NCERT Solutions Class 9 Maths Chapter 12 Quadrilaterals

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Chapter 12:
Quadrilaterals

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 51 – 84 22 61 English

Every question Step by step Exam ready
Activities, in-text questions The working shown, not Written to match how
and all exercises. just the final answer. marks are awarded.

Solutions, notes, sample papers & more at 89 pages

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 12: Quadrilaterals
Chapter 12 makes the word ‘quadrilateral’ precise and then puts parallelograms to work. You will learn
exactly which four-sided figures count as quadrilaterals (and why NOPE, CUTS or BENT do not), when a
quadrilateral is convex, and four quick tests that prove a quadrilateral is a parallelogram. Parallelograms then
prove the Midpoint Theorem, the Centroid Theorem (medians meet at a 2 : 1 point) and Varignon’s theorem,
and finally show that copies of any quadrilateral tile the plane.

TEXTBOOK BOOK PAGES
Ganita Manjari (Class 9) Part II, 51 – 84

SECTIONS QUESTIONS
22 61

MEDIUM
English

Think and Reflect — Page 51
Chapter opening: tiling the plane
THINK AND REFLECT

Q1 Can we use any given quadrilateral to tile the plane? If not, which quadrilaterals can be
used and which cannot?

Yes. Every quadrilateral, convex or non-convex, can tile the plane. (Here a quadrilateral is a
flat 4-gon whose sides do not cross, as in the rest of the chapter.)
Recipe: take a copy and turn it through 180° about the midpoint of one of its sides. The turned
copy fits exactly along that side, because the two ends of the side simply swap places. Do the
same at every side of every new copy.
It works because the four angles of any quadrilateral add up to 360°. At every corner of the
pattern four copies meet, bringing the angles 1, 2, 3 and 4 once each, so they close up with no
gap and no overlap.

Page 1 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Copies of the irregular 4-gon SOME tiling the plane (half-turn method). The yellow copy is the
starting one; the brown copies are its half-turned versions.

Why: for a rectangle or a parallelogram the tiling is just a grid of rows (Fig. 12.1).
For an irregular 4-gon the trick is to use two positions of the tile: the original
and the one turned through 180°. Round each vertex the four angles add up to 1
+ 2 + 3 + 4 = 360°, which is exactly a full turn.

The chapter returns to this in Section 12.4 (pages 69–73), where two methods are
described: turning about the midpoints of the sides (Method 1) and using the grid of
Varignon parallelograms (Method 2).

Compare: the regular pentagon cannot tile the plane, because its angle 108°
does not divide 360° (Exercise Set 12.4, Q1). Self-intersecting 4-gons such as
CUTS (Fig. 12.2) are not quadrilaterals in this sense, so they are not included in
the answer.

Page 2 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Think and Reflect — Page 52
Section 12.1 What Exactly is a Quadrilateral?
THINK AND REFLECT

Q1 Informally, a quadrilateral is a figure with four straight sides, as in the first figure ABCD
in Fig. 12.2 below. But consider the other six figures in Fig. 12.2: the five plane figures
NOPE, SILY, DART, CUTS, OPENS and the non-planar BENT. Should we call all these
figures quadrilaterals? If you answer ‘no’ for any of them, how will you define a
quadrilateral so that such a figure is excluded?

B A R
A N I
D
D C E P O S L Y T
S
U T
C O
S N B
E N
P E
T
Fig. 12.2, page 52 — ABCD, NOPE, SILY, DART (top row); CUTS, OPENS and the non-planar
BENT (bottom row; T is above the shaded plane that holds B, E and N).

No. ABCD is a quadrilateral, and so is DART (a quadrilateral with a dent). The other five should
be left out:

FIGURE WHAT GOES WRONG

NOPE E, P, O lie on one line, so the figure is really the triangle NEO.

SILY S, L, Y lie on one line, so the sides LY and YS overlap.

CUTS the sides UT and SC cross each other (self-intersecting).

OPENS it has five points and does not close up.

BENT T is not in the plane of B, E, N (non-planar).

Page 3 of 89

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Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

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Definition: take four distinct points A, B, C, D in one plane, with no three on one line. The
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ag l Page 4 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Why: three non-collinear points automatically lie in one plane, and three
segments joining them can never cross. That is why the triangle needs only one
condition. With four points all three problems can occur.

About the book: Definition 1 printed on page 53 keeps only the plane condition
and the “exactly one segment” condition, and does not repeat “no three
collinear”. With that wording NOPE would slip through: the points of OP and PE
lie on just one segment each. The Chapter Summary on page 83 does include
“no three collinear”, so read Definition 1 with that condition added, as the text
just before it says.

In-text Questions — Page 53

Page 5 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.1 What Exactly is a Quadrilateral?

Q1 Can you see how to rule out figures like OPENS?

S

O
N

P E
OPENS from Fig. 12.2, page 52.

OPENS has five points O, P, E, N, S but only four segments OP, PE, EN, NS. The path starts at O
and stops at S, so it never closes.
The definition rules it out by taking exactly four points A, B, C, D and the four segments AB,
BC, CD and DA. The last side DA brings us back to the starting point, so a quadrilateral is a
closed loop in which every vertex is an end of exactly two sides. In OPENS the points O and S
are ends of only one segment each.

Page 6 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Tip: if you joined S back to O you would get a closed figure with five sides, a
pentagon OPENS, which is still not a quadrilateral.

In-text Questions — Page 54
Section 12.1 What Exactly is a Quadrilateral?

Q1 Note that quadrilateral ABCD can also be denoted as BCDA, CDAB, DABC, DCBA,
ADCB, BADC or CBAD but not by any other sequence of vertices. (Draw ABCD and
trace the vertices in various orders to see why.)

The name of a quadrilateral lists the vertices in the order you meet them as you walk round
the boundary. So each pair of neighbouring letters, and also the last and first letters, must be
joined by a side.
You can start the walk at any of the 4 vertices and go round in either of 2 directions: 4 × 2 = 8
names.

one way round: ABCD, BCDA, CDAB, DABC

the other way: DCBA, ADCB, BADC, CBAD

Any other order, such as ACBD, puts a diagonal (AC) in the place of a side, so it describes a
different figure.

Why: four letters can be arranged in 4 × 3 × 2 × 1 = 24 orders. Each walk round
the boundary is named 8 times, so 24 ÷ 8 = 3 different 4-gons can be made
from the same four points: ABCD, ABDC and ACBD. For a convex ABCD, the 4-
gon ACBD has the two diagonals AC and BD as sides, and they cross: it is self-
intersecting. See End-of-Chapter Exercise 10.

Page 7 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Exercise Set 12.1 — Pages 55–56
Section 12.1 What Exactly is a Quadrilateral?

Q1 Let ABCD be a quadrilateral.
(i) List all sides of ABCD adjacent to side AB. List all sides opposite to AB.
(ii) List all angles of ABCD adjacent to ∠A. List all angles opposite to ∠A.
(iii) Define a pair of opposite sides and a pair of opposite angles without using the
names of the vertices.

(i) Sides adjacent to AB: BC and DA (they share the endpoints B and A). Side opposite to AB:
CD only.
(ii) Angles adjacent to ∠A: ∠B and ∠D. Angle opposite to ∠A: ∠C only.
(iii) Two sides of a quadrilateral are opposite if they have no common endpoint. Two angles are
opposite if their vertices are not adjacent, that is, not the two ends of one side (their vertices
are joined by a diagonal).

