These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2

Question 1.

ABCD is a quadrilateral in which P, Q, R, and S are midpoints of the sides AB, BC, CD, and DA (see fig. 8.29). AC is a diagonal. Show that

(i) SR || AC and SR = \(\frac {1}{2}\) AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Solution:

Given. ABCD is a quadrilateral in which P, Q, R, and S are the midpoints of side AB, BC, CD, and AD respectively.

To prove that:

(i) SR || AC and SR = \(\frac {1}{2}\) AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Proof:

(i) In ΔADC

S and R are the mid-point of side AD and DC respectively.

We know that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of it.

SR || AC and SR = \(\frac {1}{2}\) AC ……(i)

(ii) In ΔABC,

P and Q are the mid points of side AB and BC respectively.

Therefore, PQ || AC and PQ = \(\frac {1}{2}\) AC …….(ii)

From equation (i) and (ii),

PQ || SR and PQ = SR

(iii) We have

PQ = SR (Prove above)

and PQ || SR (Prove above)

We know that if one pair of opposite sides of a quadrilateral are parallel and equal then the quadrilateral is a parallelogram.

∴ PQRS is parallelogram.

Question 2.

ABCD is a rhombus and P, Q, R, and S are the midpoints of the sides AB, BC, CD, and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:

Given: Arhmbus ABCD in which P, Q, Rand S are the midpoints of sides AB, BC, CD, and DA respectively. PQ, QR, RS, and SP are joined.

To prove that: PQRS is a rectangle.

Construction: Join AC and BD.

Proof: In order to prove that PQRS is a rectangle, it is sufficient to show that it is a parallelogram whose one angle is a right angle. First, we shall prove that PQRS is a parallelogram.

In ΔABC,

P and Q are the mid points of AB and BC respectively.

∴ PQ || AC and PQ = \(\frac {1}{2}\) AC ……(i)

In ΔADC,

R and S are the mid points of CD and AD respectively.

∴ RS || AC and RS = \(\frac {1}{2}\) AC ……(ii)

From equation (i) and (ii) we have

PQ || RS and PQ = RS

Thus, PQRS is a quadrilateral such that one pair of opposite sides PQ and SR is equal and parallel. So, PQRS is a parallelogram.

PQ || AC (Opposite sides of || gm)

∴ MQ || ON

QR || BD (Opposite sides of || gm)

∴ QN || OM

Therefore, OMQN is a parallelgram.

and ∠NOM = 90°

[Diagomals of a rhombus bisects each other at a right angle]

So, ∠MQN = 90°

[Opposite angles of || gm are equal]

Therefore, PQRS is a parallelogram in which ∠Q is a right angle.

∴ PQRS is a rectangle.

Question 3.

ABCD is a rectangle and P, Q, R, and S are midpoints of the sides AB, BC, CD, and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

Given: A rectangle ABCD in which P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA respectively. PQ, QR, RS, and SP are joined.

To prove that: PQRS is a rhombus.

Construction: Join AC and BD.

Proof: In ∆ABC,

P and Q are mid points of AB and BC respectively.

∴ PQ || AC and PQ = \(\frac {1}{2}\) AC …..(i)

Similarly in ∆ACD,

SR || AC and SR = \(\frac {1}{2}\) AC ……(ii)

From equation (i) and (ii) we have

PQ || SR and PQ = SR

Therefore, PQRS is a parallelogram

∴ SR || PQ (Opposite sides of || gm PQRS)

and SR = PQ

Now, in ∆BCD,

Q and R are mid points of side BC & CD.

∴ QR || AC and QR = \(\frac {1}{2}\) BD

But, AC = BD (In rectangle diagonals are equal)

QR = \(\frac {1}{2}\) AC ……(iii)

From equation (i) and (iii) we have

PQ = QR

but PQ = SR and PS = QR

Therefore, PQ = QR = RS = PS

∴ PQRS is a rhombus.

Question 4.

ABCD is a trapezium in which AB || DC, BD is a diagonal, and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see fig. 8.30). Show that F is the midpoint of BC.

Solution:

In ∆ABD,

E is the midpoints of AD and EF || AB

∴ EF || CD

We know that a line through the midpoint of a side of a triangle parallel to another side bisects the third side. So, G is the midpoint of DB.

Now, in ∆DBC

G is the midpoint of DB and GF || CD

∴ F is the midpoint of BC.

Question 5.

In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see fig. 8.31). Show that the line segment AF and EC trisect the diagonal BD.

Solution:

We have,

ABCD is a parallelogram

∴ AB || CD

So, AE || FC ……(i)

Again, AB = CD (Opposite sides of ||gm ABCD)

∴ \(\frac {1}{2}\) AB = \(\frac {1}{2}\) CD

or, AE = FC …….(ii)

From equation (i) and (ii)

Therefore, AFCE is a parallelogram.

Now, in ΔDQC,

F is the mid point of CD,

and FP || CQ

We know that a line through the midpoint of a side of a triangle parallel to another side bisects the third side.

So, P is the midpoint of DQ

or, DP = PQ ……(i)

Similarly in ΔAPB, Q is midpoint of BP

i.e. PQ = BQ …..(ii)

From equation (i) and (ii)

PQ = DP = BQ

∴ AF and CE trisect the diagonal BD.

Question 6.

Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.

Solution:

Given: ABCD is a quadrilateral in which P, Q, R, and S are the midpoints of side AB, BC, CD, and AD respectively.

To prove that: PR & QS bisect each other. Construction: Join PQ, QR, RS, and SP and join BD.

Proof: In ΔABD,

S and P are the midpoint of side AD and AB respectively we know that the line segment joining the midpoints of any two sides of a triangle is. parallel to the third side and is half of it.

Therefore, SP || BD and SP = \(\frac {1}{2}\) BD …….(i)

Similarly,

In ΔBCD,

Q and R is the mid points of side BC & CD

∴ QR || AC and QR = \(\frac {1}{2}\) AC ……(ii)

From equation (i) and (ii)

SP || QR and SP = QR

We know that if one opposite pair of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.

∴ SPQR is a parallelogram.

So, PR and QS bisect each other.

[Because diagonal of a parallelogram bisects each other]

Question 7.

ABC is a triangle right angled at C. A line through the midpoint 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 = \(\frac {1}{2}\) AB

Solution:

Given: ABC is a right-angled triangle right angle at C. M is the midpoint of side AB and DM || BC.

To prove that:

(i) D is the mid point of AC

(ii) MD ⊥ AC

(iii) CM = MA = \(\frac {1}{2}\) AB

Construction: Join CM.

Proof: (i) In ΔABC,

M is the mid point of side AB and MD || BC

We know that a line through the midpoint of a side of a triangle parallel to another side bisects the third side.

So, D is the midpoint of AC.

(ii) We have DM || CB

∴ ∠MDC + ∠DCB = 180°

(Sum of interior angles of the same side of transveralsi)

or, ∠MDC + 90° = 180°

or, ∠MDC = 90°

Therefore, MD ⊥ AC

(iii) In ΔAMD and ΔCMD

AD = CD (Prove above)

∠ADM = ∠CDM, (Each 90°)

DM = DM (Common)

By S-A-S congruency condition

ΔAMD ≅ ΔCMD

∴ AM = CM …..(i) (By CPCT)

But AM = AB ……(ii)

(Given M is the mid point of AB)

From (i) and (ii)

CM = MA = \(\frac {1}{2}\) AB