Quadrilaterals: Class 9 Ganita Manjari · Chapter 12

Parallelograms, midpoints, medians and tilings

Ganita Manjari Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

70 min
1 simulation
45 practice questions
19 MCQs

What you will learn in Quadrilaterals

Define a quadrilateral precisely and tell convex, non-convex, self-intersecting and non-planar 4-gons apart
Prove the properties of a parallelogram and their converses, and use the five tests for a parallelogram
Prove and apply the Midpoint Theorem and its converse
Prove that the medians of a triangle meet at the centroid, two-thirds of the way along each median from its vertex
Show that the midpoints of the sides of any quadrilateral form a parallelogram, and find its sides from the diagonals
Explain, using the angle sum and half-turns, why copies of any quadrilateral tile the plane

Original companion text aligned to NCERT Mathematics Class 9, Chapter 12, pp. 51–84. Not affiliated with or endorsed by NCERT.

Floors, tiles and 4-gons

Railway platforms, school corridors and kitchen floors are usually covered with four-sided tiles fitted edge to edge. A shape with four straight sides is a quadrilateral, also called a 4-gon. Copies of one shape covering a floor with no gaps and no overlaps form a tiling; imagine the floor going on for ever and you are tiling the plane.

One case is easy: a parallelogram. Its opposite sides are equal and parallel, so a copy slid along one side fits snugly against the first. A line of such copies makes a row with straight edges at the top and bottom (Fig. 12.1). Rows then fit against rows in the same way. So copies of any parallelogram tile the plane.

3/3Rows fit against rows, so copies of the parallelogram cover the plane with no gaps.

Fig. 12.1Rows of one parallelogram cover the plane.NCERT Fig. 12.1 · p. 52

Step 1 shows a single parallelogram tile, with tick marks and arrow marks showing that its opposite sides are equal and parallel. Step 2 shows four copies slid along the direction of the bottom side to form a row whose top and bottom edges are straight lines. Step 3 shows four such rows stacked, each shifted sideways by the slant of the tile, covering the region with no gaps.

Along the way we study parallelograms closely: how to recognise one from a few measurements, and how they unlock some surprising facts about triangles and quadrilaterals. Then we return to the tile factory.

12.1 What Exactly is a Quadrilateral?

Everyone recognises a quadrilateral on sight. But try to write a rule precise enough for a computer: given four points and four segments, is the figure a quadrilateral or not? Look at the seven figures in Fig. 12.2. Each one has four straight sides. Would you call every one of them a quadrilateral?

5/5Only the ordinary shape and the dented arrowhead are left. Both are quadrilaterals.

Fig. 12.2Three rules decide which four-sided figures count as quadrilaterals.NCERT Fig. 12.2 · p. 52

Seven small figures, each made of four straight segments: an ordinary four-sided shape, a triangle with an extra corner on its base, a shape whose sides fold back over each other, a crossed bow-tie, a dented arrowhead, an open zigzag that does not close, and a card bent along a diagonal so its halves lie in different planes. Step 2 crosses out the bow-tie and the open zigzag. Step 3 also crosses out the two figures with three corners in a line. Step 4 also crosses out the bent card. Step 5 highlights the two survivors, the ordinary shape and the arrowhead, which are both quadrilaterals.

Triangles are easy: any three points not on one line give a triangle (three points on a line give only a segment). With four points $A$ , $B$ , $C$ , $D$ joined in the order $A \to B \to C \to D \to A$ , the figure can fail in three different ways.

The loop is open, or two sides cross. We want one closed loop in which two sides touch only at a corner they share. A bow-tie, whose sides cross in the middle, is called a self-intersecting 4-gon; we leave it out.
Three corners lie on one line. Then two neighbouring sides lie along that line, so the figure is really a triangle with a spare point, or two sides double back over each other. We rule this out: no three of the four points may be collinear.
The corners are not in one plane. Fold a square card along a diagonal and tilt the halves apart: the four corners now lie in two different planes. Such a figure is a non-planar 4-gon. Bent 4-gons appear only in a couple of the end-of-chapter exercises.

Putting the three conditions together gives a precise definition.

Definition

Quadrilateral

Take four different points $A$ , $B$ , $C$ , $D$ in one plane, no three of them on a line. The four segments $A B$ , $B C$ , $C D$ and $D A$ form the quadrilateral $A B C D$ when no two of these segments have any point in common other than a shared corner.

ExampleThe dented arrowhead in Fig. 12.2 is a quadrilateral; the bow-tie is not, because two of its sides cross.

Some names for the parts. In quadrilateral $A B C D$ the points $A$ , $B$ , $C$ , $D$ are its vertices (corners), and its four sides, also called edges, are $A B$ , $B C$ , $C D$ and $D A$ . The segments $A C$ and $B D$ , which join opposite corners, are its diagonals. The two ends of one side, such as $A$ and $B$ , are adjacent vertices. Two sides with a common end, such as $A B$ and $B C$ , are adjacent sides. At each vertex the two sides meeting there form an internal angle, so there are four internal angles; angles at adjacent vertices are adjacent angles. How would you describe opposite sides and opposite angles? Try Q1 of Practice 12.1.

A quadrilateral has several names. Walk once round its boundary, starting at any of the four corners and going either way, and read off the letters. Clockwise and anticlockwise you get $A B C D$ , $B C D A$ , $C D A B$ , $D A B C$ , $A D C B$ , $D C B A$ , $C B A D$ and $B A D C$ : that is $4 \times 2 = 8$ names for one shape. A name such as $A C B D$ is not allowed, because it jumps along a diagonal, and joining $A$ , $C$ , $B$ , $D$ in that order draws a different figure.

From now on, quadrilateral means a flat one whose sides do not cross, unless we say otherwise.

Convex quadrilaterals

Both survivors in Fig. 12.2 are quadrilaterals, but the arrowhead is different in one way: one of its corners is pushed inwards, making a dent. At the dent the inside of the shape wraps round the corner, so the internal angle there is more than $180^{\circ}$ (a reflex angle). A quadrilateral is convex when all four internal angles are less than $180^{\circ}$ , and non-convex when one of them is more than $180^{\circ}$ .

2/3Every angle of $P Q R S$ is below $180^{\circ}$ . In $W X Y Z$ the angle at $Z$ is $254^{\circ}$ . Both sets add up to $360^{\circ}$ .

Fig. 12.3Two tests for convexity: the angles, and the diagonals.NCERT Figs. 12.4–12.5 · p. 54

Two quadrilaterals side by side: a convex one PQRS on the left and an arrowhead WXYZ, dented at Z, on the right. Step 2 marks every internal angle: PQRS has 100, 73, 99 and 88 degrees, all less than 180; WXYZ has 27, 53 and 26 degrees and a reflex angle of 254 degrees at Z. Both sets add to 360 degrees. Step 3 draws the diagonals: in PQRS they cross at a point inside the shape; in WXYZ diagonal WY runs outside the shape below Z and the two diagonals never meet.

A second test needs no protractor: draw both diagonals. In a convex quadrilateral the diagonals cross each other inside the shape. In a non-convex one, one diagonal lies outside the shape and the two diagonals do not meet at all. Either test tells you which kind you have. Try both with the lab below.

