NCERT Solutions for Class 9 Maths Exercise 8.2

NCERT Solutions for Class 9 Maths Exercise 8.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Quadrilateral Download as PDF

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.

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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.

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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

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

Two adjacent sides are equal.

Hence, PQRS is a rhombus.

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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]

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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.

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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)

Similarly, in ADC,

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.

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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,

AD = DC [proved above]

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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