Contents

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 8.1, Ex 8.2 |

Number of Questions Solved |
18 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

### Chapter 8 Quadrilaterals Ex 8.1

Question 1.

The angles of quadrilateral are in the ratio 3: 5: 9:13. Find all the angles of the quadrilateral.

Solution.

Given, the ratio of the angles of quadrilateral are 3:5:9:13.

Let the angles of the quadrilateral are 3x, 5x, 9x and 13x.

We know that, sum of angles of a quadrilateral = 360°

∴ 3x + 5x + 9x +13x = 360°

∴ 30x = 360° ⇒ °

∴ Angles of the quadrilateral are

3x = 3 x 12 = 36°

5x = 5 x 12 = 60°

9x = 9 x 12 = 108°

and 13x =13 x 12=156°

Question 2.

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

Solution.

We have a parallelogram ABCD such that AC = BD.

In Δ ABC and Δ DCB

AC = DB [given]

AB = DC (opposite sides of a parallogram)

BC=CB (Common)

Δ ABC ≅ Δ DCB [by SAS congrunce rule]

∠ABC ≅ ∠DCB ………(i)

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

∴ ∠ABC + ∠DCB =180°

(interior opposite angles are supplementry]

From eqs. (i) and (ii), we have

∠ABC = ∠DCB =90°

i.e., ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is a rectangle.

**Hence proved.**

Question 3.

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

Solution.

We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.

∴ In Δ AOB and Δ AOD, we have

AO = AO common]

OB = OD [∵ O is the mid-point of BD]

∠AOB = ∠AOD [each = 90 °]

Δ AOB ≅ Δ AOD [by SAS congruence rule]

AB = AD [by CPCT]… (i)

Similarly, AB = BC …(ii)

BC = CD …(iii)

CD = AD …(iv)

∴ From eqs.(i), (ii), (iii) and (iv), we have AB = BC = CD = DA Hence, the quadrilateral ABCD is a rhombus.

Question 4.

Show that the diagonals of a square are equal and bisect each other at right angles.

Solution.

Given : A square PQRS in which diagonals PR and QS intersect each other at 0.

To prove : PR = SQ, PO = OR, QO = OS and PR ⊥ SQ.

Proof : In Δ POQ and Δ ROS

PQ = RS [sides of a square]

∠PQO=∠RSO [alternative interior angles]

∠QPO = ∠SRO [alternative interior angles]

⇒ Δ POQ ≅ Δ ROS [by SAS congruence rule]

⇒ PO = OR [by CPCT]

and QO = OS [by CPCT]

Thus, PR and SQ bisect each other at O.

Also, in Δ POQ and Δ ROQ,

PQ = RQ

PO = OR

OQ = OQ

⇒ ΔPOQ ≅ ΔROQ by SAS congruence rule]

⇒ ∠POQ = ∠ROQ

Also.,∠POQ + ∠ROQ = 180°

⇒ ∠POQ + ∠POQ = 180°

⇒2 ∠POQ = 180°

⇒ ∠POQ = x 180° = 90°

Thus, diagonals PR and SQ are perpendicular to each other.

Again, in Δ PQS and Δ QPR

PQ = PQ

PS = QR

∠QPS – ∠PQR

⇒ ΔPQS = ΔQPR

⇒ SQ = PR

Hence, diagonals of a square are equal and bisect each other at right angles.

Question 5.

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution.

Given : A quadrilateral ABCD in which AC = BD and AC⊥ BD such that OA = OC and OB = OD. So, ABCD is a parallelogram.

To prove : ABCD is a square.

Proof: Let AC and BD intersect at a point O.

