NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals

The NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals provide complete exercise-wise question answers with clear explanations, step-by-step proofs and labelled diagrams. These solutions cover all Exercise Sets 12.1 to 12.4, and the End-of-Chapter Exercises, including starred questions. Students can learn important concepts related to parallelograms, rhombuses, rectangles, squares, the Midpoint Theorem, Centroid Theorem and tiling patterns.

Arranged according to the textbook sequence and page numbers, the solutions make it easier to locate questions, understand the reasoning behind each answer and practise writing complete mathematical solutions. Students can also use the Class 9 Maths Ganita Manjari Chapter 12 Quadrilaterals Solutions PDF for revision and offline study.

1. Let ABCD be a quadrilateral.

(i) List all sides of ABCD adjacent to side AB. List all sides opposite to AB.

The sides adjacent to AB are AD and BC.

The side opposite to AB is CD.

(ii) List all angles of ABCD adjacent to ∠A. List all angles opposite to ∠A.

The angles adjacent to ∠A are ∠B and ∠D.

The angle opposite to ∠A is ∠C.

(iii) Define a pair of opposite sides and a pair of opposite angles without using the names of the vertices.

A pair of opposite sides of a quadrilateral are two sides that do not share a common vertex.

A pair of opposite angles are two angles that are not adjacent, that is, they do not share a common side.

3. 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:

(i) If AB ∥ DC

No. A simple quadrilateral ABCD cannot be non-convex if the opposite sides AB and DC are parallel.

Since AB and DC lie on parallel lines, the other two sides AD and BC must join the endpoints without crossing. This gives a convex quadrilateral, so no vertex can form a reflex angle.

Therefore, AB ∥ DC implies that ABCD is convex.

(ii) If AB = CD

Yes. A quadrilateral can still be non-convex even if the opposite sides AB and CD are equal.

For example, take:

A(0, 0), B(2, 0), C(1, 1), D(1, 3).

Here, AB = CD = 2 units, but C is an inward-pointing vertex, so ABCD is non-convex.

Therefore, AB = CD does not guarantee that ABCD is convex.

(iii) If ∠A = ∠C

Yes. A quadrilateral can have equal opposite angles and still be non-convex.

For example, take:

A(−2, 0), B(0, 1), C(2, 0), D(0, 3).

The figure is symmetric, so ∠A = ∠C, but B is an inward-pointing vertex and forms a reflex angle.

Therefore, ∠A = ∠C does not guarantee that ABCD is convex.

Hence:

• AB ∥ DC: ABCD must be convex.
• AB = CD: ABCD may be non-convex.
• ∠A = ∠C: ABCD may be non-convex.

Answer: 

No. A quadrilateral cannot be both self-intersecting and non-planar.

If a quadrilateral is self-intersecting, a pair of its non-adjacent sides must cross each other. Two intersecting lines always lie in the same plane. Since all four vertices lie on those two intersecting sides, all four vertices must lie in that same plane.

Therefore, a self-intersecting quadrilateral is necessarily planar.

Hence, a quadrilateral cannot be both self-intersecting and non-planar.

NCERT Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.2

NCERT Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.3

1. (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?

Answer: 

(i) Show that △PQR is congruent to △QPA and two other triangles.

Since P, Q and R are the midpoints of AB, AC and BC respectively,

AP = PB = AB/2
AQ = QC = AC/2
BR = RC = BC/2

By the Midpoint Theorem,

PQ = BC/2,
QR = AB/2,
PR = AC/2.

Now compare △PQR and △QPA:

• PQ = BC/2 = QA? Wait no, better pair by side lengths:
  • PQ = BC/2
  • QR = AB/2
  • PR = AC/2

For △QPA:

• QP = BC/2
• PA = AB/2
• QA = AC/2

Therefore,

PQ = QP, QR = PA, PR = QA.

Hence,

△PQR ≅ △QPA by SSS congruence.

Similarly,

• △PRB has sides AB/2, BC/2 and AC/2.
• △QRC also has sides AB/2, BC/2 and AC/2.

Therefore,

△PQR ≅ △QPA ≅ △PRB ≅ △QRC.

So the two other triangles are △PRB and △QRC.

(ii) Can △ABC be reconstructed from △PQR?

Yes.

Using the Midpoint Theorem,

• QR ∥ AB
• PR ∥ AC
• PQ ∥ BC

So, starting with △PQR:

1. Through P, draw a line parallel to QR.
2. Through Q, draw a line parallel to PR. Their intersection gives A.
3. Through R, draw a line parallel to PQ.
4. The line through P parallel to QR meets the line through R parallel to PQ at B.
5. The line through Q parallel to PR meets the line through R parallel to PQ at C.

Thus, the original △ABC can be reconstructed uniquely from △PQR.

Let MN meet AD at X.

Since M and N are the midpoints of AB and AC, by the Midpoint Theorem,

MN ∥ BC

As D lies on BC, we also have

MX ∥ BD

Now consider △ABD.
M is the midpoint of AB, and the line through M parallel to BD meets AD at X.

Therefore, by the converse of the Midpoint Theorem, X is the midpoint of AD.

Hence,

AX = XD

Therefore, MN bisects AD.

If D = B or D = C, the result follows directly because M or N is already the midpoint of AD.

Answer:

Let M be the midpoint of AD.

In △ADC, M is the midpoint of AD and G is the midpoint of AC. Therefore, by the Midpoint Theorem,

MG ∥ DC and MG = DC/2.

Similarly, in △ADB, M is the midpoint of AD and H is the midpoint of DB. Hence,

MH ∥ AB and MH = AB/2.

Given that

AB ∥ DC,

we get

MG ∥ AB and MH ∥ AB.

