On this ray, and on its opposite ray, cut OC = OD = 4 cm.
4. Join AD, DB, BC and CA. ADBC is the required quadrilateral.

Every one of these four figures is a rectangle, because the diagonals are equal (8 cm each) and bisect each other — Deduction 3 says the angle between them makes no difference.

APML is a square.

The two diameters are the diagonals of the quadrilateral APML, so check them against the square test.

Equal diagonals that bisect each other at right angles ⇒ the figure is a square.

Use the Carpenter’s Problem in reverse: build a rectangle first, and its corners are your right angles.

1. Let AB and CD be the two equal sticks. Find the midpoint of each by folding the thread along a stick and halving it.
2. Cross the sticks so that the two midpoints coincide at a point O. Fix them there.
3. Run the thread around the four ends A, C, B, D and back to A, pulling it taut.

ACBD is a rectangle, since its diagonals AB and CD are equal and bisect each other. So ∠C = ∠A = ∠B = ∠D = 90° — each corner of the thread outline is an exact right angle.

No. “Opposite sides parallel and equal” is a true property of rectangles, but it is not enough to define them.

A definition must let in every rectangle and keep out everything else. This one fails the second half: every parallelogram has opposite sides parallel and equal, and most parallelograms are not rectangles.

No, it is impossible. Try as many constructions as you like — the fourth angle always closes at exactly 90°.

There is no room for any other value, so such a quadrilateral cannot exist. (This also proves statement (ii) of the true/false question on page 109: a quadrilateral with three right angles must be a rectangle.)

Because of a property that holds for every quadrilateral: the sum of all its angles is 360°.

Draw one diagonal. It splits the quadrilateral into two triangles, and each triangle contributes 180°.

Once that total is fixed, three angles of 90° leave exactly 90° for the fourth. Nothing about the construction can change it.

We get 360°, and those six angles regroup exactly into the four angles of the quadrilateral.

Here ∠1 + ∠4 is the full angle of the quadrilateral at S, ∠3 + ∠6 is the full angle at M, while ∠2 and ∠5 are the whole angles at E and O. So:

The sum of all angles in any quadrilateral is 360°.

Yes, plenty. Draw two pairs of parallel lines that do not cross at right angles, and the quadrilateral they cut out has both pairs of opposite sides parallel while none of its angles is 90°.

Such quadrilaterals are called parallelograms.

With a ruler and a set-square.

1. Draw a line l. Slide the set-square along the ruler and draw a second line m parallel to l.
2. Draw a third line p crossing both, at any angle other than 90°.
3. Slide the set-square along p‘s direction and draw a fourth line q parallel to p.
4. The four crossing points are the vertices of a parallelogram.

With a compass and a ruler — copy an angle instead. Draw line l and a transversal through a point P on it. Copy the angle that l makes with the transversal, at a point Q further along it. The new arm is parallel to l, because equal corresponding angles mean parallel lines.

Yes. A rectangle has both pairs of opposite sides parallel (proved on page 90), which is exactly the parallelogram’s definition.

More precisely, a rectangle is a special kind of parallelogram — one with all its angles equal to 90°.

Angles: 30°, 150°, 30°, 150°. Sides: 4 cm, 5 cm, 4 cm, 5 cm.

Construction. Draw AB = 4 cm and AD = 5 cm with ∠A = 30° between them. Through D draw a line parallel to AB, and through B a line parallel to AD; call their meeting point C.

Angles (Deduction 6). AB ∥ CD with AD as transversal, so ∠A and ∠D are interior angles on the same side:

Yes — the opposite angles of a parallelogram are always equal. Using a letter instead of a number proves it for every parallelogram at once.

In parallelogram PEAR (vertices in order P, E, A, R), take ∠P = x.

Check: x + (180 – x) + x + (180 – x) = 360° ✓

Use ∆ABD and ∆CDB — the two halves cut off by the diagonal BD.

So the opposite sides of a parallelogram are equal — Property 1.

Yes, it is wrong. The vertices are not in corresponding order.

The correct statement is ∆ABD ≅ ∆CDB, which matches A↔C, B↔D, D↔B. Writing ∆ABD ≅ ∆CBD would match A↔C, B↔B, D↔D, and so would claim:

Neither of these is true in a general parallelogram — AB and CB are adjacent sides of different lengths, and BD does not bisect the angle at B.

No. In a general parallelogram the diagonals have different lengths.