In Δ ABO and Δ ADO, we have

BO = OD (Given)

AO = OA (Common)

∴ ∠AOB = ∠AOD = 90° (Given)

∴ ΔABO ≅ ΔADO (By SAS)

AB = AD (By CPCT)

Also,AB = DC

and AD = BC (Opposite sides of parallelogram)

∴ AB = BC = DC = AD .(i)

Again, in Δ ABC and Δ BAD, we have AB – BA AC = BD BC – AD

∴ Δ ABC = Δ BAD

∴ ∠ABC = ∠BAD…..(ii)

But ∠ABC + ∠BAD = 180°(Sum of interior angles of a parallelogram)

∴ ∠ABC = ∠BAD = 90° [From Eq. (ii)]

Thus, AB = BC = CD = DA and ∠A = 90°

∴ ABCD is a square.

Question 6.

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

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Solution.

**(i)** Given, diagonal AC of a parallelogram ABCD bisects ∠A.

i.e., ∠DAC =∠BAC = ∠BAD … (i)

Here, AB || DC and AC is a transversal.

∴ ∠DCA = ∠BAC …(ii)[pair of alternate angles]

Similarly, BC || AD and BD is a transversal.

∴ ∠BCA = ∠DAC … (iii) [pair of alternate angles]

From eqs. (i), (ii) and (iii), we get

∠DAC = ∠BCA = ∠BAC = ∠DCA .. (iv)

Now, ∠BCD = ∠BCA + ∠DCA

= ∠DAC + ∠BAC = ∠BAD

Hence, diagonal AC also bisects ∠C.

(**ii)** From eq. (iv), ∠DAC = ∠DCA

⇒ CD = DA [∵ angles opposite to equal sides are equal]

But AB = DC and AD =BC [∵ ABCD is a parallelogram]

∴ AB = BC = CD = DA

Hence, ABCD is a rhombus.

Question 7.

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Solution.

Given : ABCD is a rhombus in which AB = BC =CD = AD.

To prove :

**(i)** Diagonal AC bisects ∠A as well as ∠C.

**
(ii)** Diagonal BD bisects ∠B as well as

Construction : Join diagonals AC and BD.

Proof : ABCD is a rhombus.

∴ AB = BC = CD = AD

Also, AB || CD and AD || BC

Now,AD = CD ⇒ ∠1=∠2 …(i) [angles opposite to equal sides are equal]

Also, CD || AB [opposite sides of the parallelogram]

and AC is transversal. ∠1 = ∠3 … (ii) [alternate interior angles]

From eqs. (i) and (ii), we have

∠2 = ∠3 and ∠1 = ∠4 ⇒ AC bisects ∠C as well as ∠A.

Similarly, we prove that BD bisects ∠B as well as ∠D.

Question 8.

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 ZB as well as

Solution.

We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.

i.e. ∠1 = 4 and ∠2 = ∠3 …(i)

**(i)** Since, rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AS || CD and AC is a transversal.

∴ ∠2=∠4 … (ii) [alternate interior angles]

From eqs. (i) and (ii), we have

∠3 = ∠4 ⇒ AB = BC [∵ sides opposite to equal angles in Δ ABC are equal]

∴ AB = BC = CD = AD

⇒ ABCD is a rectangle having all of its sides equal.

∴ ABCD is a square.

**(ii)** Since, ABCD is a square, and diagonals of a square bisect the opposite

∴ BD bisects ZB as well as ZD.

Question 9.

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

(i) Δ APD = Δ CQB

(ii) AP = CQ

(iii) Δ AQB ≅ Δ CPD

(iv) AQ = CP

(v) APCQis a parallelogram.

Solution.

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.

∴AP || CQ and AQ || CP

Now, both pairs of opposite sides of quadrilateral Δ PCQ are parallel. Hence, ΔPCQ is a parallelogram. Hence proved.

Question 10.

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

(i) Δ APB ≅ Δ CQD

(ii) AP = CQ

Solution.

**(i)** In Δ APB and Δ CQD, we have

∠ APB = ∠CQD [each = 90 °]

AB = CD [opposite sides of parallelogram ABCD]

∠ABP= ∠CDQ

Δ APB = Δ CQD

**(ii)** Since, Δ APB = Δ CQD

∴ Their corresponding parts are equal.

AP = CQ

Question 11.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the figure). Show that

(i) quadrilateral ABED is a parallelogram.

