Q1 ) The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: The ratios of the angles of quadrilateral are 3: 5: 9: 13.

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

We know that, sum of angles of a quadrilateral = \(360^\circ\)

\(\Rightarrow \) 3x + 5x + 9x + 13x = \(360^\circ\)

\(\Rightarrow \) 30x = \(360^\circ\)

\(\therefore \) x = \(12^\circ\)

Thus, we get,

angles of quadrilateral

=> 3x = 3 × \(12^\circ\) = \(36^\circ\)

=>5x = 5 × \(12^\circ\) = \(60^\circ\)

=> 9x = 9 × \(12^\circ\) = \(108^\circ\)

=> 13x = 13 × \(12^\circ\) = \(156^\circ\)

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: Parallelogram is ABCD whose diagonals AC and BD are equal.

\(\Rightarrow \) AC = BD

To prove: ABCD is a rectangle.

Proof:

In \(\triangle{ABC}\) and \(\triangle{DCB}\), we have,

AC = BD ...(Given)

AB = CD

(Opposite sides of parallelogram)

BC = CB ...(Common sides)

\(\therefore \) \(\triangle{ABC}\)\(\displaystyle \cong \)\(\triangle{DCB}\)

(By SSS rule)

\(\therefore \) \(\angle{ABC}\) = \(\angle{DCB}\) ...(i)(By CPCT)

But from figure, DC || AB and transversal CB intersects them

\(\angle{ABC}\) + \(\angle{DCB}\) = \(180^\circ\)

(interior angles on same side of transversal)

\(\angle{ABC}\) + \(\angle{ABC}\) = \(180^\circ\)

(from(i))

\(\Rightarrow \) 2 \(\angle{ABC}\) = \(180^\circ\)

\(\Rightarrow \) \(\angle{ABC}\) = \(90^\circ\)

Also, \(\angle{DCB}\) = \(90^\circ\)

Thus, we can say that, ABCD is a parallelogram and some angles are \(90^\circ\).

Hence it is proved that, ABCD is a rectangle.

Q3 ) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.

\(\Rightarrow \) OA = OC and OB = OD

And also,

\(\angle{AOD} = \angle{AOB} = \angle{COD} = \angle{BOC} = 90^\circ\).

To prove: ABCD is a rhombus.

Proof:

In \(\triangle{OAB}\) and \(\triangle{ODC}\), we have,

OA = OC and OB = OD ...(given)

\(\angle{AOB}\) = \(\angle{COD}\)

(vertically opposite angles)

\(\therefore \) \(\triangle{OAB}\) \(\displaystyle \cong \) \(\triangle{OCD}\)

(By SAS rule)

\(\therefore \) AB = CD ...(i)(By CPCT)

Similarly, in \(\triangle{OAD}\) and \(\triangle{OBC}\), we have,

OA = OC and OD = OB ...(given)

\(\angle{AOD}\) = \(\angle{BOC}\)

(vertically opposite angles)

\(\therefore \) \(\triangle{OAD}\) \(\displaystyle \cong \) \(\triangle{OCB}\) ...(By SAS rule)

\(\therefore \) AD = BC ...(ii)(By CPCT)

Similarly, we can prove that,

AB = CD and

CD = BC ...(iii)

Hence, from (i), (ii) and (iii), we get,

AB = BC = AD = CD

Hence, it is proved that ABCD is a rhombus.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: A square ABCD whose diagonals AC and BD intersect at O.

To prove: Diagonals are equal and bisect each other at right angles.

i.e, AC = BD, OD = OB, OA = OC and \({AC}\perp{BD}\)

Proof: In \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,

BC = AD ...(given)

AB = BA ...(Common side)

\(\angle{ABC}\) = \(\angle{BAD}\) = \(90^\circ\)

\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SAS rule)

\(\therefore \) AC = BD ...(By CPCT)

Similarly, in \(\triangle{OAB}\) and \(\triangle{OCD}\), we have,

AB = DC ...(given)

\(\angle{OAB}\) = \(\angle{DCO}\)

(\(\because \) AB || CD and transversal AC intersect)

\(\angle{OBA}\) = \(\angle{BDC}\)

(\(\because \) AB || CD and transversal BD intersect)

\(\therefore \) \(\triangle{OAB}\) \(\displaystyle \cong \) \(\triangle{OCD}\)

(By SAS rule)

\(\therefore \) OA = OC and OB = OD

(By CPCT)

Now, in \(\triangle{AOB}\) and \(\triangle{AOD}\), we have,

OA = OD ...(proved earlier)

