**NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 – Quadrilaterals****. This Exercise contains 7 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 9 or other Chapters, you can click the link at the end of this Note.**

**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 below). AC is a diagonal. Show that :**

**a) SR || AC and SR =½ AC
**

**b) PQ = SR**

**c) PQRS is a parallelogram.**

** **

**Given **

P is the mid-point of side AB

Q is the mid-point of side BC

R is the mid-point of side CD

S is the mid-point of side DA|AC is diagonal of ABCD

** ****To prove :**

**(a) SR || AC and SR =½ AC**

**Solution :**

In ΔADC

R is the mid-point of side CD [given]

S is the mid-point of side DA [given]

Therefore

SR || AC and SR =½ AC [mid point theorem]

**(b) PQ = SR**

**Solution :**

In ΔABC

P is the mid-point of side AB [given]

Q is the mid-point of side BC [given]

Therefore

PQ || AC and PQ =½ AC [mid point theorem] (1)

Also

SR || AC and SR =½ AC [from part (a)] (2)

From (1) and (2) we conclude that

SR ||PQ || AC (3)

And , SR = PQ = ½ AC (4)

Thus PQ = SR [using (4)]

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

**Solution:**

In quadrilateral PQRS,

We have

PQ = SR [from part (b)]

PQ || SR [using (3)]

(A quadrilateral is a parallelogram if it’s opposite sides is equal and parallel.)

Therefore, 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

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

**Solution :**

In ΔABD

P is the mid-point of side AB [given]

S is the mid-point of side AD [given]

Therefore

PS || BD and PS =½ BD [By mid point theorem] (1)

In ΔBCD

Q is the mid-point of side BC [given]

R is the mid-point of side CD [given]

Therefore

QR || BD and QR =½ BD [mid point theorem] (2)

Similarly we can prove that

PQ || AC and PQ =½ AC (3)

SR || AC and SR =½ AC (4)

Therefore using (1), (2), (3) and (4) we conclude that PQRS is a parallelogram.

(A quadrilateral is a parallelogram if its opposite sides are equal & parallel)

In quadrilateral MONS

OM || SN [ using (4) AC || QR]

SM || ON [ using (1) PQ || BD]

∠MON = 90^{0 } [diagonals of a rhombus bisect perpendicularly]

We know that a quadrilateral having parallel opposite sides with one of its angles equal to 90^{0} is a rectangle.

Therefore MONS is a rectangle

Hence, ∠PSN = 90^{0
}Thus, Parallelogram PQRS has one angle as right angle and so it is a rectangle.

So, we get 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 :**

Let ABCD be a rectangle in which P, Q, R, S are the mid-points of sides AB, BC, CD and DA respectively.

PQ, QR, RS and SP are joined such that a quadrilateral PQRS is formed.

**To Prove:**

PQRS is a rhombus.

** ****Construction :**

Divide the rectangle into two triangles by joining AC.

** ****Solution :**

In ∆ABC, P and Q are the mid-points of the sides AB and BC.

Therefore, [By mid point theorem]

PQ || AC & PQ =½ AC … (a)

Similarly, in ∆ADC, [By mid point theorem]

SR || AC & SR =½ AC … (b)

From (a) and (b), we get

PQ || SR & PQ = SR …(c)

In quadrilateral PQRS, opposite sides PQ and SR are parallel and equal.[From c)]

Therefore PQRS is a parallelogram.

Also

AS = BQ [Halves of equal sides are also equal.] … (d)

In ∆APS and ∆BPQ

AP = BP [P is the mid-point of AB]

AS = BQ [from (d)]

∠PAS = ∠PBQ [Each angle is 90° as ABCD is a rectangle.]

∆APS ≅ ∆BPQ [SAS criteria of congruence]

PS = PQ …(e)

We know that if the adjacent sides of a parallelogram are equal then it is a Rhombus.