Check it yourself: a quadrilateral has exactly two pairs of opposite sides (AB,
CD and BC, DA) and two pairs of opposite angles (∠A, ∠C and ∠B, ∠D). Each
side has two adjacent sides and one opposite side.

*Q2 You have used internal angles of quadrilaterals, but they too require an exact
definition, just like how we gave one for a quadrilateral. Precisely define the internal
angle of a quadrilateral at a given vertex. Your answer should work for a non-convex
quadrilateral too. (Hint: use the opposite vertex as well.)

At vertex A of quadrilateral ABCD, the sides AB and AD make two angles: ∠DAB (less than
180°) and the reflex angle 360° − ∠DAB. The internal angle is one of them. Use the opposite
vertex C:

Page 8 of 89

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Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

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If A lies inside ∆BCD (the triangle formed by the other three vertices),
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Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

AB = CD: Yes, it can be non-convex. Example: A(0, 0), B(5, 0), C(4, 7), D(4, 2). AB = 5 and CD
= 7 − 2 = 5, and D lies inside ∆ABC.
∠A = ∠C: Yes. Example: the arrowhead A(−3, 0), B(0, 4), C(3, 0), D(0, 1). It is symmetric about
BD, so ∠A = ∠C, and D lies inside ∆ABC.

C
B

D D
A C

A B
AB = CD = 5 ∠A = ∠C

Two non-convex quadrilaterals (dent at D): left, AB = CD; right, ∠A = ∠C.

Why: the first answer is why the proof of Theorem 5 (page 59) can say that the
diagonals of ABCD meet: a quadrilateral with a pair of parallel sides is convex,
and the diagonals of a convex quadrilateral intersect.

Check it yourself: in the arrowhead, ∠A = ∠DAB is the angle between AB
(direction 3 along, 4 up) and AD (3 along, 1 up), about 53.1° − 18.4° = 34.7°. By
the mirror symmetry ∠C is also about 34.7°, while the internal angle at D is
reflex.

Page 10 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q4 Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every
possible location of point D in the plane outside these lines, decide if ABCD is self-
intersecting, non-convex, or convex. (Hint: the three lines divide the plane into 7
regions.)

The three lines cut the plane into 7 regions: the inside of ∆ABC, three regions across the sides,
and three corner regions at the vertices. In ABCD the sides are AB, BC, CD, DA, and AC is a
diagonal.

WHERE D IS ABCD IS

inside ∆ABC non-convex (dent at D)

across side AC (away from B) convex

across side AB (away from C) self-intersecting (CD crosses AB)

across side BC (away from A) self-intersecting (DA crosses BC)

corner region at A non-convex (A inside ∆BCD)

corner region at B non-convex (B inside ∆ACD)

corner region at C non-convex (C inside ∆ABD)

So 1 region gives a convex, 2 give a self-intersecting and 4 give a non-convex quadrilateral.

Page 11 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

non-convex

C
convex self-int.

non-convex

A B non-convex
self-int.
non-convex

Where D can be: the lines AB, BC, CA cut the plane into 7 regions; each is labelled with the kind
of 4-gon ABCD.

Why: across AC, the points B and D are on opposite sides of AC, and D is inside
the angle ABC, so the diagonals AC and BD cross: convex. Across AB, D and C
are on opposite sides of line AB and D is inside the angle ACB, so the side CD
must cut the side AB. In a corner region, say at A, D lies in the angle vertically
opposite to ∠BAC; then A lies inside ∆BCD, which is exactly a dent at A.

Check it yourself: with A(0, 0), B(4, 0), C(1, 3), try D(−1, 2) (convex), D(3, 3)
(self-intersecting), D(1.5, 1) (non-convex) and D(−1, −1) (non-convex, dent at A).

Page 12 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q5 Can a quadrilateral be both self-intersecting and non-planar?

No. In a self-intersecting 4-gon two sides cross, say AB and CD meet at a point E. Two lines
that meet always lie in one plane. That plane contains A and B (on line AB) and C and D (on line
CD), so all four vertices lie in one plane: the 4-gon is planar.
(Two adjacent sides, which share a vertex, can overlap only if three vertices are collinear, and
three collinear points together with a fourth point also lie in one plane.)

Why: “non-planar” means the four vertices are not in one plane, like BENT. If
two sides of BENT crossed, the two crossing lines would pull all four vertices
into a single plane, which is impossible.

Did you know? a non-planar 4-gon is also called a skew quadrilateral. The Venn
diagram on page 55 shows the skew 4-gons and the self-intersecting planar 4-
gons as separate groups for exactly this reason.

Think and Reflect — Page 59
Section 12.2 Parallelograms (Theorem 3)
THINK AND REFLECT

Q1 The angles of a quadrilateral add up to 360°. Therefore, if the opposite angles are
equal, what can we say about adjacent angles? Is the converse of your answer true?
Conclude that Theorem 3 can also be stated as follows. “If each pair of adjacent
angles in a quadrilateral ABCD …then ABCD is a parallelogram.” Fill in the blank.

If ∠A = ∠C and ∠B = ∠D, then

∠A + ∠B + ∠C + ∠D = 360°
2(∠A + ∠B) = 360° ⇒ ∠A + ∠B = 180°

similarly ∠B + ∠C = ∠C + ∠D = ∠D + ∠A = 180°

Page 13 of 89

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Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

So each pair of adjacent angles adds up to 180° (they are supplementary).
co m
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as
The converse is true. If every pair of adjacent angles adds up to 180°:
m l
m .co ag
la se
ag
∠A + ∠B = 180° = ∠B + ∠C ⇒ ∠A = ∠C

∠B + ∠C = 180° = ∠C + ∠D ⇒ ∠B = ∠D

o m
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So the two conditions say the same thing, and ag
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In-text Questions — Pagea59
Section 12.2 Parallelograms (Theorem 5)
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Q1

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se m
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Because AB ∥ DC forces ABCD to be e m
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o m Fig. 12.5). So AC
In a convex quadrilateral the two diagonals always cross each other (page 54,
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and BD meet at a point E inside ABCD, and the triangles EAB and ECDmused
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ag l Page 14 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Why: in a non-convex quadrilateral such as DART the diagonals do not meet, so
a proof that starts “let E be the point where the diagonals meet” would have
nothing to work with. That is why the book checks this step.

Exercise Set 12.2 — Page 60
Section 12.2 Parallelograms

Q1 True or false?
(i) A parallelogram with a right angle is a rectangle.
(ii) A rhombus with perpendicular diagonals is a square.
(iii) If the diagonals of a parallelogram are equal, then it is a rectangle.

(i) True. Adjacent angles of a parallelogram add up to 180°, so if ∠A = 90° then ∠B = 90°.
Opposite angles are equal, so ∠C = ∠A = 90° and ∠D = ∠B = 90°. All four angles are right
angles: a rectangle.
(ii) False. Every rhombus has perpendicular diagonals, so this describes every rhombus,
including one with angles 60° and 120°, which is not a square.
(iii) True. In parallelogram ABCD with AC = BD:

∆ABC ≅ ∆DCB (SSS: AB = DC, BC common, AC = DB)
⇒ ∠ABC = ∠DCB
∠ABC + ∠DCB = 180° (adjacent angles)
⇒ ∠ABC = ∠DCB = 90°

A parallelogram with a right angle is a rectangle, by (i).

