# NCERT Solutions for Class 9 Maths

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1.    Let in quadrilateral ABCD, ∠ A = 3x, ∠ B = 5x, ∠ C = 9x and ∠ D = 13x.

Since, sum of all the angles of a quadrilateral = 360°

∴ ∠A + ∠B + ∠C +∠D = 360°         ⇒      3x + 5x + 9x + 13x = 360°

⇒                30x = 360°                      ⇒             x = 12°

Now       ∠ A =  3x = 3 × 12 = 36°

∠ B =  5x = 5 × 12 = 60°

∠ C = 9x = 9 × 12 = 108°

And        ∠ D = 13x = 13 × 12 = 156°

Hence angles of given quadrilateral are 36°, 60°, 108° and 156°.

2.     Given: ABCD is a parallelogram with diagonal AC = diagonal BD

To prove: ABCD is a rectangle.

Proof: In triangles ABC and ABD,

AB = AB                              [Common]

AC = BD                              [Given]

AD = BC                              [opp. Sides of a ∵ gm]

∴ Δ ABC ≅ Δ BAD             [By SSS congruency]

∠ DAB = ∠ CBA                 [By C.P.C.T.] ………. (i)

But ∠ DAB + ∠ CBA = 180° …………………… (ii)

[º AD || BC and AB cuts them, the sum of the interior angles of the same side of transversal is 180° ]

From eq. (i) and (ii),

∠ DAB = ∠ CBA = 90°

Hence ABCD is a rectangle.

3.     Given: Let ABCD is a quadrilateral.

Let its diagonal AC and BD bisect each other at right angle at point O.

∴ OA = OC, OB = OD

And ∠ AOB = ∠ BOC = ∠ COD = ∠ AOD = 90°

To prove:          ABCD is a rhombus.

Proof: In Δ AOD and Δ BOC,

OA = OC                             [Given]

∠ AOD = ∠ BOC        [Given]

OB = OD            [Given]

∴ Δ AOD ≅ Δ COB [By SAS congruency]

⇒ AD = CB             [By C.P.C.T.]                …(i)

Again, In Δ AOB and Δ COD,

OA = OC             [Given]

∠ AOB = ∠ COD    [Given]

OB = OD             [Given]

∴ Δ AOB ≅ Δ COD    [By SAS congruency]

⇒ AD = CB             [By C.P.C.T.]                …(ii)

Now In Δ AOD and Δ BOC,

OA = OC                [Given]

∠ AOB = ∠ BOC       [Given]

OB = OB                [Common]

∴ Δ AOB ≅ Δ COB       [By SAS congruency]

⇒ AB = BC        [By C.P.C.T.]

From eq. (i), (ii) and (iii),

AD = BC = CD = AB

And the diagonals of quadrilateral ABCD bisect each other at right angle. Therefore ABCD is a rhombus.

4.    Given: ABCD is a square. AC and BD are its diagonals bisect each other at point O.

To prove:           AC = BD               and AC ⊥ BD at point 0.

Proof: In triangles ABC and BAD,

AB = AB                       [Common]

∠ ABC = ∠ BAD = 90°

BC = AD                        [Sides of a square]

∴ Δ ABC ≅ Δ BAD         [By SAS congruency]

⇒ AC = BD                      [By C.P.C.T.]

Now in triangles AOB and AOD,

AO = AO                       [Common]

AB = AD                       [Sides of a square]

OB = OD                       [Diagonals of a square bisect each other]

∴ Δ AOB ≅ Δ AOD              [By SSS congruency]

∠ AOB = ∠ AOD                             [By C.P.C.T.]

But ∠ AOB + ∠ AOD = 180°          [Linear pair]

∴ ∠ AOB = ∠ AOD = 90°

⇒ OA ⊥ BD or AC ⊥ BD                  Hence proved.

