NCERT Solutions for Class 9 Maths Exercise 8.1

NCERT Solutions for Class 9 Maths Exercise 8.1 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Quadrilateral Download as PDF

NCERT Solutions for Class 9 Mathematics Quadrilaterals

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

Ans. Let in quadrilateral ABCD, A = B = C = and D =

Since, sum of all the angles of a quadrilateral =

A + B + C + D =

Now A =

B =

C =

And D =

Hence angles of given quadrilateral are and

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

Ans. 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]

ABC BAD [By SSS congruency]

DAB = CBA [By C.P.C.T.] ……….(i)

But DAB + CBA = ……….(ii)

[ ADBC and AB cuts them, the sum of the interior angles of the same side of transversal is ]

From eq. (i) and (ii),

DAB = CBA =

Hence ABCD is a rectangle.

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

Ans. 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 =

To prove: ABCD is a rhombus.

Proof: In AOD and BOC,

OA = OC[Given]

AOD = BOC [Given]

OB = OD[Given]

AOD COB [By SAS congruency]

AD = CB[By C.P.C.T.] ……….(i)

Again, In AOB and COD,

OA = OC[Given]

AOB = COD [Given]

OB = OD[Given]

AOB COD [By SAS congruency]

AD = CB [By C.P.C.T.] ……….(ii)

Now In AOD and BOC,

OA = OC[Given]

AOB = BOC [Given]

OB = OB[Common]

AOB COB [By SAS congruency]

AB = BC[By C.P.C.T.] ……….(iii)

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.

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

Ans. 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 O.

Proof: In triangles ABC and BAD,

AB = AB[Common]

ABC = BAD =

BC = AD[Sides of a square]

ABC BAD [By SAS congruency]

AC = BD[By C.P.C.T.]Hence proved.

Now in triangles AOB and AOD,

AO = AO[Common]

AB = AD[Sides of a square]

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

AOB AOD [By SSS congruency]

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

But AOB + AOD = [Linear pair]

AOB = AOD =

OA BD or AC BD Hence proved.

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NCERT Solutions for Class 9 Maths Exercise 8.1

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

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

We haveAC = BD and OA = OC……….(i)

And OB = OD……….(ii)

Now OA + OC = OB + OD

OC + OC = OB + OB [Using (i) & (ii)]

2OC = 2OB

OC = OB……….(iii)

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

Now in AOB and COD,

OA = OD [proved]

AOB = COD[vertically opposite angles]

OB = OC [proved]

AOB DOC[By SAS congruency]

AB = DC [By C.P.C.T.]……….(v)

Similarly, BOC 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 ABC and BAD,

AB = BA [Common]

BC = AD [proved above]

AC = BD [Given]

ABC BAD[By SSS congruency]

ABC = BAD[By C.P.C.T.]……….(vii)

But ABC + BAD = [ABCD is a parallelogram]……….(viii)

AD BC and AB is a transversal.

ABC + ABC = [Using eq. (vii) and (viii)]

2ABC = ABC =

ABC = BAD =……….(ix)

Opposite angles of a parallelogram are equal.

But ABC = BAD =

ABC = ADC = ……….(x)

BAD = BDC =……….(xi)

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

ABC = ADC = BAD = BDC =……….(xii)

Now in AOB and BOC,

OA = OC [Given]

AOB = BOC = [Given]

OB = OB [Common]

AOB 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 with each other.

ABCD is a square.

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NCERT Solutions for Class 9 Maths Exercise 8.1

6. Diagonal AC of a parallelogram ABCD bisects A (See figure). Show that:

(i) It bisects C also.

(ii) ABCD is a rhombus.

Ans. Diagonal AC bisects 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.

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NCERT Solutions for Class 9 Maths Exercise 8.1

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

Ans. ABCD is a rhombus. Therefore, AB = BC = CD = AD

Let O be the point of bisection of diagonals.

OA = OC and OB = OD

In AOB and AOD,

OA = OA [Common]

AB = AD [Equal sides of rhombus]

OB = OD (diagonals of rhombus bisect each other]

AOB AOD [By SSS congruency]

OAD = OAB [By C.P.C.T.]

