NCERT Solutions for Chapter 8 Quadrilaterals Class 9 Maths

Answer

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Answer

We have a quadrilateral ABCD such that the diagonals AC and

BD bisect each other at right angles at O.
∴ In ΔAOB and ΔAOD, we have AO = AO  [Common]
OB = OD  [Given that O in the mid-point of BD]
∠AOB = ∠AOD [Each = 90°]
∴ ΔAOB ≌ ΔAOD   [SAS criteria]
⇒ Their corresponding parts are equal. ∴ AB = AD …(1)
Similarly,

AB = BC …(2)

BC = CD …(3)
CD = AD …(4)
∴ From (1), (2), (3) and (4),
we have AB = BC = CD = DA Thus, the quadritateral ABCD is a rhombus.

Answer

We have a square ABCD such that its diagonals AC and BD intersect at O.

(i) To prove that the diagonals are equal, i.e. AC = BD In ΔABC and ΔBAD, we have
AB = BA  [Common]
BC = AD   [Opposite sides of the square ABCD]
∠ABC = ∠BAD  [All angles of a square are equal to 90º]
∴ DABC ≌ DBAD  [SAS criteria]
⇒ Their corresponding parts are equal.
⇒ AC = BD  …(1)

(ii) To prove that ‘O’ is the mid-point of AC and BD.
∵ AD || BC and AC is a transversal.   [∵ Opposite sides of a square are parallel]
∴ ∠1 = ∠3   [Interior alternate angles]
Similarly, ∠2= ∠4  [Interior alternate angles]
Now, in ΔOAD and ΔOCB, we have AD = CB  [Opposite sides of the square ABCD]
∠1= ∠3   [Proved ∠2= ∠4 ]
∴ ΔOAD ≌ ΔOCB   [ASA criteria]
∴ Their corresponding parts are equal.
⇒ OA = OC and OD = OB ⇒ O is the mid-point of AC and BD,
i.e. the diagonals AC and BD bisect each other at O.  …(2)

(iii) To prove that AC ⊥ BD.
In ΔOBA and ΔODA, we have OB = OD  [Proved]
BA = DA  [Opposite sides of the square]
OA = OA   [Common]
∴ ΔOBA ≌ ΔODA  [SSS criteria]
⇒ Their corresponding parts are equal.
⇒ ∠AOB = ∠AOD
But ∠AOB and ∠AOD form a linear pair.
∴ ∠AOB + ∠AOD = 180º
⇒ ∠AOB = ∠AOD = 90º
⇒ AC ⊥ BD    …(3)
From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles.

Answer

We have a quadrilateral ABCD such that ‘O’ is the mid-point of AC and BD.

Also AC⊥BD.
Now, in ΔAOD and ΔAOB, we have

AO = AO  [Common]
OD = OB   [∵ O is the mid-point of BD]
∠AOD = ∠AOB  [Each = 90º]
∴ ΔAOD ≌ ΔAOB   [SAS criteria]
∴ Their corresponding parts are equal.
⇒ AD = AB   …(1)
Similarly, we have AB = BC  …(2)
BC = CD  …(3)
CD = DA  …(4)
From (1), (2), (3) and (4) we have:

AB = BC = CD = DA
∴ Quadrilateral ABCD is having all sides equal.
In ΔAOD and ΔCOB, we have

AO = CO  [Given]
OD = OB  [Given]
∠AOD = ∠COB   [Vertically opposite angles]
∴ ΔAOD ≌ ΔCOB
⇒ Their corresponding parts are equal. ⇒ ∠1= ∠2

But, they form a pair of interior alternate angles.
∴ AD || BC
Similarly, AB || DC
∴ ABCD is a parallelogram.
∵ Parallelogram having all of  its sides equal is a rhombus.
∴ ABCD is a rhombus.
Now, in ΔABC and ΔBAD, we have AC = BD  [Given]
BC = AD  [Proved]
AB = BA  [Common]
∴ ΔABC ≌ ΔBAD  [SSS criteria]
⇒ Their corresponding angles are equal.
∴ ∠ABC = ∠BAD
Since, AD || BC and AB is a transversal.
∴ ∠ABC + ∠BAD = 180º   [Interior opposite angles are supplementary]
i.e. The rhombus ABCD is having one angle equal to 90º.
Thus, ABCD is a square.

Answer

We have a parallelogram ABCD in which diagonal AC bisects ∠A.
⇒ ∠DAC = ∠BAC

(i) To prove that AC bisects ∠C.
∵ ABCD is a parallelogram.
∴ AB || DC and AC is a transversal.
∴ ∠1 = ∠3    [Alternate interior angles]   …(1)
Also, BC || AD and AC is a transversal.
∴ ∠2 = ∠4   [Alternate interior angles]    …(2)

But AC bisects ∠A    [Given]
∴ ∠1 = ∠2    …(3)
From (1), (2) and (3), we have

∠3= ∠4
⇒ AC bisects ∠C.