In the 4 cm by 5 cm parallelogram with ∠A = 30°, the diagonals come out very different — the one joining the two 30° corners (about 8.7 cm) is far longer than the one joining the two 150° corners (about 2.5 cm). Measure both in your own drawing; they are equal only when the angles are 90°.

Yes — the diagonals of a parallelogram always bisect each other, even though they need not be equal.

Deduction 8. In parallelogram EASY the diagonals meet at O. Compare ∆AOE and ∆YOS:

So O is the midpoint of both diagonals — Property 4.

Yes. The correct statement is ∆AOE ≅ ∆YOS.

OA and OS are halves of different diagonals, and there is no reason for them to be equal — in a general parallelogram they are not.

No — the angle between them can be anything, and it changes as the shape of the parallelogram changes.

Being perpendicular is what the equal sides buy you, not what parallelism buys you. That is exactly the question section 4.4 goes on to answer, and Deduction 10 settles it: the diagonals of a rhombus intersect at 90°.

No. A square is only one member of a much larger family — the rhombuses.

Draw AD and AB of the same length with any angle between them other than 90°, say 50°. Complete the figure so that all four sides match, and you get a quadrilateral with four equal sides whose angles are 50°, 130°, 50°, 130°. It is certainly not a square.

Any angle less than 180° can be used in place of 50°, so there are infinitely many rhombuses for each side length; only the one with a 90° angle is a square.

Yes, and a compass finds the fourth vertex in one step.

1. Open the compass to the length AB.
2. Keeping this as radius, draw an arc centred at B and another centred at D.
3. The arcs meet at C. Join BC and DC.

50°, 130°, 50° and 130°.

Deduction 9 first shows that in any rhombus a diagonal splits the two angles it meets into four equal parts. Apply that to ABCD, where ∠A = 50° and the diagonal BD makes four equal angles a.

By the SSS condition, using the diagonal AE as the shared side.

From the congruence, ∠G = ∠M and the angle at A is split the same way as the angle at E. Together with the two isosceles triangles GAE and MAE:

Draw one large oval for parallelograms. Inside it, two overlapping ovals — rectangles and rhombuses. The squares sit exactly in the overlap.

No, not in general. They are equal only when the rhombus happens to be a square.

In the rhombus with angles 50° and 130°, the diagonal joining the two 130° corners is much shorter than the one joining the two 50° corners. Squash a rhombus flatter and one diagonal grows while the other shrinks, though all four sides stay the same length.

Yes — always at 90°.

Why ∆GEO ≅ ∆MEO: the diagonals bisect each other (every rhombus is a parallelogram), so O is the midpoint of GM.

Deduction 10. From the congruence, ∠GOE = ∠MOE. These two angles sit along the straight line GM, so they form a linear pair:

This is Property 6: the diagonals of a rhombus intersect each other at an angle of 90°.

(i) Parallelogram PERA, ∠P = 40°

(ii) Parallelogram PQRS, ∠P = 110°

(iii) Rhombus XWVU, ∠XVU = 30°

(iv) Rhombus OIEA, ∠OEA = 20°

The one diagonal property that defines a parallelogram is: the diagonals bisect each other. So halve each length and build outwards from the crossing point.

1. Draw AB = 7 cm and mark its midpoint O, so OA = OB = 3.5 cm.
2. At O draw a ray making 140° with OB.
3. On that ray mark C with OC = 2.5 cm; on the opposite ray mark D with OD = 2.5 cm. Then CD = 5 cm and O is its midpoint.
4. Join AC, CB, BD and DA.

ACBD is the required parallelogram.

A rhombus needs its diagonals to bisect each other at right angles — and no protractor is needed, because a perpendicular bisector can be drawn with a compass.

1. Draw AB = 5 cm.
2. Construct the perpendicular bisector of AB with a compass; let it cut AB at O, so OA = OB = 2.5 cm.
3. On the perpendicular mark C and D with OC = OD = 2 cm (so CD = 4 cm).
4. Join AD, DB, BC and CA.

ADBC is the required rhombus.

You get a square.

On the geoboard the two bands are stretched between pegs so that they cross at the middle peg — that is, each band is cut in half by the other.

You get a rhombus — no longer a square, but still with four equal sides.

Extending one band by 2 cm at each end keeps its midpoint exactly where it was. So two things are unchanged and one thing changes:

Diagonals that bisect each other at right angles give a rhombus; the “equal” condition, which is what makes a rhombus a square, has been lost.

Yes. Place the two cutouts so that one side of the first lies exactly on one side of the second, with the triangles on opposite sides of that common edge.