InteractiveQuadrilateral lab

A grid with a quadrilateral ABCD whose four corners can be dragged, or moved with the arrow keys. A message below classifies the shape as convex, non-convex (naming the corner with the reflex angle), self-intersecting, or not a quadrilateral when three corners are in a line. One card lists the four sides, both diagonals, the four internal angles and their sum; another says whether opposite sides are parallel or equal and whether the diagonals bisect each other. A checkbox joins the midpoints P, Q, R and S of the sides, and buttons load a parallelogram, a trapezium, a dart and a crossed 4-gon.

Fig. 12.4Drag the corners of

A B C D

. The lab says whether it is convex, non-convex or crossed, lists its angles and runs the parallelogram tests of Section 12.2.

Table 12.1The family of 4-gons

Kind of 4-gon	Corners in one plane?	Sides cross?	An angle over 180°?	A quadrilateral here?
Non-planar	No	—	—	No
Self-intersecting	Yes	Yes	—	No
Non-convex	Yes	No	Yes, one	Yes
Convex	Yes	No	No	Yes

Practice · 5 questions

Practice 12.1: questions with answers

1
Short answerEasyNCERT Exercise Set 12.1, Q1
In quadrilateral $P Q R S$ :
1. Which sides are adjacent to side $Q R$ ? Which side is opposite it?
  Show solution
  AnswerAdjacent: $P Q$ and $R S$ . Opposite: $S P$ .
  $P Q$ shares vertex $Q$ with $Q R$ , and $R S$ shares vertex $R$ . The only side with no vertex in common with $Q R$ is $S P$ .
2. Which angles are adjacent to $∠ S$ ? Which angle is opposite it?
  Show solution
  AnswerAdjacent: $∠ R$ and $∠ P$ . Opposite: $∠ Q$ .
  $S$ is joined by sides to $R$ and $P$ , so $∠ R$ and $∠ P$ are adjacent to $∠ S$ . $Q$ is not joined to $S$ by a side, so $∠ Q$ is opposite.
3. Describe, in words that would work for any quadrilateral, when two sides are opposite and when two angles are opposite.
  Show solution
  Opposite sides: two sides that have no vertex in common. Opposite angles: the angles at two vertices that are not the two ends of any one side.
Long answerHardNCERT Exercise Set 12.1, Q2
At vertex $B$ of quadrilateral $A B C D$ , the rays $B A$ and $B C$ make two angles: one less than $180^{\circ}$ and one more, adding up to $360^{\circ}$ . Write a rule, using the other vertices, that says which of the two is the internal angle at $B$ . Your rule should also handle dented quadrilaterals.
Show solution
Look at the triangle $A C D$ formed by the other three vertices.
Rule: if $B$ lies inside $△ A C D$ , the internal angle at $B$ is the reflex angle $360^{\circ} - ∠ A B C$ ; otherwise it is the smaller angle $∠ A B C$ .
Why it works: the internal angle at $B$ is more than $180^{\circ}$ exactly when the quadrilateral is dented at $B$ , and that happens exactly when $B$ lies inside the triangle made by the other three corners (like $Z$ inside $△ W X Y$ in Fig. 12.3). In a convex quadrilateral no vertex lies inside the triangle of the other three, so every internal angle is the smaller one.
3
Short answerMediumNCERT Exercise Set 12.1, Q3
For each condition on quadrilateral $P Q R S$ , can $P Q R S$ still be non-convex (dented)?
1. $P Q ∥ S R$
  Yes, it canNo, it cannot
  Show solution
  No. Each vertex lies on one of the two parallel lines, and the whole quadrilateral lies in the strip between them. At each vertex the shape is on one side of the line through that vertex, so no internal angle can be more than $180^{\circ}$ .
2. $P S = Q R$
  Yes, it canNo, it cannot
  Show solution
  Yes. Take $P (0, 0)$ , $Q (3, 4)$ , $R (10, 0)$ , $S (8, 1)$ . Then $P S = \sqrt{64 + 1} = \sqrt{65}$ and $Q R = \sqrt{49 + 16} = \sqrt{65}$ , and $S$ lies inside $△ P Q R$ , so $P Q R S$ is dented at $S$ .
3. $∠ P = ∠ R$
  Yes, it canNo, it cannot
  Show solution
  Yes. The arrowhead $P (0, 0)$ , $Q (4, 6)$ , $R (8, 0)$ , $S (4, 2)$ is symmetric about the line $x = 4$ , so $∠ P = ∠ R$ , and it is dented at $S$ .
4. $∠ P + ∠ Q = 180^{\circ}$
  Yes, it canNo, it cannot
  Show solution
  No. Co-interior angles adding up to $180^{\circ}$ make $P S ∥ Q R$ , and then part (a) applies to that pair of sides.
4
DiagramMediumNCERT Exercise Set 12.1, Q4
Take three points $K$ , $L$ , $M$ not on one line and draw the full lines $K L$ , $L M$ and $M K$ ; they split the plane into 7 regions. A fourth point $N$ , off these lines, makes the 4-gon $K L M N$ with sides $K L$ , $L M$ , $M N$ , $N K$ . Is $K L M N$ convex, non-convex or self-intersecting when $N$ is:
1. inside $△ K L M$ ?
  Show solution
  AnswerNon-convex, dented at $N$ .
  $N$ lies inside the triangle of the other three vertices, so the angle at $N$ is reflex.
2. on the far side of $K M$ from $L$ , between the lines $L K$ and $L M$ ?
  Show solution
  AnswerConvex.
  No side crosses another, and no vertex lies inside the triangle of the others. The diagonals $K M$ and $L N$ cross, so $K L M N$ is convex.
3. on the far side of $K L$ from $M$ , between the lines $M K$ and $M L$ ?
  Show solution
  AnswerSelf-intersecting.
  Side $M N$ has to cross line $K L$ to reach $N$ , and it crosses it between $K$ and $L$ . So sides $M N$ and $K L$ cross. The same happens on the far side of $L M$ from $K$ , where $N K$ crosses $L M$ .
4. in the corner region beyond $L$ , between the extensions of $K L$ and $M L$ past $L$ ?
  Show solution
  AnswerNon-convex, dented at $L$ .
  Now $L$ lies inside $△ K M N$ , so the angle at $L$ is reflex. The corner regions beyond $K$ and beyond $M$ behave the same way.
  In all: 1 region gives a convex 4-gon, 2 give self-intersecting ones, and 4 (the inside and the three corner regions) give non-convex ones.
Short answerMediumNCERT Exercise Set 12.1, Q5
A wire is bent into a closed loop with four straight sides, and two of the sides cross. Can its four corners fail to lie in one plane? Explain.
Show solution
AnswerNo.
Only opposite sides can cross, say $A B$ and $C D$ meeting at $E$ . Two lines that meet lie in one plane, and that plane contains both lines, so it contains all four corners $A$ , $B$ , $C$ and $D$ . A self-intersecting 4-gon is therefore always flat: it cannot also be non-planar.

12.2 Parallelograms

A carpenter building a window frame can easily measure lengths, angles and diagonals, but cannot easily check that two long sides are exactly parallel. Yet the frame must be a parallelogram, a quadrilateral whose opposite sides are parallel (rectangles, rhombuses and squares are special cases). Which measurements guarantee it? Start with the facts you proved last year, collected here as one theorem.

3/4The diagonals cut each other in half at $O$ .