5.    Let ABCD be a quadrilateral in which equal diagonals AC and BD bisect each other at right angle at point O.

We have  AC = BD and OA = OC             …(i)

And          OB = OD                                     …(ii)

Now OA + OC = OB + OD

⇒ OC + OC = OB + OB                        [Using (i) and (ii)]

⇒ 20C = 20B

⇒ OC = OB                                                 … (iii)

From eq. (i), (ii) and (iii), we get, OA = OB = OC = OD   ..(iv)

Now in Δ AOB and Δ COD,

OA = OD                  [proved]

∠ AOB = ∠ COD       [vertically opposite angles]

OB = OC                   [proved]

∴ Δ AOB ≅ Δ DOC      [By SAS congruency]

AB = DC                   [By C.P.C.T.]                   …(v)

Similarly Δ BOC ≅ Δ AOD   [By SAS congruency]

⇒ BC = AD                   [By C.P.C.T.]                   …(vi)

From eq. (v) and (vi), it is concluded that ABCD is a parallelogram because opposite sides of a quadrilateral are equal.

Now in Δ ABC and Δ BAD,

AB = BA                                   [Common]

AC = BD                                   [Given]

∴ Δ ABC ≅ Δ BAD                      [By SSS congruency]

⇒ ∠ ABC = ∠ BAD                       [By C.P.C.T.]             … (vii)

But ∠ ABC + ∠ BAD = 180°            [ABCD is a parallelogram]     … (viii)

∴ AD ∵ BC and AB is a transversal.

⇒ ∠ ABC + ∠ ABC = 180°            [Using eq. (vii) and (viii)]

⇒ 2 ∠ ABC = 180°                                              ∠ ABC = 90°

∴ ∠ ABC = ∠ BAD = 90°                           … (ix)

Opposite angles of a parallelogram are equal.

But          ∠ ABC = ∠ BAD =

∴ ∠ ABC = ∠ ADC = 90°                            … (x)

∴ ∠ BAD = ∠ BDC = 90°                           … (xi)

From eq. (x) and (xi), we get

∠ ABC = ∠ ADC = ∠ BAD = ∠ BDC = 90°        … (xii)

Now in Δ AOB and Δ BOC,

OA = OC                                   [Given]

∠ AOB = ∠ BOC = 90°              [Given]

OB = OB                                   [Common]

∴ Δ AOB ≅ Δ COB                      [By SAS congruency]

⇒ AB = BC                                … (xiii)

From eq. (v), (vi) and (xiii), we get,

AB = BC = CD = AD           … (xiv)

Now, from eq. (xii) and (xiv), we have a quadrilateral whose equal diagonals bisect each other at right angle.

Also sides are equal make an angle of 90° with each other.

∴ ABCD is a square.

6.    Diagonal AC bisects L A of the parallelogram ABCD.

(i) Since AB ∵ DC and AC intersects them.

∴ ∠ 1 = ∠ 3          [Alternate angles] …………… (i)

Similarly               ∠ 2 = ∠ 4 ………………………………….. (ii)

But           ∠ 1 = ∠ 2         [Given] ………………………….. (iii)

∴ ∠ 3 = ∠ 4                       [Using eq. (i), (ii) and (iii)]

Thus        AC bisects ∠ C.

(ii) ∠ 2 = ∠ 3 = ∠ 4 = ∠ 1

⇒ AD = CD                     [Sides opposite to equal angles]

∴ AB = CD = AD = BC

Hence ABCD is a rhombus.

7.    ABCD is a rhombus. Therefore AB = BC = CD = AD

Let 0 be the point of bisection of diagonals.

∴ OA = OC and OB = OD

In Δ AOB and Δ AOD,

OA = OA          [Common]

AB = AD         [Equal sides of rhombus]

OB = OD          (diagonals of rhombus bisect each other]

∴ Δ AOB ≅ Δ AOD           [By SSS congruency]

⇒ ∠ OAD = ∠ OAB         [By C.P.C.T.]

⇒ OA bisects ∠ A ……………………………………… (i)

Similarly Δ BOC ≅ Δ DOC         [By SSS congruency]

⇒ ∠ OCB = ∠ OCD                      [By C.P.C.T.]

⇒ OC bisects ∠ C ……………………………………… (ii)

From eq. (i) and (ii), we can say that diagonal AC bisects ∠ A and ∠ C.