(ii) quadrilateral BEFC is a parallelogram.

(iii) AD || CF and AD= CF.

(iv) quadrilateral ACFD is a parallelogram.

(v) AC = DF

(vi) ΔABC ≅ ΔDEF

Solution.

Given, in Δ ABC and Δ DEF,

AB = DE,AB || DE, BC = EF and BC || EF

**(i)** Now, in quadrilateral ABED,

AB = DE and AB || DE [given]

So, ABED is a parallelogram. [∵ a pair of opposite side is equal and parallel]

**(ii)** In quadrilateral BEFC,

BC = EF and BC || EF [given]

So, BEFC is a parallelogram. [∵ a pair of opposite sides is equal and parallel]

**(iii)** Since, ABED is a parallelogram.

∴ AD || BE and AD = BE …(i)

Also, BEFC is a parallelogram.

∴ CF || BE and CF = BE …(ii)

From eqs. (i) and (ii), we get

AD || CF and AD = CF

**(iv)** In quadrilateral ACFD

∴ AD || CF and AD = CF [from part (iii)]

So, ACFD is a parallelogram.

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

**(v)** Since, ACFD is a parallelogram.

∴ AC = DF and AC 11 DF

**(vi)** Now, in Δ ABC and Δ DEF,

AB = DE [given]

BC = EF [given]

and AC = DF [from part (v)]

∴ Δ ABC ≅ Δ DEF [by SSS congruence rule]

Hence proved.

Question 12.

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

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) Δ ABC = Δ BAD

(iv) Diagonal AC = Diagonal

[Hint : Extend AB and draw a line through C parallel to DA . intersecting AB produced at E.]

Solution.

### Chapter 8 Quadrilaterals Ex 8.2

Question 1.

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

(i) SR || AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Solution.

Given : P,Q,R and S are the mid-points of the sides

AB, BC, CD and DA, respectively.

AP = BP

BQ = CQ

CR = DR

and AS = DS

To show (i) SR || AC and SR = AC 2

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Proof:

**(i)** In Δ ADC, we have S is mid-point of AD and R is mid-point of DC.

We know that, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

∴ SR || AC …….(i)

SR = AC ………….(ii)

**(ii)** Similarly, in Δ ABC, we have P is mid-point of AB and Q is the mid-point of BC.

∴ PQ || AC …………..(iii)

PQ = AC…………..(iv)

**(iii)** From eqs. (i) and (iii), we get

PQ || SI and from eq. (v), PQ = SR

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

So, PQRS is a parallelogram.

Question 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.

Solution.

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

To show : Quadrilateral PQRS is a rectangle.

Proof : By mid-point theorem.

In Δ ADC, we have S and R are the mid-points of DA and CD, respectively.

∴ SR || AC

SR = AC …………..(i)

In Δ ABC, we have P and Q are the mid-points of AB and BC, respectively.

∴ PQ || AC and PQ = AC …..(ii)

From eqs. (i) and (ii), we get PQ || SR and PQ =SR = AC

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

We know that, diagonals of a rhombus bisect each other at right angels.

∠COD = ∠EOF = 90°

Now, in ABCD, R and Q are the mid-points of CD and BC, respectively.

RQ || DB [by mid-point theorem]

⇒ RE || OF

Also,SR || AC [from eq. (i)]

⇒ FR || OE

So, OERF is a parallelogram.

∴ ∠ERF = ∠EOF = 90°[opposite angles of a parallelogram are equal]

Thus, PQRS is a parallelogram with ∠R =90°

Hence, PQRS is a rectangle.

**Hence proved.**

Question 3.

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

Solution.

Given, ABCD is a rectangle.

∴ ∠A=∠B = ∠C =∠D = 90° and AD = BC, AB = CD

Also, given P,Q,R and S are mid-points of AB, BC, CD and DA respectively.

