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NCERT Class 9 Mathematics Chapter 8 Quadrilaterals
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 9 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 9 Mathematics Chapter 8 Quadrilaterals Solutions for All Subject, You can practice these here.
Quadrilaterals
Chapter – 8
| Exercise 8.1 |
Q.1. The angles of a quadrilateral are the ratio 3:5: 9:13. Find all the angles of the quadrilateral.
Ans: Let ABCD be a quadrilateral in which
∠A: ∠B:∠C:∠D=3:5:9:13
Sum of the ratios = 3 + 5 + 9 + 13 = 30
Also, ∠ A+∠B+∠C+ ∠ D = 360⁰
| Sum of all the angles of a quadrilateral is 360⁰
Q.2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Ans: Given: In parallelogram ABCD, AC = BD
To Prove: || gm ABCD is a rectangle.
Proof: In ∆ ACB and ∆ BDA
AC = BD | Given
AB = BA | Common
BC = AD | Opposite sides of || gm ABCD
∴ ACB ≅ ∆BDA | ISSS Rule
∴ ∠ ABC = ∠ BAD …(1) c.p.c.t.
Again,∵ AD || BC | Opp. sides of || gm
ABCD and transversal AB intersects them
∴ ∠BAD+∠ABC = 180⁰…(2) Sum of consecutive interior angles on the same side of the transversal is 180⁰
From (1) and (2),
∠BAD=∠ABC = 90⁰ ∴ ∠ A = 90⁰
∴ || gm ABCD is a rectangle.
Q.3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Ans: Given: ABCD is a quadrilateral where diagonals AC and BD intersect each other at right angles at O.
To Prove: Quadrilateral ABCD is a rhombus.
Proof: In ∆AOB and ∆AOD
AO = AO | Common
OB = OD | Given
∠AOB =∠AOD | Each = 90⁰
∴ ∆AOB ≅ ∆AOD | SAS Rule
∴ AB = AD …(1) I c.p.c.t.
Similarly, we can prove that
AB = BC
BC = CD
CD = AD
In view of (1), (2), (3) and (4), we obtain
AB = BC =CD=DA
∴ Quadrilateral ABCD is a rhombus.
Q.4. Show that the diagonals of a square are equal and bisect each other at right angles.
Ans: Given: ABCD is a square
To Prove: (i) AC = BD
(ii) AC and BD bisect each other at right angles.
Proof: (i) In ∆ABC and ∆BAD,
AB = BA | Common
BC = AD | Opp. sides of square ABCD
∠ ABC = ∠ BAD | Each = 90⁰ ( ∵ ABCD is a square)
∴ ∆ ABC ≅ ∆BAD | SAS Rule
AC = BD Ic.p.c.t.
(ii) In ∆ OAD and ∆OCB
AD = CB | Opp. sides of square ABCD
∠ OAD = ∠ OCB| ∵ AD || BC and transversal
AC intersects them
∠ ODA = ∠ OBC |∵AD || BC and transversal BD intersects them
∴ ∆ OAD≅∆OCB | ASA Rule
∴ OA = OC …(1)
Similarly, we can prove that
OB = OD …(2)
In view of (1) and (2)
AC and BD bisect each other.
Again, in ∆OBA and ∆ODA
OB = OD | From (2) above
BA = DA | Opp. sides of square ABCD
OA = OA | Common
∴ ∆ OBA≅∆ ODA | SSS Rule
∴ ∠AOB =∠AOD | c.p.c.t.
But ∠AOB + ∠AOD = 180⁰ | Linear Pair Axiom
∴ ∠AOB=∠AOD = 90⁰
∴ AC and BD bisect each other at right angles.
Q.5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Ans: Given
The diagonals AC and BD of a quadrilateral ABCD and equal and bisect each other at right angles.
To Prove: Quadrilateral ABCD is a square.
Proof: In ∆OAD and ∆OCB,
OA = OC | Given
OD = OB | Given
∠ AOD = ∠ COB | Vertically Opposite angles
∴ ∆ OAD≅∆OCB | SAS Rule
∴ AD = CB | c.p.c.t.
