NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 9.2 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2

**Question 1.**

**ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). 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:** P, Q, Ft and S are mid-points of the sides.

∴ AP = PB, BQ = CQ

CR = DR and AS = DS

**(i)** In ∆ADC, we have

S is mid-point of AD and R is mid-point of the DC.

We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side.

∴ SB || AC …(i)

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

**(ii)** Similarly, in ∆ABC, we have

PQ || AC ….(iii)

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

Now, from Eqs. (i) and (iii), we get

SR = \(\frac { 1 }{ 2 }\) AC …..(v)

**(iii)** Now, from Eqs. (i) and (iii), we get

PQ || SR

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.

Hence proved.

**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 mid-points of AB, BC, CD and DA

By mid-point theorem,

∴ PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.

∴ ∠EOF = 90°

Now, RQ || BD (By mid-point theorem)

⇒ RE || OF

Also, SP|| AC [From Eq. (i)]

⇒ FR || OE

∴ OERF is a parallelogram.

So, ∠ ERF = ∠EOF = 90°

(Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with ∠R = 90°

Hence, PQRS is a rectangle.

**Question 3.**

**ABCD is a rectangle and P, Q, R ans 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 || BD and PQ = \(\frac { 1 }{ 2 }\) BD

In rectangle ABCD,

AC = BD

∴ PQ = SR …(ii)

Now, in ∆ASP and ∆BQP

AP = BP (Given)

AS = BQ (Given)

∠A = ∠B (Given)

∴ ∆ASP ≅ ∆BQP (By SAS)

∴ SP = PQ (By CPCT)…(ii)

Similarly, in ∆RDS and ∆RCQ,

SD = CQ (Given)

DR = RC (Given)

∠C=∠D (Given)

∴ ∆RDS ≅ ∆RCQ (By SAS)

∴ SR = RQ (By CPCT)…(iii)

From Eqs. (i), (ii) and (iii), it is clear that quadrilateral PQRS is a rhombus.

**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 figure). Show that F is the mid-point of BC.**

**Solution:**

**Given:** ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.

In ∆ABD, we have

EP\\AB

and E is mid-point of AD.

So, by theorem, if a line drawn through the mid-point of one side of a triangle parallel to another side bisect the third side.

∴ P is mid-point of BD.

Similarly, in ∆ BCD, we have,

PF || CD (Given)

and P is mid-point of BD.

So, by converse of mid-point theorem, F is mid-point of CB.

**Question 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.**

**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)

⇒ AE || FC and \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) DC

⇒ AF || FC and AF = FC

∴ AECF is a parallelogram.

∴ AF || FC

⇒ EQ || AP and FP || CQ

In ∆ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.

(By converse of mid-point theorem)

∴ BQ = PQ ….(i)

Again, in ∆DQC, F is the mid-point of DC and FP || CQ, so P is the mid-point of DQ. (By 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.

**Question 6.**

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

**Solution:**

Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively, i.e., AS = SD, 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 ∆ADC, S and R are mid-points of AD and CD.

We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side. (By mid-point theorem)

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

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

∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC (By mid-point theorem)…(ii)

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

PQ || SR

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

∴ Quadrilateral PQRS is a parallelogram whose diagonals are SQ and PR. Also, we know that diagonals of a parallelogram bisect each other. So, SQ and PR bisect each other.

**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**

**(iii) 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 ∆ ABC, BC || MD and M is mid-point of AB.

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

**(ii)** Since, MD || BC and CD is transversal

∴ ∠ADM = ∠ACB (Corresponding angles)

But ∠ACB = 90°

∴ ∠ADM = 90° ⇒ MD ⊥ AC

**(iii)** Now, in ∆ ADM and ∆ CDM, we have

DM = MD (Common)

AD = CD (∵ D is mid point of AC)

∴ ∠ADM = ∠MDC (Each equal to 90°)

∴ ∆ ADM = ∆ CDM (By SAS)

∴ CM = AM (By CPCT)…(i)

Also, M is mid-point of AB.

∴ AM – BM = \(\frac { 1 }{ 2 }\) AB ….(ii)

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

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

Hence proved.

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