Page 15 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Tip: correct versions of (ii) are “a rhombus with a right angle is a square” and “a
rhombus with equal diagonals is a square”.

Q2 The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and
that ABCD is a rhombus.

Given ∠DAC = ∠CAB.

AB ∥ DC, transversal AC: ∠CAB = ∠ACD (alternate angles)

AD ∥ BC, transversal AC: ∠DAC = ∠ACB (alternate angles)
∠DAC = ∠CAB ⇒ ∠ACB = ∠ACD

So AC bisects ∠C.
In ∆ABC, ∠CAB = ∠DAC = ∠ACB, so the triangle is isosceles and AB = BC (sides opposite
equal angles). In a parallelogram AB = DC and BC = AD, hence AB = BC = CD = DA: ABCD is a
rhombus.

Why: once one pair of adjacent sides of a parallelogram is equal, all four sides
are equal, because opposite sides of a parallelogram are always equal.

Page 16 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q3 The following questions examine converses of true properties. Answer them with Yes
or No. If your answer is No, what extra condition can you add so that the answer
becomes Yes?
(i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus?
(ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must
ABCD be a rhombus?
(iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a
rectangle? (In the ancient Indian study of quadrilaterals, equality of diagonals was
considered significant. Quadrilaterals were first classified according to whether their
diagonals were equal or not, before considering equality of sides.)

(i) Yes. AC bisects ∠A and ∠C. In ∆ABC and ∆ADC: ∠BAC = ∠DAC, AC is common, ∠BCA =
∠DCA, so ∆ABC ≅ ∆ADC (ASA). Hence AB = AD and CB = CD. In the same way BD bisects ∠B
and ∠D, so ∆ABD ≅ ∆CBD and AB = CB. All four sides are equal: a rhombus.
(ii) Yes. Diagonals that bisect each other make ABCD a parallelogram (Theorem 4). If they
cross at E, then ∆AEB ≅ ∆AED (SAS: AE common, ∠AEB = ∠AED = 90°, EB = ED), so AB = AD.
A parallelogram with two adjacent sides equal is a rhombus.
(iii) No. The isosceles trapezium A(0, 0), B(6, 0), C(4, 3), D(2, 3) has AC = BD = 5 but is not a
rectangle. Extra condition: the diagonals also bisect each other. Then ABCD is a parallelogram
with equal diagonals, which is a rectangle by Q1(iii).

Check it yourself: AC = √(4² + 3²) = 5 and BD = √(4² + 3²) = 5, while ∠A is not
90° (AD goes 2 along and 3 up).

Did you know? rectangles, squares and isosceles trapeziums all have equal
diagonals. That is why grouping quadrilaterals by their diagonals, as the ancient
Indian texts mentioned in the book did, puts different shapes in the same group.

Page 17 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q4 Let ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points
of the four angle bisectors form the vertices of a rectangle (see Fig. 12.11). Why did we
assume AB ≠ BC?

D C
P

Q
S

R
A B
Fig. 12.11, page 60 — parallelogram ABCD with the bisectors of its four angles.

As in Fig. 12.11, the bisectors of ∠A and ∠B meet at P, of ∠B and ∠C at Q, of ∠C and ∠D at R,
and of ∠D and ∠A at S.

In ∆APB: ∠PAB + ∠PBA = ½∠A + ½∠B = ½(∠A + ∠B) = ½ × 180° = 90°
⇒ ∠APB = 180° − 90° = 90°

The angle of PQRS at P is formed by the lines PA (which carries S) and PB (which carries Q), so
∠SPQ = ∠APB = 90°. In the same way ∆BQC, ∆CRD and ∆DSA give ∠PQR = ∠QRS = ∠RSP =
90°. A quadrilateral with all angles 90° has its opposite angles equal, so it is a parallelogram
(Theorem 3) with a right angle: PQRS is a rectangle.
Why AB ≠ BC: if AB = BC, ABCD is a rhombus. Its diagonals bisect its angles, so all four
bisectors pass through the centre and P, Q, R, S are one and the same point. There is no
rectangle.

Page 18 of 89

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Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

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m as e
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ag l Page 19 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.3.1 An Important Property of Triangles
ACTIVITY

Q1 On a piece of paper, draw and then cut out two copies of the same triangle. Keep one
copy aside. On the other, mark the midpoint of each side. (You can do this by folding
the paper to join two vertices at a time.) Draw the triangle made by the three midpoints
and cut the paper along each of these lines. Now you have four smaller triangles.
Compare them to each other and to the intact copy of the original triangle. What do
you notice? Verify that this pattern holds for triangles of other shapes.

?
? ?

Fig. 12.12, page 61 — the triangle formed by joining the midpoints of the sides. What can we
say about its sides?

All four small triangles are congruent: each can be placed exactly on any other, though the
middle one has to be turned round (through 180°) first. Each is a half-size copy of the original
triangle: its sides are half as long and its angles are the same.
Sample result (a triangle with sides 6 cm, 8 cm and 10 cm): every small triangle measured 3
cm, 4 cm and 5 cm, and the four of them laid on the intact copy covered it exactly.
The same thing happens with an acute, a right-angled and an obtuse triangle, so the pattern
does not depend on the shape. The Midpoint Theorem (page 63) explains it.

Page 20 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Why: each side of the middle triangle joins two midpoints, so by the Midpoint
Theorem it is parallel to one side of the big triangle and half as long. That makes
every small triangle have sides ½a, ½b, ½c. Exercise Set 12.3, Q1 proves the
congruences.

Did you know? since four equal small triangles fill the big one, each has exactly
one quarter of its area. Halving every length divides the area by 2 × 2 = 4.

Think and Reflect — Page 62
Section 12.3.1 An Important Property of Triangles
THINK AND REFLECT

Q1 Try to prove directly that the four smaller triangles are congruent. You will see that no
congruence test applies, because many quantities are unknown.

Call the midpoints P (on AB), Q (on AC) and R (on BC). Compare ∆APQ with ∆PBR. All we know
is AP = PB. We do not know whether AQ = PR or PQ = BR, and we know no equal angles: ∠APQ
= ∠PBR would need PQ ∥ BC, which has not been proved. One pair of equal sides is not
enough for SSS, SAS, ASA, AAS or RHS.
The middle triangle PQR is even worse: none of its sides or angles is known. So a direct proof
gets stuck. We first need a fact about a single segment joining two midpoints, namely PQ ∥ BC
and PQ = ½BC. That is the Midpoint Theorem.

Why: once the Midpoint Theorem is known, the missing facts arrive all at once:
PQ = BR = RC, QR = AP = PB and PR = AQ = QC. Then SSS applies to every pair
of the four triangles (Exercise Set 12.3, Q1). Breaking a hard problem into
smaller steps is exactly the idea the book suggests on page 62.

Page 21 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Think and Reflect — Page 63
Section 12.3.1 An Important Property of Triangles
THINK AND REFLECT

Q1 Read the statement of the Midpoint Theorem again. The following questions seem
natural to ask. What can we say about (1) a segment PQ (with P on side AB and Q on
side AC) that is parallel to the third side and has half the length, and (2) the line that
passes through the midpoint of one side and is parallel to another side? We will
address question (1) in Exercise 22 at the end of the chapter. For now, experiment and
see if you can make a guess about question (2).

