NCERT Solutions Class 9 Maths Exercise 8.2 of chapter 8-Quadrilateral - Future Study Point

# NCERT Solutions Class 9 Maths Exercise 8.2 of chapter 8- Quadrilateral

NCERT Solutions Class 9 Maths Exercise 8.2 of chapter 8-Quadrilateral are the solutions of unsolved questions of exercise 8.2 of the NCERT maths text book of class 9.All the solutions of questions are created by future study point by a step by step way for helping the students to boost their preparations for the CBSE board exams.You can also study here NCERT solutions of science and maths from class 9 -12, sample papers, solutions of previous years question papers, tips for entrance exams of government jobs, carrier in online jobs.

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pdf-NCERT Solutions of Class 9 Maths chapter 8-Quadrilateral

pdf of class 9 NCERT solutions of the chapter 7 -Triangles

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

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CBSE class 9 Notes on Lines, angles, and triangles

Addition, subtraction, and division of polynomials

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NCERT Solutions of Class 9 Science : Chapter 1 to Chapter 15

## NCERT Solutions Class 9 Maths Exercise 8.2 of chapter 8- Quadrilateral

Q1.ABCD is a quadrilateral in which P, Q, R, and S midpoints of the sides AB, BC, CD, and DA (see the given figure ).AC is a diagonal. Show that:

(i) SR ll AC, SR = 1/2(AC)

(ii) PQ = SR

(iii)PQRS is a parallelogram

Ans.

GIVEN: ABCD is a quadrilateral in which P,Q,R and S mid points of the sides AB,BC, CD and DA

TO PROVE:

(i) SR ll AC, SR = 1/2(AC)

(ii) PQ = SR

(iii)PQRS is a parallelogram

PROOF:

(i) In ฮADC, R is the mid point point of DC and S is the mid point of AD

According to mid point theorem the line segment joining the mid points of a triangle is parallel to third side and half of the third side in length.

SR ll AC

SR = 1/2(AC)

(ii) In ฮABC ,P is the mid point of AB and Q is the mid point ofย  BC ,then using mid point theorem

PQ ll AC

PQ = 1/2(AC)…(i)

SR = 1/2(AC) …(ii)[proved above in (i)]

From equation (i) and (ii)

PQ = SR

(iii) According to mid point theorem,from the figure we have

SR ll AC….(i)

PQ ll AC….(ii)

From equation (i) and (ii) PQ ll SR

PQ = SR (proved above)

According to the theorem of the parallelogram, if one pair of opposite sides of a quadrilateral are equal and parallel, then the other pair of opposite sides is also parallel and equal.

PS ll QR and PS = QR

Therefore PQRS is a parallelogram, Hence proved.

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Q2.ABCD is a rhombus and P, Q, R, and S are midpoints of the sides AB, BC, CD, and DAย  respectively. Show that quadrilateral PQRS is a rectangle.

Ans.

GIVEN: ABCD is a rhombus and P,Q,R and S are midpoints of the sides AB, BC, CD, and DA

TO PROVE: PQRS is a rectangle.

PROOF: In ฮABD

PSโฅBD (mid point theorem)

โดSU โฅ TO…..(i)

SRโฅAC (mid point theorem)

โดST โฅ OU….(ii)

From equation (i) and equation (ii),we get that OUST is a parallogram

Therefore โ  UST = โ TOU (opposite angle of parallelogram)

โ TOU = 90ยฐ(diagonal of rhombus bisect each other at 90ยฐ)

โดโ  UST = 90ยฐ

Similarly โ PQR = โ QRS = โ QPS =90ยฐ

Therefore PQRS is a rectangle, Hence proved

Q3.ABCD is a rectangle and P,Q,R and S are midpoints of the sides AB,BC,CD and DAย  respectively .Show that quadrilateral PQRS is a rhombus.

Ans.

GIVEN:ABCD is a rectangle P,Q,R and S are midpoints of the sides AB,BC,CD and DAย  respectively

TO PROVE: quadrilateral PQRS is a rhombus

PROOF:

In ฮABD, P is the mid point of AB and S is the mid point of AD

Therefore applying mid point theorem

PSโฅBD

In ฮBDC, Q is the mid point of BC and R is the mid point of DC

Therefore applying mid point theorem

QRโฅBD

In ฮADC, S is the mid point of AD and R is the mid point of DC

Therefore applying mid point theorem

SRโฅAC

In ฮABC, P is the mid point of AB and Q is the mid point of BC

Therefore applying mid point theorem

PQโฅAC

AC = BD (diagonal of rectangle)

From equation (i), (ii),(iii) and (iv), we have

PQ = QR = SR = PS

Hence PQRS is a rhombus

Q4. ABCD is a trapizium in which ABโฅ DC, BD is a diagonal andย  E is the mid point of AD. A line is drawn through E parallel to AB intersecting BC at F(see the given figure). Show that F is the mid point of BC.

