**NCERT Solutions for Class 9
Maths Chapter 8 Quadrilaterals Ex 8.2**

NCERT Solutions for Class
9 Maths Chapter 8 Quadrilaterals Ex 8.2 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 8 Quadrilaterals Ex 8.2.

**NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1****NCERT Solutions for Class 9 Maths Chapter 8****Quadrilaterals**Ex 8.2

**Ex 8.2 Class 9 Maths Question 1.**

**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 = ½ AC

**(ii)**PQ = SR

**(iii)**PQRS is a parallelogram.

**Solution:**

(i)In ∆ACD, we have

(i)

∴ S is the mid-point of AD and R is the mid-point of CD.

SR = ½ AC and SR || AC ….…(1) [By mid-point theorem]

**(ii)** In ∆ABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ = ½ AC and PQ || AC ….…(2) [By mid-point theorem]

From (1) and (2), we get

⇒ PQ = SR and PQ || SR

**(iii)** In a quadrilateral PQRS,

PQ = SR and PQ || SR [Proved
in (ii)]

∴ PQRS is a parallelogram.

**Ex 8.2 Class 9 Maths Question
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.**

**Solution.**

**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:**Quadrilateral PQRS is a rectangle.

**Proof:**By mid-point theorem.

In Î”ADC, we have S and R are the mid-points of AD and CD, respectively.

∴ SR || AC and SR = ½ AC …………..(i)

In Î”ABC, we have P and Q are the mid-points of AB and BC, respectively.

∴ PQ || AC and PQ = ½ AC ……..…..(ii)

From eqs. (i) and (ii), we get PQ || SR and PQ = SR = ½ AC

Since, a pair of opposite sides of a quadrilaterals PQRS is equal and parallel.

So, PQRS is a parallelogram.

We know that diagonals of a rhombus bisect
each other at right angles.

∠COD = ∠EOF = 90°

Now, in ABCD, R and Q are the mid-points of CD
and BC, respectively.

RQ || DB [by mid-point
theorem]

⇒ RE || OF

Also, SR || AC [from eq. (i)]

⇒ FR || OE

So, OERF is a parallelogram.

∴ ∠ERF = ∠EOF = 90°
[Opposite angles of a parallelogram are equal]

Thus, PQRS is a parallelogram with ∠R = 90°

Hence, PQRS is a rectangle.

**Hence proved.**

**Ex 8.2 Class 9 Maths Question
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.

**Solution.**

In Î”ASP and Î”BQP,

AP = BP (Given)

AS = BQ (Given)

∠A = ∠B (Each 90°)

∴ Î”ASP ≅ Î”BQP (By SAS
congruency rule)

∴ SP = PQ (By CPCT) ……(i)

In Î”PBQ and Î”RCQ,

PB = RC (Given)

BQ = CQ (Given)

∠B = ∠C (Each 90°)

∴ Î”PBQ ≅ Î”RCQ (By SAS
congruency rule)

∴ PQ = RQ (By CPCT) ……(ii)

Similarly, we can prove that RQ = SR and SR = SP (iii)

From eqs. (i), (ii) and (iii), we
get SP = PQ = QR = RS.

Therefore, PQRS is a rhombus.

**Ex 8.2 Class 9 Maths Question
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 figure). Show that F is the mid-point of BC.

**Solution.****
Given:** ABCD is a trapezium in which AB || DC, BD is a diagonal and
E is the mid-point of AD and a line is drawn through E parallel to AB
intersecting BC at F such that EF || AB.

**To**** prove:** F is the mid-point of BC.

**Proof:** Let EF intersects BD at P.

In Î”ABD, we have EP || AB [∵ EF || AB] and E is the mid-point of AD.

So, by the converse of mid-point theorem, P is
the mid-point of BD.

Similarly, in Î”BCD, we have PF || CD [∵ EF || AB and AB || CD]

and P is the mid-point of BD.

So, by the converse of mid-point
theorem, F is the mid-point of BC.

**Ex 8.2 Class 9 Maths Question
5.****
**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.

**Solution.**

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

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

**Proof:**Since ABCD is a parallelogram.

∴ AB || DC and AB = DC [opposite sides of a parallelogram]

⇒ AE || FC and ½ AB = ½ DC

⇒ AE || FC and AE = FC [∵ E and F are the mid-points of AB and CD]

Since a pair of opposite sides of a
quadrilaterals AECF is equal and parallel.

So, AECF is a parallelogram.

Then, AF || EC

⇒ AP || EQ and FP || CQ [Since opposite sides of a parallelogram
are parallel]

In Î”BAP, E is the mid-point of AB and EQ || AP,
so Q is the mid-point of BP.

[by the converse of mid-point theorem]

∴ BQ = PQ ……(i)

Again, in Î”DQC, F is the mid-point of DC and FP
|| CQ.

So, P is the mid-point of
DQ. [by the converse of
mid-point theorem]

∴ QP = DP ……(ii)

From eqs. (i) and (ii), we get BQ = PQ = PD

Hence, CE and AF trisect the diagonal BD.

**Hence proved.**

**Ex 8.2 Class 9 Maths Question
6.**Show that the line segments joining
the mid-points of the opposite sides of a quadrilateral bisect each other.

**Solution.**

i.e., AS = DS, 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 Î”ADC, S and R are the mid-points of AD and CD, respectively.

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

∴ SR || AC and SR = ½ AC ……(i) [by mid-point theorem]

Similarly, in Î”ABC, P and Q are mid-points of AB and BC, respectively.

PQ || AC and PQ = ½ AC ……(ii) [by mid-point theorem]

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

PQ || SR and PQ = SR = ½ AC

So, PQRS is a parallelogram whose diagonals are SQ
and PR.

We know that diagonals of a parallelogram bisect
each other.

So, SQ and PR bisect each other at O, i.e., SO = OQ
and PO = OR.

**Hence proved.**

**Ex
8.2 Class 9 Maths Question 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 = ½ AB

**Solution.**

∠C = 90° and M is the mid-point of AB. Also, DM || BC

**(i)** In Î”ABC, BC || MD and M is the mid-point
of AB.

∴ D is the mid-point of AC. (By converse of mid-point theorem)

**(ii)**
Since MD || BC and CD is
transversal.
.

∠ADM = ∠ACB (Corresponding
angles)

But ∠ACB = 90°

⇒ ∠ADM = 90°

⇒ MD ⊥ AC

**(iii)**
Now, in Î”ADM and Î”CDM, we have

DM = MD
(Common)

AD = CD
(D is the mid-point of AC)

∠ADM = ∠CDM (Each 90°)

Therefore, Î”ADM ≅ Î”CDM

∴ CM = AM …….(i)

Also, M is the mid-point of AB.

AM = BM = ½ AB
…….(ii)

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

CM = AM = ½ AB

**Hence proved.**

**NCERT Solutions for Maths Class 10**