Since both MG and MH pass through M and are parallel to AB, they lie on the same straight line. Thus, G, M and H are collinear.

Therefore,

GH ∥ AB.

Why is AB ≠ CD assumed?

If AB = CD and AB ∥ CD, then ABCD is a parallelogram. The diagonals of a parallelogram bisect each other, so the midpoint G of AC and the midpoint H of BD would be the same point.

Thus,

G = H,

and GH would have zero length, so it would not form a proper segment parallel to AB.

Hence, the condition AB ≠ CD ensures that G and H are distinct points.

(i) Reconstructing a congruent quadrilateral A′B′C′D′

Since P, Q, R and S are the midpoints of AB, BC, CD and DA, respectively, we can rebuild the vertices successively.

1. On the copied parallelogram PQRS, construct a point A′ so that △SPA′ is congruent to the original △SPA. Choose the corresponding position of A′ relative to SP.
2. Draw the line A′P and extend it beyond P. Mark B′ so that
   A′P = PB′.
   Thus, P is the midpoint of A′B′.
3. Extend B′Q beyond Q and choose C′ such that
   B′Q = QC′.
   Hence, Q is the midpoint of B′C′.
4. Extend C′R beyond R and locate D′ so that
   C′R = RD′.
   Thus, R is the midpoint of C′D′.
5. Join D′ to A′.

This gives a quadrilateral A′B′C′D′ corresponding to ABCD.

(ii) Why is A′B′C′D′ congruent to ABCD?

By construction,

△SPA′ ≅ △SPA.

Also, B′ is obtained from A′ by a half-turn about P, just as B is obtained from A about P. In the same way,

• C′ is obtained from B′ by a half-turn about Q,
• D′ is obtained from C′ by a half-turn about R.

Thus the same sequence of midpoint constructions used in ABCD is reproduced in A′B′C′D′, so corresponding lengths and angles are preserved.

It remains to check that S is also the midpoint of A′D′.

Since Q and R are the midpoints of B′C′ and C′D′, respectively, the Midpoint Theorem in △B′C′D′ gives

QR ∥ B′D′ and QR = B′D′/2.

But PQRS is a parallelogram, so

PS ∥ QR and PS = QR.

Therefore,

PS ∥ B′D′ and PS = B′D′/2.

Now P is the midpoint of A′B′. Hence the line through P parallel to B′D′ meets A′D′ at its midpoint. That point is S. Therefore,

A′, S and D′ are collinear and SA′ = SD′.

So P, Q, R and S are exactly the midpoints of the four sides of A′B′C′D′, just as in ABCD. Hence,

A′B′C′D′ ≅ ABCD.

Different valid placements of A′ can produce translated, rotated or reflected copies, but all are congruent to ABCD.

(iii) If PQRS is a square, show that AC and BD are equal and perpendicular. Prove the converse.

By the Midpoint Theorem,

PQ ∥ AC and PQ = AC/2,

while

QR ∥ BD and QR = BD/2.

If PQRS is a square, then

PQ = QR and PQ ⟂ QR.

Therefore,

AC/2 = BD/2, so AC = BD,

and since PQ ∥ AC and QR ∥ BD,

AC ⟂ BD.

Hence, the diagonals AC and BD are equal and perpendicular.

Conversely, suppose

AC = BD and AC ⟂ BD.

Then

PQ = AC/2 = BD/2 = QR,

and, because PQ ∥ AC and QR ∥ BD,

PQ ⟂ QR.

Thus the Varignon parallelogram PQRS has two equal adjacent sides and a right angle. Therefore, PQRS is a square.

Hence,

PQRS is a square if and only if AC and BD are equal and perpendicular.

NCERT Class 9 Maths Ganita Manjari Chapter 12 Exercise Set 12.4

A regular pentagon cannot tile the plane because its interior angles do not fit exactly around a point.

For a regular pentagon, each interior angle is

108°.

To tile the plane without gaps or overlaps, the angles meeting at a vertex must add up to exactly 360°.

But:

3 × 108° = 324°, leaving a gap of 36°,

while

4 × 108° = 432°, which is more than 360°.

So no whole number of regular pentagons can meet at a point and completely fill the space.

Therefore, the plane cannot be tiled using regular pentagons alone.

Irregular pentagons can tile the plane because their angles need not all be equal and can be arranged so that the angles around each vertex add up to 360°.

2. Draw a non-convex 4-gon DART. Show how we can tile the plane with copies of DART. Both methods that we discussed earlier will work. Which do you prefer?

Answer: 

One example of a non-convex quadrilateral DART can be obtained by taking

D(1, 1), A(0, 3), R(0, 0) and T(3, 0),

and joining the points in the order

D – A – R – T – D.

Here D lies inside triangle ART, so the interior angle at D is greater than 180°. Hence, DART is a non-convex quadrilateral.

To tile the plane, we can use either of the following methods:

Method 1: Half-turn method

Take one copy of DART. Rotate a congruent copy through 180° about the midpoint of one of its sides. The rotated copy fits with the first one along that side. Continue the same process along the exposed sides. Repeating these half-turns produces a pattern of congruent copies that covers the plane without gaps or overlaps.

Method 2: Varignon parallelogram method

Construct the Varignon parallelogram of DART by joining the midpoints of its four sides. Copies of DART can then be arranged according to the repeating parallelogram pattern. Copies with the same orientation are placed at corresponding positions, while the remaining spaces are filled using half-turned copies.

Both constructions give a tiling of the plane even though DART is non-convex.

Preference: I prefer the half-turn method, because every new tile can be positioned simply by rotating the previous tile through 180° about the midpoint of a suitable side.

NCERT Class 9 Maths Ganita Manjari Chapter 12 End-of-Chapter Exercises