NCERT Solutions for Class 9 Maths Chapter 8 – Quadrilaterals

Here we provide NCERT Solutions for Class 9 Maths Chapter 8 – Quadrilaterals for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Class 9 Maths solution Chapter 8 pdf. Now you will get step by step solution to each question. 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

(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

(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
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.

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

Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.

(iii)  Since ABED and BEFC are parallelograms.
⇒ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)

Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel) Thus, AD || CF

(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.

(vi) In ΔABC and ΔDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, ΔABC ≅ ΔDEF by SSS congruence condition.
12. ABCD is a trapezium in which AB || CD and

AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iv) diagonal AC = diagonal BD

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

Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
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

AB = AB (Common)
∠DBA = ∠CBA
Thus, ΔABC ≅ ΔBAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

Page No: 150

Exercise 8.2

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.

Given,
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.

Given,
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)
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.

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.
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,

ABCD is a parallelogram
Therefor, AB || CD

also, AE || FC
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.

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

(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)
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

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