Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8

Quadrilaterals is a chapter in geometry that focuses on polygons with four sides. It builds upon the concepts of lines, angles, and triangles.

Key Topics

• Basic concepts: Definition of a quadrilateral, its properties, and different types (angle sum property, types of quadrilaterals).
• Properties of Parallelograms: Specific properties of parallelograms, including opposite sides, angles, and diagonals.
• Conditions for a Quadrilateral to be a Parallelogram: Various conditions that guarantee a quadrilateral to be a parallelogram.
• Rectangles, Rhombuses, and Squares: Properties and relationships between these special types of parallelograms.
• Kite: Definition and properties of a kite.

Core Ideas

• Identifying the necessary conditions for a quadrilateral to be a parallelogram.
• Applying the properties of quadrilaterals to solve geometric problems.

This chapter provides a foundation for understanding more complex geometric shapes and their properties.

NCERT Solutions for Class 9 Maths Chapter 8

Exercise 8.1

1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Ans : 

In ∆ABC and ∆DCB,

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency]

⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)

Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we have

∠ABC = ∠DCB = 90°

∴ ABCD is a rectangle.

2. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Ans : 

∴ In ∆AOB and ∆AOD, we have

AO = AO [Common]

∠AOB = ∠AOD [Each 90]

∴ ∆AQB ≅ ∆AOD [By,SAS congruency

∴ AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

3. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Ans : 

(i) AC bisects angle C

• Since ABCD is a parallelogram, AB is parallel to DC and AD is parallel to BC.
• Therefore, angle DAC is equal to angle BCA (alternate interior angles).
• Similarly, angle BAC is equal to angle DCA (alternate interior angles).
• But angle DAC is equal to angle BAC (given).
• Hence, angle BCA is equal to angle DCA.

Therefore, AC bisects angle C.

(ii) ABCD is a rhombus

• We know that AC bisects both angle A and angle C.
• This means triangle ABC is congruent to triangle ADC (ASA congruence).
• Therefore, AB = CD and BC = AD (corresponding parts of congruent triangles).
• Since ABCD is a parallelogram, 
• AB = CD and AD = BC.
• Combining the above equalities, we get AB = BC = CD = AD.

Therefore, ABCD is a rhombus.

4. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

(i) ABCD is a square

(ii) diagonal BD bisects ∠B as well as ∠D.

Ans : 

Proof:

i) ABCD is a square

• Since AC bisects angle A, angle BAC = angle DAC.
• Since ABCD is a rectangle, 
• angle A and angle C are right angles.
• Therefore, angle BAC = angle DAC = 45°.
• Similarly, angle BCA = angle DCA = 45°.
• In triangle ABC, angle ABC = 180° – (angle BAC + angle BCA) = 180° – 90° = 90°.
• So, all angles of ABCD are 90°, and it is a rectangle with all sides equal.
• Hence, ABCD is a square.

ii) 

Since ABCD is a square, all its sides are equal and all its angles are right angles.

• Consider triangles ABD and CBD.
• AB = BC (sides of a square)
• AD = CD (sides of a square)
• BD is common to both triangles.
• Therefore, by SSS congruence, triangle ABD is congruent to triangle CBD.
• Hence, angle ABD = angle CBD and angle ADB = angle CDB.

Therefore, diagonal BD bisects both angle B and angle D.

5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that

Ans : 

Proof:

(i) ΔAPD ≅ ΔCQB

• Given: ABCD is a parallelogram, DP = BQ.
• Proof:
  • In triangles APD and CQB,
    • AD = BC (opposite sides of a parallelogram)
    • ∠ADP = ∠CBQ (alternate interior angles as AD || BC and BD is the transversal)
    • DP = BQ (given)
  • Therefore, by SAS congruence, ΔAPD ≅ ΔCQB.

(ii) AP = CQ

• Since ΔAPD ≅ ΔCQB (proved in (i)), their corresponding parts are equal.
• Therefore, AP = CQ.