AB = AD

(sides of square)

AO = OA

(Commom side)

\(\therefore \) \(\triangle{AOB}\) \(\displaystyle \cong \) \(\triangle{AOD}\)

(By SSS rule)

\(\therefore \) \(\angle{AOB}\) = \(\angle{AOD}\) ...(By CPCT)

\(\angle{AOB}\) + \(\angle{AOD}\) = \(180^\circ\)

(linear pair axiom)

\(\angle{AOB}\) = \(\angle{AOD}\) = \(90^\circ\)

\(\therefore \) \(AO\perp{BD}\)

\(\Rightarrow \) \(AC\perp{BD}\).

Also, AC = BD, OA = OC, OB = OD and \(AC\perp{BD}\)

Hence, it is proved that diagonals are equal and bisect each other at right angles.

Q5 ) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: A quadrilateral ABCD in which AC = BD and AC \(\perp{BD}\) such that OA = OC
and OB = OD. So, ABCD is a parallelogram.

To prove: ABCD is a square.

Proof: Let AC and BD intersect at a point O.

In \(\triangle{ABO}\) and \(\triangle{ADO}\), we have,

BO = OD ...(given)

AO = OA ...(Common side)

\(\angle{AOB}\) = \(\angle{AOD}\) = \(90^\circ\) ...(given)

\(\therefore \) \(\triangle{ABO}\) \(\displaystyle \cong \) \(\triangle{ADO}\)

(By SAS rule)

\(\therefore \) AB = AD ...(By CPCT)

Also, AB = DC and AD = BC

(opposite sides of parallelogram)

\(\therefore \) AB = BC = DC = AD ...(i)

Similarly, in \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,

AC = BD ...(given)

AB = BA ...(Common side)

BC = AD ...(from(i))

\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

(By SSS rule)

\(\therefore \) \(\angle{ABC}\) = \(\angle{BAD}\) ...(ii)(By CPCT)

But \(\angle{ABC}\) + \(\angle{BAD}\) = \(180^\circ\) >br> (linear pair axiom)

\(\angle{ABC}\) = \(\angle{BAD}\) = \(90^\circ\) ...(from (ii))

\(\therefore \) AB = BC = CD = DA, and \(\angle{A}\) = \(90^\circ\)

Hence, it is proved that, ABCD is a square.

Q6 )
Diagonal AC of a parallelogram ABCD bisects \(\angle{A}\) (see figure). Show that

(i) It bisects \(\angle{C}\) also,

(ii) ABCD is a rhombus.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given, diagonal AC of a parallelogram ABCD bisects \(\angle{A}\).

\(\therefore \) \(\angle{DAC}\) = \(\angle{BAC}\) = \(\frac{1}{2} \) \(\angle{BAD}\) ...(i)

Here, AB || CD and AC is the transversal.

\(\angle{DCA}\) = \(\angle{CAB}\) ...(ii)(alternate angles)

\(\angle{BCA}\) = \(\angle{DAC}\) ...(iii)(alternate angles)

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

\(\angle{DAC}\) = \(\angle{BAC}\) = \(\angle{DCA}\) = \(\angle{BCA}\)

Now, \(\angle{BCD}\) = \(\angle{BCA}\) + \(\angle{DCA}\)

\(\Rightarrow \) \(\angle{BCD}\)= \(\angle{DAC}\) + \(\angle{CAB}\)

\(\therefore \) \(\angle{BCD}\) = \(\angle{BAD}\)

Thus, we can say that, diagonal AC bisects \(\angle{C}\).

Now, in \(\triangle{OAD}\) and \(\triangle{OCD}\), we have,

OA = OC

(since, diagonals bisects each other)

DO = OD ...(Common side)

\(\angle{AOD}\) = \(\angle{COD}\) = \(90^\circ\)

\(\therefore \) \(\triangle{OAD}\) \(\displaystyle \cong \) \(\triangle{OCD}\)

(By SAS rule)

\(\therefore\) AD = CD ...(By CPCT)

Now, AB = CD and AD = BC

(sides of parallelogram)

\(\therefore \) AB = CD = AD = BC

Hence it is proved that, ABCD is a rhombus.

Q7 ) ABCD is a rhombus. Show that diagonal AC bisects \(\angle{A}\) as well as \(\angle{C}\) and diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is a rhombus.

i.e., AD = AB = BC = CD ... (i)

To prove: (i) Diagonal AC bisect \(\angle{A}\) as well as \(\angle{C}\).