∠ODA = ∠ OBC | c.p.c.t.
∴ ∠ODA =∠OBC
∴ AD || BC
Now, AD = CB and AD || CB
∴ Quadrilateral ABCD is a || gm.
In ∆AOB and ∆AOD
AO = AO | Common
OB = OD | Given
∠AOB = ∠ AOD | Each = 90⁰ (Given)
∴ ∆ AOB ≅ ∆AOD |SAS Rule
∴ AB = AD
Now, ABCD is a parallelogram and AB = AD
∴ ABCD is a rhombus.
Again, in ∆ ABC and ∆BAD,
AC=BD | Given
BC=AD | ∵ ABCD is a rhombus
AB = BA | Common
∴ ∆ABC ≅ ∆BAD | SSS Rule
∴ ∠ABC≅∠ BAD | c.p.c.t.
∵ AD || BC | Opp. sides of || gm ABCD
and transversal AB intersects them.
∴ ∠ABC+∠BAD = 180⁰
| Sum of consecutive interior angles on the same side of the transversal is 180⁰
∴ ∠ABC = ∠BAD = 90⁰
Similarly, ∠BCD = ∠ ADC = 90⁰
∴ ABCD is a square.
Q.6. Diagnosal AC of a parallelogram ABCD bisects ∠A. Show that:
(i) it bisects ∠C also
(ii) ABCD is a rhombus
Ans: Given: Diagonal AC of a parallelogram ABCD bisects ∠A.
To Prove: (i) it bisects ∠C also.
(ii) ABCD is a rhombus.
Proof: (i) In ∆ADC and ∆ CВА,
AD=BC | Opp. sides of || gm ABCD
CA = CA | Opp. sides of ||gm ABCD
∴ ∆ADCA ≅∆СВА | SSS Rule
∴ ∠ACD = ∠CAB | c.p.c.t.
and ∠DAC = ∠BCA | c.p.c.t.
But ∠CAB=∠DAC | Given
∴ ∠ACD = ∠ BCA
∴ AC bisects ∠C also
(ii) From above,
∠ACD = ∠CAD
∴ AD = CD | Opposite sides of equal angles of a triangle are equal
∴ AB = BC = CD = DA |∵ ABCD is a || gm
∴ ABCD is a rhombus.
Q.7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D
Ans: Given: ABCD is a rhombus.
To Prove: (i) Diagonal AC bisects ∠A as well as ∠C
(ii) Diagonal BD bisects ∠B as well as ∠D
Proof: ∵ ABCD is a rhombus
∴ AD=CD
∴ ∠DAC = ∠DCA …(1)
| Angles opposite to equal sides of a triangle are equal
Also, CD || AB
and transversal AC intersects them
∴ ∠DCA = ∠BCA ..(2) | Alt. Int. ∠S
Friv (1) and (2)
∠DCA = ∠BCA
⇒ AC bisects ∠C
Similarly AC bisects ∠A
(ii) Proceeding similarly as in (i) above, we can prove that BD bi-sects ∠B as well as ∠D.
Q.8. ABCD is a rectangle in which diagonal AC bisects ∠A. Show that (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.
Ans: Given: ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C
To Prove: (i) ABCD is a square.
(ii) diagonal BD bisects ∠B as well as ∠D.
Proof: (i) ∵ AB || DC
and transversal AC intersects them
∴ ∠ACD = ∠CAB || Alt. Int. ∠S
But ∠CAB=∠CAD
∴ ∠ACD=∠CAD
∴ AD = CD || Sides opposite to equal angles of a triangle are equal
∴ ABCD is a square
(ii) In ∆ BDA and ∆DBC
BD = DB | Common
DA = BC | Sides of a square ABCD
AB = DC | Sides of a square ABCD
∴ ∆BDA ≅ ∆DBC | SSS Rule
∴ ∠ABD = ∠ CDB | c.p.c.t.
