Ganita Manjari · Class 9 · Part II · Chapter 12

Ganita Manjari Class 9 Maths Chapter 12 Quadrilaterals Exercise 12.1 Solutions

All 5 questions on defining a quadrilateral: adjacent and opposite parts, internal angles, convex or non-convex, and why a 4-gon cannot be both crossing and non-planar.

Last updated 6 October 2026

5Questions
1Diagrams
Ch 12Quadrilaterals

How to answer every question

Use the definition A quadrilateral has 4 distinct points, no three collinear, and its sides meet only at vertices.
Draw it Sketch the figure, and a dent or a crossing becomes visible.
Test an example Use coordinates to check an idea.
Prove or give a counter-example One valid example shows a statement can happen; one failure shows it cannot always.

Adjacent: sharing an endpoint (sides) or joined by a side (vertices).

Convex: every internal angle is less than 180°, or equivalently the diagonals intersect.

Self-intersecting: two sides cross each other.

Basic definitions

Chapter 12 Exercise 12.1 Question 1 Solution : Let ABCD be a quadrilateral. (i) List all sides adjacent to AB and all sides opposite to AB. (ii) List all angles adjacent to ∠A and all angles opposite to ∠A. (iii) Define a pair of opposite sides and a pair of opposite angles without using the names of vertices.

Answer: sides, angles, definitions

Part (i) · Sides

Two sides are adjacent if they share an endpoint. AB has endpoints A and B. The side at B is BC and the side at A is DA.

Side	Sides adjacent to it	Side opposite to it
AB	BC and DA	CD
BC	AB and CD	DA

CD shares no endpoint with AB, so CD is opposite to AB.

Part (ii) · Angles

Angles at adjacent vertices (joined by a side) are adjacent. A is joined to B by AB and to D by DA.

Angle	Adjacent angles	Opposite angle
∠A	∠B and ∠D	∠C

C is the only vertex not joined to A by a side, so ∠C is opposite to ∠A.

Part (iii) · Definitions without names

Opposite sides: two sides of a quadrilateral that have no common endpoint.
Opposite angles: two internal angles whose vertices are not the endpoints of one side (they are the ends of a diagonal).

(i) Adjacent to AB: BC and DA; opposite to AB: CD. (ii) Adjacent to ∠A: ∠B and ∠D; opposite: ∠C. (iii) Opposite sides share no endpoint; opposite angles are at the two ends of a diagonal.

Chapter 12 Exercise 12.1 Question 2 Solution : Define precisely the internal angle of a quadrilateral at a given vertex. Your definition must work for a non-convex quadrilateral too. (Hint: use the opposite vertex as well.)

StarredAnswer: use the diagonal from the vertex

The problem

The two sides at A, AB and AD, split the plane into two angles, one smaller than 180° and one larger, and the two add up to 360°. For a convex quadrilateral the smaller one is internal. For a non-convex one, at the dent the internal angle is the larger one.

The definition

Join A to the opposite vertex C. Look at the diagonal AC.

If AC lies inside the quadrilateral, the internal angle at A is ∠BAC + ∠CAD. This can be more than 180°.
If AC lies outside the quadrilateral, the internal angle at A is the angle ∠BAD between AB and AD that is less than 180°.

Check on the non-convex 4-gon DART

In DART the dent is at D, and the diagonal DR lies inside. So the internal angle at D is ∠ADR + ∠RDT, which is more than 180°. At A the other diagonal AT lies outside, so the internal angle at A is just ∠DAR, which is less than 180°.

Every other vertex works the same way, and the four internal angles add up to 360°.

The internal angle at A is ∠BAC + ∠CAD if the diagonal AC lies inside the quadrilateral, and the angle between the two sides that is less than 180° otherwise.

Convex or not

Chapter 12 Exercise 12.1 Question 3 Solution : In a quadrilateral ABCD, suppose AB ∥ DC. Can ABCD be non-convex? What if we instead assume AB = CD? What if we instead assume ∠A = ∠C?