(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Proof: Let AC and BD are the diagonals of rhombus ABCD.

In \(\triangle{ABC}\) and \(\triangle{ADC}\), we have,

AD = AB ...(Given)

AC = CA ...(Common side)

CD = BC ...(From eq. (i))

\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ADC}\) ...(By SSS rule)

Thus, \(\angle{DAC}\) = \(\angle{BAC}\) ...(By CPCT)

Also, \(\angle{DCA}\) = \(\angle{BCA}\)

Also, \(\angle{DAC}\) = \(\angle{DCA}\)

And \(\angle{BAC}\) = \(\angle{BCA}\)

This shows that, Diagonal AC bisect \(\angle{A}\) as well as \(\angle{C}\).

Now, in \(\triangle{BDC}\) and \(\triangle{BDA}\), we have,

AB = BC ...(Given)

BD = BD ...(Common side)

AD = CD ...(Given)

\(\therefore \) \(\triangle{BDC}\) \(\displaystyle \cong \) \(\triangle{BDA}\) ...(By SSS rule)

Thus, \(\angle{BDA}\) = \(\angle{BDC}\) ...(By CPCT)

Also, \(\angle{DBA}\) = \(\angle{DBC}\)

Also, \(\angle{BDA}\) = \(\angle{DBA}\)

And \(\angle{BDC}\) = \(\angle{DBC}\)

This shows that, Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Hence, proved.

Q8 )
ABCD is a rectangle in which diagonal AC bisects \(\angle{A}\) as well as \(\angle{C}\). Show that:

(i) ABCD is a square

(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is a rectangle. AB = CD and BC = AD ...(i)

To prove: (i) ABCD is a square. i.e., AB = BC = CD = DA

(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Proof: In \(\triangle{ADC}\) and \(\triangle{ABC}\), we have,

\(\angle{DAC}\) = \(\angle{BAC}\)

(Since, AB || DC and AC is transversal that intersects)

Similarly, \(\angle{DCA}\) = \(\angle{BCA}\)

AC = CA ...(Common side)

\(\therefore \) \(\triangle{ADC}\) \(\displaystyle \cong \) \(\triangle{ABC}\) ...(By ASA rule)

\(\therefore \)AD = AB ...(By CPCT)

Also, CD = BC ...(ii)

Thus, from eq. (i) and(ii), we get,

AB = BC = AD = CD

Hence, it is proved that ABCD is a square.

Now, in \(\triangle{AOB}\) and \(\triangle{COB}\), we have,

AB = BC ...(Given)

BO = OB ...(Common side)

OA = OC

(Since,diagonal bisectS each other)

\(\therefore \) \(\triangle{AOB}\) \(\displaystyle \cong \) \(\triangle{COB}\) ...(By SSS rule)

\(\therefore \) \(\angle{OBA}\) = \(\angle{OBC}\) ...(By CPCT)

This shows that, Diagonal BD bisects \(\angle{B}\).

Similarly, Now, in \(\triangle{AOD}\) and \(\triangle{COD}\), we have,

AD = CD ...(Given)

OD = DO ...(Common side)

OA = OC

(Since,diagonal bisectS each other)

\(\therefore \) \(\triangle{AOD}\) \(\displaystyle \cong \) \(\triangle{COD}\) ...(By SSS rule)

\(\therefore \) \(\angle{ADO}\) = \(\angle{CDO}\) ...(By CPCT)

Hence, it is proved that, Diagonal BD bisects \(\angle{D}\), too.

Q9 )
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that

i) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\)

ii) AP = CQ

iii) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\)

iv) AQ = CP

v) APCQ is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given, ABCD is a parallelogram and P and Q are lie on BD such that DP = BQ ...(i)

i) Now,in \(\triangle{APD}\) and \(\triangle{CQB}\), we have,

DP = BQ ...(Given)

AD = BC

(Opposite sides are equal in parallelogram)

\(\angle{ADP}\) = \(\angle{QBC}\)

(Since, AD || BC and BD is a transversal)

\(\therefore\) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\) ...(By SAS rule)

ii)

\(\because \) \(\triangle{APD}\) \(\displaystyle \cong \) \(\triangle{CQB}\)

\(\therefore \) AP = CQ ...(By CPCT)

iii) Here, Now, in \(\triangle{AQB}\) and \(\triangle{CPD}\), we have,

DP = BQ ...(Given)

AB = CD

(Opposite sides are equal in parallelogram)

