NCERT Solutions Class 8 Maths Chapter 4 Quadrilaterals

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 8 · M AT H S

NCERT Solutions

Chapter 4: Quadrilaterals

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

82 – 111 15 86 English

Solutions, notes, sample papers & more at 73 pages

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 4: Quadrilaterals
Chapter 4 of Ganita Prakash Part I studies four-sided figures — rectangles, squares, parallelograms,
rhombuses, kites and trapeziums. The whole chapter is built on one habit: after you notice a property, you
deduce it using congruent triangles, parallel-line angles and the 360° angle sum, instead of trusting
measurement alone.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) 82 – 111

SECTIONS QUESTIONS

15 86

MEDIUM

English

In-text Questions — Page 82

Page 1 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Introduction

Q1 Observe the following figures. Figs. (i), (ii), and (iii) are quadrilaterals, and the
others are not. Why?

(i) (ii) (iii)

(iv) (v)

The five figures (i) to (v) printed on page 82. In (i), (ii) and (iii) the angles between the
sides are marked.

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.
Check each figure against the two conditions a quadrilateral must satisfy:

it is closed — you can trace it and return to the starting point;
it is made of four straight sides that meet only at their endpoints.

Figures (iv) and (v) fail the second condition. Both 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.

Page 2 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Why it matters: the word quadrilateral comes from the Latin quadri (four) and
latus (side). The definition is what lets us talk about “the four angles of the figure” at
all — in (i), (ii) and (iii) each pair of sides that meet at a vertex makes exactly one
interior angle, so the figure has four well-defined angles. A figure whose boundary
contains a curve has no such set of four angles.

In-text Questions — Pages 83–85
4.1 Rectangles and Squares — A Carpenter's Problem

MATH TALK

Q1 Are there other ways to define a rectangle?

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

DEFINITION WHERE IT COMES FROM

All angles are 90° and opposite sides are the familiar one, stated first
equal

The diagonals are equal and bisect each the answer to the Carpenter's Problem
other

All angles are 90° Deduction 4 — the “opposite sides equal” part turns out to follow
on its own

A definition is a test: a figure is a rectangle exactly when it passes the test. Two different tests
can be equally good, provided each one lets in the same figures and keeps out the same figures.

Tip: the third definition is the shortest, so the book finally adopts it — a rectangle is
a quadrilateral in which the angles are all 90°. Everything else about a rectangle is
then a theorem, not part of the definition.

Page 3 of 73

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Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Why it happens: the angle between the two strips is not fixed. Deduction 3 shows
that any angle at all gives a rectangle, as long as the strips are equal and cross at
their midpoints. So the carpenter has one free choice — she can swing the strips to
make the rectangle long and thin or nearly square — and the figure stays a
rectangle.

Did you know? This is a real workshop method. Carpenters in Europe use it to
square up a frame, and farmers in Mozambique use it to set out a rectangular base
for a house.

Q3 What is the length of the other diagonal?

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:

AB = CD (opposite sides of a rectangle)

∠BAD = ∠CDA = 90° (angles of a rectangle)

AD = DA (common side)

So ∆ADC ≅ ∆DAB by the SAS condition

Hence AC = BD (corresponding parts of congruent triangles)

Why it happens: each diagonal is the hypotenuse of a right triangle whose legs are
a length and a breadth of the rectangle. Both diagonals sit on right triangles with
the same two legs, so the two hypotenuses must match. The congruence argument
says exactly this, without needing to compute any length.

Check it yourself: draw three very different rectangles and measure both diagonals
of each. They agree every time — but measuring three cases is not a proof, which is
why Deduction 1 is written out.

Page 5 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q4 What is the point of intersection of the two diagonals?

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.

AB = CD (opposite sides)
∠AOB = ∠COD (vertically opposite angles)

Now for ∠1 = ∠OAB and ∠2 = ∠OCD, write ∠3 = ∠DBC:

∠B = 90°, so ∠3 + ∠1 = 90°

In ∆BCD: ∠3 + ∠2 + 90° = 180°, so ∠3 + ∠2 = 90°

Hence ∠1 = ∠2 = 90° – ∠3

So ∆AOB ≅ ∆COD by the AAS condition

Therefore OA = OC and OB = OD

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

Why it happens: ∠1 and ∠2 are both the “left-over” part of a right angle after
taking away the same angle ∠3 — one inside the corner at B, the other inside
triangle BCD. Two angles that are each 90° – ∠3 have to be equal. That single
equality is what unlocks the congruence.

Tip: to bisect something is to cut it into two equal parts. Saying “the diagonals bisect
each other” is shorter than saying “each diagonal passes through the midpoint of
the other”, and means the same thing.

Q5 What should the angle be between the diagonals?

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.

Page 6 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

x = 90° → all four sides equal → a square

x close to 0° or 180° → a long, thin rectangle

Why it happens: the four endpoints all lie at the same distance from the crossing
point O (half of the common diagonal length), so they lie on a circle centred at O.
Turning one strip about O just slides two of the points around that circle — the four
points stay on the circle, and the figure stays a rectangle.

Q6 Can the following equalities be used to establish that ∆AOD ≅ ∆COB? AO = CO
(proved above); ∠AOB = ∠COD (vertically opposite angles); AD = CB

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:

AO = CO, ∠AOD = ∠COB (vertically opposite), OD = OB

⇒ ∆AOD ≅ ∆COB by SAS

Why it matters: a congruence condition is a rule about which parts match, not just
about how many equalities you can list. SAS needs the angle to sit between the two
sides. Listing an angle from somewhere else in the figure proves nothing about
these two triangles.

Q7 Let us check what quadrilateral we get if we draw the two diagonals such that their
lengths are equal, they bisect each other and have an arbitrary angle, say 60°,
between them. Can you find all the remaining angles?

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

Page 7 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Given ∠AOB = 60°

∠COD = 60° (vertically opposite)

∠AOD = 180° – 60° = 120° (linear pair)
∠BOC = 120° (vertically opposite)

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

TRIANGLE APEX ANGLE AT O BASE ANGLES

∆AOB 60° 60°, 60°

∆COD 60° 60°, 60°

∆AOD 120° 30°, 30°

∆BOC 120° 30°, 30°

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.

Why it happens: OA = OB makes ∆AOB isosceles, so its base angles are equal — that
is the only reason the numbers come out. Each vertex angle of ABCD is one base
angle from a 60° triangle plus one base angle from a 120° triangle, and 60 + 30 = 90
every time.

Q8 In ∆AOB, since OA = OB, the angles opposite them are equal, say a. Can you find the
value of a?

a = 60°.

In ∆AOB: a + a + 60 = 180 (interior angles of a triangle)

2a = 180 – 60 = 120

a = 60

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

Page 8 of 73

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Class 8 Maths Chapter 4 Quadrilaterals
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In-text Questions — Pages 86–87
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Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Why it happens: half of one diagonal, half of the other, and the angle between them
completely fix a triangle. The two triangles on opposite sides of O are built from
exactly the same three pieces, so the third sides — which are the opposite sides of
ABCD — must match.

With all angles 90° and opposite sides equal, ABCD satisfies the book's first definition of a
rectangle.

Q3 Will ABCD remain a rectangle if the angles between the diagonals are changed? Can
we generalise this?

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.