(ii) To prove ABCD is a rhombus.
In DABC, we have

∠1= ∠4    [∵ ∠1 = ∠2 = ∠4]
⇒ BC = AB   [Sides opposite to equal angles are equal]   …(4)
Similarly,

AD = DC   …(5)
But ABCD is a parallelogram   [Given]
∴ AB = DC   [Opposite sides of a parallelogram]   …(6)
From (4), (5) and (6), we have

AB = BC = CD = DA
Thus, ABCD is a rhombus.

Answer

ABCD is a rhombus.

∴ AB = BC = CD = AD
Also, AB || CD an AD || BC
Now, AD = CD ⇒ ∠1 = ∠2    …(1)
[Angles opposite to equal sides are equal]
Also, CD || AB   [Opposite sides of the parallelogram]

and AC is AC is transversal.
∴ ∠1= ∠3   [Alternate interior angles]   …(2)
From (1) and (2), 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.

Answer

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

i.e. ∠ 1= ∠ 4 and ∠ 2 = ∠ 3   ...(1)

(i) Since, rectangle is a parallelogram.
∴ ABCD is a parallelogram.
⇒ AB || CD and AC is a transversal.
∴ ∠2= ∠4   [Alternate interior angles] ...(2)
From (1) and (2), we have

∠3= ∠4
⇒ AB = BC   [∵ Sides opposite to equal angles in D 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 angles.
∴ BD bisects ∠B as well as ∠D.

Answer

We have parallelogram ABCD. BD is a diagonal and ‘P’ and ‘Q’ are such that
PD = QB   [Given]

(i) To prove that ΔAPD ≌ ΔCQB
∵ AD || BC and BD is a transversal.    [∵ ABCD is a parallelogram.]
∴ ∠ADB = ∠CBD   [Interior alternate angles]
⇒ ∠ADP = ∠CBQ
Now, in ΔAPD and ΔCQB, we have
AD = CB   [Opposite side of the parallelogram]
P D = QB   [Given]
∠CBQ = ∠ADP  [Proved]
∴ Using SAS criteria, we have ΔAPD ≌ ΔCQB

(ii) To prove that AP = CQ

Since, ΔAPD ≌ ΔCQB  [Proved]
∴ Their corresponding parts are equal.

⇒ AP = CQ

(iii) To prove that ΔAQB ≌ ΔCPD.
∵ AB || CD and BD is a transversal.   [∵ ABCD is a parallelogram.]
∴ ∠ABD = ∠CDB
⇒ ∠ABQ = ∠CDP

Now, in ΔAQB and ΔCPD,
we have QB = PD  [Given]
∠ABQ = ∠CDP  [Proved]
AB = CD  [Opposite sides of parallelogram ABCD]
∴ ΔAQB ≌ ΔCPD  [SAS criteria]

(iv) To prove that AQ = CP.
Since, ΔAQB ≌ ΔCPD   [Proved]
∴ Their corresponding parts are equal.
⇒ AQ = CP.

(v) To prove that APCQ is a parallelogram.
Let us join AC.
Since, the diagonals of a || gm bisect each other

∴ AO = CO  …(1)
and BO = DO
⇒ (BO ∠ BQ) = (DO ∠ DP)   [∵ BQ = DP (Given)]
⇒ QO = PO  …(2)
Now, in quadrilateral APCQ,

We have, AO = CO and QO = PO
[from (1) and (2)]
i.e. AC and QP bisect each other at O.
⇒ APCQ is a parallelogram.

Answer

(i) In ΔAPB and ΔCQD, we have
∠APB = ∠CQD [90º each]

AB = CD  [Opposite sides of parallelogram ABCD]
∠ABP = ∠CDQ   [AB || CD and AB is a transversal]
⇒ Using AAS criteria, we have ΔAPB ≌ ΔCQD

(ii) Since, ΔAPB ≌ ΔCQD  [Proved]
∴ Their corresponding parts are equal.
⇒ AP = CQ

Answer

(i) To prove that ABED is a parallelogram.
Since “A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length.” Now AB = DE  [Given]
AB | | DE  [Given]
i.e. ABED is a quadrilateral in which a pair of opposite sides (AB and DE) is parallel and of equal length.
∴ ABED is a parallelogram.

(ii) To prove that BECF is a parallelogram.
∵ BC = EF   [Given]
and BC || EF   [Given]
i.e. BECF is a quadrilateral in which a pair of opposite sides (BC and EF) is parallel and of equal length.
∴ BECF is a parallelogram.

