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# NCERT Solutions for Class 9 Maths Exercise 8.1

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NCERT solutions for Class 9 Maths Quadrilateral

## 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

###### 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]

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.

###### 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]

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.

###### 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]

BC = AD[Sides of a square]

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.

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]

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

AB = BA [Common]

AC = BD [Given]

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

AD BC and AB is a transversal.

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

2ABC = ABC =

Opposite angles of a parallelogram are equal.

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

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.

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.

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]

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.

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)

Also A = B = C = D =

1 = 2 and 3 = 4

[AC bisects A and C (given)]

AC = AC [Common]

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

Hence ABCD is a square.

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)

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

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]

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.]

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.

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.]

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.

(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.

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 ABC and DEF,

AB = DE [Given]

BC = EF [Given]

AC = DF [Proved]

ABC DEF[By SSS congruency]

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

(iv) Diagonal AC = Diagonal BD

Ans. Given: ABCD is a trapezium.

AB CD and AD = BC

To prove: (i) A = B

(ii) C = D

(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]

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

AB = AB [Common]

1 = 2 [Proved]

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

## NCERT Solutions for Class 9 Maths Exercise 8.1

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