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# 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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## 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 ΔADC, 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 ΔABC ,P is the mid point of AB and Q is the mid point of  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 DA  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 ΔABD

PS∥BD (mid point theorem)

∴SU ∥ TO…..(i)

SR∥AC (mid point theorem)

∴ST ∥ OU….(ii)

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

Therefore ∠ UST = ∠TOU (opposite angle of parallelogram)

∠TOU = 90°(diagonal of rhombus bisect each other at 90°)

∴∠ UST = 90°

Similarly ∠PQR = ∠QRS = ∠QPS =90°

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 DA  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 DA  respectively

TO PROVE: quadrilateral PQRS is a rhombus

PROOF:

In ΔABD, P is the mid point of AB and S is the mid point of AD

Therefore applying mid point theorem

PS∥BD

$\boldsymbol{PS=\frac{1}{2}BD....\left ( i \right )}$

In ΔBDC, Q is the mid point of BC and R is the mid point of DC

Therefore applying mid point theorem

QR∥BD

$\boldsymbol{QR=\frac{1}{2}BD....\left ( ii \right )}$

In ΔADC, S is the mid point of AD and R is the mid point of DC

Therefore applying mid point theorem

SR∥AC

$\boldsymbol{SR=\frac{1}{2}AC}$

$\boldsymbol{SR=\frac{1}{2}BD....\left ( iii \right )}$

In ΔABC, P is the mid point of AB and Q is the mid point of BC

Therefore applying mid point theorem

PQ∥AC

$\boldsymbol{PQ=\frac{1}{2}AC}$

AC = BD (diagonal of rectangle)

$\boldsymbol{PQ=\frac{1}{2}BD....\left ( iv \right )}$

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

Ans.

GIVEN: AB∥ DC

E is the mid point of AD

EF∥ AB

TO PROVE:F is the mid point of BC

PROOF: Let EF intersects diagonal BD at G

In ΔABD

EF∥ AB

∴EG ∥ AB

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

In ΔBDC

GF ∥ DC (DC∥ AB∥EF)

According to converse of mid point theorem ,If G is the mid point of BD and GF ∥ 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 of  the sides AB and CD  respectively (see the figure). Show that the line segments AF and EC trisect the diagonal BD.

Ans.

GIVEN:ABCD is a parallelogram

E is mid point of  the side AB

F is mid point of  the side CD

TO PROVE: DP = PQ = BQ

PROOF: In ABCD  parallelogram

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

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

AB = DC  (opposite sides of parallelogram )

AE = DC/2….(ii)

From the equations (i) and (ii)

CF = AE

CF ∥ 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 ∥ CE

Therefore in ΔDQC,F is the mid point of DC

PF ∥ CQ ( since AF ∥ CE proved above)

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

∴DP = PQ….(i)

Similarly ,we can consider ΔAPB 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 ΔABD, P is the mid point of AB and S is the mid point of AD

Therefore applying mid point theorem

PS∥BD

$\boldsymbol{PS=\frac{1}{2}BD....\left ( i \right )}$

n ΔBDC, Q is the mid point of BC and R is the mid point of DC

Therefore applying mid point theorem

QR∥BD

$\boldsymbol{QR=\frac{1}{2}BD....\left ( ii \right )}$

From (i) and (ii)

PS = QR

PS ∥ QR (since PS ∥ BD and QR ∥ 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 ∥ 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 ⊥ AC

$\boldsymbol{\left ( iii \right )CM=MA=\frac{1}{2}AB}$

Ans.

GIVEN: ΔABC in which ∠C= 90°

M is mid point of hypotenuse AB

BC∥ 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 BC∥ MD, then D will also the mid point of AC.

(ii) MD ⊥ AC

Since BC∥ MD (given)

∴∠MDC + ∠C = 180° (sum of co-interior angles)

∠MDC + 90°= 180°

∠MDC = 180° – 90° = 90°

Therefore MD ⊥ AC

$\boldsymbol{\left ( iii \right )CM=MA=\frac{1}{2}AB}$

In ΔAMD and ΔCMD ,we have

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

DM = DM (common)

∠ADM = ∠CDM = 90° [proved above in (ii)]

ΔADM ≅ ΔCDM(SAS rule of congruency of triangles)

CM = MA (by CPCT)

Since, M is the mid point of AB

$\boldsymbol\therefore {CM=\frac{1}{2}AB}$

Therefore

$\boldsymbol {CM=MA=\frac{1}{2}AB}$

Hence proved

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

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