Guess for (2): the line through the midpoint of one side, drawn parallel to a second side,
passes through the midpoint of the third side, and the part of it inside the triangle is half as
long as the second side.
Sample experiment: in a triangle with BC = 8 cm, draw (with set-squares) the line through the
midpoint P of AB parallel to BC. It meets AC at Q, and measuring gives AQ = QC and PQ = 4 cm.
Other triangles give the same result. This is Theorem 7, proved on page 64.
Guess for (1): such a segment PQ must join the midpoints of AB and AC. (This is proved in
Exercise 22(i).)

Why: for (2): there is only one line through P parallel to BC. The segment joining
P to the midpoint of AC is parallel to BC (Midpoint Theorem), so it must be that
very line. For (1): as a segment parallel to BC slides from A down to BC, its
length grows steadily from 0 to BC, so it has length ½BC at exactly one position,
the one through the midpoints.

In-text Questions — Page 64

Page 22 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.3.1 Converse of the Midpoint Theorem

Q1 Take ∆ABC and suppose the line through midpoint P of AB, drawn parallel to BC, meets
AC at point Q. The line parallel to BA through C meets line PQ at R. BCRP is a
parallelogram, so CR = BP. BP = PA and ∠AQP = ∠CQR (vertically opposite angles).
Combining with CR ‖ BA, ∆APQ ≅ ∆CRQ. (Why?)

A

P Q
R

B C
Fig. 12.15, page 64 — P is the midpoint of AB, PQR ∥ BC, and CR ∥ BA.

In ∆APQ and ∆CRQ:

AP = CR (since AP = BP and BP = CR)
∠AQP = ∠CQR (vertically opposite angles)
∠PAQ = ∠RCQ (alternate angles: CR ∥ BA, transversal AC)

Two angles and a side not between them are equal: AP is opposite ∠AQP, and CR is opposite
the matching angle ∠CQR. So ∆APQ ≅ ∆CRQ by the AAS test (A ↔ C, P ↔ R, Q ↔ Q).

Page 23 of 89

as e
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

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m as e
.co

ag l Page 24 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.3.2 A Surprising Property of Medians

Q1 Medians CP and BQ of ∆ABC meet at M; Y and X are the midpoints of CM and BM.
Using the Midpoint Theorem in ∆ABC and in ∆MBC we get that PQ and XY are parallel
to BC and hence to each other. Moreover PQ = XY = BC/2. So ∆MPQ ≅ ∆MYX by ASA
(How?). Therefore MQ = MX = XB and MP = MY = YC.

A

P Q
M
X Y

B C
Fig. 12.18, page 65 — medians CP and BQ meet at M; X and Y are the midpoints of BM and
CM.

PQ = YX (both equal ½BC)

∠MPQ = ∠MYX (alternate angles: PQ ∥ XY, transversal PY; P, M, Y lie on CP)
∠MQP = ∠MXY (alternate angles: PQ ∥ XY, transversal QX; Q, M, X lie on BQ)

The equal sides PQ and YX lie between the equal angles, so ∆MPQ ≅ ∆MYX by ASA, with P ↔
Y and Q ↔ X.
Hence MQ = MX and MP = MY. Since X and Y are midpoints, MX = XB and MY = YC. So BM =
2MQ and CM = 2MP.

Page 25 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Why: the two triangles sit on opposite sides of M like a bow-tie: PQ is above M
and XY below it, both parallel to BC. Lines through M (CP and BQ) cut both
parallel segments, which is why the angles are alternate angles.

Q2 What about the third median AR? We can repeat the argument with medians AR and
BQ! Let N be the point at which AR and BQ intersect. Now the same reasoning gives BN
: NQ = 2 : 1 = AN : NR. (Use Fig. 12.19 if you wish to repeat the reasoning.)

A

P Q
N

B R C
Fig. 12.19, page 66 — medians AR and BQ meet at N. The two unlabelled dots are the
midpoints of AN and BN.

Let K and L be the midpoints of AN and BN (the two unlabelled dots).

∆ABC: Q, R are midpoints of CA, CB ⇒ QR ∥ AB and QR = ½AB
∆NAB: K, L are midpoints of NA, NB ⇒ KL ∥ AB and KL = ½AB

So QR ∥ KL and QR = KL

Page 26 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

In ∆NQR and ∆NLK: QR = LK, ∠NQR = ∠NLK and ∠NRQ = ∠NKL (alternate angles, as QR ∥ KL).
So ∆NQR ≅ ∆NLK (ASA). Hence NQ = NL = LB and NR = NK = KA, which gives BN : NQ = 2 : 1
and AN : NR = 2 : 1.

Why: this is the same argument as in Q1 with the triangle ‘turned’: Q, N, L lie on
BQ and R, N, K lie on AR. Since BQ is divided in the ratio 2 : 1 by both M and N,
they are the same point, and that is the Centroid Theorem.

Think and Reflect — Page 66

Page 27 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.3.2 A Surprising Property of Medians
THINK AND REFLECT

Q1 Give another proof of the Centroid Theorem using a clever construction that we will
describe only partly and ask you to complete. Once again let M be the intersection
point of the two medians CP and BQ. We want to show that the intersection point of
line AM with side BC is the midpoint of BC. For this, extend AM up to a carefully chosen
point S such that BSCM becomes a parallelogram. (Hint: To arrange MC ‖ BS, first note
that PM and MC are along the same line. Now where should M be on segment AS?
Does your placement of S also prove MB ‖ CS?)

A

P Q
M

B X C

S
Fig. 12.20, page 66 — medians CP and BQ meet at M; AM is extended through X (on BC) to S.

Choose S on AM extended so that MS = AM, that is, M is the midpoint of AS.

Page 28 of 89

ase
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

com
m.
∆ABS: P, M are midpoints of AB, AS ⇒ PM ∥ BS and PM = ½BS
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medians pass through M. ag
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c m = MS =
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agl

co m
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m as e
.co

ag l Page 29 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Sections 12.3.1 and 12.3.2 (Midpoint Theorem, Centroid, Varignon parallelogram)

Q1 (i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of ∆ABC, show that
∆PQR is congruent to ∆QPA and to two other triangles which you should identify.
(ii) Suppose someone erases ∆ABC, leaving only ∆PQR on the paper. Can you
reconstruct ∆ABC from ∆PQR?

A

P Q

B R C
Fig. 12.22, page 67 — P, Q, R are the midpoints of AB, AC and BC.

(i) By the Midpoint Theorem, used three times:

PQ = ½BC = BR = RC
QR = ½AB = AP = PB
RP = ½AC = AQ = QC

So ∆PQR ≅ ∆QPA by SSS (PQ = QP, QR = PA, RP = AQ). In the same way ∆PQR ≅ ∆RBP (PQ =
RB, QR = BP, RP = PR) and ∆PQR ≅ ∆CRQ (PQ = CR, QR = RQ, RP = QC). All four small triangles
are congruent.
(ii) Yes. Through each vertex of ∆PQR draw the line parallel to the opposite side: through P
parallel to QR, through Q parallel to PR, through R parallel to PQ. These three lines meet in pairs
at A, B and C. This works because side AB passes through P and is parallel to QR (Midpoint

Page 30 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Theorem in ∆ABC), and there is only one such line. The same is true for AC and BC.

