# NCERT Solutions for Class 9 Maths Exercise 8.2

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NCERT solutions for Class 9 Maths Quadrilateral

## NCERT Solutions for Class 9 Mathematics Quadrilaterals

###### 1. ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (See figure). AC is a diagonal. Show that:

(i) SR AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Ans. In ABC, P is the mid-point of AB and Q is the mid-point of BC.

Then PQ AC and PQ = AC

(i) In ACD, R is the mid-point of CD and S is the mid-point of AD.

Then SR AC and SR = AC

(ii) Since PQ = AC and SR = AC

Therefore, PQ = SR

(iii) Since PQ AC and SR AC

Therefore, PQ SR [two lines parallel to given line are parallel to each other]

Now PQ = SR and PQ SR

Therefore, PQRS is a parallelogram.

###### 2. ABCD is a rhombus and P, Q, R, S are mid-points of AB, BC, CD and DA respectively. Prove that quadrilateral PQRS is a rectangle.

Ans. Given: P, Q, R and S are the mid-points of respective sides AB, BC, CD and DA of rhombus. PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Construction: Join A and C.

Proof: In ABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ AC and PQ = AC ……….(i)

In ADC, R is the mid-point of CD and S is the mid-point of AD.

SR AC and SR = AC ……….(ii)

From eq. (i) and (ii), PQ SR and PQ = SR

PQRS is a parallelogram.

Now ABCD is a rhombus. [Given]

AB = BC

AB = BC PB = BQ

1 = 2 [Angles opposite to equal sides are equal]

Now in triangles APS and CQR, we have,

AP = CQ [P and Q are the mid-points of AB and BC and AB = BC]

Similarly, AS = CR and PS = QR [Opposite sides of a parallelogram]

APS CQR [By SSS congreuancy]

3 = 4 [By C.P.C.T.]

Now we have 1 + SPQ + 3 =

And 2 + PQR + 4 = [Linear pairs]

1 + SPQ + 3 = 2 + PQR + 4

Since 1 = 2 and 3 = 4 [Proved above]

SPQ = PQR ……….(iii)

Now PQRS is a parallelogram [Proved above]

SPQ + PQR = ……….(iv) [Interior angles]

Using eq. (iii) and (iv),

SPQ + SPQ = 2SPQ =

SPQ =

Hence PQRS is a rectangle.

NCERT Solutions for Class 9 Maths Exercise 8.2

###### 3. ABCD is a rectangle 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 rhombus.

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

To prove: PQRS is a rhombus.

Construction: Join AC.

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

PQ AC and PQ = AC ……….(i)

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

SR AC and SR = AC ……….(ii)

From eq. (i) and (ii), PQ SR and PQ = SR ……….(iii)

PQRS is a parallelogram.

Now ABCD is a rectangle. [Given]

AD = BC AS = BQ ……….(iv)

In triangles APS and BPQ,

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

PAS = PBQ [Each ]

And AS = BQ [From eq. (iv)]

APS BPQ [By SAS congruency]

PS = PQ [By C.P.C.T.] ………(v)

From eq. (iii) and (v), we get that PQRS is a parallelogram.

PS = PQ

Hence, PQRS is a rhombus.

NCERT Solutions for Class 9 Maths Exercise 8.2

###### 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.

Ans. Let diagonal BD intersect line EF at point P.

In DAB,

E is the mid-point of AD and EP AB [ EF AB (given) P is the part of EF]

P is the mid-point of other side, BD of DAB.

[A line drawn through the mid-point of one side of a triangle, parallel to another side intersects the third side at the mid-point]

Now in BCD,

P is the mid-point of BD and PF DC [ EF AB (given) and AB DC (given)]

EF DC and PF is a part of EF.

F is the mid-point of other side, BC of BCD. [Converse of mid-point of theorem]

NCERT Solutions for Class 9 Maths Exercise 8.2

###### 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.

Ans. Since E and F are the mid-points of AB and CD respectively.

AE = AB and CF = CD……….(i)

But ABCD is a parallelogram.

AB = CD and AB DC

AB = CD and AB DC

AE = FC and AE FC [From eq. (i)]

AECF is a parallelogram.

FA CE FP CQ [FP is a part of FA and CQ is a part of CE] ………(ii)

###### Since the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.

In DCQ, F is the mid-point of CD and FP CQ

P is the mid-point of DQ.

DP = PQ ……….(iii)

Similarly, In ABP, E is the mid-point of AB and EQ AP

Q is the mid-point of BP.

BQ = PQ ……….(iv)

From eq. (iii) and (iv),

DP = PQ = BQ ………(v)

Now BD = BQ + PQ + DP = BQ + BQ + BQ = 3BQ

BQ = BD ……….(vi)

From eq. (v) and (vi),

DP = PQ = BQ = BD

Points P and Q trisects BD.

So AF and CE trisects BD.

NCERT Solutions for Class 9 Maths Exercise 8.2

###### 6. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.

Ans. Given: A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.

To prove: EG and FH bisect each other.

Construction: Join AC, EF, FG, GH and HE.

Proof: In ABC, E and F are the mid-points of respective sides AB and BC.

EF AC and EF AC ……….(i)

G and H are the mid-points of respective sides CD and AD.

HG AC and HG AC ……….(ii)

From eq. (i) and (ii),

EF HG and EF = HG

EFGH is a parallelogram.

Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.

NCERT Solutions for Class 9 Maths Exercise 8.2

###### 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.

Ans. (i) In ABC, M is the mid-point of AB [Given]

MD BC

AD = DC [Converse of mid-point theorem]

Thus D is the mid-point of AC.

(ii) BC (given) consider AC as a transversal.

1 = C [Corresponding angles]

1 = [C = ]

Thus MD AC.

###### (iii) In AMD and CMD,

1 = 2 = [proved above]

MD = MD [common]

AMD CMD [By SAS congruency]

AM = CM [By C.P.C.T.] ……….(i)

Given that M is the mid-point of AB.

AM = AB ……….(ii)

From eq. (i) and (ii),

CM = AM = AB

## NCERT Solutions for Class 9 Maths Exercise 8.2

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