CG Board Class 9 Mathematics Chapter 11 Solutions: Quadrilaterals (चतुर्भुज) – 2026-27

A quadrilateral is a closed two-dimensional geometric figure enclosed by four line segments in a plane, formed by joining four points in an order such that no three points are collinear. It consists of four sides, four vertices, four interior angles, and two diagonals.

Examples of quadrilateral-shaped surfaces in a classroom/school:

1. Top surface of a study table or desk.
2. Door panel.
3. Notice board / Display board.
4. Floor tiles and wall tiles.
5. Teacher’s table surface.

A rectangle is a four-sided closed figure formed by four line segments. Since any polygon bounded by four line segments is classified as a quadrilateral, a rectangle meets all basic definitions of a quadrilateral (having 4 sides, 4 vertices, and 4 interior angles). Hence, a rectangle is a special quadrilateral where opposite sides are parallel and equal, and each angle measures 90°.

Definition of a trapezium: A quadrilateral in which at least/only one pair of opposite sides is parallel is called a trapezium.

• Figures (iv), (v), (viii), and (ix) are not trapeziums:
• Fig. (iv) & (v): Neither pair of opposite sides is parallel.
• Fig. (viii): It is a kite where adjacent pairs of sides are equal, but no pair of opposite sides is parallel.
• Fig. (ix): None of the opposite side pairs are parallel.
• Figures (i), (ii), (iii), (vi), and (vii) are trapeziums because they have at least one pair of parallel opposite sides.

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

• Rhombus: Parallelogram (both pairs of opposite sides are parallel and all sides are equal).
• Rectangle: Parallelogram (both pairs of opposite sides are parallel and all angles are 90°).
• Square: Parallelogram (both pairs of opposite sides are parallel, all sides are equal, and all angles are 90°).
• Trapezium: Not a parallelogram in general (only one pair of opposite sides is parallel).

Yes.** A trapezium requires at least one pair of opposite sides to be parallel. Since a parallelogram has both pairs of opposite sides parallel, it automatically satisfies the condition of having at least one pair of parallel sides. Therefore, every parallelogram is also a trapezium.

No, not necessarily.** A rectangle has all interior angles equal to 90°, but its adjacent sides may not be equal. A square requires all four sides to be equal in addition to all angles being 90°. A rectangle is a square only if all four of its sides are equal.

No, such a parallelogram is not possible.**

In any quadrilateral, the sum of interior angles is 360°. If three angles are 90°, the fourth angle must be 360° – (90° + 90° + 90°) = 90°. A parallelogram with all four angles equal to 90° is, by definition, a rectangle. Hence, it is impossible to have such a parallelogram that is not a rectangle.

Let the measures of the angles be 3x, 5x, 7x, and 9x.

By angle sum property of a quadrilateral, sum of interior angles = 360°.

3x + 5x + 7x + 9x = 360°

24x = 360°

x = 360° / 24 = 15°

Therefore, the angles are:

• First angle = 3 × 15° = 45°
• Second angle = 5 × 15° = 75°
• Third angle = 7 × 15° = 105°
• Fourth angle = 9 × 15° = 135°

Let each angle of the quadrilateral be x.

By angle sum property of a quadrilateral,

x + x + x + x = 360°

4x = 360°

x = 360° / 4 = 90°

Thus, the measure of each angle is 90°.

Given:** ABCD is a rhombus in which diagonals AC = BD.

To Prove: ABCD is a square.

Proof:

In △ABC and △DCB:

• AB = DC (Sides of a rhombus are equal)
• BC = CB (Common side)
• AC = BD (Given)

By SSS congruency rule, △ABC ≅ △DCB.

Therefore, ∠ABC = ∠DCB (by C.P.C.T.).

Since AB || DC and BC is a transversal:

∠ABC + ∠DCB = 180° (Interior angles on same side of transversal)

∠ABC + ∠ABC = 180° → 2∠ABC = 180° → ∠ABC = 90°.