Tip: the order of letters records which vertices match. In ∆PQR ≅ ∆RBP, P ↔ R,
Q ↔ B, R ↔ P, so the side PQ (= ½BC) matches RB (= ½BC).

Check it yourself: draw any triangle PQR and the three parallels. Measure: P, Q,
R should come out as the midpoints of the sides of the big triangle.

Q2 In ∆ABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC.
Show that MN bisects AD.

A

M N

B D C
Fig. 12.23, page 68 — M and N are the midpoints of AB and AC; D is any point on BC.

Let MN meet AD at O.

Page 31 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

∆ABC: M, N are midpoints of AB, AC ⇒ MN ∥ BC (Midpoint Theorem)

So MO ∥ BD
∆ABD: M is the midpoint of AB and MO ∥ BD ⇒ O is the midpoint of AD (Theorem 7)

So MN cuts AD at its midpoint: MN bisects AD, wherever D is on BC.

Why: the same argument works in ∆ADC with N: NO ∥ DC, so again O is the
midpoint of AD. As a bonus, MO = ½BD and ON = ½DC, which add up to MN =
½BC, as they should.

Page 32 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q3 In a quadrilateral ABCD, suppose AB ‖ DC and AB ≠ CD. Suppose G and H are the
midpoints of AC and BD respectively. Prove that GH ‖ AB. (Why did we assume AB ≠
CD?)

B
A

G
H

C

D
Fig. 12.24, page 68 — AB ∥ DC; G and H are the midpoints of the diagonals AC and BD.

Let M be the midpoint of AD.

∆ADC: M, G are midpoints of AD, AC ⇒ MG ∥ DC
∆DAB: M, H are midpoints of DA, DB ⇒ MH ∥ AB

AB ∥ DC ⇒ MG and MH are both parallel to AB and pass through M

There is only one line through M parallel to AB, so M, G and H all lie on it. Hence GH ∥ AB.
Why AB ≠ CD: if also AB = CD, then ABCD has a pair of equal and parallel sides, so it is a
parallelogram (Theorem 5). Its diagonals bisect each other, so G and H are the same point.
Then “GH” is a single point, not a segment, and GH ∥ AB means nothing.

Page 33 of 89

as e
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

co m
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.cknow? since MG = ½DC and MH = ½AB, the length GH a g= ½(DC − AB)
m
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Q4 Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R
and S respectively.
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c
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co m
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m as e
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ag l Page 34 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q5 Suppose PQRS is the Varignon parallelogram of ABCD.

A P B A′ P B′

S Q S Q

C C′
D R D′ R
Fig. 12.25, page 68 — left: ABCD with its Varignon parallelogram PQRS (dashed); right: the
same PQRS with a dotted copy A′B′C′D′.

(i) Copy only PQRS on another paper. Show how you will recreate a congruent copy
A′B′C′D′ of ABCD from PQRS. This will be relevant when we return to study tilings at
the end of this chapter. (Hint: How will you place vertex A′? How will you place B′, C′
and D′?)
(ii) There are multiple ways to construct A′B′C′D′ in (i). Justify why the quadrilateral
A′B′C′D′ you constructed is congruent to ABCD. You may need to show why S is
collinear with the points A′ and D′ that you constructed, and similarly for P, Q and R.
(Hint: Use congruence of triangles, for example ∆SDR ≅ ∆SD′R.)
(iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove
the converse.

(i) Copy the corner triangles of ABCD outside PQRS. With compasses place A′ outside SP so
that A′S = AS and A′P = AP. Place B′, C′, D′ in the same way from ∆PBQ, ∆QCR, ∆RDS. Join
A′B′C′D′.
(ii) By SSS, ∆SA′P ≅ ∆SAP and ∆SD′R ≅ ∆SDR. So

∠A′SP + ∠PSR + ∠RSD′ = ∠ASP + ∠PSR + ∠RSD = 180° (A, S, D collinear)

⇒ A′, S, D′ are collinear and A′D′ = AS + SD = AD

In the same way at P, Q and R: A′B′ = AB, B′C′ = BC, C′D′ = CD. The angles match too, e.g.
∠D′A′B′ = ∠SA′P = ∠SAP = ∠DAB. So A′B′C′D′ ≅ ABCD.
(iii) Midpoint Theorem: PQ ∥ AC, PQ = ½AC and QR ∥ BD, QR = ½BD.

Page 35 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

PQRS square ⇒ PQ = QR, PQ ⊥ QR ⇒ AC = BD and AC ⊥ BD
AC = BD, AC ⊥ BD ⇒ PQ = QR, ∠PQR = 90° ⇒ PQRS is a square

Why the last step holds: PQRS is already a parallelogram (Theorem 9). Two
equal adjacent sides make it a rhombus, and a rhombus with a right angle is a
square.

Another way for (i): place A′ as above. Then extend A′P beyond P by its own
length to get B′ (so P is the midpoint of A′B′), extend B′Q by its own length to get
C′, and extend C′R by its own length to get D′. By Theorem 9 the midpoint of
D′A′ forms a parallelogram with P, Q, R; there is only one such point, so it is S.
You never need to check S separately.

Did you know? PQRS alone does not fix ABCD. If you put A′ at a different place
and use the ‘extend by its own length’ method, you still get a quadrilateral
whose Varignon parallelogram is PQRS, but a different one. That is why the hint
asks how to place A′: it is the one free choice. Method 2 of the tiling section
(page 72) uses exactly this idea.

Think and Reflect — Page 70

Page 36 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.4 Tiling the Plane Using Any 4-gon
THINK AND REFLECT

Q1 Suppose we have a tiling of the plane. Consider any vertex. As we go around this
vertex and consider the angles made by consecutive lines, the total of these angles
must be 360°. This suggests an idea. What if we take 4 copies of SOME and fit them
together around a common point so that each angle is used once as we go around?

O
S
2
1

3
4
E M
Fig. 12.26, page 69 — the irregular 4-gon SOME with its angles labelled 1, 2, 3, 4.

The four copies fit exactly round the point, with no gap and no overlap there, because ∠1 +
∠2 + ∠3 + ∠4 = 360°, the angle sum of a quadrilateral.
The angles can be put round the point in different orders. Start with angle 1; the other three can
follow in 6 orders, and reading each order clockwise instead of anticlockwise pairs these up.
That leaves 3 different patterns, shown below (as in Fig. 12.27).

Page 37 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

1 2 1 4 1 4
3
34 3 2 2

(a) (b) (c)
Three ways to fit four copies round a point, each angle used once (as in Fig. 12.27). The blue
numbers name the angle of each copy at the red point.

Each pattern is a possible start for a tiling, because it already fills the full turn round one vertex.

Check it yourself: in the SOME drawn here the angles are about 120°, 96°, 80°
and 64°, and 120° + 96° + 80° + 64° = 360°.

Why: in a tiling, every vertex is surrounded completely by corners of tiles, so
the corner angles there must make one full turn, 360°. A quadrilateral provides
exactly 360° with its four angles, which is why four copies, one angle each, are
the natural choice.