In isosceles ∆AOB (OA = OB), base angles a:

a + a + x = 180 ⇒ 2a = 180 – x ⇒ a = 90 – x⁄2

In isosceles ∆AOD (OA = OD), base angles b:

b + b + (180 – x) = 180 ⇒ 2b = x ⇒ b = x⁄2

Each angle of ABCD = a + b = (90 – x⁄2) + x⁄2 = 90

The x cancels. That is the whole point: the answer does not depend on x, so no matter what
the angle between the diagonals is, the quadrilateral is a rectangle.

Why it happens: at each vertex, one half-angle grows exactly as fast as the other
shrinks. Open the diagonals by a little and a falls by half that amount while b rises by
the same half. Their sum can never move away from 90°.

Check it yourself: put x = 60 into the formulas: a = 90 – 30 = 60 and b = 30, matching
the worked case on page 85 exactly.

Page 10 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q4 Since we know that ∆AOB is isosceles, we can denote the measures of both of its
base angles by a. What is the value of a (in degrees) in terms of x?

a = 90 – x⁄2 degrees.

a + a + x = 180 (sum of the interior angles of a triangle)
2a = 180 – x

a = (180 – x)⁄2 = 90 – x⁄2

X A = 90 – X/2 B = X/2 A+B

40° 70° 20° 90°

60° 60° 30° 90°

90° 45° 45° 90°

140° 20° 70° 90°

Why algebra is used here: a numerical case such as x = 60° checks one rectangle.
Writing the answer in terms of x checks all of them at once — the letter stands for
every possible angle the two strips could make.

Q5 What can we say about AB and CD, and AD and BC?

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

∆AOB ≅ ∆COD ⇒ AB = CD

∆AOD ≅ ∆COB ⇒ AD = CB

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.

Page 11 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Tip: this is what makes the second definition legitimate — a rectangle is a
quadrilateral whose diagonals are equal and bisect each other. Deductions 1
and 2 prove every rectangle passes this test; Deduction 3 proves everything that
passes it is a rectangle.

In-text Questions — Pages 89–90
4.1 Rectangles and Squares — Deduction 4

Q1 In the earlier definition, we stated that a rectangle has (a) opposite sides of equal
length, and (b) all angles equal to 90°. Would we be wrong if we just define a
rectangle as a quadrilateral in which all the angles are 90°?

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.

Why this is worth doing: a good definition should carry no dead weight. If part of a
definition can be proved from the rest, it belongs among the properties, not in the
definition. Keeping the definition minimal also makes it easier to test a figure — you
only have to check four angles.

Tip: three right angles are already enough, since the angle sum of a quadrilateral is
360° and 360 – 270 = 90. The book uses all four in the definition because it is simpler
to state.

Q2 If you think that this definition is incomplete, try constructing a quadrilateral in
which the angles are all 90° but the opposite sides are not equal. Are you able to
construct such a quadrilateral?

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

Page 12 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

CD = AB and AD = BC

The moment you make CD longer or shorter than AB, the fourth corner refuses to be a right
angle. You have only three free choices (two lengths and the starting direction); the fourth side
is decided for you.

Why it happens: after three 90° turns you are facing back along your original
direction. The figure can only close if the two sides you drew in that direction are
equal, and likewise for the other pair. Deduction 4 turns this observation into a proof
using congruent triangles.

Q3 Join BD. ∆BAD and ∆DCB seem congruent. Can we justify this claim? Two equalities
can be directly seen in the triangles. What can we say about ∠1 and ∠2?

∠1 = ∠2, and that is exactly the missing piece — the same argument as in Deduction 2.
Let ABCD have all four angles 90°, and join BD. Write ∠1 = ∠ADB, ∠2 = ∠CBD and ∠3 = ∠DBA.

Since ∠B = 90°: ∠3 + ∠2 = 90°

In ∆ABD: ∠3 + ∠1 + 90° = 180°, so ∠3 + ∠1 = 90°

Hence ∠1 = ∠2 = 90° – ∠3
Now compare ∆BAD and ∆DCB:

∠A = ∠C = 90°

∠1 = ∠2 (just proved)

BD = DB (common side)

⇒ ∆BAD ≅ ∆DCB by the AAS condition

⇒ AD = CB and DC = BA

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

Page 13 of 73

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Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

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Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

∠A + ∠B = 90° + 90° = 180°

∠A and ∠B are interior angles on the same side of the transversal AB

When such a pair adds to 180°, the two lines are parallel

⇒ AD ∥ BC

Why it happens: if AD and BC were not parallel they would meet somewhere,
forming a triangle with AB. That triangle would already contain 90° + 90° = 180° at
two of its vertices, leaving nothing for the third — impossible. So they never meet.

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

Q6 Can you similarly show that AB is parallel to DC (AB||DC)?

Yes — repeat the argument with a different transversal.

Take AD as the transversal cutting AB and DC

∠A + ∠D = 90° + 90° = 180°

These are interior angles on the same side of AD

⇒ AB ∥ DC

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°.

Tip: the same transversal argument works at any of the four sides, because every
angle of a rectangle is 90°. You may pick whichever pair of angles is most
convenient.

In-text Questions — Pages 90–93

Page 15 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

4.1 A Special Rectangle — the Square

Q1 In the quadrilaterals below, are there any non-rectangles?

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

FIGURE SIDES ALL ANGLES 90°? RECTANGLE?

(i) 5 cm, 2 cm, 5 cm, 2 cm Yes Yes

(ii) 3.6 cm, 6 cm, 3.6 cm, 6 cm Yes Yes

(iii) 5 cm, 1 cm, 5 cm, 1 cm Yes Yes

(iv) 4 cm, 4 cm, 4 cm, 4 cm Yes Yes — and also a square

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 (ii) and (iii) are) changes nothing —
angles do not depend on how a figure is turned.

Why (iv) counts: a square passes the rectangle test — all its angles are 90°. It is a
special rectangle, one whose sides happen to be all equal. So every square is a
rectangle, but not every rectangle is a square, exactly as a Malayali is an Indian
while not every Indian is a Malayali.

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

Q2 Let us consider the Carpenter's Problem again. If the wooden strips have to be
placed such that the thread passing through their endpoints forms a square, what
must be done?

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

Page 16 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Equal + bisecting → all angles 90° (a rectangle)

+ crossing at 90° → all four sides equal as well

⇒ a square

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.

Try This: to build a square of diagonal 8 cm, draw an 8 cm segment, mark its
midpoint O, draw the perpendicular to it at O, and cut off 4 cm on each side of O.
Join the four endpoints in order.

Q3 What more needs to be done to get equal sidelengths as well? Can this be achieved
by properly choosing the angle between the diagonals?

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:

x = 180 – x ⇒ 2x = 180 ⇒ x = 90°

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

Why it happens: the four triangles AOB, BOC, COD, DOA all have two sides equal to
half a diagonal. They differ only in the angle between those two sides. Make that
angle the same in all four and the triangles become congruent by SAS — so their
third sides, which are the four sides of the quadrilateral, become equal too.

Q4 To find the angle formed by the diagonals, what are the two triangles we should
consider for congruence? Can this be used to find the angles ∠BOA and ∠BOC
formed by the diagonals?

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

Page 17 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

BA = BC (sides of the square)

OA = OC (the diagonals bisect each other)

BO = BO (common side)
⇒ ∆BOA ≅ ∆BOC by the SSS condition

So ∠BOA = ∠BOC (corresponding parts)

But ∠BOA + ∠BOC = 180° (they form a straight angle along AC)

⇒ ∠BOA = ∠BOC = 90°

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

Why it happens: B is the same distance from A and from C, so B lies on the
perpendicular bisector of AC. O is the midpoint of AC. So BO is the perpendicular
bisector of AC — which is another way of saying the diagonals meet at 90°.