Since ABCD is a rhombus with one angle 90°, all its angles are 90° and all sides are equal.

Hence, ABCD is a square.

Given:** ABCD is a rhombus whose diagonals AC and BD intersect at O.

To Prove: AC ⊥ BD (i.e., ∠AOB = 90°).

Proof:

In △AOB and △BOC:

• AO = CO (Diagonals of a parallelogram bisect each other)
• OB = OB (Common side)
• AB = BC (Sides of a rhombus are equal)

By SSS congruency rule, △AOB ≅ △BOC.

Therefore, ∠AOB = ∠BOC (by C.P.C.T.).

Also, ∠AOB + ∠BOC = 180° (Linear pair).

∠AOB + ∠AOB = 180° → 2∠AOB = 180° → ∠AOB = 90°.

Hence, the diagonals of a rhombus are perpendicular to each other.

Proof:** Let ABCD be a rectangle where ∠ABC = ∠DCB = 90°.

In △ABC and △DCB:

• AB = DC (Opposite sides of a rectangle/parallelogram)
• ∠ABC = ∠DCB = 90°
• BC = CB (Common side)

By SAS congruency rule, △ABC ≅ △DCB.

Therefore, AC = BD (by C.P.C.T.).

Thus, the diagonals of a rectangle are equal in length.

Let ABCD be a square.

1. Equal diagonals: In △ABC and △DCB, AB = DC, ∠B = ∠C = 90°, BC = BC. By SAS, △ABC ≅ △DCB → AC = BD.
2. Bisect each other: Since a square is a parallelogram, its diagonals AC and BD bisect each other at O (AO = OC, BO = OD).
3. Bisect at right angles: In △AOB and △BOC, AO = CO, OB = OB, AB = BC. By SSS, △AOB ≅ △BOC → ∠AOB = ∠BOC = 90°.

Hence proved.

Theorem 11.8 (Mid-Point Theorem):** The line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.

• Converse of Theorem 11.8: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.
• Verification / Proof:

In △ABC, let D be the mid-point of AB, and line DE is drawn parallel to BC meeting AC at E. Draw CF || BA meeting DE extended at F.

In △ADE and △CFE:

• ∠DAE = ∠FCE (Alternate interior angles, AB || CF)
• ∠AED = ∠CEF (Vertically opposite angles)
• AD = CF (since BDFC is a parallelogram, BD = CF and AD = BD)

By ASA congruency, △ADE ≅ △CFE → AE = EC (by C.P.C.T.).

Hence, E is the mid-point of AC.

Given:** ABCD is a parallelogram. X is the mid-point of AB and Y is the mid-point of CD.

To Prove: AXCY is a parallelogram.

Proof:

Since ABCD is a parallelogram:

• AB || CD → AX || CY (Parts of parallel lines are parallel)
• AB = CD (Opposite sides of a parallelogram)

both sides by 1/2 का गुणा करने पर:

(1/2) AB = (1/2) CD

Since X and Y are mid-points of AB and CD respectively:

AX = (1/2) AB and CY = (1/2) CD

Therefore, AX = CY.

In quadrilateral AXCY, one pair of opposite sides (AX and CY) is both parallel and equal (AX || CY and AX = CY).

By Theorem 11.7, AXCY is a parallelogram.

Hence proved.

Given:** AB || DC intersected by transversal l at P and R. Let angle bisectors of interior angles ∠APR, ∠BPR, ∠CRP, ∠DRP intersect to form quadrilateral PQRS.

To Prove: PQRS is a rectangle.

Proof:

1. ∠APR + ∠BPR = 180° (Linear pair)

(1/2) ∠APR + (1/2) ∠BPR = 90° → ∠SPR + ∠QPR = 90° → ∠SPQ = 90°.