(iii) ΔAQB ≅ ΔCPD

• Proof:
  • In triangles AQB and CPD,
    • AB = CD 
    • ∠ABQ = ∠CDP (alternate interior angles as AB || CD and BD is the transversal)
    • BQ = DP (given)
  • Therefore, by SAS congruence, ΔAQB ≅ ΔCPD.

(iv) AQ = CP

• Since ΔAQB ≅ ΔCPD (proved in (iii)), their corresponding parts are equal.
• Therefore, AQ = CP.

(v) APCQ is a parallelogram

• From (ii) and (iv), we have AP = CQ and AQ = CP.
• Therefore, APCQ is a parallelogram.

6. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

Ans : 

Proof:

(i) In △APB and △CQD,

∠APB = ∠CQD = 90° (Each 90∘)

AB = CD 

∠ABP = ∠CDQ (Alternate interior angles for AB∥CD)

∴ △APB ≅ △CQD (By AAS congruency)   

(ii) AP = CQ (By CPCT)

Therefore, AP = CQ.

7. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that

(i )∠A=∠B

(ii )∠C=∠D

(iii) ∆ABC ≅ ∆BAD

(iv) diagonal AC = diagonal BD

Ans : 

(i) ∠A = ∠B

• Since AB is parallel to CD, AD is a transversal.
• Therefore, ∠A and ∠B are alternate interior angles.
• Hence, ∠A = ∠B.

(ii) ∠C = ∠D

• Similarly, since AB is parallel to CD, BC is a transversal.
• Therefore, ∠C and ∠D are alternate interior angles.
• Hence, ∠C = ∠D.

(iii) ΔABC ≅ ΔBAD

• In triangles ABC and BAD,
  • AB is common.
  • BC = AD (given)
  • ∠A = ∠B (proved in (i))
• Therefore, by the SAS congruence rule, ΔABC ≅ ΔBAD.

(iv) Diagonal AC = diagonal BD

• Since ΔABC ≅ ΔBAD, their corresponding parts are equal.
• Therefore, AC = BD.

Exercise 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that

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

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Ans : 

Proof:

(i) Consider triangle ADC.

• By the mid-point theorem, SR || AC and SR = 1/2 AC.

(ii) PQ = SR

• Similarly, consider triangle ABC.
• By the mid-point theorem, PQ || AC and PQ = 1/2 AC.
• From (i) and (ii), we have SR || AC and PQ || AC.
• Therefore, SR || PQ.
• Also, SR = 1/2 AC and PQ = 1/2 AC, so SR = PQ.

(iii) PQRS is a parallelogram

• From (ii), we have PQ || SR and PQ = SR.
• Therefore, PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.

Ans : 

1. PQRS is a Parallelogram:

Using the mid-point theorem, we can show that PQ is parallel to SR and PQ = SR. 

Similarly, PS is parallel to QR 

and PS = QR.

Since one pair of opposite sides is parallel and equal, PQRS is a parallelogram.

2. Diagonals of PQRS are equal:

In rhombus ABCD, AC = BD (diagonals of a rhombus are equal).

Since P, Q, R, and S are midpoints, PR = 1/2 AC and QS = 1/2 BD.

Therefore, PR = QS.

3. Conclusion:

A parallelogram with equal diagonals is a rectangle.

Since PQRS is a parallelogram with equal diagonals PR and QS, it is a rectangle.

Hence, PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.

Ans :

1. PQRS is a Parallelogram:

• Using the mid-point theorem, we can show that PQ is parallel to SR and PQ = SR. 
• Similarly, PS is parallel to QR 
• and PS = QR.
• Since one pair of opposite sides is parallel and equal, PQRS is a parallelogram.

2. Diagonals of PQRS are equal and perpendicular:

• In rectangle ABCD, AC = BD (diagonals of a rectangle are equal).
• Since P, Q, R, and S are midpoints, PR = 1/2 AC and QS = 1/2 BD.
• Therefore, PR = QS.
• Also, in a rectangle, the diagonals intersect at right angles. Therefore, PR and QS are perpendicular to each other.

3. Conclusion:

• A parallelogram with equal and perpendicular diagonals is a rhombus.
• Since PQRS is a parallelogram with equal and perpendicular diagonals PR and QS, it is a rhombus.