In-text Questions — Pages 70–71

Page 38 of 89

as e
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

Section 12.4 Tiling the Plane Using Any 4-gon
co m
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ommultiple ways of doing this, as shown in Fig. 12.27. Can
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blue numbers name the angle of each copy at the red point.
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Yes: pattern (c) can be continued into a tiling. In (c) the angles 1, 2, 3, 4 come round the point
in order, and every two neighbouring copies share a whole side of the same length
c o m (for
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agl
se m
com g l a
m . a
e
ag las

co m
m .
m la se
.co ag
a s em
agl c
m .
m a s e
. co agl
asem
agl

co m
m .
m as e
.co

ag l Page 39 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

The tiling of Fig. 12.28, page 70, redrawn: green copies are SOME slid along its diagonals, brown
copies are SOME turned through 180°. The yellow copy is the starting one.

In patterns (a) and (b) some neighbouring copies meet along sides of different lengths, so the
copies do not line up edge to edge and it is hard to see how to go on.

Why: a side can be shared completely only by two copies whose touching
sides have the same length. In (c) the copies touching along a ray are the
original SOME and its half-turn, and a half-turn about the midpoint of a side
gives back the same side. In (a) and (b) the touching sides are different sides of
SOME, such as SO against ME, which have different lengths.

Page 40 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q2 How can we understand the figure (Fig. 12.28)? There seems to be a repeating pattern.
(1) Can you precisely describe a procedure to draw the pattern so that someone can
draw the tiling on their own, based only on your description? (2) Can you justify why
your procedure works?

Fig. 12.28, page 70 — copies of SOME tiling the plane (redrawn).

(1) Procedure (Method 1 of the book):
1. Draw SOME and mark the midpoints of its four sides.
2. For one side, draw the copy of SOME turned through 180° about that side’s midpoint: each
vertex X goes to the point X′ for which the midpoint lies halfway between X and X′. The new
copy shares that whole side with SOME.
3. Do step 2 for every side of every copy you have drawn, skipping copies that are already
there.

Page 41 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

O = M′ E′
S

S′
E M = O′
Method 1: the second copy is the first one turned through 180° about the midpoint of OM (red dot),
so the edge OM is shared.

(2) Why it works. A half-turn about the midpoint of a side swaps the two ends of that side, so
two copies share the whole side, with no gap or overlap along it. Round every vertex the four
copies bring the angles 1, 2, 3, 4 once each, which add up to 360°, so they close up exactly.

Why: a full proof needs three facts: matching along edges, matching round
vertices, and no clash between copies reached by different routes. The third
holds because two half-turns one after the other amount to a slide, so every
copy is SOME or its half-turned version, slid along the diagonals; this is
explained in the next question.

Method 2: draw the grid of Varignon parallelograms of SOME (page 72). Place
copies of SOME on every second cell in each direction (the shaded cells of Fig.
12.32), each with its side-midpoints at the corners of its cell. The gaps left
between them are half-turned copies of SOME.

Page 42 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q3 Now repeat this procedure at each edge of every copy of SOME that you add. Two
steps of a possible continuation are shown in Fig. 12.30. The first step (Fig. 12.30 A)
adds 4 copies around the starting copy and the second step (Fig. 12.30 B) adds 4 more
copies. Note two interesting things about the second step: (1) each new copy can be
obtained by rotating any one of its neighbours, and both ways give the same result. (2)
The new copies fit perfectly. Can you explain these facts by reasoning? This is needed
to prove that the method works!

S O

E M

Like Fig. 12.30 B, page 71: SOME (yellow), the 4 copies of step 1 (brown, each turned 180°
about the midpoint of a side of SOME) and the 4 corner copies of step 2 (green).

Use one fact: a half-turn about one point followed by a half-turn about another point is a slide,
in which every point moves the same distance in the same direction.
(1) The new corner copy at vertex O can be made by turning either brown neighbour (the copy
on OM or the copy on SO) about the midpoint of its other side at O. Each route is two half-turns
of SOME, and both give the same slide: SOME slid along its diagonal from E to O. So both
routes give the same copy. The other corners work the same way.
(2) The corner copy has its angle 4 at O. Round O we now have angle 2 (SOME), angle 3 (copy
on OM), angle 1 (copy on SO) and angle 4: 360° in total, so it fills the gap exactly. Its two sides
at O are the sides of the two brown neighbours, so the edges match too.

Page 43 of 89

as e
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

co m
m.

o m l ase
Why:mwrite ag Z sends any
.c points as coordinate pairs. A half-turn about the point
la se X to Z + Z − X (add the coordinates). The brown copy on OM sends X to O
ag
point
+ M − X, and its other side at O runs from O to O + M − E, with midpoint Z =

com
½(2O + M − E). Turning about Z sends O + M − X to 2O + M − E − (O + M − X) =
m .
X + (O − E). The other route, through the
e
copy on SO, gives 2O + S − E − (S + O ag
s the same slide by O − E, which carries E to
− X) = X + (O − E) as well. Bothlaare
O. a g

c o m
.
m is SOME
e
Did you know? this is why the whole tiling repeats: every green copy
s
.
slid by m steps along the two diagonals EO and SM, andgevery
cowhole a la brown copy
a semthe half-turned SOME slid in the same way.
is
ag l

m a s
e m .co agl
Q4 laspossibilities shown in Fig. 12.27 occurs in the tiling?
ag
Do you see which of the three

co m
m .
as e
. com 1 2 1 4 agl 1 4
se m
3
a 34 3 2 2
agl
sem
(a) com (b) (c) g l a
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asways of fitting four copies of SOME round a point (redrawn). The
g l
Fig. 12.27, page 70 — three
blueanumbers name the angle of each copy at the red point.

co m
m .
m la se
.co ag

s emThe third one, (c). In the tiling, the four copies round any vertex bring the angles 1, 2, 3, 4 in
ag la order, going round the point in the opposite sense to the way 1, 2, 3, 4 go round the tile itself.
The copies round the point alternate between the starting position and the half-turned position,
m .c
m
like the alternating colours of Fig. 12.28.
o l a se
That is why neighbouring copiesm .c whole sides: the copies showing angles 1 and 2 meet ag
e
share
las 2 and 3 along OM, and so on.
ag
along a full side SO, those showing

co m
m .
m as e
.co

ag l Page 44 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Check it yourself: in the tiling drawn for Q1, pick any vertex and read the angle
numbers round it. You meet 1, 2, 3, 4 in order (in one direction or the other),
never 1, 3, 2, 4 or 1, 3, 4, 2.

A Cutting-edge Discovery — Page 74

Page 45 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.4 Tiling the Plane Using Any 4-gon
ACTIVITY

Q1 The following shape, the first of its kind, that leads to aperiodic tilings was discovered
only recently in 2023 by a team of four mathematicians (Smith, Myers, Kaplan and
Goodman-Strauss). Cut out 15 identical copies of the following shape (known as the
‘hat’) and start to tile with it. Do you see any pattern?

The ‘hat’, page 74 — the first single shape found (in 2023) that tiles the plane only
aperiodically.

The hats fit together snugly, with no gaps and no overlaps, and you can keep adding pieces.
But you do not find a block that simply repeats, the way the rows of a brick wall or the SOME
tiling repeat. Small groups of hats arranged in the same way do turn up again and again, but
they never line up into a regular grid. You will also find that a few of the hats have to be turned
over (used as mirror images) to make the pieces fit.
This is what aperiodic means: the hat covers the whole plane, but no tiling by it can be slid
onto itself.