1. Similarly, ∠SRQ = 90°, ∠PSR = 90°, and ∠PQR = 90°.
2. Alternate interior angles are equal:

∠APR = ∠CRP → (1/2) ∠APR = (1/2) ∠CRP → ∠SPR = ∠PRQ.

Since these are alternate interior angles for lines PS and RQ with transversal PR, PS || RQ.

1. Similarly, PQ || SR.
2. Thus, PQRS is a parallelogram with each angle equal to 90°.

Hence, PQRS is a rectangle.

Given:** △ABC is isosceles with AB = AC. AD bisects exterior ∠PAC (i.e. ∠PAD = ∠DAC). CD || BA.

To Prove: (i) ∠DAC = ∠BCA, (ii) ABCD is a parallelogram.

Proof:

(i) In △ABC, AB = AC → ∠BCA = ∠ABC (Angles opposite to equal sides).

Exterior angle ∠PAC = ∠ABC + ∠BCA = ∠BCA + ∠BCA = 2 ∠BCA.

Since AD bisects ∠PAC, ∠PAC = 2 ∠DAC.

Therefore, 2 ∠DAC = 2 ∠BCA → ∠DAC = ∠BCA.

(ii) From part (i), ∠DAC = ∠BCA. These are alternate interior angles for lines AD and BC with transversal AC.

Therefore, AD || BC.

Also, CD || BA is given.

Since both pairs of opposite sides are parallel (AD || BC and CD || BA), ABCD is a parallelogram.

Given:** ABCD is a parallelogram. AP ⊥ BD and CQ ⊥ BD.

To Prove: (i) △APB ≅ △CQD, (ii) AP = CQ.

Proof:

(i) In △APB and △CQD:

• ∠APB = ∠CQD = 90° (Since AP ⊥ BD and CQ ⊥ BD)
• ∠ABP = ∠CDQ (Alternate interior angles, AB || CD with transversal BD)
• AB = CD (Opposite sides of parallelogram ABCD)

By AAS congruency rule, △APB ≅ △CQD.

(ii) Since △APB ≅ △CQD:

AP = CQ (by C.P.C.T.).

Given:** ABCD is a rectangle where diagonal AC bisects ∠A and ∠C (∠DAC = ∠BAC and ∠DCA = ∠BCA).

To Prove: (i) ABCD is a square, (ii) Diagonal BD bisects ∠B and ∠D.

Proof:

(i) Since ABCD is a rectangle, AB || CD and AD || BC, and all angles are 90°.

∠DAC = ∠BCA (Alternate interior angles, AD || BC).

Given ∠DAC = ∠BAC → ∠BAC = ∠BCA.

In △ABC, sides opposite to equal angles are equal → AB = BC.

Since adjacent sides of rectangle ABCD are equal (AB = BC = CD = DA), ABCD is a square.

(ii) Since ABCD is a square, AB = AD and BC = CD.

In △ABD, AB = AD → ∠ABD = ∠ADB.

Since AB || CD, ∠ABD = ∠CDB → ∠ADB = ∠CDB (BD bisects ∠D).

Similarly, ∠ABD = ∠CBD (BD bisects ∠B).

Hence, diagonal BD bisects both angles B and D.

Given:** AB = DE and AB || DE; BC = EF and BC || EF.

To Prove: AC = DF and AC || DF.

Proof:

1. In quadrilateral ABED: AB = DE and AB || DE.

By Theorem 11.7, ABED is a parallelogram.

Therefore, AD = BE and AD || BE …..(1)

1. In quadrilateral BCFE: BC = EF and BC || EF.

By Theorem 11.7, BCFE is a parallelogram.

Therefore, CF = BE and CF || BE …..(2)

1. From (1) and (2):

AD = CF and AD || CF.

1. Thus, in quadrilateral ACFD, one pair of opposite sides (AD and CF) is equal and parallel.

By Theorem 11.7, ACFD is a parallelogram.