But ∠CDB = ∠CBD | ∵ CB = CD (Sides of a square ABCD)
∴ ∠ ABD = ∠ CBD
∴ BD bisects ∠ B
Now, ∠ ABD = ∠ CBD
∠ABD = ∠ ADB | ∵ AB=AD
∠CBD =∠CDB | ∵ CB=CD
∴ ∠ ADB = ∠ CDB
∴ BD bisects ∠ D
Q.9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that:
(i) ∆ APD≅∆CQB
(ii) AP = CQ
(iii) ∆ AQB ≅ ∆CPD
(iv) AQ = CP
(v) APQC is a parallelogram.
Ans: Given: In parallelogram ABCD, two points P and Q are on di- agonal BD such that DP = BQ
Construction: Join AC to intersect BD at O.
Proof: (i) In ∆APD and ∆CQB | From (ii)
PD = QB | Given
AD = BC | Opp. sides of || gm ABCD
∴ ∆ APD≅∆ CQB | SSS Rule
(ii) ∵ APCQ is a || gm | Proved in (i) above
AP = CO | Op. sides of a || gm are equal
(iii) In ∆AQB and ∆ CPD,
AQ = CP | From (iii)
QB = PD | Given
AB = CD | Opp. sides of || gm ABCD
∴ ∆ AQB ≅ ∆CPD | SSS Rule
(iv) APQC is a || gm | Proved in (i) above
∴ AQ = CP | Opp. sides of a || gm are equal
(v) ∵ The diagonals of a parallelogram bisect each other.
∴ OB = OD
∴ OB-BQ=OD-DP | ∵ BQ = DP (given)
∴ OQ=OP …(1)
Also, OA = OC…(2) | Diagonals of a || gm bisect each other
In view of (1) and (2), APQC is a parallelogram.
Q.10. ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD respectively. Show that:
(i) ΔΑΡΒ ≅ ∆ CQD
(ii) AP=CQ
Ans: Given: ABCD is a parallelo- gram and AP and
CQ are perpendiculars from verti- ces A and Con diagonal BD respec tively.
To Prove: (i) ∆ APB ≅ ∆ CQD
(ii) AP = CQ
Proof: (i) In ∆APB and ∆ CQD,
AB = CD | Opp. sides || gm ABCD
∠ ABP = ∠ CDQ | ∵ AB || DC and transversal BD intersects them
∠APB = ∠CQD | Each = 90⁰
∴ ΔΑΡΒ ≅ Δ CQD | AAS Rule
(ii) ∵ ∆ APB≅∆CQD | Proved above in (i)
∴ AP = CQ | c.p.c.t.
Q.11. In ∆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 respec-tively. 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: Given: In ∆ 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.
To Prove: (i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BECF is a parallelogram
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC=DF
(vi) ∆ ABC ≅∆ DEF
Proof: (i) In quadrilateral ABED,
AB = DE and AB || DE |Given
∴ quadrilateral ABED is a parallelogram.
|∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length
(ii) In quadrilateral BECF,
BC = EF and BC || EF | Given
∴ quadrilateral BECF is a parallelogram.
| ∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length
(iii) ∵ ABED is a parallelogram
| Proved in (i)
∴ AD || BE and AD = BE …(1)
| ∵ Opposite sides of a || gm are parallel and equal
∵ BEFC is a parallelogram | Proved in (ii)
∴ BE|| CF and BE = CF …(2)
| ∵ Opposite sides of a || gm are parallel and equal
From (1) and (2), we obtains AD || CF and AD = CF
(iv) In quadrilateral ACFD
AD || CF and AD = CF | From (iii)
∴ quadrilateral ACFD is a parallelogram.