Page 46 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Did you know? the hat is made of 8 identical kites. Before 2023, the known
aperiodic tilings, such as Roger Penrose’s from the 1970s, needed at least two
different shapes. Later in 2023 the same four mathematicians found the
‘spectre’, a related shape that tiles only aperiodically even without turning any
piece over.

Your own answer: with only 15 pieces you will see a patch, not the whole plane.
Proving that the pattern never repeats needs advanced mathematics; your
observation is that no repeating block shows up.

Exercise Set 12.4 — Pages 74–75

Page 47 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Section 12.4 Tiling the Plane Using Any 4-gon

Q1 Justify why the plane cannot be tiled with a regular pentagon. (Hint: Read the first 3
sentences of ‘Think and Reflect’ in the section on tiling.) (There are many ways to tile
the plane using a suitable irregular pentagon. The most recent method was found in
2015.)

C
c
d B

D b
e
a
E A
The irregular pentagon printed with Exercise Set 12.4, Q1 (drawn to scale).

ANGLES SIDES

A = 60° a=1

B = 135° b = 1/2

C = 105° c = 1/(√2(√3 − 1))

D = 90° d = 1/2

E = 150° e = 1/2

Picture beside the pentagon (page 74)
A coloured pattern in which copies of this irregular pentagon, in navy, tan and
grey, cover the page without gaps, repeating in rows.

Page 48 of 89

ase
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

com
m.

m pentagon is as e
.co
Each angle of a regular
ag l
a s em
gl of angles) ÷ 5 = 540° ÷ 5 = 108°
a(sum
mto 360°.
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m .c ag
la se there, we need 108° × k = 360°, so k = 360 ÷ 108
If the vertex is a corner of every pentagon
= 3⅓, not a whole number. Three agpentagons leave a gap of 36°; four would overlap (432° >
360°).
If the vertex lies on a side of one pentagon (a straight angle, 180°), the cornersom
. c there must
fill the other 180°: 108° × k = 180° gives k = 1⅔, again impossible.
m em
as
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o a regular pentagon cannot fit round a vertex: the planeacannot
So copies.cof
m
ase pentagons.
agl
regular

s
m a
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se 3 triangles by the two diagonals from one
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l
g to 3 × 180° = 540° (End-of-Chapter Exercise 21).
vertex, so its angles addaup

c o m
.
m 105°, 90°
e
Compare: the irregular pentagon in the figure has angles 60°, 135°,
s
. com
and 150° (sum 540°). Some of them do combine to 360°,gfor
a la example 60° +
a s em150° + 150° and 90° + 135° + 135°, so copies can close up round a vertex, and
agl the pentagon tiles the plane as the picture shows.

se m
com g l a
m . a
e
las
g DART. Show how we can tile the plane with copies of DART.
Q2 Draw a non-convex a4-gon
Both methods that we discussed earlier will work. Which do you prefer?
co m
m .
o
m la se
gstill add up to 360°, with the
.c DART as in Fig. 12.2, with the dent at D. Its internal angles
a
m
se reflex angle at D.
Take
la
ag c
Method 1: turn DART through 180° about the midpoint of each of its sides to get four
m .
a s e
com agl
neighbours, and repeat for every new copy. Each turned copy shares a whole side with its
neighbour, and round every vertex .the
m
angles at D, A, R and T meet once each (360°). The
se a
agl
result:

co m
m .
m as e
.co

ag l Page 49 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

A tiling by the non-convex 4-gon DART made with Method 1 (half-turns about the midpoints of the
edges). The yellow copy is the starting one.

Method 2: draw the grid of Varignon parallelograms of DART and place DART on every second
cell in each direction, as in Fig. 12.32. The gaps are again copies of DART, turned through 180°.
Preference (sample answer): Method 1: it needs only cut-outs turned about midpoints, with no
construction lines. Method 2 is neater on paper, as the grid shows in advance where every
copy goes.

Why: neither method uses convexity. Method 1 uses only “the angles add up to
360°” and “a half-turn about a midpoint gives back the same side”. Method 2
uses Theorem 9, which holds for non-convex 4-gons too (End-of-Chapter
Exercise 13).

Page 50 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Your own answer: either preference is fine if you give a reason, for example
speed with cut-outs or accuracy with a ruler and grid.

End-of-Chapter Exercises — Pages 75–83
Whole chapter

Q1 Using a fact about parallelograms, show how to tile the plane using any given triangle.
(Hint: Can you use the parallelogram tiling in the introduction?)

Take any triangle ABC and a copy of it turned through 180° about the midpoint M of side BC.
The copy has vertices C, B and A′, where M is the midpoint of AA′.
In the quadrilateral ABA′C the diagonals AA′ and BC bisect each other at M, so it is a
parallelogram (Theorem 4). By the introduction (Fig. 12.1), copies of any parallelogram tile the
plane in slanted rows. Now draw the diagonal BC in every parallelogram: each one splits back
into the triangle and its turned copy. So copies of the triangle tile the plane.

Any triangle (yellow) and its copy turned 180° about the midpoint of a side make a
parallelogram; the parallelograms tile the plane.

Page 51 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Why: round every vertex of this tiling, six triangles meet, and they bring each of
the angles A, B, C twice: 2(A + B + C) = 2 × 180° = 360°.

Q2 Mark the midpoint of the line drawn on the paper (see Fig. 12.33), given that the
horizontal lines are equally spaced. Justify your answer.

Fig. 12.33, page 75 — a line drawn across equally spaced horizontal lines of a ruled sheet.

The red line starts on the 2nd ruled line from the top and ends on the 6th, so it crosses 4 equal
gaps. Its midpoint is the point where it crosses the 4th ruled line from the top, the middle one
of the five lines it meets.

Page 52 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Justification. Call the ends X (on line 2) and Y (on line 6), and let the red line cross line 4 at Z.
From X draw the perpendicular to the ruled lines; let it meet line 4 at U and line 6 at V. The lines
are equally spaced, so XU = UV: U is the midpoint of XV. In ∆XVY, line 4 passes through the
midpoint U of XV and is parallel to VY, which lies on line 6. By the converse of the Midpoint
Theorem (Theorem 7), line 4 bisects XY. So Z is the midpoint of XY.

Did you know? the same reasoning shows that equally spaced parallel lines cut
any slanting line into equal pieces. A ruled notebook is therefore a handy tool
for dividing a strip of paper into 2, 3, 4, … equal parts.

Q3 You know that the sum of angles of a quadrilateral is 360°, even for a non-convex
quadrilateral. (Recall the proof.) Now consider a self-intersecting quadrilateral ABCD,
where AB and CD intersect at point E. Show that ∠A + ∠B + ∠C + ∠D < 360°. Can you
construct ABCD such that ∠A + ∠B + ∠C + ∠D = 2°?

AB and CD cross at E, so the figure is two triangles, ∆EBC and ∆EAD, meeting at E like a bow-
tie. Since E lies on AB and on CD:

∠A = ∠EAD, ∠D = ∠EDA (angles of ∆EAD)
∠B = ∠EBC, ∠C = ∠ECB (angles of ∆EBC)
∠A + ∠D = 180° − ∠AED, ∠B + ∠C = 180° − ∠BEC
∠AED = ∠BEC (vertically opposite)
⇒ ∠A + ∠B + ∠C + ∠D = 360° − 2∠BEC < 360°

Yes, the sum can be 2°. We need 360° − 2∠BEC = 2°, so ∠BEC = 179°. Draw two lines through
E that make an angle of 179°. Mark B and C on the arms of this 179° angle with EB = EC = 5 cm,
and A and D on the opposite rays with EA = ED = 5 cm. Join AB, BC, CD, DA. Each of ∆EBC and
∆EAD is isosceles with apex angle 179°, so each of ∠A, ∠B, ∠C, ∠D is 0.5°, and the sum is 2°.