Every side is 8 cm, so any side of one triangle fits any side of the other. The common edge disappears into the interior and becomes a diagonal, and the four outer edges — 8 cm each — become the sides of a quadrilateral.

It is a rhombus with side 8 cm and angles 60°, 120°, 60°, 120°.

It is not a square, since 60° ≠ 90°.

A join is possible along any pair of edges of the same length, and for each such pair the second triangle may be flipped over or turned round. That gives three genuinely different quadrilaterals.

The book shows the first two of these.

First figure — joined along the 6 cm edge: a rhombus of side 8 cm.

Second figure — joined along an 8 cm edge, one triangle given a half turn: a parallelogram with sides 8 cm and 6 cm.

Since all three sides are different, a join is only possible along matching edges — 6 with 6, 9 with 9, or 12 with 12. For each of the three edges there are two placements (a half turn or a mirror image), so there are six ways in all.

Two clear families appear.

The three half-turn joins give parallelograms. A half turn about the midpoint of the shared edge reverses all directions, so each remaining side becomes parallel to its partner:

The three mirror joins give kites, because reflection produces two adjacent pairs of equal sides:

Kite ABCD has AB = BC and CD = DA. Let the diagonals meet at O.

Step 1 — the diagonal BD splits the kite into two congruent triangles.

That is part (i): BD bisects both ∠ABC and ∠ADC.

Step 2 — now compare ∆AOB and ∆COB.

So BD bisects AC and is perpendicular to it — part (ii).

Yes. In trapezium PQRS with PQ ∥ SR, each slanting side is a transversal cutting the two parallel sides, so the two angles at its ends are interior angles on the same side.

So measure the two base angles ∠S and ∠R, and then simply subtract:

For example, if you measure ∠S = 70° and ∠R = 55°, then ∠P = 110° and ∠Q = 125°. Check: 70 + 55 + 110 + 125 = 360° ✓

Construction.

1. Draw UV, the longer parallel side.
2. Draw a line parallel to UV at the height you want.
3. Mark X and W on that line so that UX = VW. The easiest way is to keep the figure symmetric about the perpendicular bisector of UV — set X and W the same distance in from each end.
4. Join UX and VW.

Finding the rest without measuring. Measure ∠U once. Then:

XWZY is a rectangle, and that is what produces the congruent triangles.

Drop perpendiculars XY and WZ from X and W onto UV.

Being a rectangle, its opposite sides are equal, so XY = WZ — the two perpendiculars have the same length.

By the RHS condition (right angle–hypotenuse–side), using the rectangle XWZY established above.

This is Property 2: in an isosceles trapezium, the angles opposite to the equal sides are equal.

All four sides are 4 cm; the angles are 60°, 120°, 60° and 120°. The figure is a rhombus.

Join the two cutouts along a full side. That side becomes a diagonal of length 4 cm, and the four remaining edges become the sides.

Use the kite’s diagonal property proved on page 105: one diagonal is the perpendicular bisector of the other.

1. Draw PQ = 6 cm — this will be the diagonal that gets bisected.
2. Construct the perpendicular bisector of PQ with a compass; let it meet PQ at T, so PT = TQ = 3 cm.
3. On the perpendicular, mark R and S on opposite sides of PQ with RS = 8 cm — but not with T as the midpoint. For instance take TR = 2 cm and TS = 6 cm.
4. Join PR, RQ, QS and SP.

PRQS is the required kite.

(i) The top and bottom sides are parallel, so the two angles at the ends of each slanting side add up to 180°.

(ii) The slanting sides are marked equal, so this is an isosceles trapezium; the 100° angle is at one end of the shorter parallel side.

(i) The rhombus (including the square): a kite has two adjacent pairs of equal sides and a parallelogram has equal opposite sides, so all four sides are equal.

(ii) Yes — the square. A square has AB = BC and CD = DA, so it meets the book’s definition of a kite (page 105), and all its angles are 90°, so it is a rectangle.

(iii) No. Every rhombus is a kite, but not every kite is a rhombus.

So the correct relationship is rhombuses ⊂ kites — the set of rhombuses lies wholly inside the set of kites.

∠IOD = 30°.

In the figure, PAIR is a rectangle with R at the bottom-left and I at the bottom-right, O lies on the side AI, and RODS is a second rectangle hanging off RO. The marked 30° is ∠ORI, the angle the segment RO makes with the side RI.

Use the diagonal property of a square: equal diagonals bisecting each other at 90°. A compass gives the right angle, so no protractor is needed.