Fig. 12.5One parallelogram, four kinds of evidence.NCERT Fig. 12.6 · p. 57

Parallelogram ABCD, with AB along the bottom and DC along the top. Step 1 draws diagonal AC, shades the two triangles it makes and marks AB equal to DC and AD equal to BC with tick marks, with AC labelled common. Step 2 marks the angles: 54 degrees at A and C, and 126 degrees at B and D. Step 3 draws both diagonals meeting at O, with tick marks showing OA equal to OC and OB equal to OD. Step 4 highlights only AB and DC, marked as parallel and equal.

Theorem 1Properties of a parallelogram

In a parallelogram $A B C D$ : (a) the opposite sides are equal; (b) the opposite angles are equal; (c) the diagonals bisect each other.

(a) Diagonal $A C$ cuts the parallelogram into $△ A B C$ and $△ C D A$ (step 1 of Fig. 12.5). As $A B ∥ D C$ , the alternate angles $∠ B A C$ and $∠ D C A$ are equal; as $A D ∥ B C$ , $∠ B C A = ∠ D A C$ . Side $A C$ is common, so $△ A B C ≅ △ C D A$ (ASA). Hence $A B = C D$ and $B C = D A$ .

(b) Side $A B$ is a transversal of the parallel lines $A D$ and $B C$ , so the co-interior angles give $∠ A + ∠ B = 180^{\circ}$ . In the same way $∠ B + ∠ C = 180^{\circ}$ . Comparing the two, $∠ A = ∠ C$ ; likewise $∠ B = ∠ D$ (step 2).

(c) Let the diagonals meet at $O$ . In $△ O A B$ and $△ O C D$ : $A B = C D$ by (a), $∠ O A B = ∠ O C D$ and $∠ O B A = ∠ O D C$ (alternate angles). The triangles are congruent (ASA), so $O A = O C$ and $O B = O D$ (step 3).

Theorem 1 goes from parallelogram to measurements. The carpenter needs the opposite direction, from measurements to parallelogram. So write each part as if … then … and swap the two halves to get its converse, as you did in Chapter 9.

Table 12.2Theorem 1 and its converses

Part	If a quadrilateral is a parallelogram, then …	Converse: a quadrilateral is a parallelogram if …
(a)	its opposite sides are equal	both pairs of opposite sides are equal
(b)	its opposite angles are equal	both pairs of opposite angles are equal
(c)	its diagonals bisect each other	its diagonals bisect each other

A true statement can have a false converse, so each of these needs a proof of its own. (Use the lab to test them first: all three survive.) One natural plan is to take the old proof and walk it in reverse, but every reversed step must be checked.

Theorem 2Equal opposite sides

If both pairs of opposite sides of a quadrilateral are equal, it is a parallelogram.

Let $A B C D$ have $A B = C D$ and $B C = D A$ , and draw diagonal $A C$ . Then $△ A B C ≅ △ C D A$ by SSS, with $A C$ common. Matching angles are equal: $∠ B A C = ∠ D C A$ and $∠ B C A = ∠ D A C$ .

$B$ and $D$ lie on opposite sides of line $A C$ . (On the same side, the congruent triangles would force sides $B C$ and $D A$ to cross. Try it in Fig. 12.6.) So the equal angles are alternate angles for the transversal $A C$ , which gives $A B ∥ D C$ and $B C ∥ A D$ .

Notice the reversal: part (a) of Theorem 1 went from parallel sides, through ASA, to equal sides; this proof travels from equal sides, through SSS, back to parallel sides.

SimulationFind the fourth corner

A coordinate grid with fixed points A(-4, -1), B(-2, 3) and C(4, -1) and a draggable point D that snaps to whole-number positions. The four sides AB, BC, CD and DA are drawn with their lengths shown, so the learner can move D until CD equals AB and DA equals BC, and compare the two positions where that happens.

Fig. 12.6

A

B

and

C

are fixed. Drag

D

and watch the four side lengths.

Try this

Drag $D$ until $C D$ matches $A B$ and $D A$ matches $B C$ . There are two such points: find both.
At one of them $A B C D$ is a parallelogram. What goes wrong at the other?
At the parallelogram position, compare the step from $C$ to $D$ with the step from $B$ to $A$ .

Theorem 3Equal opposite angles

If both pairs of opposite angles of a quadrilateral are equal, it is a parallelogram.

The proof shows something more. In a quadrilateral, both pairs of opposite angles are equal and every two adjacent angles add up to $180^{\circ}$ say the same thing. The first gives the second as above; and if $∠ A + ∠ B = 180^{\circ} = ∠ B + ∠ C$ , then $∠ A = ∠ C$ . So Theorem 3 can be restated in terms of adjacent angles (see the quick check below).

Theorem 4Diagonals that bisect each other

A quadrilateral whose diagonals cut each other in half is a parallelogram.

Theorem 5One pair equal and parallel

If one pair of opposite sides of a quadrilateral is both equal and parallel, it is a parallelogram.

Suppose $A B ∥ D C$ and $A B = D C$ (step 4 of Fig. 12.5), and draw diagonal $A C$ . A quadrilateral with a pair of parallel sides is convex (Q3 of Practice 12.1), so $B$ and $D$ lie on opposite sides of $A C$ , and the alternate angles $∠ B A C$ and $∠ D C A$ are equal.

So $△ A B C ≅ △ C D A$ by SAS ( $A B = C D$ , $∠ B A C = ∠ D C A$ , $A C$ common). Hence $∠ B C A = ∠ D A C$ . These are alternate angles for the transversal $A C$ , so $B C ∥ A D$ , and $A B C D$ is a parallelogram.

Example 12.1

Angles that force a parallelogram

In quadrilateral $P Q R S$ , $∠ P = ∠ R = (2 x + 10)^{\circ}$ and $∠ Q = ∠ S = (3 x - 20)^{\circ}$ . Find $x$ and the four angles. Is $P Q R S$ a parallelogram?

Try it yourself first. Then reveal the solution one step at a time.

Example 12.2

A parallelogram inside a parallelogram

$A B C D$ is a parallelogram. Point $P$ is on side $A B$ and point $Q$ is on side $C D$ , with $A P = C Q$ . Prove that $P B Q D$ is a parallelogram.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 4 questions