Now in Δ AOB and Δ BOC,

OB = OB          [Common]

AB = BC         [Equal sides of rhombus]

OA = OC           (diagonals of rhombus bisect each other]

∴ Δ AOB ≅ Δ COB          [By SSS congruency]

⇒ ∠ OBA = ∠ OBC         [By C.P.C.T.]

⇒ OB bisects ∠ B ……………………………………… (iii)

Similarly Δ AOD ≅ Δ COD         [By SSS congruency]

⇒ ∠ ODA = ∠ ODC                     [By C.P.C.T.]

⇒ BD bisects Z D …………………………………….. (iv)

From eq. (iii) and (iv), we can say that diagonal BD bisects ∠ B and ∠ D.

8.    ABCD is a rectangle. Therefore AB = DC …………………………….. (i)

Also        ∠ A = ∠ B = ∠ C = ∠ D = 90°

(i) In Δ ABC and Δ ADC

∠ 1 = ∠ 2 and ∠ 3 = ∠ 4

[AC bisects ∠ A and ∠ C (given)] AC = AC       [Common]

∴ Δ ABC ≅ Δ ADC                      [By ASA congruency]

⇒ AB = AD …………………………………… (ii)

From eq. (i) and (ii), AB = BC = CD = AD Hence ABCD is a square.

(ii)       In Δ ABC and Δ ADC

AB = BA           [Since ABCD is a square]

AD = DC          [Since ABCD is a square]

BD = BD           [Common]

∴ Δ ABD ≅ Δ CBD          [By SSS congruency]

⇒ ∠ ABD = ∠ CBD          [By C.P.C.T.] …………………………… (iii)

And        ∠ ADB = ∠ CDB                     [By C.P.C.T.]

From eq. (iii) and (iv), it is clear that diagonal BD bisects both ∠ B and ∠ D.

9.    (i) In Δ APD and Δ CQB,

DP = BQ                         [Given]

∠ ADP = ∠ QBC            [Alternate angles (AD ∵ BC and BD is transversal)]

AD = CB                         [Opposite sides of parallelogram]

∴ Δ APD ≅ Δ CQB            [By SAS congruency]

(ii) Since Δ APD ≅ Δ CQB

⇒ AP = CQ                        [By C.P.C.T.]

(iii) In Δ AQB and Δ CPD,

BQ = DP                        [Given]

∠ ABQ = ∠ PDC           [Alternate angles (AB CD and BD is transversal)]

AB = CD                        [Opposite sides of parallelogram]

∴ Δ AQB ≅ Δ CPD           [By SAS congruency]

(iv) Since Δ AQB ≅ Δ CPD

⇒ AQ = CP                       [By C.P.C.T.]

AP = CQ                      [proved in part (i)]

AQ = CP                      [proved in part (iv)]

Since opposite sides of quadrilateral APCQ are equal. Hence APCQ is a parallelogram.

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

To prove: (i) Δ APB ≅ Δ CQD (ii) AP = CQ

Proof: (i) In Δ APB and Δ CQD,

∠ 1 = ∠ 2                       [Alternate interior angles]

AB = CD                       [Opposite sides of a parallelogram are equal]

∠ APB = ∠ CQD = 90°

∴ Δ APB ≅ Δ CQD          [By ASA Congruency]

(ii) Since          Δ APB ≅ Δ CQD

∴ AP = CQ          [By C.P.C.T.]

11.

(i) In Δ ABC and Δ DEF

AB = DE                             [Given]

And AB ∵ DE                     [Given]

∴ ABED is a parallelogram.

(ii) In Δ ABC and Δ DEF

BC = EF                           [Given]

And BC ∵ EF                   [Given]

∴ BEFC is a parallelogram.

(iii) As ABED is a parallelogram.

Also BEFC is a parallelogram.

∴ CF ∵ BE and CF = BE                            … (ii)

From (i) and (ii), we get

⇒ ACFD is a parallelogram.

(v) As ACFD is a parallelogram.