\(\angle{ABQ}\) = \(\angle{CDP}\)

(Since, AB || CD and BD is a transversal)

\(\therefore \) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\) ...(By SAS rule)

iv)

\(\because \) \(\triangle{AQB}\) \(\displaystyle \cong \) \(\triangle{CPD}\)

\(\therefore \) AQ = CP...(By CPCT)

v) Now,in \(\triangle{APQ}\) and \(\triangle{PCQ}\), we have,

AQ = CP (From part iv))

AP = CQ (From part ii))

PQ = QP ...(Common side)

\(\therefore \) \(\triangle{APQ}\) \(\displaystyle \cong \) \(\triangle{PCQ}\) ...(By SSS rule)

\(\therefore \) \(\angle{APQ}\) = \(\angle{PQC}\)

And \(\angle{AQP}\) = \(\angle{CPQ}\)

(Vertically Opposite angles)

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.

\(\therefore \) AP || CQ and AQ || CP

Now, both pairs of opposite sides of quadrilateral APCQ are parallel.

Hence, it is proved that APCQ is a parallelogram.

Q10 )
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\)

(ii) AP = CQ

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.

\(\because \) AB || CD and BD is a transversal, we get,

\(\angle{CDB}\) = \(\angle{DBA}\) ...(i)

Now, in \(\triangle{APB}\) and \(\triangle{CQD}\), we have,

CD = AB ...(Sides of parallelogram)

\(\angle{CQD}\) = \(\angle{APB}\) = \(90^\circ\) ...(Given)

\(\angle{CDQ}\) = \(\angle{ABP}\) ...(From Eq. (i))

\(\therefore \) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\) ...(By ASA rule)

\(\therefore \) AP = CQ ...(By CPCT)

Hence, it is proved.

Q11 )
In \(\triangle{ABC}\) and \(\triangle{DEF}\), 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) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\)

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: In \(\triangle{ABC}\) and \(\triangle{DEF}\), AB = DE, AB || DE, BC = EF and BC || EF

(i) Now, in quadrilateral ABED,

AB = DE and AB || DE ...(Given)

Since, a pair of opposite sides is equal and parallel

Therefore, ABED is a parallelogram.

(ii) In quadrilateral BEFC,

BC = EF and BC || EF ...(Given)

Since, a pair of opposite sides is equal and parallel

Therefore, BEFC is a parallelogram.

(iii)Since, ABED is a parallelogram,

AD || BE and AD = BE ...(i)

Also, BEFC in a parallelogram,

CF || BE and CF = BE ...(ii)

Thus, from Eq. (i) and (ii), we get,

AD || CF and AD = CF ...(From part (iii))

Therefore, ACFD is a parallelogram.

(v)Since, ACFD is a parallelogram.

we get, AC = DF and AC || DF

(vi)Now, in \(\triangle{ABC}\) and \(\triangle{DEF}\),

AB = DE ...(Given)

BC = EF ...(Given)

and AC = DF ...(From part (v))

Therefore, \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\) ...(By SSS test)

Q12 )
ABCD is a trapezium in which AB || CD and AD = BC (see figure).

Show that

i) \(\angle{A}\) = \(\angle{B}\)

ii) \(\angle{C}\) = \(\angle{D}\)

iii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

iv) Diagonal AC = Diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is trapezium.

AB || CD and AD = BC

Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.

Now, ADCE is a parallelogram.

\(\therefore \) AD || CE and AD = CE

But AD = BC

\(\therefore \) AD = BC = CE

i) We know that, \(\angle{A}\) + \(\angle{E}\) = \(180^\circ\)

(Since, interior angles on the same side of the transversal )

\(\Rightarrow \) \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\)

\(\therefore \) \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\)

(Since, BC = EC)

Also, \(\angle{ABC}\) = \(180^\circ\) - \(\angle{CBE}\)

(Since, ABE is straight line)

\(\angle{ABC}\) = \(180^\circ\) - \(180^\circ\) + \(\angle{A}\)

\(\angle{B}\) = \(\angle{A}\)

Hence, it is proved.

ii)Now, \(\angle{A}\) + \(\angle{D}\) = \(180^\circ\)

(Since, interior angles on the same side of the transversal)

\(\Rightarrow \) \(\angle{D}\) = \(180^\circ\) - \(\angle{A}\)

\(\Rightarrow \) \(\angle{D}\) = \(180^\circ\) - \(\angle{B}\)

(from eq.(i)) ...(ii)

Also, \(\angle{C}\) + \(\angle{B}\) = \(180^\circ\) ...180 (Since, interior angles on
the same side of the transversal BC)

\(\Rightarrow \) \(\angle{C}\) = \(180^\circ\) - \(\angle{B}\) ...(iii)

from Eq. (ii) and (iii), we get,

\(\angle{C}\) = \(\angle{D}\)

Hence, proved.

iii)Now, in \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,

AD = BC ...(Given)

\(\angle{A}\) = \(\angle{B}\) ...(From Eq.(i))

AB = BA ...(Common side)

\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SAS rule)

Hence, proved.

iv)

\(\because \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)

\(\therefore \) AC = BD

Hence, it is proved.