Tip: the same argument works in any rhombus, which is why Property 6 on page 102
says the diagonals of a rhombus meet at 90°. A square is just a rhombus that is also
a rectangle.

Q5 Using this fact, construct a square with a diagonal of length 8 cm.

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.

Page 18 of 73

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Class 8 Maths Chapter 4 Quadrilaterals
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Diagonals: AC = BD = 8 cm (equal)
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They bisect each other at O and meet at 90°
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⇒ ABCD is a square with each side = √(4² + 4²) = 4√2 ≈ 5.66 cm
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WHAT IT USED DOES A SQUARE HAVE
IT?

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1 — diagonals equal AB = CD, ∠BAD = ∠CDA = 90°, AD common Yes

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— diagonals bisect each AB = CD, vertically opposite angles, ∠1 = ∠2 Yes

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

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ag ∠1, ∠2,
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Q7 There is one more special property of a square. What are the measures of
∠3, and ∠4? See if you can reason and/or experiment to figure this out! ... Similarly,
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All four are 45°.

co m
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m as e
.co


a g l Page 19 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

In ∆ADC: ∠1 + ∠3 + 90 = 180 ⇒ ∠1 + ∠3 = 90

AD = DC (sides of a square) ⇒ ∠1 = ∠3

So 2∠1 = 90 ⇒ ∠1 = ∠3 = 45°

In ∆ABC: AB = BC and ∠B = 90°

⇒ ∠2 = ∠4 = 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.

Why it happens: a diagonal of a square splits it into two isosceles right triangles. In
an isosceles triangle the base angles are equal, and here they must share the 90° left
over from the right angle at the vertex — so each gets exactly half.

Tip: this is the same 45° that appeared in Deduction 3 when x = 90: a = 90 – 45 = 45
and b = 45. The two routes agree, which is a good sign that both are right.

Figure it Out — Page 94
4.1 Rectangles and Squares

MATH TALK TRY THIS

Q1 Find all the other angles inside the following rectangles.

Figure (i) — rectangle ABCD with A bottom-left, B bottom-right, C top-right, D top-left, both
diagonals drawn, and ∠CAB = 30°.
Because the diagonals of a rectangle are equal and bisect each other, OA = OB = OC = OD, so
each of the four triangles around O is isosceles.

Page 20 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

∠CAB = 30° (given) ⇒ ∠DAC = 90 – 30 = 60°

∆AOB is isosceles ⇒ ∠ABD = ∠BAC = 30° ⇒ ∠DBC = 60°

In ∆ABC: ∠ACB = 180 – 90 – 30 = 60° ⇒ ∠ACD = 30°

In ∆ABD: ∠ADB = 180 – 90 – 30 = 60° ⇒ ∠BDC = 30°

At O: ∠AOB = 180 – 30 – 30 = 120° = ∠COD

∠AOD = ∠BOC = 180 – 120 = 60°

D C
30° 30°
60° 60°

120°
O
60° 60°

120°
60° 60°

30° 30°
A B

Figure (i): every angle inside the rectangle, starting from the given ∠CAB = 30°.

Figure (ii) — rectangle PQRS with P bottom-left, Q top-left, R top-right, S bottom-right, both
diagonals drawn, and ∠QOR = 110° at the crossing point O.

Page 21 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

∠POS = 110° (vertically opposite to ∠QOR)

∠QOP = ∠ROS = 180 – 110 = 70° (linear pairs)

∆QOR isosceles (OQ = OR): ∠OQR = ∠ORQ = (180 – 110) ÷ 2 = 35°

∆POS isosceles (OP = OS): ∠OPS = ∠OSP = 35°

∆POQ isosceles (OP = OQ): ∠OPQ = ∠OQP = (180 – 70) ÷ 2 = 55°

∆ROS isosceles (OR = OS): ∠ORS = ∠OSR = 55°

Check at each vertex: 35° + 55° = 90° ✓

Why the isosceles triangles appear: the diagonals of a rectangle are equal and
bisect each other, so all four half-diagonals OP, OQ, OR, OS have the same length.
Every triangle formed at O therefore has two equal sides, and its base angles must
be equal. That is what turns one given angle into all twelve.

Q2 Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each
other, and intersect at an angle of (i) 30° (ii) 40° (iii) 90° (iv) 140°

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.

Page 22 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

ANGLE AT HALF-ANGLES AT EACH VERTEX (90 – X/2, FIGURE OBTAINED
O X/2)

(i) 30° 75° and 15° a long, thin rectangle

(ii) 40° 70° and 20° rectangle

(iii) 90° 45° and 45° a square

(iv) 140° 20° and 70° rectangle (same shape as (ii),
turned)

Why (ii) and (iv) give the same shape: 40° and 140° are supplementary, and the
four angles at O are always x, x, 180 – x, 180 – x. Choosing 140° simply swaps which
pair of angles is the smaller one, so you get the same rectangle standing the other
way round.

Check it yourself: measure the four sides in case (iii). They should all come to about
5.7 cm, since each side is √(4² + 4²) = 4√2 cm.

Q3 Consider a circle with centre O. Line segments PL and AM are two perpendicular
diameters of the circle. What is the figure APML? Reason and/or experiment to
figure this out.

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

PL = AM — every diameter of a circle has the same length ⇒ diagonals equal

Both pass through the centre O, and O is the midpoint of each ⇒ diagonals bisect each

other

PL ⊥ AM (given) ⇒ diagonals meet at 90°

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

Page 23 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: all four vertices lie on the circle, so each is at distance r from O. The

m l a se
four triangles round O are therefore congruent right isosceles triangles with legs r,
o
so all four.csides of APML equal r√2 and all four angles equal 45°a+g45° = 90°.
a s em
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aTry This: draw a circle of radius 3 cm and two perpendicular diameters. Measure a

co m
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side of APML — it should be about 4.2 cm, since 3√2 ≈ 4.24.
m .
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We have seen how to get 90° using paper folding. Now, suppose we do not have any
m
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paper but two sticks of equal length, and a thread. How do we make an exact 90°
using these?
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se m
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a Use the Carpenter's Problem in reverse: build a rectangle first, and its corners are your right
s
angles.
m a
m .co agl
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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.
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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.

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∠C = ∠A m .
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ACBD is a rectangle, since its diagonals AB and CD are equal and bisect each other. So
= ∠B = ∠m
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D = 90° — each corner of the thread outline is an exact rightaangle.
. c a g
s e m
a gla Why it is exact and not approximate: nothing here depends on judging an angle
m
by eye. The two conditions you can control physically — sticks of equal length,

a se
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crossing at their midpoints — are precisely the conditions Deduction 3 proves are
enough to force all four angles to.90°. a g
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Another way: use the isosceles triangle instead. Tie the thread to make a triangle

c o m
with two equal sides on the sticks; the line from the apex to the midpoint of the base
.
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is perpendicular to the base, because it is the perpendicular bisector.
m a
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m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 24 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q5 We saw that one of the properties of a rectangle is that its opposite sides are
parallel. Can this be chosen as a definition of a rectangle? In other words, is every
quadrilateral that has opposite sides parallel and equal, a rectangle?

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.