Page 53 of 89

ase
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

co m
m.

o m l ase
.c proof: for a non-convex quadrilateral, draw the diagonal
a g from the
m
se with the reflex angle. It lies inside and splits the 4-gon into two triangles,
Recall the
la
ag
vertex
whose angles add up to 2 × 180° = 360°.

. com ag
s em 4-gon is not fixed: it depends on the
Why: the angle sum of a self-intersecting
a
l crossing angle gives a sum close to 360°; a
angle at the crossing point. aAgsmall
crossing angle close to 180° (a very flat bow-tie) gives a sum close to 0°.
co m
em.
m l as
.co ag
a semIn a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP =
agl
Q4
BQ (see Fig. 12.34). Show that APCQ is a parallelogram.
m a s
m .co agl
la se
ag C
D co m
m .
as e
. com agl
a se m P
agl
se m
com Q g l a
m . a
e
aglas
B
co m
m .
m A la se
.co ag
a s em
agl
Fig. 12.34, page 75 — P and Q on the diagonal BD with DP = BQ.
.c
s e m
m a
. co agl
sem
laof ABCD meet at O. The diagonals of a parallelogram bisect each
Let the diagonals AC and gBD
a
other, so OA = OC and OB = OD.

co m
m .
m as e
.co

ag l Page 54 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

OP = OD − DP
OQ = OB − BQ
OD = OB and DP = BQ ⇒ OP = OQ

So in the quadrilateral APCQ the diagonals AC and PQ bisect each other at O. By Theorem 4,
APCQ is a parallelogram.

Another way: ∆ADP ≅ ∆CBQ (SAS: AD = CB, ∠ADP = ∠CBQ as alternate
angles, DP = BQ), so AP = CQ. In the same way ∆ABQ ≅ ∆CDP gives AQ = CP.
Both pairs of opposite sides are equal, so APCQ is a parallelogram (Theorem 2).

Check it yourself: if DP is longer than OD, P lies beyond O. Then OP = DP − OD
= BQ − OB = OQ, and the proof still works.

Page 55 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q5 A right-triangle shaped cutout of a paper is folded such that point A touches point B
(Fig. 12.35). Show that the crease line can be used to find the midpoint of not only AB
but also that of AC.

Step 1 Step 2
A A

B C B C
Fig. 12.35, page 76 — a right-triangle cut-out (right angle at B) folded so that A touches B.

When A is folded onto B, the part of the paper on one side of the crease lands exactly on the
other part. So the crease meets AB at a point M with MA = MB, the midpoint of AB, and the two
angles that the crease makes with AB at M are equal; since they add up to 180°, each is 90°.
The crease is the perpendicular bisector of AB.
The right angle of the triangle is at B, so BC ⊥ AB too. Two lines perpendicular to the same line
are parallel: the crease ∥ BC.
So the crease is the line through the midpoint M of AB parallel to BC. By Theorem 7 it bisects
the third side: it meets AC at its midpoint N. The ends of the crease mark both midpoints.

Did you know? Theorem 7 also says MN = ½BC: the crease is exactly half as
long as the base.

Page 56 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

Q6 You saw how to use the Midpoint Theorem to divide a given triangle into 4 congruent
triangles. Can we divide a triangle into 3 congruent triangles? This exercise shows us
how to do that and more, provided we are allowed to cut and reassemble.
(i) Draw medians AP, BQ and CR of ∆ABC, meeting in a common point M. Cut along
each median to get 6 triangles. Show with justification how to assemble the 6 pieces
into 3 congruent triangles! (Hint: Align ∆MPB and ∆MPC along equal sides PB and PC,
matching P with itself and B with C.) Prove that you get a triangle. What are its side
lengths?
*(ii) Repeat the procedure with each of the 3 assembled triangles you got in (i). Show
that each of the resulting 9 triangles has sides AB/3, BC/3 and AC/3. (The procedure
and the result above were unnoticed till 2014, when they were discovered by Lee
Sallows, an amateur mathematician!)
(iii) Naturally, we can next ask about the possibility of cutting a triangle and
reassembling it into 2 congruent triangles. You were asked this question in the chapter
Measuring Space: Perimeter and Area. Can you solve this now? (Hint: Use median AP
in ∆ABC and then a median of ∆APB.)

(i) Turn ∆MPB through 180° about P: B lands on C (P is the midpoint of BC) and M lands on M′.

∠M′PC + ∠MPC = ∠MPB + ∠MPC = 180° (B, P, C collinear) ⇒ M, P, M′ collinear

So ∆MPC and the moved piece make one triangle MCM′, with sides (Centroid Theorem)

MM′ = 2MP = ⅔AP, M′C = MB = ⅔BQ, MC = ⅔CR

The pieces at Q and R pair up the same way, so the three triangles are congruent (SSS).
(ii) ∆MCM′ has medians ½BC, ½AB, ½AC (for example CP = ½BC, as P is the midpoint of MM′).
Applying (i) to it gives sides ⅔ × ½ = ⅓ of those of ∆ABC: 9 triangles with sides AB/3, BC/3,
AC/3.
(iii) Cut along AP, then along the median PN of ∆APB. Turn ∆PNA through 180° about N (A → B,
P → P′). ∆PBP′ has PB = PC, BP′ = PA, PP′ = 2PN = AC, so ∆PBP′ ≅ ∆APC (SSS).

Page 57 of 89

Class 9 Maths Chapter 12 Quadrilaterals AglaSem · NCERT Solutions

A

M

B P C

M′
△MPB (orange) turned 180° about P lands on △M′PC (dashed): with △MPC it makes the
triangle MCM′.

Why: for (ii), call the triangle T = MCM′ and find its three medians. (1) P is the
midpoint of MM′, so the median from C is CP = ½BC. (2) Let K be the midpoint
of MC. Since AM = 2MP = MM′, M is the midpoint of AM′; since MK = ½MC =
MR, M is also the midpoint of RK. The diagonals of ARM′K bisect each other, so
it is a parallelogram and the median M′K = AR = ½AB. (3) Let L be the midpoint
of CM′. BMCM′ is a parallelogram (its diagonals BC and MM′ bisect each other
at P), so CM′ ∥ BM and CL = ½CM′ = ½BM = MQ. So MQ and LC are equal and
parallel, MQCL is a parallelogram (Theorem 5), and the median ML = QC = ½AC.

Page 58 of 89

as e
Class 9 Maths Chapter 12 Quadrilaterals
ag l AglaSem · NCERT Solutions

. com
Check it yourself: areas agree. The 3 triangles of (i) share the area of ∆ABC,
m a s
and each has sides ⅔ of the medians. The 9 triangles of (ii) are copiesemof ∆ABC
co⅓, each with area ⅑ of ∆ABC, and 9 × ⅑ = 1.
scaled .by
m agl
la se
ag

com
m . ag
la se
ag

co m
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m l as
m .co ag
la se
ag
m a s
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ag

co m
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a se m
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co m
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co m
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ag l Page 59 of 89