Therefore, AC = DF and AC || DF.

Hence proved.

Given:** ABCD is a trapezium with AB || DC. E is the mid-point of AD. Line EF || AB meeting BC at F.

To Prove: F is the mid-point of BC.

Proof:

Join diagonal AC intersecting EF at point O.

In △ADC:

• E is the mid-point of AD (Given).
• EO || DC (Since EF || AB and AB || DC → EF || DC).

By Converse of Mid-Point Theorem (Theorem 11.8), O is the mid-point of AC.

In △ABC:

• O is the mid-point of AC (Proved above).
• OF || AB (Since EF || AB).

By Converse of Mid-Point Theorem (Theorem 11.8), F is the mid-point of BC.

Hence proved.

Given:** ABCD is a rhombus. P, Q, R, S are mid-points of AB, BC, CD, DA respectively.

To Prove: Quadrilateral PQRS is a rectangle.

Proof:

Join diagonals AC and BD. Diagonals of a rhombus intersect at right angles (AC ⊥ BD, ∠AOB = 90°).

In △ABC, P and Q are mid-points of AB and BC.

By Mid-Point Theorem (Theorem 11.8):

PQ || AC and PQ = (1/2) AC …..(1)

In △ADC, S and R are mid-points of AD and CD.

By Mid-Point Theorem (Theorem 11.8):

SR || AC and SR = (1/2) AC …..(2)

From (1) and (2), PQ || SR and PQ = SR.

Thus, PQRS is a parallelogram.

Similarly, in △ABD, PS || BD.

Since AC ⊥ BD, and PQ || AC and PS || BD, the sides PQ and PS are perpendicular (∠SPQ = 90°).

A parallelogram with one interior angle equal to 90° is a rectangle.

Hence, quadrilateral PQRS is a rectangle.

Given:** ABCD is a rectangle. P, Q, R, S are mid-points of sides AB, BC, CD, DA respectively.

To Prove: Quadrilateral PQRS is a rhombus.

Proof:

Join diagonals AC and BD. In a rectangle, diagonals are equal (AC = BD).

1. In △ABC, P and Q are mid-points of AB and BC.

By Mid-Point Theorem: PQ = (1/2) AC and PQ || AC.

1. In △ADC, S and R are mid-points of AD and CD.

By Mid-Point Theorem: SR = (1/2) AC and SR || AC.

1. In △ABD, S and P are mid-points of AD and AB.

By Mid-Point Theorem: SP = (1/2) BD and SP || BD.

1. In △BCD, R and Q are mid-points of CD and BC.

By Mid-Point Theorem: RQ = (1/2) BD and RQ || BD.

Since AC = BD, we have:

(1/2) AC = (1/2) BD → PQ = SR = SP = RQ.

A quadrilateral in which all four sides are equal is a rhombus.

Hence, PQRS is a rhombus.

Given:** ABCD is a quadrilateral. P, Q, R, S are mid-points of AB, BC, CD, DA. AC is a diagonal.

To Prove:

(1) SR || AC and SR = (1/2) AC

(2) PQ = SR

(3) PQRS is a parallelogram

Proof:

(1) In △ADC:

S is the mid-point of AD and R is the mid-point of CD (Given).

By Mid-Point Theorem (Theorem 11.8), the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.

Therefore, SR || AC and SR = (1/2) AC.

(2) In △ABC:

P is the mid-point of AB and Q is the mid-point of BC (Given).

By Mid-Point Theorem (Theorem 11.8):

PQ || AC and PQ = (1/2) AC.

Since SR = (1/2) AC and PQ = (1/2) AC, we get PQ = SR.

(3) From parts (1) and (2):

SR || AC and PQ || AC → SR || PQ.

Also, PQ = SR.

In quadrilateral PQRS, one pair of opposite sides (PQ and SR) is both equal and parallel.

By Theorem 11.7, PQRS is a parallelogram.

Hence proved.