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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
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)
[ AD
BC 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]
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.
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.
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)]
2
ABC =
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.
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]
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.
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.
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 (AD
BC 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 (AB
CD 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.
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.
(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]
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
NCERT Solutions for Class 9 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 9 Maths have total 15 chapters. 9 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 9 solutions PDF and Maths ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.
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