Parallelogram with ∠A = 30°, sides 4 cm and 5 cm

Opposite sides parallel ✓ opposite sides equal ✓

But ∠A = 30° ≠ 90° ⇒ not a rectangle

Why the angles must be mentioned: parallelism fixes the directions of the sides but
not the angle between the two directions. Push the top of a rectangle sideways and
the sides stay parallel and equal while every angle changes — you get a “leaning”
parallelogram. Only the condition “all angles are 90°” stops this from happening.

Tip: compare with the definitions that do work: “all angles 90°”, or “diagonals equal
and bisect each other”. In a general parallelogram the diagonals bisect each other
but are not equal — which is exactly what the failed definition leaves out.

In-text Questions — Pages 94–95
4.2 Angles in a Quadrilateral

Q1 Is it possible to construct a quadrilateral with three angles equal to 90° and the
fourth angle not equal to 90°?

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

Page 25 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Angle sum of a quadrilateral = 360°

Three right angles use up 90 × 3 = 270°

Fourth angle = 360 – 270 = 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.)

Tip: this is why the book's shortest definition of a rectangle could have said “three
angles are 90°” instead of four. Both tests admit exactly the same figures.

Q2 But why not?

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°.

180° + 180° = 360°

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

Why this is the real reason: failing at ten attempts only tells you the task is hard.
The angle-sum property tells you it is impossible — no attempt, however careful, can
ever succeed. That is the difference between a conjecture from experiment and a
proof.

Q3 Consider a quadrilateral SOME. Draw a diagonal SM. We get two triangles ∆SEM and
∆SOM. What do we get when we add all six angles?

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

Page 26 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

In ∆SEM: ∠1 + ∠2 + ∠3 = 180°

In ∆SOM: ∠4 + ∠5 + ∠6 = 180°

Adding: ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° + 180° = 360°

Regroup: (∠1 + ∠4) + (∠3 + ∠6) + ∠2 + ∠5 = 360°

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°.

E

2 M
3
6

1
4
S 5

O
diagonal SM splits SOME into two
triangles

One diagonal turns any quadrilateral into two triangles, so its angle sum is 180° + 180° = 360°.

Why the diagonal has to be inside: for this regrouping to work, the diagonal must
lie inside the figure so that ∠1 and ∠4 really are the two parts of the angle at S. In a
concave quadrilateral one of the two diagonals falls outside — but the other one still
lies inside, so the result survives (see Question 10 of the last Figure it Out).

In-text Questions — Pages 95–99

Page 27 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

4.3 More Quadrilaterals with Parallel Opposite Sides

Q1 Rectangles (and therefore squares) have parallel opposite sides. Are there
quadrilaterals that have parallel opposite sides that are not rectangles?

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.

Parallelogram: a quadrilateral in which opposite sides are parallel.

Why rectangles do not exhaust the list: being parallel is about direction only. Two
families of parallel lines can meet at any angle at all; 90° is just one choice out of
infinitely many. Every other choice gives a parallelogram that is not a rectangle.

Tip: so the set of parallelograms is larger than the set of rectangles. In the Venn
diagram, the rectangle oval sits entirely inside the parallelogram oval.

Q2 Construct such a figure by recalling how parallel lines can be constructed using a
ruler and a set-square, or a compass and a ruler.

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.

Page 28 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

co m
m.
Why the copied-angle method works: it is the converse of the corresponding-

m l a se
angles property. If a transversal makes equal corresponding angles with two lines,
o
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those lines.c can never meet — so they are parallel. The set-squareagmethod is the
l a seidea in one motion: sliding keeps the angle with the ruler constant.
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same

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Q3 Is a rectangle a parallelogram?

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

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the parallelogram's definition.
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a g with all its angles equal
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to 90°.
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STATEMENT TRUE?
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Every rectangle is a parallelogram Yes

Every parallelogram is a rectangle
agl No

Every square is a parallelogram Yes
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Tip: in the Venn diagram this becomes three nested regions — square inside

a s erectangle
l
inside parallelogram. Reading “inside” as “every ... is a ...” keeps the whole
a g family straight.
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as adjacent sides of lengths 4 cm and 5 cm, and an angle of
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Draw a parallelogram with
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the lengths of the remaining sides?

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m . 30°, 150°, 30°, 150°. Sides: 4 cm, 5 cm, 4 cm, 5 cm. a
a s e Angles:

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line parallel to AB, and through B a line parallel to AD; call their meeting point C.

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

m .
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same side:

g l as
a

co m
m .
m ase
.co


a g l Page 29 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

∠A + ∠D = 180° ⇒ ∠D = 180 – 30 = 150°

∠A + ∠B = 180° ⇒ ∠B = 150°

∠C + ∠D = 180° ⇒ ∠C = 30°
Check: 30 + 150 + 30 + 150 = 360° ✓

Sides (Deduction 7). Compare ∆ABD and ∆CDB: the angles marked with one arc are equal
(opposite angles of the parallelogram), the angles marked with two arcs are equal (alternate
angles, since AD ∥ BC with BD as transversal), and BD is common.

∆ABD ≅ ∆CDB (AAS)

⇒ AD = CB = 5 cm and AB = CD = 4 cm

Why adjacent angles must add to 180°: the two sides you started from are cut by a
transversal that joins two parallel sides. Interior angles on the same side of a
transversal always add to a straight angle — so knowing one angle of a
parallelogram tells you all four.

Q5 What about the opposite angles? Will they be equal in all parallelograms? If yes,
how can we be sure? Let us take one of the angles to be x. What are the other
angles?

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.

∠P + ∠R = 180° ⇒ ∠R = 180 – x

∠A + ∠R = 180° ⇒ ∠A = 180 – (180 – x) = 180 – 180 + x = x

So ∠P = ∠A = x

Similarly ∠R = ∠E = 180 – x

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

Page 30 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Why the two supplements cancel: ∠A is supplementary to ∠R, and ∠R is
supplementary to ∠P. Taking the supplement twice brings you back where you
started, so ∠A must equal ∠P. The x's cancel in the algebra for exactly that reason.

Tip: this gives Property 3 — in a parallelogram, adjacent angles add up to 180° and
opposite angles are equal. One measured angle is therefore enough to fill in all four.

Q6 Deduction 7 — What can we say about the sides of a parallelogram? Can we again
use congruence to show this? Which two triangles can be considered for this?

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

∠A = ∠C (opposite angles of a parallelogram, from Deduction 6)

∠ADB = ∠CBD (alternate angles, since AD ∥ BC and BD is a transversal)

BD = DB (common side)

⇒ ∆ABD ≅ ∆CDB by the AAS condition

⇒ AD = CB and AB = CD

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

Why a diagonal is the natural tool: the sides you want to compare are not in the
same triangle until you draw one. The diagonal creates two triangles that between
them contain all four sides, and the parallel lines immediately hand you a pair of
equal alternate angles.

Tip: notice the logical order — Deduction 6 (angles) is used inside Deduction 7
(sides). Once a property is proved, it becomes a tool for the next proof.

Q7 Is it wrong to write ∆ABD ≅ ∆CBD? Why?

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

Page 31 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

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:

AB = CB and ∠ABD = ∠CBD

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.

Why order is not a formality: the whole value of a congruence is the list of equal
parts you can read off it (CPCT). If the letters are in the wrong order, that list is a list
of false statements, and any later step built on it collapses.

Q8 Are the diagonals of a parallelogram always equal? Check with the parallelogram
that you have constructed.

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 across the 150° corners is far longer than the one across the 30° corners. Measure both in
your own drawing and you will find they never agree unless the angles are 90°.

PARALLELOGRAM DIAGONALS EQUAL?

