NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 8.2 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## 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 the mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that: D

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

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Solution.

Given : P,Q,R, and S are the mid-points of the sides

AB, BC, CD, and DA, respectively.

AP = BP

BQ = CQ

CR = DR

and AS = DS

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

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Proof:

**(i)** In Δ ADC, we have S is mid-point of AD and R is mid-point of DC.

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

∴ SR || AC …….(i)

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

**(ii)** Similarly, in Δ ABC, we have P is mid-point of AB and Q is the mid-point of BC.

∴ PQ || AC …………..(iii)

PQ = \(\frac { 1 }{ 2 } \) AC…………..(iv)

**(iii)** From eqs. (i) and (iii), we get

PQ || SI and from Eq. (v), PQ = SR

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

So, 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 show: Quadrilateral PQRS is a rectangle.

Proof: By mid-point theorem.

In Δ ADC, we have S and R are the mid-points of DA and CD, respectively.

∴ SR || AC

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

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

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

From eqs. (i) and (ii), we get PQ || SR and PQ =SR = \(\frac { 1 }{ 2 } \) 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.

Given, ABCD is a rectangle.

∴ ∠A=∠B = ∠C =∠D = 90° and AD = BC, AB = CD

Also, given P,Q,R and S are mid-points of AB, BC, CD and DA respectively.

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

AP = BP (Given)

AS= BQ (Given)

∠A= ∠B (Given)

∴ Δ ASP ≅ Δ BQP (By SAS)

∴ SP = PQ (By CPCT) …(iv)

SD = CQ (Given)

DR = RC (Given)

∠C = ∠D (Given)

Δ RDS = Δ RCQ (By SAS)

SR = RQ (By CPCT) …(v)

From Eqs. (iii), (iv) and (v), it is clear that quadrilateral 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 fig.). 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 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 Δ ABD, we have EP || AB [∵ EF || AB] and E is mid-point of AD.

So, by the converse of mid-point theorem.

We get, P is mid-point of BD.

Similarly, in ABCD, we have PF || CD [∵ EF || AB and AB || CD]

and P is mid-point of BD.

So, by the converse of mid-point theorem, F is 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 fig.). Show that the line segments AF and EC trisect the diagonal BD.

Solution.

Given: ABCD is a parallelogram and E, 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]

⇒ AF || FC and \(\frac { 1 }{ 2 } \) AP = \(\frac { 1 }{ 2 } \) DC

⇒ AF || FC and AF = 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 || FC

⇒ AP || FQ and FP || CQ

Since opposite sides of a parallelogram are parallel.

In Δ 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 Δ 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.

Let ABCD be a quadrilateral and P, Q, R and S be the D mid-points of the sides AB, BC, CD, and DA, respectively.

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 Δ ADC, S and R are 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 = \(\frac { 1 }{ 2 } \)AC …(i) [by mid-point theorem]

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

PQ || AC and PQ = \(\frac { 1 }{ 2 } \)AC [by mid-point theorem]

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

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

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

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

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

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

Solution.

Given, ABC is a right angled triangle.

∠C = 90° and M is the mid-point of AB.

Also, DM || BC

**(i)**In Δ ABC, BC || MD and Mis mid-point of AB.

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

Since, MD || BC and CD is transversal. .

∠ADM = ∠ACB (Corresponding angles)

But ∠ACB = 90°

∠ADM = 90° => MD ⊥ AC

Now, in AADM and A CDM, we have

DM = MD

AD = CD

∠ADM = ∠MDC

AADM = ACDM

CM = AM

Also, M is mid-point of AB.

AM = BM = \(\frac { 1 }{ 2 } \) AB

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

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

Hence proved.

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