(iii) To prove that AD || CF and AD = CF ∵ ABED is a parallelogram. [Proved]
∴ Its opposite sides are parallel and equal.
⇒ AD || BE and AD = BE  …(1)
Also BEFC is a parallelogram.   [Proved]
∴ BE || CF and BE = CF   [∵ Opposite sides of a parallelogram are parallel and equal]         …(2)
From (1) and (2), we have

AD || CF and AD = CF

(iv) To prove that ACFD is a parallelogram.
∵ AD || CF  [Proved] and

AD = CF  [Proved]
i.e. In quadrilateral ACFD, one pair of opposite sides (AD and CF) is parallel and of equal length.
∴ Quadrilateral ACFD is a parallelogram.

(v) To prove that AC = DE.
∵ ACFD is a parallelogram.   [Proved]
∴ Its opposite sides are parallel and of equal length. i.e. AC = DF

(vi) To prove that ΔABC ≌ ΔDEF In D ABC and DDEF, we have:
AB = DE  [Opposite sides of a parallelogram]
BC = EF   [Opposite sides of a parallelogram]
AC = DF   [Proved]
∴ Using SSS criteria, we have ΔABC ≌ ΔDEF.

Answer

We have AB || CD and AD = BC

(i) To prove that ∠A = ∠B.
Produce AB to E and draw CE || AD.
∴ AB || DC
⇒ AE || DC   [Given]
Also AD || CE  [Construction]
∴ AECD is a parallelogram.
⇒ AD = CE   [opposite sides of the parallelogram AECD]
But AD = BC  [Given]
∴ BC = CE
Now, in ΔBCE, we have

BC = CE
⇒ ∠CBE = ∠CEB    …(1)
[∵ Angles opposite to equal sides of a triangle are equal]
Also, ∠ABC + ∠CBE = 180°  [Linear pair] ...(2)
and ∠A + ∠CEB = 180°  [∵ Adjacent angles of a parallelogram are supplementary] …(3)
From (2) and (3), we get

∠ABC + ∠CBE = ∠A + ∠CEB
But ∠CBE = ∠CEB    [Using (1)]
∴ ∠ABC = ∠A
or ∠B= ∠A
or ∠A= ∠B

(ii) To prove that ∠C = ∠D.
AB || CD and AD is a transversal.
∴ ∠A + ∠D = 180º   [Sum of interior opposite angles]
Similarly, ∠B + ∠C = 180º
⇒ ∠A + ∠D= ∠B + ∠C
But ∠A= ∠B  [Proved]
∴ ∠C= ∠D

(iii) To prove ΔABC ≌ ΔBAD

In ΔABC and ΔBAD, we have

AB = BA   [Common]
BC = AD   [Given]
∠ABC = ∠BAD    [Proved]
∴ ΔABC ≌ ΔBAD  [Using SAS criteria]
(iv) To prove that diagonal AC = diagonal BD ∵
ΔABC ≌ ΔBAD   [Proved]
∴ Their corresponding parts are equal.
⇒ the diagonal AC = the diagonal BD.

Answer

We have P as the mid-point of AB, Q as the mid-point of BC, R as the mid-point of CD, S as the mid-point of DA, and AC as the diagonal of quadrilateral ABCD.

(i) To prove that SR =(1/2) AC and SR || AC.
In ΔACD, we have
S as the mid-point of AD, R as the mid-point of CD.
∵ The line segment joining the mid-point of any two sides of a triangle is parallel to the third side and half of it.
∴ SR = (1/2)AC and SR || AC

(ii) To prove that PQ = SR.
In ΔABC, we have P is the mid-point of AB, Q is the mid-point of BC.
∴ PQ = (1/2) AC   …(1)
Also, SR = (1/2) AC   [Proved]  …(2)
From (1) and (2),

PQ = SR

(iii) To prove that PQRS is a parallelogram.
In ΔABC, P and Q are the mid-points of AB and BC.
∴ PQ = (1/2)AC and PQ || AC   ...(3)
In ΔACD, S and R are the mid-points of DA and CD.
∴ SR = (1/2)AC and SR || AC   …(4)
From (3) and (4), we get
PQ =(1/2)AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
i.e. One pair of opposite sides in quadrilateral PQRS is equal and parallel.
∴ PQRS is a parallelogram.

Answer

We have P as the mid-point of AB, Q as the midpoint of BC, R as the mid-point of CD, S as the mid-point of DS.