General parallelogram No

Rhombus No (unless it is a square)

Rectangle Yes

Square Yes

Why equal diagonals mean a rectangle: Deduction 3 showed that equal diagonals
which bisect each other force all four angles to be 90°. Since the diagonals of every
parallelogram bisect each other, adding “equal” is enough to turn a parallelogram
into a rectangle. So the diagonals of a parallelogram are equal precisely when it is a
rectangle.

Page 32 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q9 Do they bisect each other (do they intersect at their midpoints)? Reason and/or
experiment to figure this out.

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:

AE = YS (opposite sides of the parallelogram)

∠AEO = ∠YSO (alternate angles, AE ∥ YS)

∠EAO = ∠SYO (alternate angles)

⇒ ∆AOE ≅ ∆YOS by the ASA condition
⇒ OA = OY and OE = OS

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

Why this works without equal diagonals: the congruence uses only one pair of
equal sides and the alternate angles that parallelism guarantees. It says the two
halves of the figure across O are copies of each other, so each diagonal is cut into
two equal pieces — but it says nothing about comparing one diagonal with the
other.

Q10 Is it wrong to write ∆AOE ≅ ∆SOY? Why?

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

CORRECT: ∆AOE ≅ ∆YOS WRONG: ∆AOE ≅ ∆SOY
A ↔ Y, O ↔ O, E ↔ S A ↔ S, O ↔ O, E ↔ Y

gives OA = OY, OE = OS ✓ would give OA = OS, OE = OY ✗

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.

Page 33 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

co m
m.
Tip: to get the order right, look at which vertex of one triangle plays the same role as

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com with Y. l
which vertex of the other. A is opposite the side EO; the vertex opposite SO is Y, so A
. a g
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must be paired
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Do the diagonals of a parallelogram intersect at a particular angle?
.
Q11

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

co m
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parallelogram changes.

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QUADRILATERAL ANGLE BETWEEN THE DIAGONALS
a g l
se m
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General parallelogram any value; not fixed

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Rectangle any value (still not fixed)

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Rhombus always 90°
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Square a

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Being perpendicular is what the equal sides buy you, not what parallelism buys you. That is

s e m it: the diagonals of
exactly the question section 4.4 goes on to answer, and Deduction 10 settles

. com intersect at 90°.
a rhombus
a gla
a s em
agl Why sides control this angle: if all four sides are equal, each vertex is the same
distance from the two vertices next to it, so every vertex lies on the perpendicular
se m
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bisector of a diagonal. That forces the diagonals to be perpendicular. Unequal sides
m
ase
put the vertices off those bisectors, and the angle is free again.

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In-text Questions — Pages 99–102
m .
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4.4 Quadrilaterals with Equal Sidelengths
a g l
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agl Are squares the only quadrilaterals that have equal sidelengths? Let us explore this
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question through construction.

m a s
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as

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

com
m .
m ase
.co


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Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

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.

Rhombus: a quadrilateral in which all the sides have the same length.

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.

Why four equal sides do not fix the shape: a four-bar linkage with equal rods is
not rigid — you can push it over into a leaning shape without changing any side
length. A triangle made of three rods cannot be deformed this way, which is why SSS
is a congruence condition for triangles but there is no “SSSS” for quadrilaterals.

Page 35 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q2 Can we complete this quadrilateral so that all its sides are of the same length? Mark
a point C whose distance from B and D is equal to AB (or AD).

D

50°
A B

Page 99 — the two equal sides AD and AB drawn with an angle of 50° between them. The
tick on AD marks the point D at the same distance from A as B.

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.

Page 36 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

AB = AD (drawn equal)

BC = AB (radius of the arc from B)

DC = AB (radius of the arc from D)

⇒ AB = BC = CD = DA — a rhombus

Why the arcs must meet: every point on the arc centred at B is at distance AB from
B, and every point on the arc centred at D is at distance AB from D. Their crossing
point is the one place that is the correct distance from both — so it is the only
possible position for C on that side of BD.

Tip: the same construction with a starting angle of 90° gives a square, so the square
really is just the rhombus you get for one particular choice of angle.

Q3 What are the other angles of the rhombus ABCD that we have constructed? Reason
and/or experiment to figure this out.

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.

In ∆ADB: a + a + 50 = 180

2a = 130 ⇒ a = 65°

∠B = ∠D = a + a = 65 + 65 = 130°

∠C = ∠A = 50°

Check: 50 + 130 + 50 + 130 = 360° ✓

There is a second, quicker route. Every rhombus is a parallelogram, so adjacent angles are
supplementary:

∠A = ∠C = 50° and ∠D = ∠B = 180 – 50 = 130°

Page 37 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Why both routes agree: they must — a rhombus is a parallelogram (its opposite
sides are parallel, as the equal alternate angles in Deduction 9 show), so the
parallelogram rules apply. Meeting the same answer by two independent arguments
is a good check on both.

Q4 It can be seen that ∆GAE ≅ ∆MAE (How?)

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

GA = MA (all sides of rhombus GAME are equal)

GE = ME (same reason)

AE = AE (common side)

⇒ ∆GAE ≅ ∆MAE by SSS

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:

GE = GA ⇒ a = d ME = MA ⇒ b = c

congruence ⇒ a = b and c = d

⇒a=b=c=d

Why this gives Property 5: a = b says the diagonal AE cuts the angle at A into two
equal halves, and c = d says the same at E. That is exactly what “the diagonals of a
rhombus bisect its angles” means. It also gives EM ∥ GA and GE ∥ AM, since the equal
parts are alternate angles — so every rhombus is a parallelogram.

Page 38 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

co m
m.
So a rhombus is a parallelogram, and a rectangle is also a parallelogram. How can
e
Q5

com
this be represented using a Venn diagram? Where will the set of squares occur in
g l as
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this diagram?

a s
a gl

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

as e
a g l
Parallelograms
co m
e m.
m l as
m .co a g
l a se
a g
Rectangles
m Rhombuses a s
m .co agl
l a se
a g Squares

co m
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m as e
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se m
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A square is exactly a quadrilateral that is both a rectangle and a rhombus.

a se
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m
ase
REGION WHAT LIVES THERE

Rectangle only agl all angles 90°, sides not all equal

co m
.
Rhombus only all sides equal, angles not 90°

e m
m l as
.co a g
Overlap squares — all angles 90° and all sides equal

a s em Outside both, inside the big oval
agl
a plain leaning parallelogram

.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 39 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Why the squares fall in the overlap: a square is a rectangle (all angles 90°) and it is
also a rhombus (all sides equal). Conversely, anything in the overlap has all angles
90° and all sides equal, which is precisely the definition of a square. So the overlap is
not merely where squares happen to be — it is the set of squares.

Q6 Are the diagonals of a rhombus equal?

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.

Diagonals equal + bisect each other ⇒ all angles 90° (Deduction 3)

A rhombus with all angles 90° = a square

Why the sides cannot control the diagonals: a rhombus is a flexible linkage. The
four rods fix the sides but leave one degree of freedom — the angle — and the two
diagonals move in opposite directions as that angle changes. Only at 90° do they
cross.

Tip: the properties a rhombus does have are Property 4 (diagonals bisect each
other), Property 5 (they bisect the angles) and Property 6 (they meet at 90°). “Equal
length” is not on the list.

Q7 Do the diagonals of a rhombus intersect at any particular angle? In the rhombus
GAME, we have ∆GEO ≅ ∆MEO (why?)