∴ AC = DF

(vi) In Δ ABC and Δ DEF,

AB = DE                                  [Given]

BC = EF                                   [Given]

AC = DF                                  [Proved]

∴ Δ ABC ≅ Δ DEF                 [By SSS congruency]

12.  Given: ABCD is a trapezium.

AB II CD and AD = BC

To prove:   (i) ∠ A= ∠ B

(ii) ∠ C= ∠ D

(iii) Δ ABC ≅ Δ BAD

(iv) Diag. AC = Diag. BD

Construction: Draw CE II AD and extend

AB to intersect CE at E.

Proof:

(i) As AECD is a parallelogram.         [By construction]

∴ BC = EC

⇒ ∠ 3 = ∠ 4                                      [Angles opposite to equal sides are equal]

Now ∠ 1 + ∠ 4 = 180°                    [Interior angles]

And ∠ 2 + ∠ 3 = 180°                    [Linear pair]

⇒ ∠ 1 + ∠ 4 = ∠ 2 + ∠ 3

⇒ ∠ 1 = ∠ 2                               [∵ ∠ 3 = ∠ 4]

⇒ ∠ A = ∠ B

(ii) ∠ 3 = ∠ C                             [Alternate interior angles]

And ∠ D = ∠ 4                           [Opposite angles of a parallelogram]

But  ∠ 3 = ∠ 4                            [Δ BCE is an isosceles triangle]

∴ ∠ C = ∠ D

(iii) In Δ ABC and Δ BAD,

AB = AB                                             [Common]

∠ 1 = ∠ 2                                             [Proved]

∴ Δ ABC ≅ Δ BAD                             [By SAS congruency]

⇒ AC = BD                                         [By C.P.C.T.]

13.  In Δ ABC, P is the mid-point of AB and Q is the mid-point of BC.

Then PQ || AC and PQ =  AC

(i) In Δ ACD, R is the mid-point of CD and S is the mid-point of AD.

Then SR || AC and SR =  AC

(ii) Since PQ =  AC and SR =  AC

Therefore            PQ = SR

(iii) Since PQ || AC and SR || AC

Therefore    PQ || SR     [two lines parallel to given line are parallel to each other]

Now PQ = SR and PQ || SR

Therefore PQRS is a parallelogram.

14.  Given: P, Q, R and S are the mid-points of respective sides AB, BC, CD and DA of rhombus. PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Construction: Join A and C.

Proof: In Δ ABC, P is the mid-point of AB and Q is the mid­point of BC.

∴ PQ || AC and PQ =  AC                     …(i)

In Δ ADC, R is the mid-point of CD and S is the mid-point of AD.

∴ SR || AC and SR =  AC                     … (ii)

From eq. (i) and (ii), PQ || SR and PQ = SR

∴ PQRS is a parallelogram.

Now ABCD is a rhombus.                                            [Given]

∴ AB = BC

AB = BC                           ⇒                                  PB = BQ

∴ ∠ 1 = ∠ 2                     [Angles opposite to equal sides are equal]

Now in triangles APS and CQR, we have,

AP = CQ                        [P and Q are the mid-points of AB and BC and AB = BC]

Similarly                AS = CR and PS = QR  [Opposite sides of a parallelogram]

∴ Δ APS ≅ Δ CQR                                      [By SSS congreuancy]

⇒ ∠ 3 = ∠ 4                                             [By C.P.C.T.]

Now we have                ∠ 1 + ∠ SPQ + ∠ 3 = 180°

And                               ∠ 2 + ∠ PQR + ∠ 4 = 180°                  [Linear pairs]

∴ ∠ 1 + ∠ SPQ + ∠ 3 = ∠ 2 + ∠ PQR + ∠ 4

Since ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4                         [Proved above]

∴ ∠ SPQ = ∠ PQR                                            … (iii)

Now PQRS is a parallelogram                          [Proved above]

∴ ∠ SPQ + ∠PQR = 180°                               … (iv)        [Interior angles]

Using eq. (iii) and (iv),

∠ SPQ + ∠ SPQ = 180°                ⇒                    2 ∠ SPQ = 180°

⇒ ∠ SPQ = 90°

Hence PQRS is a rectangle.

15.  Given: A rectangle ABCD in which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.

To prove:          PQRS is a rhombus.

Construction: Join AC.

Proof: In Δ ABC, P and Q are the mid-points of sides AB, BC respectively.

