NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

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

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 × \(12^\circ\) = \(156^\circ\)

Q2 ) If the diagonals of a parallelogram are equal, then show that it is a rectangle.

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.

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.

Q4 ) Show that the diagonals of a square are equal and bisect each other at right angles.

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.

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.

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 \(\angle{D}\).

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 \(\angle{D}\).

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.

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

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) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{DEF}\)

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].

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.