Yes — always at 90°.
Why ∆GEO ≅ ∆MEO: the diagonals bisect each other (every rhombus is a parallelogram), so O is
the midpoint of GM.

Page 40 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

GE = ME (sides of the rhombus)

GO = MO (O is the midpoint of GM)

EO = EO (common side)

⇒ ∆GEO ≅ ∆MEO by SSS

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

∠GOE + ∠MOE = 180°

2 × ∠GOE = 180° ⇒ ∠GOE = 90°

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

Why it happens: E is equidistant from G and M, and so is O. Two points that are
both equidistant from G and M determine the perpendicular bisector of GM — and
EO is that line. Perpendicular bisector means exactly “through the midpoint, at right
angles”.

Figure it Out — Page 102
4.4 Quadrilaterals with Equal Sidelengths

Q1 Find the remaining angles in the following quadrilaterals.

(i) Parallelogram PERA — vertices P (bottom-left), E (bottom-right), A (top-right), R (top-left); PE
∥ RA and PR ∥ EA; ∠P = 40°.

∠A = ∠P = 40° (opposite angles of a parallelogram)

∠E = 180 – 40 = 140° (adjacent angles are supplementary)

∠R = ∠E = 140°

Check: 40 + 140 + 40 + 140 = 360° ✓

(ii) Parallelogram PQRS — P (bottom-left), Q (bottom-right), R (top-right), S (top-left); SR ∥ PQ
and SP ∥ RQ; ∠P = 110°.

Page 41 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

∠R = ∠P = 110° (opposite angles)

∠Q = 180 – 110 = 70° and ∠S = 70°

Check: 110 + 70 + 110 + 70 = 360° ✓

(iii) Rhombus XWVU — all four sides carry a tick, the diagonal XV is drawn, and ∠XVW = 30°.

A diagonal of a rhombus bisects its angles, so

∠XVU = ∠XVW = 30° ⇒ ∠UVW = 30 + 30 = 60°

∠X = ∠UVW = 60° (opposite angles), split by XV into 30° + 30°

∠U = 180 – 60 = 120° and ∠W = ∠U = 120°

Check: 60 + 120 + 60 + 120 = 360° ✓

(iv) Rhombus OIEA — all four sides ticked, the diagonal OE is drawn, and ∠OEI = 20°.

∠OEA = ∠OEI = 20° (the diagonal bisects ∠E)

∠AOE = ∠EOI = 20° (the diagonal bisects ∠O too, and ∠O = ∠E)

∠E = ∠AEI = 20 + 20 = 40° and ∠O = 40°

∠A = ∠I = 180 – 40 = 140°

Check: 40 + 140 + 40 + 140 = 360° ✓

Why one angle is enough every time: in a parallelogram, opposite angles are
equal and adjacent ones add to 180°, so a single angle fixes all four. In a rhombus
you get more: a diagonal cuts each of the two angles it meets into two equal halves,
so even a half-angle like the 30° in (iii) determines the whole figure.

Q2 Using the diagonal properties, construct a parallelogram whose diagonals are of
lengths 7 cm and 5 cm, and intersect at an angle of 140°.

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.

Page 42 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

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.

Diagonals AB = 7 cm and CD = 5 cm, unequal ⇒ not a rectangle

Angle at O = 140° ≠ 90° ⇒ not a rhombus

Diagonals bisect each other ⇒ a genuine parallelogram

Why bisecting is enough: if OA = OB and OC = OD, then ∆AOC ≅ ∆BOD by SAS
(vertically opposite angles at O). So AC = BD, and the equal alternate angles make AC
∥ BD. The same argument on the other pair gives AD ∥ CB — both pairs of opposite
sides parallel, which is the definition.

Q3 Using the diagonal properties, construct a rhombus whose diagonals are of lengths
4 cm and 5 cm.

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.

Each side = √(2.5² + 2²) = √(6.25 + 4) = √10.25 ≈ 3.2 cm

All four sides are equal because all four triangles round O are congruent right triangles with

legs 2.5 cm and 2 cm.

Page 43 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

co m
m.
Why perpendicular bisecting forces equal sides: the four triangles AOC, COB,

m l a se
BOD, DOA all have legs 2.5 cm and 2 cm with a right angle between them, so they
o g
.c by SAS. Their hypotenuses — the four sides of theaquadrilateral
m
are congruent —

l a setherefore be equal.
ag
must

o m
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Check it yourself: measure the four sides; each should be a little over 3.2 cm.
Measure the angle at O; it should readm .
a s e exactly 90°.

agl

co m
m.
Geoboard Activity — Page 103
m as e
.co l
4.5 Playing with Quadrilaterals — Geoboard Activity

a g
a s em
a glQ1 Place two rubber bands perpendicular to each other, forming diagonals of equal
length. Join the ends. What is the quadrilateral that you get? Justify your answer.

m a s
m .co agl
se

g l a
a
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.
co m
m .
m as e
.co a g l
Diagonals equal ✓ + bisect each other ✓ ⇒ all four angles are 90° (Deduction 3)
s m
eDiagonals
gl a meet at 90° ✓ ⇒ all four sides are equal (Deduction 5)
a
m
⇒ a square
a se
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e m
g l
Why all four sides come out as equal: the four small triangles round the centre each
a length (half a band) with a right angle between them. So
have two legs of the same
they are congruent by SAS, and their hypotenuses — the four sides of the figure —
co m
must be equal.
m .
m as e
.co a g l
s e m Check it yourself: if each band spans 6 units, then each half is 3 units and each side
agla of the square measures √(3² + 3²) = 3√2 ≈ 4.24 units.
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 44 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q2 Extend one of the diagonals on both sides by 2 cm. What quadrilateral will you get
now? Justify your answer.

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:

CONDITION BEFORE AFTER EXTENDING

Diagonals bisect each other Yes Yes — the midpoint did not move

Diagonals perpendicular Yes Yes — the directions did not change

Diagonals equal Yes No — one is now 4 cm longer

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.

Why the sides stay equal: call the half-lengths p and q. Each of the four triangles at
the centre is right-angled with legs p and q, so every side of the quadrilateral
measures √(p² + q²). Making p larger changes that common value but keeps all four
sides identical.

Tip: this is the geoboard version of the fact that a square is a rhombus with equal
diagonals. Stretch one diagonal and the square relaxes into a leaning rhombus.

Joining Triangles — Pages 104–105
4.5 Playing with Quadrilaterals — Joining Triangles

Q1 Take two cardboard cutouts of an equilateral triangle of sidelength 8 cm. Can you
join them to get a quadrilateral?

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.

Page 45 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

2 triangles × 3 sides = 6 edges

2 edges are used up as the join ⇒ 6 – 2 = 4 sides ✓

Why they must be joined along a whole edge: if you slide one triangle so that only
part of an edge touches, the outline gains extra corners and stops being four-sided.
The edges must match end to end, which is why equal side lengths matter.

Q2 What type of a quadrilateral is this? Justify your answer.

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

All four outer edges are sides of equilateral triangles ⇒ all equal 8 cm ⇒ rhombus

Angles: at the two ends of the common edge, two 60° angles meet:

60° + 60° = 120°

At the other two vertices the angle is a single triangle angle: 60°

Check: 60 + 120 + 60 + 120 = 360° ✓

It is not a square, since 60° ≠ 90°.

Why it cannot be anything else: four equal sides is precisely the definition of a
rhombus. The diagonal that used to be the join is 8 cm — equal to a side — so it
makes two equilateral triangles, which is what fixes the angles at 60° and 120°.