We have to prove that PQRS is a rectangle.
Let us join AC.
∵ In ΔABC, P and Q are the mid-points of AB and BC.
∴ PQ =(1/2)AC and PQ || AC   …(1)
Also in ΔADC, R and S are the mid-points of CD and DA.
∴ SR = (1/2)AC and SR || AC
From (1) and (2), we get
PQ =(1/2) AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
i.e. One pair of opposite sides of quadrilateral PQRS is equal and parallel.
∴ PQRS is a parallelogram.
Now, in ΔERC and ΔEQC,

∠1= ∠2   [∵ The diagonal of a rhombus bisects the opposite angles]
CR = CQ   [Each is equal to(1/2) of a side of rhombus]
CE = CE   [Common]
∴ ΔERC ≌ ΔEQC   [SAS criteria]
⇒ ∠3= ∠4  [c.p.c.t.]
But ∠3 + ∠4 = 180º  [Linear pair]
⇒ ∠3= ∠4 = 90°
But ∠5= ∠3   [Vertically opposite angles]
∴ ∠5 = 90º PQ || AC
⇒ PQ || EF
∴ PQEF is a quadrilateral having a pair of opposite sides parallel and one of the angles is 90º.
∴ PQEF is a rectangle.
⇒ ∠RQP = 90º
∴ One angle of parallelogram PQRS is 90º.
Thus, PQRS is a rectangle.

Answer

In a rectangle ABCD, P is the mid-point of AB, Q is the midpoint of BC, R is the mid-point of CD, S is the mid-point of DA AC is the diagonal.

Now, in ΔABC,
PQ =(1/2)AC and PQ || AC  [Mid-point theorem]  …(1)
Similarly, in ΔACD,
SR =(1/2)AC and SR || AC   …(2)
From (1) and (2), we get

PQ = SR and PQ || SR

Similarly, by joining BD,
we have PS = QR and PS || QR
i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.
∴ PQRS is a parallelogram.
Now, in ΔPAS and ΔPBQ,

∠A= ∠B   [Each = 90º]
AP = BP  [Each = (1/2)AB]
AS = BQ    [Each =(1/2) of opposite sides of a rectangle]
∴ ΔPAS ≌ ΔPBQ   [SAS criteria]
∴ Their corresponding parts are equal.
⇒ PS = PQ

Also PS = QR   [Proved]
and PQ = SR    [Proved]
∴ PQ = QR = RS = SP
i.e. PQRS is a parallelogram having all of its sides equal.
⇒ PQRS is a rhombus.

Answer

In trapezium ABCD, AB || DC. E is the mid-point of AD.
EF is drawn parallel to AB.

We have to prove that F is the mid-point of BC.
Join BD.
In DDAB,
∵ E is the mid-point of AD   [Given]
and EG || AB   [∵ EF || AB]
∴ Using the converse of mid-point theorem, we get that G is the mid-point BD.
Again in DBDC,
∵ G is the mid-point of BD  [Proved]
GF || DC   [∵ AB || DC and EF || AB and GF is a part of EF]
∴ Using the converse of the mid-point theorem, we get that F is the mid-point of BC.

Answer

We have ABCD is a parallelogram such that: E is the mid-point of AB and F is the mid-point of CD.

Let us join the opposite vertices B and D.
Since, the opposite sides of a parallelogram are parallel and equal.
∴ AB || DC ⇒ AE || FC   …(1)
Also, AB = DC
or (1/2) AB =(1/2)DC ⇒ AE = FC     …(2)
From (1) and (2),
we can say that AECF is quadrilateral having a pair of the opposite sides as parallel and equal.
∴ AEFC is a parallelogram.
⇒ AE || CF
Now, in DDBC,

F is the mid-point of DC   [Given]
and FP || CQ    [∵ AF || CE]
⇒ P is the mid-point of DQ    [Converse of mid-point theorem]
⇒ DP = PQ    …(3)
Similarly, in DBAP,

BQ = PQ   …(4)
∴ From (3) and (4), we have
DP = PQ = BQ
⇒ The line segments AF and EC trisect the diagonal BD.

Answer

Answer

We have a triangle ABC, such that ∠C = 90º M is the mid-point of AB and MD || BC

(i) To prove that D is the mid-point of AC.
In ΔACB, we have M as the mid-point of AB.   [Given]
MD || BC  [Given]
∴ Using the converse of mid-point theorem, D is the mid-point of AC.

(ii) To prove that MD ⊥ AC.
Since, MD || BC   [Given] and AC is a transversal.]
∴ ∠ MDA = ∠BCA   [Corresponding angles]
But ∠BCA = 90º   [Given]
∴ ∠MDA = 90º ⇒ MD ⊥ AC.

(iii) To prove that CM = MA =(1/2) AB
In ΔADM and ΔCDM, we have
∠ADM = ∠CDM   [Each = 90º]
MD = MD   [Common]
AD = CD  [∵ M is the mid-point of AC (Proved)]
∴ ΔADM ≌ ΔCOM   [SAS criteria]
⇒ MA = MC  [c.p.c.t.]  …(1)
∵ M is the mid-point AB.   [Given]
∴ MA =(1/2)AB  …(2)
From (1) and (2), we have
CM = MA = (1/2) AB