Practice 12.2: questions with answers

1
Short answerEasyNCERT Exercise Set 12.2, Q1
True or false?
1. A parallelogram with two equal adjacent sides is a rhombus.
  TrueFalse
  Show answer
  Answer True
  Opposite sides of a parallelogram are equal, so if two adjacent sides are equal, all four are.
2. A parallelogram whose diagonals bisect each other is a rectangle.
  TrueFalse
  Show answer
  Answer False
  The diagonals of every parallelogram bisect each other (Theorem 1), so this tells you nothing new.
3. A rhombus with equal diagonals is a square.
  TrueFalse
  Show answer
  Answer True
  A rhombus is a parallelogram. In parallelogram $A B C D$ with $A C = B D$ : $△ A B C ≅ △ B A D$ (SSS: $A B$ common, $B C = A D$ , $A C = B D$ ), so $∠ A B C = ∠ B A D$ . These co-interior angles add up to $180^{\circ}$ , so each is $90^{\circ}$ : a parallelogram with equal diagonals is a rectangle. A rhombus with right angles is a square.
4. A rectangle with perpendicular diagonals is a square.
  TrueFalse
  Show answer
  Answer True
  The diagonals bisect each other at right angles, so the four triangles round their meeting point are congruent (SAS) and all four sides are equal.
ProofMediumNCERT Exercise Set 12.2, Q2
In parallelogram $K L M N$ , $∠ L K M = ∠ N K M = 35^{\circ}$ . Find $∠ K M L$ and $∠ K M N$ , show that $K L M N$ is a rhombus, and find all four of its angles.
Show solution
Answer $∠ K M L = ∠ K M N = 35^{\circ}$ . The angles are $70^{\circ}, 110^{\circ}, 70^{\circ}, 110^{\circ}$
As $K N ∥ L M$ , $∠ K M L = ∠ N K M = 35^{\circ}$ (alternate angles). As $K L ∥ N M$ , $∠ K M N = ∠ L K M = 35^{\circ}$ . So diagonal $K M$ bisects $∠ M$ as well as $∠ K$ .
In $△ K L M$ the angles at $K$ and $M$ are both $35^{\circ}$ , so the sides opposite them are equal: $L M = K L$ . Two adjacent sides of the parallelogram are equal, so all four are, and $K L M N$ is a rhombus.
$∠ K = ∠ M = 35^{\circ} + 35^{\circ} = 70^{\circ}$ , and $∠ L = ∠ N = 180^{\circ} - 70^{\circ} = 110^{\circ}$ .
3
Short answerMediumNCERT Exercise Set 12.2, Q3
Answer Yes or No. If No, add one condition that makes the answer Yes.
1. Diagonal $W Y$ of quadrilateral $W X Y Z$ bisects both $∠ W$ and $∠ Y$ . Must $W X Y Z$ be a rhombus?
  Show solution
  AnswerNo. Add: diagonal $X Z$ also bisects $∠ X$ and $∠ Z$ .
  In $△ W X Y$ and $△ W Z Y$ , $W Y$ is common, $∠ X W Y = ∠ Z W Y$ and $∠ X Y W = ∠ Z Y W$ , so the triangles are congruent (ASA): $W X = W Z$ and $Y X = Y Z$ . That is a kite, and a kite need not have four equal sides.
  If diagonal $X Z$ also bisects $∠ X$ and $∠ Z$ , the same argument gives $X W = X Y$ and $Z W = Z Y$ . Then all four sides are equal: a rhombus.
2. The diagonals of a quadrilateral cross at right angles. Must it be a rhombus?
  Show solution
  AnswerNo. Add: the diagonals also bisect each other.
  A kite has perpendicular diagonals, but its sides come in two different pairs. If the diagonals also bisect each other, the quadrilateral is a parallelogram (Theorem 4), and the four right triangles round the centre are congruent (SAS), so all sides are equal: a rhombus.
3. A mason checks the outline of a room by measuring both diagonals, and they are equal. Must the outline be a rectangle?
  Show solution
  AnswerNo. Add: both pairs of opposite sides are equal too.
  An isosceles trapezium also has equal diagonals. If the opposite sides are also equal, the outline is a parallelogram (Theorem 2), and a parallelogram with equal diagonals is a rectangle (Q1(c)). Masons check opposite sides as well as diagonals for exactly this reason.
ProofHardNCERT Exercise Set 12.2, Q4
A park $P Q R S$ is a parallelogram with $P Q = 40$ m and $Q R = 25$ m. Paths run along the bisectors of its four corner angles and cross in pairs at four points. Show that these four points form a rectangle. What would happen if the park were a rhombus?
Show solution
Adjacent angles such as $∠ P$ and $∠ Q$ add up to $180^{\circ}$ , so their halves add up to $90^{\circ}$ . In the triangle formed by side $P Q$ and the two bisectors, the third angle is therefore $180^{\circ} - 90^{\circ} = 90^{\circ}$ : the bisectors from $P$ and $Q$ meet at right angles.
The same holds for each pair of adjacent corners, so the quadrilateral formed by the four meeting points has four right angles: it is a rectangle. Its four corners are separate points because $P Q \neq Q R$ .
In a rhombus each angle bisector is a diagonal. The bisectors from $P$ and $R$ lie on one line, as do those from $Q$ and $S$ , so all four meet at the single point where the diagonals cross: there is no rectangle.

12.3 Applications of Parallelograms

Push a hinged parallelogram out of shape and its opposite sides stay parallel, a property engineers put to work in lamps and drawing tools (see the box). For us the bigger prize lies inside geometry: parallelograms let us prove facts about triangles that are hard to reach directly. (Physics will show you one more use, combining two forces by completing a parallelogram.)

12.3.1 An Important Property of Triangles

ActivityNCERT Activity: four triangles · p. 61

Four triangles from one

You’ll need

squared paper
a ruler and a pencil
tracing paper

Predict first

You join the midpoints of the three sides of a triangle, making four small triangles. How will they compare?

The middle one is larger than the corner onesAll four are identical in shape and sizeOnly the three corner pieces matchThey have equal areas but different shapes

On squared paper, plot $A (2, 10)$ , $B (0, 0)$ and $C (12, 2)$ and draw $△ A B C$ .
Mark the midpoint of each side. Every coordinate is even, so each midpoint lands exactly on a grid point: count squares to find it.
Join the three midpoints. Trace one of the four small triangles and lay the tracing on each of the others, turning it if needed.
Count the steps: going from the midpoint of $A B$ to the midpoint of $A C$ , how many squares across and up? Going from $B$ to $C$ ?

Table 12.3

Segment	Squares across	Squares up
Midpoint of AB to midpoint of AC
B to C

Your entries are saved automatically.

2/4Join the midpoints: four triangles, numbered 1 to 4.

Fig. 12.7Joining the midpoints cuts any triangle into four identical pieces.NCERT Fig. 12.12 · p. 61

A scalene triangle. Step 1 marks the midpoint of each side, with tick marks showing equal halves. Step 2 joins the midpoints, splitting the triangle into four pieces numbered 1 to 4, with piece 4 in the middle, upside down. Step 3 pulls the pieces apart: pieces 1, 2 and 3 point the same way. Step 4 gives piece 4 a half-turn and moves it beside the others, and all four pieces look identical.

Counting squares worked for one triangle; a proof must work for all. The step counts point the way. Take $△ A B C$ with $M$ the midpoint of $A B$ and $N$ the midpoint of $A C$ , and prove the fact about the single join $M N$ : it is parallel to $B C$ and half as long. Applied to all three joins, that fact will give the four congruent triangles (Q1 of Practice 12.3).

4/4 $M B C K$ has $C K$ equal and parallel to $B M$ , so it is a parallelogram: $M K ∥ B C$ and $M K = B C$ .

Fig. 12.8Doubling

M N

turns the problem into a parallelogram.NCERT Fig. 12.14 · p. 62

Triangle ABC with M the midpoint of AB and N the midpoint of AC, joined by an orange segment MN. Step 2 extends MN beyond N by its own length to a point K, with tick marks showing MN equal to NK, and joins C to K. Step 3 shades triangles AMN and CKN, with tick marks showing AN equal to NC and AM equal to CK. Step 4 shades the parallelogram MBCK, with arrow marks showing MN parallel to BC and BM parallel to CK.