Q3 Take two cardboard cutouts of an isosceles triangle with sidelengths 8 cm, 8 cm,
and 6 cm. What are the different ways they can be joined to get a quadrilateral?

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.

Page 46 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

JOIN HOW THE SECOND PIECE IS RESULTING QUADRILATERAL
ALONG PLACED SIDES

the 6 cm edge mirror image (or half turn — both give the 8, 8, 8, 8 rhombus
same figure here)

an 8 cm edge half turn about the midpoint of the edge 8, 6, 8, 6 parallelogram

an 8 cm edge mirror image across the edge 8, 6, 6, 8 kite

The book shows the first two of these.

Why the mirror and the half turn agree on the 6 cm edge: the triangle is isosceles
with the 6 cm side as its base, so it is already symmetric about the perpendicular
bisector of that base. Flipping it makes no difference to the outline. On an 8 cm edge
there is no such symmetry, so the two placements give different figures.

Q4 What quadrilaterals are these? Justify your answers.

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

All four outer edges are the 8 cm sides ⇒ four equal sides ⇒ rhombus

The 6 cm join becomes a diagonal, and it bisects the two angles it meets

(each half-triangle is isosceles, so the diagonal is a line of symmetry)

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

Top and bottom edges: 6 cm each Left and right edges: 8 cm each

Opposite sides equal ⇒ parallelogram (Question 9 of the next Figure it Out proves this

implication)

More directly: a half turn about the midpoint of the shared 8 cm edge sends each side to a
parallel copy pointing the opposite way, so opposite sides are parallel. That is the definition of a
parallelogram.

Page 47 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Why a half turn always gives a parallelogram: rotating a triangle by 180° about
the midpoint of one side reverses every direction. The image of the 6 cm side
therefore runs parallel to the original 6 cm side, and the image of the third side runs
parallel to that third side — both pairs of opposite sides come out parallel.

Q5 Take two cardboard cutouts of a scalene triangle with sides 6 cm, 9 cm, and 12 cm.
What are the different ways they can be joined to get a quadrilateral?

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.

JOIN ALONG HALF TURN GIVES SIDES MIRROR GIVES SIDES

6 cm 9, 12, 9, 12 9, 12, 12, 9

9 cm 6, 12, 6, 12 6, 12, 12, 6

12 cm 6, 9, 6, 9 6, 9, 9, 6

Tip: the mirror joins may need care. Reflecting across the longest edge can push the
reflected vertex so that the outline caves in — you then get a concave quadrilateral
rather than a convex one. Cut the pieces out and try each of the six before deciding.

Q6 Are you able to identify the different quadrilaterals that are obtained by joining the
triangles? Justify your answer whenever you identify a quadrilateral.

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:

join along 6 cm ⇒ parallelogram with sides 9 cm and 12 cm
join along 9 cm ⇒ parallelogram with sides 6 cm and 12 cm

join along 12 cm ⇒ parallelogram with sides 6 cm and 9 cm

Page 48 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

The three mirror joins give kites (when the outline stays convex), because reflection produces
co m
two adjacent pairs of equal sides:
e m.
m l as
m .co a g
l a se
join along 12 cm ⇒ sides 6, 9, 9, 6 ⇒ a kite with diagonal 12 cm
a g
This last one is the figure the book uses on page 105 to introduce the kite. Its adjacent sides are

com
. ag
equal in pairs — 6 cm with 6 cm at the top, 9 cm with 9 cm at the bottom.

se m
l a
ag rotation gives a parallelogram: a reflection keeps
Why reflection gives a kite and
the shared edge as a line of symmetry, so the two sides on either side of it are mirror

. c om
partners — adjacent equal pairs, which is a kite. A half turn instead sends each side

s
to the opposite corner, so the equal pairs end up opposite each othere m— which is a
. com
parallelogram.
a gla
a s em
agl Check it yourself: whichever way you join them, the four angles must total 360°,

m a s
agl
because the two triangles contribute 180° each.

m.co
l a se
a g
In-text Questions — Pages 105–107
co m
.
4.6 Kite and Trapezium

se m
o m l a
g ∠ABC and ∠ADC, (ii)
Q1 .cProperty 1: In the kite, show that the diagonal BD (i) bisects
a
e m
aglas bisects the diagonal AC, that is, AO = OC, and is perpendicular to it. Hint: Is ∆AOB ≅
∆COB?

se m
com g l a
m . a
ase
agl
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.

co m
m .
e
AB = CB (given)
m l as
m.coAD = CD (given)
a g
l a se
ag BD = BD (common)

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⇒ ∆ABD ≅ ∆CBD by SSS
s e m
m a
⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB
m . co agl
l a se
agboth ∠ABC and ∠ADC.
That is part (i): BD bisects

m
Step 2 — now compare ∆AOB and ∆COB.

. co
e m
m l as
.co a g
Page 49 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

AB = CB (given)

∠ABO = ∠CBO (just proved)

BO = BO (common)
⇒ ∆AOB ≅ ∆COB by SAS

⇒ AO = OC and ∠AOB = ∠COB

But ∠AOB + ∠COB = 180° (linear pair along AC)

⇒ ∠AOB = ∠COB = 90°

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

Why only one diagonal has these powers: B and D are each equidistant from A
and C, so both lie on the perpendicular bisector of AC — and BD is therefore that
perpendicular bisector. Nothing similar holds for A and C, which is why AC does not
bisect BD in general.

Tip: in a rhombus both pairs of adjacent sides are equal, so both diagonals get these
properties. That is exactly why the rhombus's diagonals bisect each other, bisect the
angles and meet at 90°.

Page 50 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q2 Construct a trapezium. Measure the base angles (marked in the figure). Can you
find the remaining angles without measuring them?

P Q

S R

Page 106 — trapezium PQRS with PQ ∥ SR. The two base angles, at S and at R, are the
marked ones.

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.

Property 1: ∠S + ∠P = 180° and ∠R + ∠Q = 180°

So measure the two base angles ∠P and ∠Q, and then simply subtract:

∠S = 180° – ∠P ∠R = 180° – ∠Q

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

Why only one pair of parallel sides is needed: the co-interior-angle rule applies to
each leg separately, and each leg touches both parallel sides. The non-parallel pair
contributes nothing extra — which is why a trapezium's opposite angles are not
equal in general, unlike a parallelogram's.

Page 51 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Check it yourself: after computing ∠R and ∠S, measure them with a protractor.
They should agree to within a degree — measurement error, not a flaw in the
reasoning.

Q3 How do we construct an isosceles trapezium? Construct an isosceles trapezium
UVWX, with UV||XW. Measure ∠U. Can you find the remaining angles without
measuring them?

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:

∠V = ∠U (angles opposite the equal sides — Property 2)

∠X = 180° – ∠U (co-interior angles along the leg UX)

∠W = 180° – ∠V = ∠X

If ∠U = 72°, then ∠V = 72°, ∠X = ∠W = 108°, and 72 + 72 + 108 + 108 = 360° ✓

Why one measurement suffices: the trapezium supplies one relation (co-interior
angles add to 180°) and the equal legs supply a second (∠U = ∠V). Two relations
plus the 360° total leave only one unknown, so a single measured angle pins
everything down.

Q4 Does it appear that the angles opposite to the equal sides — ∠U and ∠V — are also
equal? Can we find congruent triangles here? Consider line segments XY and WZ
perpendicular to UV. What type of quadrilateral is XWZY?

XWZY is a rectangle, and that is what produces the congruent triangles.
Drop perpendiculars XY and WZ from X and W onto UV.