∵ A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length
(v) ∵ ACFD is a parallelogram
| Proved in (iv)
∴ AC || DF and AC = DF
| In a parallelogram opposite sides and parallel and of equal length
(vi) In ∆ ABC and ∆ DEF,
AB = DE | ∵ ABED is a parallelogram
BC = EF | ∵ BEFC is a parallelogram
AC = DF | Proved in (v)
∴ Δ ABC≅∆DEF | SSS Rule
Q.12. ABCD is a trapezium in which AB || CD and AD = BC. Show that:
(i) ∠ A = ∠ B
(ii) ∠ C =∠D
(iii) ∆ ABC ≅ ∆ BAD
(iv) diagonal AC = diagonal BD
[Hing. Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Ans: Given: ABCD is a trapezium in which AB || CD and AD = BC
Construction: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.
Proof: (i) AB || CD | Given
and AD || EC | By construction
∴ AECD is a parallelogram
| Aquadrillelogram if a pair of opposite sides is parallel and is of equal length
∴ AD = EF | Opp. sides of a || gm are equal
But AD = BC | Given
∴ EC = BC
∴ ∠CBE =∠CEB …(1)
| Angles of opposite to equal sides of a triangle are equal
∠ B+∠CBE = 180⁰ …(2)
| Linear Pair Axiom
∵ AD ||FC
| By construction and transversal AE intersects them
∴ ∠ A+∠CEB = 180⁰ …(3)
| The sum of consecutive interior angles on the same side of the transversal is 180⁰
From (2) and (3),
∠ B+∠CBE =∠A+ ∠ CEB
But ∠ CBE =∠CEB | From (1)
∴ ∠B = ∠ A or ∠ A = ∠ B
(ii) ∵ AB ||BD
∴ ∠ A + ∠ D = 180⁰
| The sum of consecutive interior angles on the same side of the transversal is 180⁰
and ∠ B+ ∠ C = 180⁰
∴ ∠A +∠D= ∠ B+ ∠ C
But ∠ A = ∠ B | Proved in (i)
∴ ∠ D =∠C or∠C =∠D
(iii) In ∆ABC and ∆BAD,
AB = BA | Common
BC = AD | Given
∠ ABC = ∠ BAD | From (i)
∴ ∆ ABC≅∆BAD | SAS Rule
(iv) ∵ ∆ ABC ≅ ∆ BAD | From (iii) above
∴ AC = BD | c.p.c.t.
| Exercise 8.2 |
Q.1. ABCD is a quadrilateral in which P, Q R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Ans: Given: ABCD is a quadrilateral in which P, Q, R and S are mid- points of the sides AB, BC, CD and DA, AC is a diagonal.
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Proof: In ∆ DAC
∵ Is the mid-point of DA and R is the midpoint of DC
| Mid-point theorem
(ii) In ∆BAC
∵ P is the midpoint of AB and Q is the midpoint of BC
| Mid-point theorem
∴ PQ = SR
(iii) PQ ||AC | From (i)
SR || AC | From (ii)
∴ PQ|| SR | Two lines parallel to the same line are parallel to each other
Also, PQ = SR | From (ii)
∴ PQRS is a parallelogram |A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length
Q.2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Ans: Proof: In triangles RDS and PBQ
DS = QB | Half of opposite sides of || gm ABCD which are equal
DR = PB | Half of opposite sides of || gm ABCD which are equal
∠SDR = ∠QBP | Opposite ∠S of || gm ABCD which are equal
∴ ∆ARDS≅ ∆ PBQ | SAS Axiom
∴ SR = PQ | c.p.c.t.
In triangles RCQ and PAS,
RC = AP | Half of opposite sides of || gm ABCD which are equal
CQ=AS | Half of opposite sides of || gm ABCD which are equal
∠ CRQ = ∠ PAS | Opposite ∠S of || gm ABCD which are equal
∴ ∆ CRCQ ≅ ΔΡΑS | SAS Axiom
∴ RQ = SP | c.p.c.t.
∴ In PQRS,
SR = PQ and RQ = SP
∴ PQRS is a parallelogram,
In ∆ CDB,
∵ R and Q are the mid-points of DC and CB respectively.