Hence, PQRS is a rhombus.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.

Ans : 

Step 1: Introduction of Point G

Let the line EF intersect BD at point G.

Step 2: Applying the Midpoint Theorem

• In triangle ABD, E is the midpoint of AD and EG is parallel to AB. Therefore, by the converse of the mid-point theorem, G is the midpoint of BD.

Step 3: Applying the Midpoint Theorem Again

• In triangle BCD, GF is parallel to CD (as EF is parallel to AB, and AB is parallel to CD) and G is the midpoint of BD. Therefore, by the converse of the mid-point theorem, F is the midpoint of BC.

Conclusion:

Hence, F is the midpoint of BC.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Ans : 

Since, the opposite sides of a parallelogram are parallel and equal.

∴ AB || DC

⇒ AE || FC …(1)

and AB = DC

⇒ 1/2AB = 1/2DC

⇒ AE = FC …(2)

From (1) and (2), we have

AE || PC and AE = PC

∴ ∆ECF is a parallelogram.

Now, in ∆DQC, we have F is the mid-point of DC and FP || CQ

[∵ AF || CE]

⇒ DP = PQ …(3)

⇒ BQ = PQ …(4)

[By converse of mid-point theorem]

∴ From (3) and (4), we have

DP = PQ = BQ

So, the line segments AF and EC trisect the diagonal BD.

6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

(i) D is the mid-point of AC

(ii) MD ⊥ AC

(iii) CM = MA = 1/2AB

Ans : 

Proof:

i) D is the mid-point of AC

• Since MD is parallel to BC and M is the midpoint of AB, by the converse of the mid-point theorem, D is the midpoint of AC.

ii) MD ⊥ AC

• Since MD is parallel to BC and AC is a transversal, the corresponding angles ∠MDC and ∠ACB are equal.
• As ∠ACB is a right angle (given), ∠MDC is also a right angle.
• Therefore, MD is perpendicular to AC.

iii) CM = MA = 1/2 AB

• Since M is the midpoint of AB, MA = MB = 1/2 AB.
• From part (i), 
• In triangles AMD and CMD,
  • AM = CM (proved above)
  • AD = CD (proved in part (i))
  • ∠AMD = ∠CMD (both right angles)
• Therefore, by the SAS congruence rule, triangle AMD is congruent to triangle CMD.
• Hence, MD = CM.
• Since MA = MB and CM = MD, we have CM = MA = 1/2 AB.

NCERT Solutions for Class 9 Maths Chapter 8

FAQ’s

NCERT Solutions for Class 9 Maths Chapter 8 What is a quadrilateral and how is it classified?

A quadrilateral is a polygon with four sides, four angles, and four vertices. It is classified based on its sides and angles into types such as parallelogram, rectangle, rhombus, square, trapezium, and kite.

What are NCERT Solutions for Class 9 Maths Chapter 8 the conditions for a quadrilateral to be a parallelogram?

A quadrilateral is a parallelogram if:

• Both pairs of opposite sides are equal, or
• Both pairs of opposite sides are parallel, or
• One pair of opposite sides is both equal and parallel, or
• The diagonals bisect each other.

How NCERT Solutions for Class 9 Maths Chapter 8 can we prove that a parallelogram is a rectangle?

If the diagonals of a parallelogram are equal, then it is a rectangle. This is because equal diagonals imply all angles are right angles.

What NCERT Solutions for Class 9 Maths Chapter 8 is the difference between a rhombus and a square?

Both rhombus and square have all sides equal, but:

• In a rhombus, the angles are not necessarily 90°.
• In a square, all sides are equal and all angles are 90°.

What NCERT Solutions for Class 9 Maths Chapter 8 does the mid-point theorem state and how is it used in quadrilaterals?

The mid-point theorem states that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
In quadrilaterals, it helps prove properties like:

• Midpoints of sides of a parallelogram form another parallelogram.
• In a rectangle, midpoints form a rhombus.
• In a rhombus, midpoints form a rectangle.