In This Post we are providing Chapter 8 Quadrilaterals NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Quadrilaterals Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Quadrilaterals NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
Exercise 8.1
1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Answer
Let x be the common ratio between the angles.
Sum of the interior angles of the quadrilateral = 360°
Now,
3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = 12°
Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer
Given,
AC = BD
To show,
To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
AC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD by SSS congruence condition.
∠A = ∠B (by CPCT)
also,
∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Thus ABCD is a rectangle.
3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Answer
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given,
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show,
ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
Answer
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show,
AC = BD, AO = OC and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
∠ABC = ∠BAD = 90°
AC = AD (Given)
Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.
Thus, AC = BD by CPCT. Therefore, diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.
Thus, AO = CO by CPCT. (Diagonal bisect each other.)
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.
also, ∠AOB = ∠COB
∠AOB + ∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at right angles)
5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer
Given,
Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.
To prove,
Quadrilateral ABCD is a square.
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.
Thus, AB = CD by CPCT. — (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB || CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.
Thus, AD = CD by CPCT. — (ii)
also,
AD = BC and AD = CD
⇒ AD = BC = CD = AB — (ii)
also, ∠ADC = ∠BCD by CPCT.
and ∠ADC + ∠BCD = 180° (co-interior angles)
⇒ 2∠ADC = 180°
⇒ ∠ADC = 90° — (iii)
One of the interior ang is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.
6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Answer
(i)
In ΔADC and ΔCBA,
AD = CB (Opposite sides of a ||gm)
DC = BA (Opposite sides of a ||gm)
AC = CA (Common)
Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.
Thus,
∠ACD = ∠CAB by CPCT
and ∠CAB = ∠CAD (Given)
⇒ ∠ACD = ∠BCA
Thus, AC bisects ∠C also.
(ii) ∠ACD = ∠CAD (Proved)
⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a ||gm)
Thus, ABCD is a rhombus.
7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Answer
Let ABCD is a rhombus and AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
⇒ ∠DAC = ∠BCA (Alternate interior angles)
⇒ ∠DCA = ∠BCA
Therefore, AC bisects ∠C.
Similarly, we can prove that diagonal AC bisects ∠A.
Also, by preceding above method we can prove that diagonal BD bisects ∠B as well as ∠D.
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 ∠B as well as ∠D.
Answer
(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
Therefore, AB = BC = CD = AD
Thus, ABCD is a square.
(ii) In ΔBCD,
BC = CD
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
also, ∠CDB = ∠ABD (Alternate interior angles)
⇒ ∠CBD = ∠ABD
Thus, BD bisects ∠B
Now,
∠CBD = ∠ADB
⇒ ∠CDB = ∠ADB
Thus, BD bisects ∠D
Page No: 147
9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Answer
(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a ||gm)
Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.
(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.
(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = BCCD (Opposite sides of a ||gm)
Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.
(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.
(v) From (ii) and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.
10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Answer
(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.
(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.
11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).
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) ΔABC ≅ ΔDEF.
Answer
⇒ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)
(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.
(v) AC || DF and AC = DF because ACFD is a parallelogram.
12. ABCD is a trapezium in which AB || CD and
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC = diagonal BD
Answer
Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefor, BC = CE
⇒ ∠CBE = ∠CEB
also,
∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° (Linear pair)
⇒ ∠A = ∠B
(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of transversal)
⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)
⇒ ∠D = ∠C
(iii) In ΔABC and ΔBAD,
AB = AB (Common)
∠DBA = ∠CBA
AD = BC (Given)
Thus, ΔABC ≅ ΔBAD by SAS congruence condition.
(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.
1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that :
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
(i) In ΔDAC,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SR || AC and SR = 1/2 AC
(ii) In ΔBAC,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQ || AC and PQ = 1/2 AC
also, SR = 1/2 AC
Thus, PQ = SR
(iii) SR || AC – from (i)
and, PQ || AC – from (ii)
⇒ SR || PQ – from (i) and (ii)
also, PQ = SR
Thus, PQRS is a parallelogram.
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.
Answer
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
To Prove,
PQRS is a rectangle.
Construction,
AC and BD are joined.
Proof,
In ΔDRS and ΔBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.
RS = PQ by CPCT — (i)
In ΔQCR and ΔSAP,
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.
RQ = SP by CPCT — (ii)
Now,
In ΔCDB,
R and Q are the mid points of CD and BC respectively.
⇒ QR || BD
also,
P and S are the mid points of AD and AB respectively.
⇒ PS || BD
⇒ QR || PS
Thus, PQRS is a parallelogram.
also, ∠PQR = 90°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
Thus, PQRS is a rectangle.
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.
Answer
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
Construction,
AC and BD are joined.
To Prove,
PQRS is a rhombus.
Proof,
In ΔABC
P and Q are the mid-points of AB and BC respectively
Thus, PQ || AC and PQ = 1/2 AC (Mid point theorem) — (i)
In ΔADC,
SR || AC and SR = 1/2 AC (Mid point theorem) — (ii)
So, PQ || SR and PQ = SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PS || QR and PS = QR (Opposite sides of parallelogram) — (iii)
Now,
In ΔBCD,
Q and R are mid points of side BC and CD respectively.
Thus, QR || BD and QR = 1/2 BD (Mid point theorem) — (iv)
AC = BD (Diagonals of a rectangle are equal) — (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
So, PQRS is a rhombus.
4. 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 Fig. 8.30). Show that F is the mid-point of BC.
Answer
Given,
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
To prove,
F is the mid-point of BC.
Proof,
BD intersected EF at G.
In ΔBAD,
E is the mid point of AD and also EG || AB.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In ΔBDC,
G is the mid point of BD and also GF || AB || DC.
Thus, F is the mid point of BC (Converse of mid point theorem)
Page No: 151
5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.
Given,
ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,
Therefor, AB || CD
Now,
AB = CD (Opposite sides of parallelogram ABCD)
⇒ 1/2 AB = 1/2 CD
⇒ AE = FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AF || EC (Opposite sides of a parallelogram)
Now,
In ΔDQC,
F is mid point of side DC and FP || CQ (as AF || EC).
P is the mid-point of DQ (Converse of mid-point theorem)
⇒ DP = PQ — (i)
Similarly,
In APB,
E is mid point of side AB and EQ || AP (as AF || EC).
Q is the mid-point of PB (Converse of mid-point theorem)
⇒ PQ = QB — (ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.
6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Answer
Now,
In ΔACD,
R and S are the mid points of CD and DA respectively.
Thus, SR || AC.
Similarly we can show that,
PQ || AC
PS || BD
QR || BD
Thus, PQRS is parallelogram.
PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.
7. 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 ⊥ AC
(iii) CM = MA = 1/2 AB
Answer
(i) In ΔACB,
M is the mid point of AB and MD || BC
Thus, D is the mid point of AC (Converse of mid point theorem)
(ii) ∠ACB = ∠ADM (Corresponding angles)
also, ∠ACB = 90°
Thus, ∠ADM = 90° and MD ⊥ AC
(iii) In ΔAMD and ΔCMD,
AD = CD (D is the midpoint of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (common)
Thus, ΔAMD ≅ ΔCMD by SAS congruence condition.
AM = CM by CPCT
also, AM = 1/2 AB (M is mid point of AB)
Hence, CM = MA = 1/2 AB