Theorem 6The Midpoint Theorem

In any triangle, the segment between the midpoints of two sides runs parallel to the remaining side and is exactly half as long.

Construction. Extend $M N$ beyond $N$ to a point $K$ with $N K = M N$ , and join $C K$ (step 2 of Fig. 12.8).

In $△ A M N$ and $△ C K N$ : $A N = C N$ ( $N$ is the midpoint of $A C$ ), $M N = K N$ (by construction) and $∠ A N M = ∠ C N K$ (vertically opposite). So $△ A M N ≅ △ C K N$ (SAS). Hence $C K = A M$ , and $∠ N C K = ∠ N A M$ ; these are alternate angles for the transversal $A C$ , so $C K ∥ A B$ .

Now $C K = A M = M B$ and $C K ∥ M B$ . So quadrilateral $M B C K$ has a pair of opposite sides that are equal and parallel, and by Theorem 5 it is a parallelogram.

Therefore $M K ∥ B C$ and $M K = B C$ . Since $M N$ is half of $M K$ , we get $M N ∥ B C$ and $M N = \frac{1}{2} B C$ .

Example 12.3

Half the gap

A step-ladder stands on level ground like the letter A. A rope tied to the midpoints of its two legs stops them sliding apart. The feet are $1.3$ m apart. How long is the rope, and which way does it run?

Try it yourself first. Then reveal the solution one step at a time.

Now reverse the roles. Start from the midpoint $M$ of $A B$ alone, and draw the line through $M$ parallel to $B C$ . Where does it cross $A C$ ? In the activity's triangle, the parallel through $(1, 5)$ moves 6 across for every 1 up, so it reaches $(7, 6)$ , exactly the midpoint of $A C$ . (Other ways of turning the Midpoint Theorem round are explored in Q22 of the end-of-chapter exercises.)

3/3There is only one line through $M$ parallel to $B C$ , so the two lines are the same and $Q = N$ .

Fig. 12.9Only one line through

M

is parallel to

B C

.NCERT Fig. 12.16 · p. 64

Triangle ABC with M the midpoint of AB. Step 1 draws a dashed line through M parallel to BC, meeting AC at a point Q marked with a question mark. Step 2 marks the midpoint N of AC and joins MN, with tick marks for the equal halves of AB and AC. Step 3 shows that the dashed line and MN are the same line, so Q and N are the same point.

Theorem 7Converse of the Midpoint Theorem

In a triangle, take the midpoint of one side and draw the line through it parallel to a second side. This line meets the third side at its midpoint, and the part of it inside the triangle is half as long as the side it is parallel to.

Let $M$ be the midpoint of $A B$ , and let the line through $M$ parallel to $B C$ meet $A C$ at $Q$ . Set $Q$ aside for a moment and mark the midpoint $N$ of $A C$ (Fig. 12.9).

By the Midpoint Theorem, $M N ∥ B C$ and $M N = \frac{1}{2} B C$ . But through $M$ there is only one line parallel to $B C$ . So line $M N$ and line $M Q$ are the same line, and they cross $A C$ at the same point: $Q = N$ . Hence $Q$ is the midpoint of $A C$ , and $M Q = M N = \frac{1}{2} B C$ .

You can also prove Theorem 7 by running the proof of Theorem 6 backwards: see Q15(b) of the end-of-chapter exercises.

12.3.2 A Surprising Property of Medians

Where should you hold a cardboard triangle so that it balances flat on one fingertip? The answer comes from its medians: the three segments joining each vertex to the midpoint of the opposite side. In $△ A B C$ with midpoints $D$ , $E$ , $F$ of $B C$ , $C A$ and $A B$ , the medians are $A D$ , $B E$ and $C F$ . When three or more lines pass through one common point they are called concurrent, and the activity below suggests the medians always are.

Try thisNCERT Fig. 12.17 · p. 65

Balance a triangle

You’ll need

a sheet of stiff card
a ruler, a pencil and scissors

Predict first

Where on median

A D

do you expect the three medians to meet?

At the midpoint of

A D

One-third of the way from

A

D

Two-thirds of the way from

A

D

At a different place in every triangle

Cut a triangle about 20 cm across out of the card.
Find the midpoint of each side and draw the three medians. Do they meet at one point? Call it $G$ .
Measure each median, and the part of it from the vertex to $G$ . Fill in the table.
Try to balance the triangle flat on the tip of a pencil placed under $G$ .

Table 12.4

Median	Whole length (cm)	Vertex to G (cm)	Ratio of the two parts
AD
BE
CF

Your entries are saved automatically.

The claim is that the medians are concurrent and that their common point cuts every median into two parts in the ratio $2 : 1$ . How can a theorem about halves produce thirds? Mark the midpoints of the two longer parts; if those midpoints and the meeting point cut each median into three equal pieces, we are done. Begin with just two medians.

4/4The third median $A D$ passes through $G$ too, and $A G = 2 G D$ .

Fig. 12.10A parallelogram hidden inside the triangle cuts each median into thirds.NCERT Fig. 12.18 · p. 65

Triangle ABC with midpoints D, E and F of BC, CA and AB. Step 1 draws medians BE and CF meeting at G. Step 2 marks X and Y, the midpoints of BG and CG, and shades the parallelogram FEYX, with FE and XY both parallel to BC. Step 3 marks each median cut into three equal parts: BX, XG and GE equal, and CY, YG and GF equal. Step 4 adds the third median AD, which passes through G, with the note AG = 2 GD.

Theorem 8The Centroid Theorem

The three medians of a triangle pass through one point, which splits every median in the ratio $2 : 1$ , the longer part being next to the vertex. This point is the centroid of the triangle.

Let medians $B E$ and $C F$ of $△ A B C$ meet at $G$ , and let $X$ and $Y$ be the midpoints of $B G$ and $C G$ .

In $△ A B C$ , $F$ and $E$ are the midpoints of $A B$ and $A C$ , so $F E ∥ B C$ and $F E = \frac{1}{2} B C$ (Theorem 6). In $△ G B C$ , $X$ and $Y$ are the midpoints of $G B$ and $G C$ , so $X Y ∥ B C$ and $X Y = \frac{1}{2} B C$ .

So $F E$ and $X Y$ are equal and parallel, and $F E Y X$ is a parallelogram (Theorem 5). Its diagonals $F Y$ and $E X$ lie along $C F$ and $B E$ , so they meet at $G$ ; and they bisect each other (Theorem 1), so $G E = G X$ and $G F = G Y$ .

Together with $B X = X G$ and $C Y = Y G$ , this gives $B X = X G = G E$ and $C Y = Y G = G F$ . So $B G = 2 G E$ and $C G = 2 G F$ : $G$ divides both medians in the ratio $2 : 1$ .

Now repeat the argument with medians $A D$ and $B E$ . They meet at a point $G^{'}$ with $B G^{'} = 2 G^{'} E$ . Only one point of segment $B E$ divides it in the ratio $2 : 1$ from $B$ , so $G^{'} = G$ . Hence the third median also passes through $G$ , and $A G = 2 G D$ .

Example 12.4

A pump at the centroid

A triangular village pond has a fountain pump at its centroid $G$ . The median from the bathing steps at corner $A$ to the midpoint $D$ of the opposite bank is $27$ m long. How far is the pump from $A$ , and from $D$ ? The median from corner $B$ has its part $B G = 14$ m; how long is that whole median $B E$ ?