∴ PQ || AC and PQ =  AC                            … (i)

In Δ ADC, R and S are the mid-points of sides CD, AD respectively.

∴ SR || AC and SR =  AC                             … (ii)

From eq. (i) and (ii),                   PQ || SR and PQ = SR            … (iii)

∴ PQRS is a parallelogram.

Now ABCD is a rectangle.             [Given]

AD =  BC                     ⇒                  AS = BQ                   … (iv)

In triangles APS and BPQ,

AP = BP                                         [P is the mid-point of AB]

∠ PAS = ∠ PBQ                            [Each 90° ]

And AS = BQ                                [From eq. (iv)]

∴ Δ APS ≅ Δ BPQ                        [By SAS congruency]

⇒ PS = PQ                                     [By C.P.C.T.]          …(v)

From eq. (iii) and (v), we get that PQRS is a parallelogram.

⇒ PS = PQ

⇒ Two adjacent sides are equal.

Hence, PQRS is a rhombus.

16.  Let diagonal BD intersect line EF at point P.

In Δ DAB,

E is the mid-point of AD and EP || AB           [∵ EF || AB (given) P is the part of EF]

∴ P is the mid-point of other side, BD of Δ DAB.

[A line drawn through the mid-point of one side of a triangle, parallel to another side intersects the third side at the mid-point]

Now in Δ BCD,

P is the mid-point of BD and PF || DC     [∵ EF || AB (given) and AB || DC (given)]

∴ EF || DC and PF is a part of EF.

∴ F is the mid-point of other side, BC of Δ BCD. [Converse of mid-point of theorem]

17.  Since E and F are the mid-points of AB and CD respectively.

∴ AE =  AB and CF =  CD             … (i)

But ABCD is a parallelogram.

∴ AB = CD and AB || DC

AB =  CD and AB || DC

⇒ AE = FC and AE || FC                     [From eq. (i)]

∴ AECF is a parallelogram.

⇒ FA || CE      ⇒     FP || CQ       [FP is is a part of FA and CQ is a part of CE]   …(ii)

Since the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.

In Δ DCQ, F is the mid-point of CD and    ⇒     FP || CQ

∴ P is the mid-point of DQ.

⇒ DP = PQ                                 … (iii)

Similarly, In Δ ABP, E is the mid-point of AB and           ⇒          EQ || AP

∴ Q is the mid-point of BP.

⇒ BQ = PQ                                                                … (iv)

From eq. (iii) and (iv),

DP = PQ = BQ                                                       … (v)

Now BD = BQ + PQ + DP = BQ + BQ + BQ = 3BQ

⇒ BQ =   BD                                                            … (vi)

From eq. (v) and (vi),

DP = PQ = BQ =  BD

⇒ Points P and Q trisects BD.

So AF and CE trisects BD.

18.  Given: A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.

To prove:           EG and FH bisect each other.

Construction: Join AC, EF, FG, GH and HE.

Proof: In Δ ABC, E and F are the mid-points of respective sides AB and BC.

∴ EF || AC and EF  AC                                   … (i)

G and H are the mid-points of respective sides CD and AD.

∴ HG || AC and HG  AC                                 … (ii)

From eq. (i) and (ii),

EF || HG and EF = HG

∴ EFGH is a parallelogram.

Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.

19.

(i) In Δ ABC, M is the mid-point of AB                              [Given]

MD || BC

∴ AD = DC                               [Converse of mid-point theorem]

Thus D is the mid-point of AC.

(ii) l || BC (given) consider AC as a transversal.

∴ ∠ 1 = ∠ C                         [Corresponding angles]

⇒ ∠ 1 = 90°                        [∠ C = 90°]

Thus MD ⊥ AC.

(iii) In Δ AMD and Δ CMD,

∠ 1 = ∠ 2 = 90°                        [proved above]

MD = MD                                 [common]

∴ Δ AMD ≅ Δ CMD                [By SAS congruency]

⇒ AM = CM                             [By C.P.C.T.]                   … (i)

Given that M is the mid-point of AB.

∴ AM =  AB                                 … (ii)

From eq. (i) and (ii),

CM = AM =  AB.

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