Ans.

GIVEN: ABโฅ DC

E is the mid point of AD

EFโฅ AB

TO PROVE:F is the mid point of BC

PROOF: Let EF intersects diagonal BD at G

In ฮABD

EFโฅ AB

โดEG โฅ AB

According to converse of mid point theorem ,If E is the mid point of AD and EG โฅ AB, then G will also the mid point of BD

In ฮBDC

GF โฅ DC (DCโฅ ABโฅEF)

According to converse of mid point theorem ,If G is the mid point of BD and GF โฅ DC, then F will also the mid point of BC

Hence proved

NCERT Solutions Class 9 Maths Exercise 8.2 of chapter 8- Quadrilateral

Q5. In a parallelogram ABCD, E and F are mid point ofย  the sides AB and CDย  respectively (see the figure). Show that the line segments AF and EC trisect the diagonal BD.ย

Ans.

GIVEN:ABCD is a parallelogram

E is mid point ofย  the side AB

F is mid point ofย  the side CD

TO PROVE: DP = PQ = BQ

PROOF: In ABCDย  parallelogram

CF = DC/2….(i)(F is mid point ofย  the side DC)

AE = AB/2(E is mid point ofย  the side AB)

AB = DCย  (opposite sides of parallelogram )

AE = DC/2….(ii)

From the equations (i) and (ii)

CF = AE

CF โฅ AE

If a pair of opposite sides of a quadrilateral is equal and parallel then another pair of opposite sides is also equal and parallel

AF โฅ CE

Therefore in ฮDQC,F is the mid point of DC

PF โฅ CQ ( since AF โฅ CE proved above)

According to converse of mid point theorem ,If F is the mid point of DC and PF โฅ CQ, then P will also the mid point of DQ.

โดDP = PQ….(i)

Similarly ,we can consider ฮAPB and can prove

PQ = BQ…..(ii)

From equation (i) and (ii)

DP = PQ = BQ

Hence proved

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Q6. Show that line segments joining the mid points of the opposite side of a quadrilateral bisect each other.

Ans.

P ,Q,R and S are the mid points of the sides AB, BC, DC and AD respectively.

TO PROVE: PR and QS bisect each other

PROOF:

In ฮABD, P is the mid point of AB and S is the mid point of AD

Therefore applying mid point theorem

PSโฅBD

n ฮBDC, Q is the mid point of BC and R is the mid point of DC

Therefore applying mid point theorem

QRโฅBD

From (i) and (ii)

PS = QR

PS โฅ QR (since PS โฅ BD and QR โฅ BD)

If a pair of opposite sides of a quadrilateral is equal and parallel then another pair of opposite sides is also equal and parallel

Therefore SR = PQ , SR โฅ PQ

Now it is clear that PQRS is a parallelogram

Since diagonal of parallelogram bisect each other

Therefore PR and QS bisect each other, Hence proved

Q7. 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 โฅ AC

Ans.

GIVEN: ฮABC in which โ C= 90ยฐ

M is mid point of hypotenuse AB

BCโฅ MD

TO PROVE:

(i) D is the mid point of AC

According to converse of mid point theorem ,If M is the mid point of AB and BCโฅ MD, then D will also the mid point of AC.

(ii) MD โฅ AC

Since BCโฅ MD (given)

โดโ MDC + โ C = 180ยฐ (sum of co-interior angles)

โ MDC + 90ยฐ= 180ยฐ

โ MDC = 180ยฐ – 90ยฐ = 90ยฐ

Therefore MD โฅ AC

In ฮAMD and ฮCMD ,we have

AD = CD [D is mid point of AC ,proved above in (i)]

DM = DM (common)

โ ADM = โ CDM = 90ยฐ [proved above in (ii)]

ฮADM โ ฮCDM(SAS rule of congruency of triangles)

CM = MA (by CPCT)

Since, M is the mid point of AB

Therefore

Hence proved

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

### NCERT Solutions for class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions for class 9 scienceย

 Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers

### NCERT Solutions for class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

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### NCERT Solutions for Class 10 Science

 Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world Chapter 4- Carbon and its Compounds Chapter 12- Electricity Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current Chapter 6- Life Process Chapter 14-Sources of Energy Chapter 7-Control and Coordination Chapter 15-Environment Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

### NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem ย Chapter 16- Probability

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### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

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