Answer: No, Yes, Yes

Case 1 · AB ∥ DC: cannot be non-convexNot possible

Since ABCD does not cross itself, B and C are on the same side of line AD. Then ∠A and ∠D are interior angles on the same side of the transversal AD, so ∠A + ∠D = 180°. In the same way ∠B + ∠C = 180°.

So every internal angle is less than 180°, which means ABCD is convex. (This is also why the diagonals must intersect in Theorem 5.)

Case 2 · AB = CD: can be non-convexPossible

Take A(0, 0), B(4, 0), C(3, 1), D(0.6, 4.2). Then AB = 4 and CD = √(2.4² + 3.2²) = 4. The point C lies inside triangle ABD, so ABCD has a dent at C.

Equal sides alone do not force the shape.

Case 3 · ∠A = ∠C: can be non-convexPossible

Take the arrowhead A(−2, 0), B(0, −1), C(2, 0), D(0, −3). It is symmetric about the line BD, so ∠A = ∠C (about 29.7° each). The point B lies inside triangle ACD, so the dent is at B, where the internal angle is more than 180°.

(i) If AB ∥ DC, ABCD is always convex. (ii) If AB = CD, it can be non-convex. (iii) If ∠A = ∠C, it can be non-convex.

Chapter 12 Exercise 12.1 Question 4 Solution : Take three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible position of D outside these lines, decide whether ABCD is self-intersecting, non-convex or convex. (Hint: the three lines divide the plane into 7 regions.)

Answer: 1 convex, 2 self-intersecting, 4 non-convex

The seven regions

The triangle itself is one region. Three regions touch a side of the triangle, and three regions touch only a vertex. The sides of ABCD are AB, BC, CD and DA, so the point D joins to C and to A.

Q4 · the seven regions and the kind of 4-gon ABCD in each

Reading the result

Region of D	What ABCD is	Why
1. Inside ∆ABC	non-convex	D is a dent (reflex angle at D)
2. Across AB (opposite C)	self-intersecting	BC and DA cross
3. Across BC (opposite A)	self-intersecting	AB and CD cross
4. Across CA (opposite B)	convex	the diagonals AC and BD intersect
5. Beyond C	non-convex	C becomes a dent
6. Beyond A	non-convex	A becomes a dent
7. Beyond B	non-convex	B becomes a dent

I checked this by testing many points in each region.

Convex only in region 4 (across CA). Self-intersecting in regions 2 and 3. Non-convex in region 1 and the three vertex regions 5, 6, 7.

Chapter 12 Exercise 12.1 Question 5 Solution : Can a quadrilateral be both self-intersecting and non-planar?

Answer: No

Why notImpossible

A self-intersecting quadrilateral has two sides that cross, say AB and CD cross at a point E. Two lines that meet at a point lie in one plane. So A, B, C and D all lie in the plane of those two lines.

Therefore the quadrilateral is planar. The same works if BC and DA are the sides that cross.

No. If two sides cross, the four vertices lie in the plane of those two crossing lines, so the quadrilateral is planar.

Answers at a glance

Each answer is checked with a figure or coordinates.

Question	What is asked	Key idea	Answer
Q1	Adjacent and opposite	Shared endpoint or not	Adjacent to AB: BC, DA; opposite: CD
Q2	Internal angle at a vertex	Look at the diagonal from the vertex	∠BAC + ∠CAD if AC is inside
Q3	AB ∥ DC / AB = CD / ∠A = ∠C	Co-interior angles; examples	Convex; can be non-convex; can be non-convex
Q4	Position of D	Seven regions	1 convex, 2 self-intersecting, 4 non-convex
Q5	Crossing and non-planar	Crossing lines are coplanar	No

Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 12, Exercise Set 12.1. The solutions, explanations and diagrams on this page are our own working.

How to answer every question

Basic definitions

Convex or not

Answers at a glance

More Chapter 12 Quadrilaterals solutions

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