Try it yourself first. Then reveal the solution one step at a time.

Midpoints of a quadrilateral

Midpoints hold one more surprise. In the lab, tick Join the midpoints and drag $A$ , $B$ , $C$ , $D$ anywhere: make the shape long and thin, or dented. The shape $P Q R S$ inside, joining the midpoints $P$ , $Q$ , $R$ , $S$ of $A B$ , $B C$ , $C D$ , $D A$ , keeps both pairs of opposite sides parallel the whole time.

4/4Both pairs of opposite sides of $P Q R S$ are parallel: it is a parallelogram.

Fig. 12.11Two diagonals, four uses of the Midpoint Theorem.NCERT Fig. 12.21 · p. 67

A general quadrilateral ABCD with the midpoints P, Q, R and S of its sides. Step 2 draws diagonal AC dashed and highlights PQ and SR, both parallel to AC. Step 3 draws diagonal BD dashed and highlights QR and PS, both parallel to BD. Step 4 shades PQRS, a parallelogram.

Theorem 9Midpoint Theorem for Quadrilaterals

Join the midpoints of the sides of any quadrilateral in order: the result is always a parallelogram, called its Varignon parallelogram.

Example 12.5

Perimeter from the diagonals

The diagonals of quadrilateral $A B C D$ are $A C = 10$ cm and $B D = 14$ cm. Find the perimeter of its Varignon parallelogram $P Q R S$ . Then say what special parallelogram $P Q R S$ is when the diagonals of a quadrilateral are (a) perpendicular, (b) equal.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 5 questions

Practice 12.3: questions with answers

1
ProofMediumNCERT Exercise Set 12.3, Q1
In $△ X Y Z$ , the points $L$ , $M$ , $N$ are the midpoints of $Y Z$ , $Z X$ and $X Y$ .
1. Show that $△ L M N$ is congruent to $△ X N M$ and to two other triangles. Name them.
  Show solution
  Answer $△ X N M$ , $△ N Y L$ and $△ M L Z$ .
  By the Midpoint Theorem, $M N = \frac{1}{2} Y Z$ , $L M = \frac{1}{2} X Y$ and $L N = \frac{1}{2} X Z$ .
  In $△ X N M$ : $X N = \frac{1}{2} X Y = L M$ , $X M = \frac{1}{2} X Z = L N$ and $N M$ is common, so $△ X N M ≅ △ L M N$ (SSS). In the same way $△ N Y L$ and $△ M L Z$ have sides $\frac{1}{2} X Y$ , $\frac{1}{2} Y Z$ and $\frac{1}{2} X Z$ , so they are congruent to $△ L M N$ too. This proves what the tracing showed.
2. An old drawing of $△ X Y Z$ has faded, and only the three joins $L M$ , $M N$ , $N L$ can still be seen. How can you redraw $△ X Y Z$ exactly?
  Show solution
  Through each vertex of $△ L M N$ draw the line parallel to the opposite side. These three lines meet in pairs at $X$ , $Y$ and $Z$ .
  This works because, by the Midpoint Theorem, side $Y Z$ passes through $L$ and is parallel to $M N$ , and through $L$ there is only one line parallel to $M N$ . The same holds for the other two sides.
ProofMediumNCERT Exercise Set 12.3, Q2
A rope is stretched across a triangular field $P Q R$ from the midpoint $X$ of side $P Q$ to the midpoint $Y$ of side $P R$ . Show that every straight path from $P$ to a point $T$ on the far side $Q R$ is cut in half by the rope.
Show solution
Let $X Y$ meet $P T$ at $U$ . By the Midpoint Theorem $X Y ∥ Q R$ , so $X U ∥ Q T$ .
In $△ P Q T$ , the line $X U$ passes through the midpoint $X$ of $P Q$ and is parallel to $Q T$ . By the converse of the Midpoint Theorem it bisects $P T$ : $U$ is the midpoint of $P T$ .
3
ProofHardNCERT Exercise Set 12.3, Q3
In trapezium $A B C D$ , $A B ∥ D C$ , $A B = 14$ cm and $D C = 8$ cm. $G$ and $H$ are the midpoints of the diagonals $A C$ and $B D$ .
1. Prove that $G H ∥ A B$ .
  Show solution
  Let $M$ be the midpoint of $A D$ . By Theorem 6 in $△ A D C$ , $M G ∥ D C$ and $M G = \frac{1}{2} D C$ ; in $△ D A B$ , $M H ∥ A B$ and $M H = \frac{1}{2} A B$ .
  Since $D C ∥ A B$ , both $M G$ and $M H$ are lines through $M$ parallel to $A B$ , so they are the same line. Hence $M$ , $G$ , $H$ lie on one line parallel to $A B$ , and $G H ∥ A B$ .
2. Find $G H$ in cm.
  cm
  Show answer
  Answer 3 cm
  $G H = M H - M G = \frac{1}{2} (14) - \frac{1}{2} (8) = 7 - 4 = 3$ cm.
3. Why must $A B \neq D C$ for this question to make sense?
  Show solution
  If $A B = D C$ , then $A B C D$ is a parallelogram (Theorem 5), its diagonals bisect each other, and $G$ and $H$ are the same point. There would be no segment $G H$ .
4
ProofMediumNCERT Exercise Set 12.3, Q4
$P$ , $Q$ , $R$ , $S$ are the midpoints of sides $A B$ , $B C$ , $C D$ , $D A$ of quadrilateral $A B C D$ .
1. Prove that $P R$ and $Q S$ cut each other in half.
  Show solution
  $P R$ and $Q S$ are the diagonals of the parallelogram $P Q R S$ (Theorem 9), and the diagonals of a parallelogram bisect each other.
2. If $A C = B D = 12$ cm, find the length of each side of $P Q R S$ .
  cm
  Show answer
  Answer 6 cm
  Each side is half a diagonal: $\frac{1}{2} (12) = 6$ cm. All four sides are equal, so $P Q R S$ is a rhombus, and its diagonals $P R$ and $Q S$ are perpendicular.
3. Conversely, if $P R ⊥ Q S$ , must $A C = B D$ ?
  Show solution
  AnswerYes.
  $P Q R S$ is a parallelogram whose diagonals bisect each other at right angles, so it is a rhombus (Practice 12.2, Q3(b)). Then $P Q = Q R$ , that is, $\frac{1}{2} A C = \frac{1}{2} B D$ , so $A C = B D$ .
5
ProofHardNCERT Exercise Set 12.3, Q5
$P Q R S$ is the Varignon parallelogram of quadrilateral $A B C D$ ( $P$ on $A B$ , $Q$ on $B C$ , $R$ on $C D$ , $S$ on $D A$ ).
1. You are given $P Q R S$ and the position of vertex $A$ only. Show how to rebuild $B$ , $C$ and $D$ .
  Show solution
  Extend $A P$ beyond $P$ by its own length to get $B$ (a half-turn of $A$ about $P$ ). In the same way, $C$ is the half-turn of $B$ about $Q$ , and $D$ is the half-turn of $C$ about $R$ .
2. Explain why $S$ turns out to be the midpoint of $D A$ in the rebuilt figure.
  Show solution
  Let $S^{'}$ be the midpoint of $D A$ in the rebuilt quadrilateral. By Theorem 9, $P Q R S^{'}$ is a parallelogram. But $P Q R S$ is also a parallelogram, and $P$ , $Q$ , $R$ fix the fourth vertex of a parallelogram uniquely. So $S^{'} = S$ .
3. Does $P Q R S$ alone fix the shape of $A B C D$ ?
  Show solution
  AnswerNo.
  Starting from a different point $A$ gives a different quadrilateral with the same Varignon parallelogram. You need one vertex as well. (This matters for Method 2 of tiling, where a tile is placed on a grid cell.)
4. If $P Q R S$ is a square, show that $A C$ and $B D$ are equal and perpendicular. Is the converse true?
  Show solution
  AnswerYes, the converse is true.
  $P Q = \frac{1}{2} A C$ and $Q R = \frac{1}{2} B D$ , so $P Q = Q R$ gives $A C = B D$ . Also $P Q ∥ A C$ and $Q R ∥ B D$ , so $P Q ⊥ Q R$ gives $A C ⊥ B D$ .
  Conversely, if $A C = B D$ and $A C ⊥ B D$ , the parallelogram $P Q R S$ has two equal adjacent sides and a right angle, so it is a square.