∴ RQ ||DB ⇒ RF||EO
Similarly, RE || FO ∴ OFRE is a || gm ∴∠ R= ∠ EOF = 90⁰
| ∵ Opposite ∠ S of a || gm are equal and diagonals of a rhombus intersect at 90⁰
Thus PQRS is a rectangle
Q.3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Ans: Given: ABCD is a rectangle. P, Q, R and S are the mid-points of AB, BC, CD and DA respectively, PQ, QR, RS and SP are joined.
To prove: Quadrilateral PQRS is a rhombus.
Construction: Joined AC
Proof: In A ABC,
∴ P and Q are the mid-points of AB
and BC respectively.
In ∆ ADC,
S and R are the mid-points of AD and DC respectively.
…(2)
From (1) and (2), PQ || SR and PQ = SR
∴ Quadrilateral PQRS is a parallelogram …(3)
In rectangle ABCD
AD = BC | Opposite sides
| Halves of equals are equal
⇒ AS = BQ
In ∆ APS and ∆ BPQ,
AP = BP | ∵ P is the midpoint of AB
AS = BQ | Proved above
∠ PAS =∠PBO | Each = 90⁰
∴ ∆APS = ∆BPQ | SAS Axiom
∴ PS = PQ …(4) | c.p.c.t.
In view of (3) and (4), PQRS is a rhombus.
Q.4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersect- in BC at F. Show that F is the midpoint of BC.
Ans: Given: ABCD is a trapezium in which AB || DC, BD is a diago- nal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F.
To Prove: F is the midpoint of BC.
Proof: Let BD intersect EF at G.
In ∆ DAB,
∵ E is the mid-point of DA and EG || AB
∴ G is the midpoint of DB
| By converse of mid-point theorem Again, in ∆ BDC,
∵ G is the mid-point of BD and GF || AB || DC
∴ F is the midpoint of BC. | By converse of mid-point theorem.
Q.5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the segments AF and EC trisect the diagonal BD.
Ans: Given: In a parallelogram ABCD, E and the mid-points of sides AB and CD respectively.
To Prove: Line segments AF and EC trisect the diagonal BD.
Proof: ∵ AB || DC | Opposite sides of || gm ABCD
∴ AE || FC …(1)
∵ AB= DC | Opposite sides of || gm ABCD
| Half of equals are equal
⇒ AE=CF …(2)
In view of (1) and (2),
AECF is a parallelogram
|A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length
∴ EC|| AF …(3) | Opposite sides of || gm AECF
In ∆ DBC,
∵ F is the midpoint of DC
and FP || CQ |EC|| AF
∴ P is the mid-point of DQ | By converse of mid-point theorem
⇒ DP = PQ …(4)
Similarly, in ∆BAP,
BQ = PQ …(5)
From (4) and (5), we obtain
DP = PQ = BQ
⇒ Line segments AF and EC trisect the diagonal BD.
Q.6. ABC is a triangle right angled at C, A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the midpoint of AC
Ans: Given: ABC is a triangle right angled at C.
A line through the midpoint M of hypotenuse
AB and parallel to BC intersects AC at D.
To Prove: (i) D is the midpoint of AC.
(ii) MD ⊥ AC
Proof: (i) In ∆ ACB
∵ M is the mid-point of AB and MD || BC
∴ D is the midpoint of AC. | By converse of mid-point theorem
(ii) ∵ MD || BC and AC intersects them
∴ ∠ ADM = ∠ACB | Corresponding angles
But ∠ACB = 90⁰ | Given
∴ ∠ADM = 90⁰
⇒ MD⊥AC
(iii) Now, ∠ ADM + ∠ CDM = 180⁰ | Linear Pair Axiom
Ans: ∠ ADM = ∠ CDM = 90⁰
In ∆ ADM and ∆ CDM,
AD = CD | D is the midpoint of AC
∠ADM = ∠ CDM | Each = 90⁰
DM = DM | Common
∴ ∆ ADM ≅ ∆ CDM | SAS Rule
∴ MA = MC | c.p.c.t.
But M is the midpoint of AB