Page 52 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

XW ∥ UV (given), so YZ is a transversal

a = 180° – ∠XYZ = 180° – 90° = 90°

b = 180° – ∠WZY = 180° – 90° = 90°

So all four angles of XWZY are 90° ⇒ XWZY is a rectangle

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

In ∆UXY and ∆VWZ:

UX = VW (equal legs of the isosceles trapezium)

XY = WZ (opposite sides of rectangle XWZY)

∠UYX = ∠VZW = 90°

⇒ the triangles are congruent (RHS) ⇒ ∠U = ∠V

Why the perpendiculars are the key idea: ∠U and ∠V sit at opposite ends of the
figure with nothing joining them. Dropping the two perpendiculars manufactures a
rectangle in the middle, and the rectangle hands over the one equality (XY = WZ)
needed to make the two end triangles congruent.

Q5 Now, it can be shown that ∆UXY ≅ ∆VWZ. (How?)

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

∠XYU = ∠WZV = 90° (XY and WZ were drawn perpendicular to UV)

UX = VW (hypotenuses — the equal non-parallel sides)

XY = WZ (one leg each, opposite sides of rectangle XWZY)

⇒ ∆UXY ≅ ∆VWZ

Hence ∠U = ∠V and also UY = VZ

Page 53 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

This is Property 2: in an isosceles trapezium, the angles opposite to the equal sides are
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equal.
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Why UY = VZ matters too: it says the two slanting ends stick out by the same
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symmetric — it has a line of symmetry through the midpoints of UV and XW.

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Tip: the same figure shows why the diagonals of an isosceles trapezium are equal:
agby SAS, since UX = VW, ∠U = ∠V and UV is common.
∆UXV and ∆VWU are congruent
So XV = WU.

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it Out — Pages 107–109
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4.6 Kite and Trapezium

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TRY THIS
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Find all the sides and the angles of the quadrilateral obtained by joining two
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equilateral triangles with sides 4 cm.

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All four sides are 4 cm; the angles are 60°, 120°, 60° and 120°. The figure is a rhombus.
.c two cutouts along a full side. That side becomes a diagonal
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Sides: 4 cm, 4 cm, 4 cm, 4 cm ⇒ a rhombus
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At the two ends of the join, two 60° angles come together:

60° + 60° = 120° (twice)
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Why it is not a square: a square needs 90° angles. Here the angles are decided by agl
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a g l Page 54 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Tip: the shorter diagonal is 4 cm (the join). The longer one measures 2 × (4 × √3 ∕ 2) =
4√3 ≈ 6.93 cm, since each equilateral triangle has height (√3∕2) × 4.

Q2 Construct a kite whose diagonals are of lengths 6 cm and 8 cm.

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.

PR = QR = √(3² + 2²) = √13 ≈ 3.6 cm

PS = QS = √(3² + 6²) = √45 ≈ 6.7 cm

Two adjacent pairs of equal sides ⇒ a kite ✓

Why R and S must be placed unequally: if you took TR = TS = 4 cm, the second
diagonal would also be bisected and you would get a rhombus — a special kite.
Making TR ≠ TS keeps the two pairs of sides different, giving the ordinary kite shape.
Many different kites have diagonals 6 cm and 8 cm; each choice of TR gives another
one.

Q3 Find the remaining angles in the following trapeziums —

Trapezium (i). The top and bottom sides carry arrowheads, so they are parallel. The two marked
angles, 135° and 105°, are at the ends of the shorter (bottom) side. Each slanting side is a
transversal, so the angles at its two ends add to 180°.

Page 55 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Above the 135° corner: 180 – 135 = 45°

Above the 105° corner: 180 – 105 = 75°

Check: 135 + 105 + 45 + 75 = 360° ✓

Trapezium (ii). Here the two slanting sides carry tick marks, so they are equal — this is an
isosceles trapezium. The 100° angle lies between one equal side and the shorter parallel side.

Other end of that same leg: 180 – 100 = 80°

Property 2: the angles on each parallel side are equal, so

the other angle on the short parallel side = 100°

the other angle on the long parallel side = 80°

Check: 100 + 100 + 80 + 80 = 360° ✓

45° 75°
100° 100°

135° 105°
80° 80°
(i)
(ii)

Given angles in red/blue, computed angles in green. Both figures total 360°.

Why (ii) needs the tick marks: in an ordinary trapezium the two given angles could
be anything, and 100° alone would only fix the angle at the other end of that leg. The
equal legs supply the second relation — Property 2 — which is what lets a single
measurement determine all four angles.

Page 56 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

Q4 Draw a Venn diagram showing the set of parallelograms, kites, rhombuses,
rectangles, and squares. Then, answer the following questions — (i) What is the
quadrilateral that is both a kite and a parallelogram? (ii) Can there be a
quadrilateral that is both a kite and a rectangle? (iii) Is every kite a rhombus? If not,
what is the correct relationship between these two types of quadrilaterals?

Parallelograms

Rectangles Rhombuses

Squares

Kites

Rectangles and rhombuses both sit inside parallelograms; their overlap is the squares. The
rhombuses also sit entirely inside the kites, which spill outside the parallelograms.

(i) The rhombus. A quadrilateral that is both a kite and a parallelogram must have two adjacent
pairs of equal sides (kite) and equal opposite sides (parallelogram) — and that forces all four
sides to be equal.

Kite: AB = BC and CD = DA

Parallelogram: AB = CD and BC = DA

Together: AB = BC = CD = DA ⇒ a rhombus (the square is the special case)

(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.

Page 57 of 73

Class 8 Maths Chapter 4 Quadrilaterals AglaSem · NCERT Solutions

A note on the answer key: the answers printed at the end of the chapter give “No”
here. That would only be right if a kite were required to have its two pairs of different
lengths. But part (i) of this very question expects a rhombus to count as a kite, so the
same reading must let a square count as a kite — and a square is certainly a
rectangle. The consistent answer is yes, the square.

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

Rhombus: AB = BC = CD = DA ⇒ certainly AB = BC and CD = DA ⇒ a kite ✓

Kite with AB = BC = 6 cm and CD = DA = 9 cm ⇒ not a rhombus ✗

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

Q5 If PAIR and RODS are two rectangles, find ∠IOD.

∠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.

In ∆ORI: ∠RIO = 90° (angle of rectangle PAIR at I, since O lies on AI)

∠IRO = 30° (given)

⇒ ∠ROI = 180 – 90 – 30 = 60°

RODS is a rectangle, so ∠ROD = 90°

I and D lie on the same side of RO, and ∠ROI + ∠IOD = ∠ROD

⇒ ∠IOD = 90 – 60 = 30°

Page 58 of 73

as e
Class 8 Maths Chapter 4 Quadrilaterals
a g l AglaSem · NCERT Solutions

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Rectangle PAIR in grey, rectangle RODS in green, with RO shared between the

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parts of right angles measured from the same line RO — one at R inside ∆ORI, the
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the size of the drawing.

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Construct a square with diagonal 6 cm without using a protractor.

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Use the diagonal property of a square: equal diagonals bisecting each other at 90°. A compass
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gives the right angle, so no protractor is needed.
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1. Draw AB = 6 cm.
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2. With centre A and any radius more than half of AB, draw arcs above and below AB. Repeat
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with centre B and the same radius. Join the two crossing points — this line is the

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perpendicular bisector of AB, and it meets AB at its midpoint O.

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Page 59 of 73