Q1 )
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC,
CD and DA (see figure). AC is a diagonal. Show that

i) SR || AC and SR = (1/2) AC

ii) PQ = SR

iii) PQRS is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: P, Q, R, and S are mid-points of the sides.

Therefore, AP = PB, BQ = CQ
CR = DR and AS = DS

i) Now, in \(\triangle{ADC}\), we have,

S is the midpoint of AD and R is the mid point of CD.

As, we know that,

By midpoint theorem, the line segment joining the mid points of two sides of a triangle is parallel to the third side.

Thus, we can say that, SR || AC ...(i)

and SR = (\(\frac{1}{2} \) ) AC ...(ii)

ii) Similarly, now, in \(\triangle{ABC}\), we have,

PQ || AC ...(iii)

and PQ = (\(\frac{1}{2} \) ) AC ...(iv)

Now, from (ii) and (iv), we get,

SR = PQ = (\(\frac{1}{2} \) ) AC ...(v)

Now, from (i), (iii) and (v), we get,

PQ || SR and PQ = SR.

\(\therefore \) PQRS is a parallelogram

(Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel)

Hence, proved.

Q2 ) 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.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in \(\triangle{ADC}\), we have,

Thus, SR || AC ...(i)

and SR = (\(\frac{1}{2} \) ) AC ...(ii)

Similarly, in \(\triangle{ABC}\), we have,

PQ || AC ...(iii)

and PQ = (\(\frac{1}{2} \) ) AC ...(iv)

from (i), (ii), (iii) and (iv), we get,

SR = PQ = (\(\frac{1}{2} \) ) AC and PQ || SR

\(\therefore \) Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.

\(\therefore \) \(\angle{EOF}\) = \(90^\circ\).

Now, By midpoint theorem, we have,

RQ || BD

Thus, RE || OF

As SR || AC ...(from (i))

Thus, FR || OE

\(\therefore \) OERF is a parallelogram

So, \(\angle{ERF}\) = \(\angle{EOF} = 90° \)

(since, Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with \(\angle{R}= 90° \)

Hence, it is proved that PQRS is a rectangle.

Q3 )
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.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is a rectangle.

\(\Rightarrow \) \(\angle{A}\) = \(\angle{B}\) = \(\angle{C}\) = \(\angle{D}\) = \(90^\circ\) and

AB = CD and BC = AD.

Also, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Therefore, by midpoint theorem,

PQ || BD and PQ = \(\frac{1}{2} \) BD and

SR || AC and SR = \(\frac{1}{2} \) AC

In rectangle ABCD,

AC = BD

\(\therefore \) PQ = SR ...(i)

Now, in \(\triangle{ASP}\) and \(\triangle{BQP}\), we have,

AP = BP ...(Given)

AS = BQ ...(Given)

\(\angle{A}\) = \(\angle{B}\) ...(Given)

\(\therefore \) \(\triangle{ASP}\) \(\displaystyle \cong \) \(\triangle{BQP}\) ...(By SAS rule)

\(\therefore \) SP = BQ ...(ii)(By CPCT)

Similarly, in \(\triangle{RDS}\) and \(\triangle{RCQ}\), we have,

SD = CQ ...(Given)

DR = RC ...(Given)

\(\angle{C}\) = \(\angle{D}\) ...(Given)

\(\therefore \) \(\triangle{RDS}\) \(\displaystyle \cong \) \(\triangle{RCQ}\) ...(By SAS rule)

\(\therefore \) SR = RQ ...(iii)(By CPCT)

Thus, from eq. (i), (ii) and (iii), we can say that,

The quadrilateral PQRS is a rhombus is proved.

Q4 )
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of
AD.

A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.

In \(\triangle{ABD}\), we have,

EP || AB and E is mid-point of AD.

So, by theorem, if a line drawn through the
mid-point of one side of a triangle is parallel to another side, then it bisects the third side.