∴ PQ || AC and PQ = AC ……(i)

AP = BP (Given)

AS= BQ (Given)

∠A= ∠B (Given)

∴ Δ ASP ≅ Δ BQP (By SAS)

∴ SP = PQ (By CPCT) …(iv)

SD = CQ (Given)

DR = RC (Given)

∠C = ∠D (Given)

Δ RDS = Δ RCQ (By SAS)

SR = RQ (By CPCT) …(v)

From Eqs. (iii), (iv) and (v), it is clear that quadrilateral PQRS is a rhombus.

Question 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 fig.). Show that F is ‘ the mid-point of BC.

Solution.

Given : ABCD is a trapezium in which AB || DC, BD is a diagonal and E is mid-point of AD and a line is drawn through E parallel to AB intersecting BC at F such that EF || AB. To prove : F is the mid-point of BC.

Proof : Let EF intersects BD at P.

In Δ ABD, we have EP || AB [∵ EF || AB] and E is mid-point of AD.

So, by converse of mid-point theorem.

We get, P is mid-point of BD.

Similarly, in ABCD, we have PF || CD [∵ EF || AB and AB || CD]

and P is mid-point of BD.

So, by converse of mid-point theorem, F is mid-point of BC.

Question 5.

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

(see fig.). Show that the line segments AF and EC trisect the diagonal BD.

Solution.

Given : ABCD is a parallelogram and E, F are the mid-points of sides AB and CD, respectively.

To prove : Line segments AF and EC trisect the diagonal BD.

Proof : Since, ABCD is a parallelogram.

∴ AB || DC and AB = DC [opposite sides of a parallelogram]

⇒ AF || FC and AP = DC

⇒ AF || FC and AF = FC

[∴ E and F are the mid-points of AB and CD] Since, a pair of opposite sides of a quadrilaterals AECF is equal and parallel.

So, AECF is a parallelogram.

Then, AF || FC

⇒ AP || FQ and FP || CQ

Since, opposite sides of a parallelogram are parallel.

In Δ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.

[by converse of mid-point theorem]

∴ BQ = PQ …(i)

Again, in Δ DQC, F is the mid-point of DC and FP || CQ.

So, P is the mid-point of DQ. [by converse of mid-point theorem]

∴ QP = DP …(ii)

From eqs. (i) and (ii), we get BQ = PQ = PD

Hence, CE and AF trisect the diagonal BD.

Hence proved.

Question 6.*
*Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution.

Let ABCD be a quadrilateral and P, Q, R and S be the D mid-points of the sides AB, BC, CD and DA, respectively.

i.e., AS = DS, AP = BP, BQ = CQ and CR = DR

We have to show that PR and SQ bisect each other,

i.e., SO = OQ and PO = OR

Now, in Δ ADC, S and R are mid-points of AD and CD, respectively.

We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.

∴ SR || AC and SR = AC …(i) [by mid-point theorem]

Similarly, in Δ ABC, P and Q are mid-points of AB and BC, respectively.

PQ || AC and PQ = AC [by mid-point theorem]

From eqs. (i) and (ii), we get

PQ || SP and PQ = SR = AC

So, PQRS is a parallelogram whose diagonals are SQ and also, we know that diagonals of parallelogram bisect each other. So, SQ and PR bisect each other at O,

i.e. SO = OQ and PO = OR.

[Since, a pair of opposite sides of a quadrilaterals PQRS is equal and parallel.]

Hence proved.

Question 7.*
*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

CM = MA = AB

Solution.

Given, ABC is a right angled triangle.

∠C = 90° and M is the mid-point of AB.

Also, DM || BC

**(i)**In Δ ABC, BC || MD and Mis mid-point of AB.

∴ D is the mid-point of AC. (By converse of mid-point theorem)

Since, MD || BC and CD is transversal. .

∠ADM = ∠ACB (Corresponding angles)

But ∠ACB = 90°

∠ADM = 90° => MD ⊥ AC

Now, in AADM and A CDM, we have

DM = MD

AD = CD

∠ADM = ∠MDC

AADM = ACDM

CM = AM

Also, M is mid-point of AB.

AM = BM = AB

From Eq. (i) and (ii), we get

CM = AM = AB

Hence proved.

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