1. Draw AB = 6 cm.
2. With centre A and any radius more than half of AB, draw arcs above and below AB. Repeat with centre B and the same radius. Join the two crossing points — this line is the perpendicular bisector of AB, and it meets AB at its midpoint O.
3. On this perpendicular mark C and D with OC = OD = 3 cm, one on each side.
4. Join AC, CB, BD and DA.

ACBD is the required square.

UVWX is a square.

Geometric reasoning. The four corner triangles each have two sides equal to half the side of CASE with a right angle between them, so they are congruent (SAS) and UV = UX = XW = WV.

Each corner triangle is right-angled and isosceles, so its base angles are 45°. At U:

By construction. In a 6 cm square, each inner side measures about 4.2 cm and each inner angle 90°.

Other inner squares (Figure b). Going round in one direction, mark a point on each side at the same distance d from a corner and join them; the corner triangles are again congruent, so this is a square for every d.

Yes, it must be a square.

By construction. Draw AB = 5 cm and, at A, a perpendicular AD = 5 cm using compass arcs. With centre B and centre D draw arcs of radius 5 cm; they meet at C. Measure ∠B, ∠C and ∠D — each reads 90°, and BC = CD = 5 cm.

It is a parallelogram.

Let ABCD have AB = CD and BC = AD. Draw the diagonal AC.

Now read those equal angles as alternate angles across the transversal AC:

Both pairs of opposite sides are parallel, so ABCD is a parallelogram by definition.

Yes — 360°, provided the angle at the caved-in vertex D is measured as the reflex angle, the one on the inside of the figure.

The diagonal AC now lies outside the figure, so use the other diagonal, BD, which does lie inside.

By construction. Measure ∠A, ∠B, ∠C and the reflex angle at D (360° – ∠ADC); they add up to 360°.

(i) False. Equal diagonals that bisect each other give a rectangle; a 6 cm × 3 cm rectangle is not a square.

(ii) True. The fourth angle = 360° – 3 × 90° = 90°, so all four angles are right angles.

(iii) True. If OA = OC and OB = OD, then ∆AOB ≅ ∆COD (SAS), so ∠BAO = ∠DCO and AB ∥ CD; similarly AD ∥ BC.

(iv) False. A kite also has perpendicular diagonals, but it need not be a rhombus.

(v) True. If ∠A = ∠C = x and ∠B = ∠D = y, then 2(x + y) = 360°, so x + y = 180°. Adjacent angles are supplementary, so both pairs of opposite sides are parallel.

(vi) True. Each angle = 360° ÷ 4 = 90°, so it is a rectangle.

(vii) False. An isosceles trapezium has only one pair of parallel sides; its equal sides are not parallel.

Bring one edge exactly onto the opposite edge and press the crease flat.

The crease you make is the perpendicular bisector of the two side edges: it passes through their midpoints, and it meets them at 90° because the two halves lie exactly on top of each other.

Fold the half-sheet in half again, this time across the first crease.

You now have a quarter sheet with two creases meeting at right angles, and their crossing point is the centre of the original sheet. Every corner of the quarter sheet is a right angle, and one of those corners is the centre point.

Fold the centre corner over along a straight line joining a point on one edge to a point on the other. The corner turns down as a small triangle.

Because you are folding through four layers of paper at once, this single crease is copied into all four quarters when you open the sheet out.

The four copies of the triangular crease meet up to form a rhombus centred at the middle of the sheet.

If your triangular fold happened to cut off equal lengths on the two edges, the rhombus is a square; otherwise it is a leaning rhombus.

The picture shows a family of nested rhombuses, one inside the other, all sharing the same centre.

Make several triangular folds at the centre corner instead of one — folding the corner over a small amount, opening it, folding it over a larger amount, opening again, and so on.

1. Fold the sheet into a quarter, as before.
2. At the centre corner, fold a small triangle and press. Unfold.
3. Fold a slightly larger triangle at the same corner, parallel to the first crease. Unfold.
4. Repeat with steadily larger triangles.
5. Open out the whole sheet.

Each fold gives one rhombus, and folds made parallel to one another give rhombuses of the same shape sitting neatly inside each other.

Make the triangular fold at the centre corner a 45° fold — that is, fold so that the crease cuts off equal lengths along the two edges meeting at the centre.

The easiest way to be exact: bring one of the two creases exactly onto the other at the centre corner. That fold bisects the right angle, so it makes 45° with each — and it automatically cuts off equal lengths.