From (c) and (e), we conclude that 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. below). Show that F is the mid-point of BC.**

**Given :** A trapezium ABCD with AB || DC, E is the mid-point of AD and EF|| AB.

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

**Solution :**

Since,

AB || DC and EF || AB [given]

Therefore,

AB||EF||DC

From fig.,

In ∆ADB

AB||EO [AB||EF, given]

DE = EA [E is the midpoint of DA]

So, DO = OB [By converse of midpoint theorem]

From fig.,

In ∆DBC

DC||OF [DC||EF, given]

DO = OB [proved above]

Therefore,

CF = FB [By converse of midpoint theorem]

Thus, we conclude that F is the mid-point of BC.

Hence proved.

**5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see fig. below). Show that the line segments AF and EC trisect the diagonal BD. **

** **

**Given : **A parallelogram ABCD, in which E and F are mid-points of sides AB and DC respectively.

So, AE =½ AB

CF =½ DC

**To Prove :** DP = PQ = QB=1/3 BD

**Solution :**

We have

AE =½ AB [given]

CF =½ DC [given]

Also

AB = DC and AB || DC [opposite sides of a parallelogram are equal & parallel]

Therefore,

AE = CF and AE || CF

Hence the quadrilateral AECF is a parallelogram as one pair of opposite sides is parallel and equal.

In ∆BAP,

E is the mid-point of AB [given]

EQ || AP [AECF is a parallelogram]

Therefore,

Q is mid-point of PB [By converse of mid-point theorem]

Or, PQ = QB (1)

Similarly,

In ∆DQC,

P is the mid-point of DQ [By converse of midpoint theorem]

Or, DP = PQ (2)

Using (1) and (2), we have

DP = PQ = QB=1/3 BD

Or, 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.**

**Given :** ABCD is a quadrilateral in which PR and QS are the line segments joining the mid-points of opposite sides.

** **** **

**To Prove :** PR and QS bisect each other.

**Construction : **Join PQ, QR, RS, SP and AC.

**Solution : **

In ∆ABC, P and Q are mid-points of AB and BC respectively.

PQ =½ AC and PQ || AC (1) [By midpoint theorem]

In ∆ADC, S and R are mid-points of AD and CD respectively.

SR =½ AC and SR || AC (2) [By midpoint theorem]

From (1) and (2), we get

PQ = SR and PQ || SR

We know that a quadrilateral is a parallelogram if it’s one pair of opposite sides is equal and parallel.

Therefore, PQRS is a parallelogram.

Now, PR and QS are diagonals of the parallelogram PQRS.

Therefore, PR and QS bisect each other. [Diagonal of a parallelogram bisect each other]

Hence proved.

** **

**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**

** ****Given :**

A triangle ABC, in which ∠C = 90° and M is the mid-point of AB and BC || DM.

**To Prove :**

(i) D is the mid-point of AC

(ii) DM ⊥ AC

(iii) CM = MA =½ AB

**Construction :** Join CM.** **

**Solution : **

**(i)** In ∆ABC,

M is the mid-point of AB. [Given]

BC || DM [Given]

D is the mid-point of AC [By converse of mid-point theorem]

**(ii)** ∠ADM = ∠ACB [Pair of corresponding angles]

But ∠ACB = 90° [Given]

So, ∠ADM = 90°

Also,

∠ADM + ∠CDM = 180° [Linear pair]

Therefore, ∠CDM = 90°

Hence, MD ⊥ AC

Proved.

**(iii)** AD = DC (1)[ ∵ D is the mid-point of AC part(i)]

In ∆ADM and ∆CMD, we have

∠ADM = ∠CDM [Each angle is 90°]

AD = DC [From part (i)]

DM = DM [Common]

∆ADM ≅ ∆CDM [SAS criteria of congruence]

Thus,

CM = MA (2) [By C.P.C.T.]

Since M is mid-point of AB(given),

MA =½ AB (3)

So, CM = MA =½ AB [From (2) and (3)]

Hence Proved.

**NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 – Quadrilaterals, has been designed by the NCERT to test the knowledge of the student on the following topics:-**

- The Mid-point Theorem

– The line segment joining the mid-points of two sides of a triangle is parallel to the third side

– The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

**The next Exercise for** **NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.1 – Areas of Parallelograms And Triangles can be accessed by clicking here.**

** Maths – NCERT Solutions Class 9**

**NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 – Quadrilaterals**

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