Therefore, P is the midpoint of BD.

Similarly, in \(\triangle{BCD}\),

we have, PF || CD,

Therefore by converse of mid point theorem, F is mid-point of BD.

Q5 )
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD
respectively (see figure).

Show that the line segments AF and EC trisect the diagonal BD.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD, respectively.

To Prove: Line segments AF and EC trisect the diagonal BD.

Proof:

\(\because \) ABCD is a parallelogram,

\(\therefore \) AB || DC and AB = DC

(since, Opposite sides of a parallelogram)

Thus, AE || FC and (\(\frac{1}{2} \)) AB = (\(\frac{1}{2} \)) CD

\(\Rightarrow \) AE = FC

\(\therefore \) AECF is a parallelogram.

Thus, AF || EC

And hence, EQ || AP and FP || CQ.

In \(\triangle{BAP}\), E is the mid-point of AB and EQ || AP,

So, By converse of mid-point theorem,

Q is the mid-point of BP.

\(\Rightarrow \) BQ = PQ ...(i)

Similarly, in \(\triangle{DQC}\), F is the mid-point of DC and FP || CQ,

So, P is the mid-point of DQ.

\(\Rightarrow \) DP = PQ ...(ii)

From Equations (i) and (ii) , we get,

BQ = PQ = DP

Hence, it is proved that CE and AF trisect the diagonal BD.

Q6 ) Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.

\(\Rightarrow \) AS = SD, AP = BP, BQ = CQ and CR = DR.

We have to show that: PR and SQ bisect each other i.e., SO = OQ and PO = OR.

Now, in \(\triangle{ADC}\), S and R are mid-point of AD and CD.

We know that, the line segment joining the mid-points of two sides of a triangle is
parallel to the third side.

Thus, by midpoint theorem,

SR || AC and SR = (\(\frac{1}{2} \)) AC ...(i)

Similarly, in \(\triangle{ABC}\), P and Q are mid-point of AB and BC.

Thus, by midpoint theorem,

PQ || AC and PQ = (\(\frac{1}{2} \) ) AC ...(ii)

From Eq. (i) and (ii), we get,

PQ || SR and PQ = SR = (\(\frac{1}{2} \) ) AC

Therefore, Quadrilateral PQRS is a parallelogram whose diagonals SQ are PR.

Also, we know that diagonals of a parallelogram bisect each other. So, and bisect each other.

Hence, it is proved that PR and SQ bisect each other.

Q7 )
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse
AB and parallel to BC intersects AC at D.

Show that

i) D is the mid-point of AC

ii) MD is perpendicular to AC

iii) CM = MA = (\(\frac{1}{2} \)) AB

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Answer :

Given: \(\triangle{ABC}\) is a right angled triangle \(\Rightarrow \) \(\angle{C}\) = \(90^\circ\)

and M is the mid-point of AB.

Also, DM || BC

i) In \(\triangle{ABC}\), M is the mid-point of AB and BC || MD,

So, By converse of mid-point theorem,

D is the mid-point of AC.

\(\Rightarrow \) AD = CD ...(i)

ii) Since, BC || MD and CD is transversal.

\(\therefore \) \(\angle{ADM}\) = \(\angle{ACB}\)

(Corresponding angles)

But, \(\angle{C}\) = \(90^\circ\)

\(\therefore \) \(\angle{ADM}\) = \(90^\circ\)

Hence, proved that MD is perpendicular to AC.

iii) Now, in \(\triangle{ADM}\) and \(\triangle{ACM}\), we have,

AD = CD ...(from (i))

DM = MD ...(Common side)

\(\angle{ADM}\) = \(\angle{MDC}\) = \(90^\circ\) ...(Proved)

\(\therefore \) \(\triangle{ADM}\) \(\displaystyle \cong \) \(\triangle{CDM}\) ...(By SAS rule)

\(\therefore\) CM = AM ...(ii)(By CPCT)

Also, M is the mid point of AB.

\(\Rightarrow \) AM = BM = (\(\frac{1}{2} \) ) AB ...(iii)

Thus, from (ii) and (iii),

CM = AM = (\(\frac{1}{2} \) ) AB

Hence, proved.

There are total 19 questions present in ncert solutions for class 9 maths chapter 8 quadrilaterals

There are total 5 long question/answers in ncert solutions for class 9 maths chapter 8 quadrilaterals

There are total 2 exercise present in ncert solutions for class 9 maths chapter 8 quadrilaterals