12.4 Tiling the Plane Using Any 4-gon

Look for tilings and you find them everywhere: the cells of a beehive, the stone lattice of a Mughal jaali screen, the woven squares of a charpai, the grid under a rangoli. Some use one shape and some use several. The tile factory's question is about one shape: can copies of any quadrilateral tile the plane?

ActivityNCERT Figs. 12.26–12.27 · p. 69

Spin and trace

You’ll need

thin card and a large sheet of paper
a ruler, a pencil and scissors
a drawing pin and a piece of thick cardboard

Predict first

Number the corners of a lopsided 4-gon 1, 2, 3, 4. If four copies fit snugly round one point, which corners meet there?

Four copies of corner 1The four largest corners availableCorners 1, 2, 3 and 4, one of eachNo four copies can fit round a point

Cut a 4-gon about 6 cm across out of card, with no equal sides and no parallel sides. Number its corners 1, 2, 3, 4 and mark the midpoint of each edge.
Lay a large sheet on a piece of thick cardboard. Put the template in the middle, trace round it and copy the corner numbers into the tracing.
Push a pin through the midpoint of one edge, spin the template half a turn about the pin, and trace it again with its numbers.
Repeat from every edge of every tracing. Look closely at a point where four tracings meet.

Why should the four corners fit? Round any point where tiles meet, the angles must fill a full turn of $360^{\circ}$ , and the four angles of a quadrilateral add up to exactly that. The angle sum is the first clue.

Method 1: half-turns. This is exactly what the pin did. A turn of $180^{\circ}$ (a half-turn) about the midpoint of an edge swaps the edge's two ends. So the turned copy sits on the far side of that edge and shares all of it (Fig. 12.12). Applying half-turns at every edge of every new copy spreads copies over the whole plane.

4/4At every corner, angles 1, 2, 3 and 4 meet once each and fill $360^{\circ}$ .

Fig. 12.12Half-turns about the midpoints of the sides place every new copy.NCERT Figs. 12.29–12.30 · p. 71

An irregular convex 4-gon in blue with corners numbered 1 to 4. Step 1 shows a copy lying on top of it and a dot at the midpoint of its right-hand side. Step 2 turns the copy through 180 degrees about the dot, so it lies beside the tile, sharing that side, with its numbers reversed. Step 3 adds copies turned about the midpoints of the other three sides, surrounding the tile. Step 4 adds one more blue copy and circles a corner where angles 1, 2, 3 and 4 meet once each.

Why does the method never jam or overlap? Pair a tile with its half-turned neighbour: together they form a six-sided shape whose opposite sides are equal and parallel. Such hexagons slot together in rows, like the cells of a beehive. So every blue copy is a slide of the first tile, and every orange copy is a slide of its turned partner. Q23 of the end-of-chapter exercises fills in the details.

InteractiveTile with any 4-gon

A tiling of the screen by copies of one 4-gon: blue copies are slides of the thick tile and orange copies are half-turns of it about the midpoint of one side. The four corners of the thick tile can be dragged, or moved with the arrow keys, and the tiling redraws at once. Numbers 1 to 4 in every copy show the four angles meeting at each vertex. A checkbox draws the lines through the midpoints of the sides, which form a grid of identical parallelograms, and buttons load a general 4-gon, a dart and a kite. A message says whether the tile is convex, dented or crossed, and warns that a crossed 4-gon gives overlaps instead of a tiling.

Fig. 12.13Drag the corners of the thick tile and the whole tiling follows. Try a dart, then try to make the sides cross.

Method 2: start from a grid. Draw a grid made of copies of the tile's Varignon parallelogram, like the rows of parallelograms at the start of the chapter. Choose cells spaced like the blue ones in Fig. 12.14, skipping one cell along each row and each column. On each chosen cell place a tile with its side midpoints on the cell's corners. The placed tiles touch only at corners, and every gap between them is a half-turned copy of the tile. Why should this work? Tick Midpoint grid in the tiler: the Varignon parallelograms of the half-turn tiling line up into exactly such a grid.

3/3Each gap is a half-turned copy of the tile (orange), and the tiling is complete.

Fig. 12.14The Varignon parallelograms of the tiles form a grid.NCERT Figs. 12.31–12.32 · p. 72

A grid of dashed lines forming identical parallelograms. Step 1 shades chosen cells in blue, spaced one cell apart in both directions. Step 2 places a blue 4-gon on each chosen cell, with the midpoints of its sides at the corners of the cell; neighbouring tiles touch only at corners. Step 3 fills each gap with an orange half-turned copy, completing the tiling.

Neither method needs the tile to be convex: a dented 4-gon tiles in exactly the same way (Q2 of Practice 12.4). A crossed 4-gon is different, because its copies overlap. So the answer to our opening question is yes: copies of any quadrilateral can tile the plane. Not every shape tiles, though; you will meet one that cannot in Practice 12.4.

Practice · 2 questions

Practice 12.4: questions with answers

Short answerMediumNCERT Exercise Set 12.4, Q1
Can copies of a regular octagon tile the plane on their own? Explain using angles. Which regular polygons can?
Show solution
AnswerNo. Only the equilateral triangle, the square and the regular hexagon can.
Walking round a regular octagon you turn $360^{\circ} \div 8 = 45^{\circ}$ at each corner, so each internal angle is $180^{\circ} - 45^{\circ} = 135^{\circ}$ .
Where tiles meet at a point, their angles must add up to $360^{\circ}$ ; if the point lies on the edge of one tile, the others must fill $180^{\circ}$ . But $360 \div 135$ and $180 \div 135$ are not whole numbers, so corners of $135^{\circ}$ can never fill the space round a point. Floors that use octagons always need small squares as well.
A regular polygon tiles on its own only if its angle divides $360^{\circ}$ : the equilateral triangle ( $60^{\circ}$ ), the square ( $90^{\circ}$ ) and the regular hexagon ( $120^{\circ}$ ). A regular pentagon ( $108^{\circ}$ ) fails, and any polygon with 7 or more sides has an angle between $120^{\circ}$ and $180^{\circ}$ , so it fails too.
ActivityMediumNCERT Exercise Set 12.4, Q2
Write a step-by-step recipe that a friend could follow to tile a floor with copies of a dented 4-gon $W X Y Z$ (the dent is at $Z$ ) using half-turns. Then say how the grid method would begin, and which of the two you would rather use.
Show solution
Method 1: give a copy a half-turn about the midpoint of each side, and repeat at every new edge. The reflex corner $Z$ is filled by corners of neighbouring copies; round every point the four angles still add up to $360^{\circ}$ .
Method 2: join the midpoints of the sides to get the Varignon parallelogram of $W X Y Z$ , repeat it to make a grid, and place a copy on every other cell in each direction, with its side midpoints on the cell's corners. The gaps are half-turned copies.
Either is fine: Method 1 needs only midpoints, while Method 2 needs the grid first. Check your tiling with the dart in the tiler.