OA bisects A ……….(i)

Similarly, BOC 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 AOB and BOC,

OB = OB [Common]

AB = BC [Equal sides of rhombus]

OA = OC (diagonals of rhombus bisect each other]

AOB COB [By SSS congruency]

OBA = OBC [By C.P.C.T.]

OB bisects B ……….(iii)

Similarly, AOD COD [By SSS congruency]

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

BD bisects D ……….(iv)

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

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NCERT Solutions for Class 9 Maths Exercise 8.1

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 both B as well as D.

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

And BC = AD

Also A = B = C = D =

(i) In ABC and ADC

1 = 2 and 3 = 4

[AC bisects A and C (given)]

AC = AC [Common]

ABC ADC[By ASA congruency]

AB = AD ……….(ii)

From eq. (i) and (ii), AB = BC = CD = AD

Hence ABCD is a square.

(ii) In ABC and ADC

AB = BA [Since ABCD is a square]

AD = DC [Since ABCD is a square]

BD = BD [Common]

ABD CBD [By SSS congruency]

ABD = CBD [By C.P.C.T.]……….(iii)

And ADB = CDB[By C.P.C.T.]……….(iv)

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

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NCERT Solutions for Class 9 Maths Exercise 8.1

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

(i) APD CQB

(ii) AP = CQ

(iii) AQB CPD

(iv) AQ = CP

(v) APCQ is a parallelogram.

Ans. (i) In APD and CQB,

DP = BQ[Given]

ADP = QBC [Alternate angles (ADBC and BD is transversal)]

AD = CB [Opposite sides of parallelogram]

APD CQB [By SAS congruency]

(ii) Since APD CQB

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

(iii) In AQB and CPD,

BQ = DP[Given]

ABQ = PDC [Alternate angles (ABCD and BD is transversal)]

AB = CD[Opposite sides of parallelogram]

AQB CPD [By SAS congruency]

(iv) Since AQB CPD

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

(v) In quadrilateral APCQ,

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.

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NCERT Solutions for Class 9 Maths Exercise 8.1

10. ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See figure). Show that:

(i) APB CQD

(ii) AP = CQ

Ans. Given: ABCD is a parallelogram. AP BD and CQ BD

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

Proof: (i) In APB and CQD,

1 = 2[Alternate interior angles]

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

APB = CQD =

APB CQD [By ASA Congruency]

(ii) Since APB CQD

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

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NCERT Solutions for Class 9 Maths Exercise 8.1

11. An ABC and DEF, AB = DE, AB DE, BC = EF and BC EF. Vertices A, B and C are joined to vertices D, E and F respectively (See 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) ABC DEF

Ans. (i) In ABC and DEF

AB = DE[Given]

And AB DE[Given]

ABED is a parallelogram.

(ii) In ABC and DEF

BC = EF[Given]

And BC EF[Given]

BEFC is a parallelogram.

(iii) As ABED is a parallelogram.

AD BE and AD = BE ……….(i)

Also BEFC is a parallelogram.

CF BE and CF = BE……….(ii)

From (i) and (ii), we get

AD CF and AD = CF

(iv) As AD CF and AD = CF

ACFD is a parallelogram.

(v) As ACFD is a parallelogram.

AC = DF

(vi) In ABC and DEF,

AB = DE [Given]

BC = EF [Given]

AC = DF [Proved]

ABC DEF[By SSS congruency]

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NCERT Solutions for Class 9 Maths Exercise 8.1

12. 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. Given: ABCD is a trapezium.

AB CD and AD = BC

To prove: (i) A = B

(ii) C = D

(iii) ABC BAD

(iv) Diag. AC = Diag. BD

Construction: Draw CE AD and extend

AB to intersect CE at E.

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

AD = EC

But AD = BC [Given]

BC = EC

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

Now 1 + 4 = [Interior angles]

And 2 + 3 = [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 [BCE is an isosceles triangle]

C = D

(iii) In ABC and BAD,

AB = AB [Common]

1 = 2 [Proved]

AD = BC [Given]

ABC BAD[By SAS congruency]

(iv) We had observed that,

ABC BAD

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

NCERT Solutions for Class 9 Maths Exercise 8.1

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