Recap

Quadrilaterals: key points and summary

Four points in a plane, no three on a line, form quadrilateral $A B C D$ when the sides $A B$ , $B C$ , $C D$ , $D A$ meet only at shared corners (Section 12.1).
A quadrilateral is convex when every internal angle is less than $180^{\circ}$ ; equivalently, its diagonals cross (Convex quadrilaterals).
A parallelogram has equal opposite sides, equal opposite angles and diagonals that bisect each other (Theorem 1).
Each of these converses is true: equal opposite sides, equal opposite angles, or bisecting diagonals make a quadrilateral a parallelogram (Theorem 2, Theorem 3, Theorem 4).
One pair of opposite sides that is both equal and parallel is enough to make a parallelogram (Theorem 5).
Midpoint Theorem: in a triangle, the segment between the midpoints of two sides runs parallel to the remaining side and is half its length (Theorem 6).
Converse: draw the parallel to one side through the midpoint of another side, and it cuts the third side in half (Theorem 7).
The three medians meet at the centroid, which lies two-thirds of the way along each median from its vertex (Theorem 8).
The midpoints of the sides of any quadrilateral form a parallelogram with sides half the diagonals (Theorem 9).
Copies of any quadrilateral tile the plane: half-turns about the midpoints of the sides place the copies, and the four angles meeting at each corner add up to $360^{\circ}$ (Section 12.4).

End-of-chapter exercises

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 12.1, Q1	Practice 12.1 · Q1
NCERT Exercise Set 12.1, Q2	Practice 12.1 · Q2
NCERT Exercise Set 12.1, Q3	Practice 12.1 · Q3
NCERT Exercise Set 12.1, Q4	Practice 12.1 · Q4
NCERT Exercise Set 12.1, Q5	Practice 12.1 · Q5
NCERT Exercise Set 12.2, Q1	Practice 12.2 · Q1
NCERT Exercise Set 12.2, Q2	Practice 12.2 · Q2
NCERT Exercise Set 12.2, Q3	Practice 12.2 · Q3
NCERT Exercise Set 12.2, Q4	Practice 12.2 · Q4
NCERT Exercise Set 12.3, Q1	Practice 12.3 · Q1
NCERT Exercise Set 12.3, Q2	Practice 12.3 · Q2
NCERT Exercise Set 12.3, Q3	Practice 12.3 · Q3
NCERT Exercise Set 12.3, Q4	Practice 12.3 · Q4
NCERT Exercise Set 12.3, Q5	Practice 12.3 · Q5
NCERT Exercise Set 12.4, Q1	Practice 12.4 · Q1
NCERT Exercise Set 12.4, Q2	Practice 12.4 · Q2
NCERT End-of-chapter Exercises, Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises, Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises, Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises, Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises, Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises, Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises, Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises, Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises, Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises, Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises, Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises, Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises, Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Exercises, Q15	End-of-chapter exercises · Q15
NCERT End-of-chapter Exercises, Q16	End-of-chapter exercises · Q16
NCERT End-of-chapter Exercises, Q17	End-of-chapter exercises · Q17
NCERT End-of-chapter Exercises, Q18	End-of-chapter exercises · Q18
NCERT End-of-chapter Exercises, Q19	End-of-chapter exercises · Q19
NCERT End-of-chapter Exercises, Q20	End-of-chapter exercises · Q20
NCERT End-of-chapter Exercises, Q21	End-of-chapter exercises · Q21
NCERT End-of-chapter Exercises, Q22	End-of-chapter exercises · Q22
NCERT End-of-chapter Exercises, Q23	End-of-chapter exercises · Q23
NCERT End-of-chapter Exercises, Q24	End-of-chapter exercises · Q24

Quick check: MCQs on Quadrilaterals

Convex quadrilaterals4 questions
12.2 Parallelograms4 questions
12.3.1 An Important Property of Triangles4 questions
Midpoints of a quadrilateral4 questions
12.4 Tiling the Plane Using Any 4-gon3 questions

Glossary · 12

Key terms (English–Hindi)

centroidकेंद्रकkendrak: The point where the three medians of a triangle meet, two-thirds of the way along each median from its vertex.First seen in the chapter →
concurrentसंगामीsangami: Lines that all pass through one common point.First seen in the chapter →
converseविलोमvilom: The statement if B then A made from if A then B. It needs its own proof.First seen in the chapter →
convex quadrilateralउत्तल चतुर्भुजuttal chaturbhuj: A quadrilateral whose internal angles are all less than $180^{\circ}$ . A non-convex one has a dent, where one angle is more than $180^{\circ}$ .First seen in the chapter →

diagonalविकर्णvikarn: A segment joining two vertices of a polygon that are not ends of the same side.First seen in the chapter →

medianमाध्यिकाmadhyika: A segment joining a vertex of a triangle to the midpoint of the opposite side.First seen in the chapter →

non-planar 4-gon: A four-sided figure whose four corners do not all lie in one plane, like a card folded along a diagonal.First seen in the chapter →

parallelogramसमांतर चतुर्भुजsamantar chaturbhuj: A quadrilateral whose opposite sides are parallel.First seen in the chapter →

quadrilateralचतुर्भुजchaturbhuj: A flat figure made of four segments joining four points, no three on a line, in which the sides meet only at shared corners.First seen in the chapter →

self-intersecting 4-gon: A four-sided figure in which two sides cross each other; it is not counted as a quadrilateral here.First seen in the chapter →

tiling: A covering of the plane by copies of one or more shapes, with no gaps and no overlaps.First seen in the chapter →

Varignon parallelogram: The parallelogram formed by joining the midpoints of the sides of a quadrilateral in order.First seen in the chapter →

Quadrilaterals: Class 9 Ganita Manjari · Chapter 12

What you will learn in Quadrilaterals

Floors, tiles and 4-gons

12.1 What Exactly is a Quadrilateral?

Convex quadrilaterals

Practice 12.1: questions with answers

12.2 Parallelograms

Practice 12.2: questions with answers

12.3 Applications of Parallelograms

12.3.1 An Important Property of Triangles

12.3.2 A Surprising Property of Medians

Midpoints of a quadrilateral

Practice 12.3: questions with answers

12.4 Tiling the Plane Using Any 4-gon

Practice 12.4: questions with answers

Quadrilaterals: key points and summary

End-of-chapter exercises

Practice questions with answers (NCERT “End-of-chapter exercises”)

Find your NCERT exercise question

Quick check: MCQs on Quadrilaterals

Key terms (English–Hindi)

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