In this page we have *NCERT book Solutions for Class 9th Maths Chapter 8 Quadrilaterals* for
EXERCISE 8.2 . Hope you like them and do not forget to like , social share
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ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that

(i) SR || AC and SR =1/2 AC

(ii) PQ = SR

(iii) PQRS is a parallelogram

(i) In Δ ACD

S is the mid point of AD

R is the mid point of CD

By Mid point theorem

SR || AC and SR =1/2 AC --(X)

(ii)

In Δ ABC

P is the mid point of AB

Q is the mid point of BC

By Mid point theorem

PQ || AC and PQ =1/2 AC --(Y)

From equation X and Y

PQ = SR

and PQ || SR||AC

(iii) In Quadrilaterals PQRS

PQ=SR

and PQ||SR

Hence PQRS is a parallelogram

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

The figure is shown as below

To Prove: quadrilateral PQRS is a rectangle

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR

PQ ||SR

Hence PQSR is a parallelogram

Now in Δ BCD

Q and R are mid points of sides BC and DC

By Mid point theorem

BD || QR

or

QM || OL

Also we already proved that

PQ ||SR

or LQ ||OM

In LQOM

QM || OL

LQ ||OM

Therefore LQOM is a parallelogram

Now,we know that opposite angles in the parallelogram are equal

∠ LQM = ∠ LOM

Now ∠ LOM=90 as diagonals bisects at right angle in Rhombus

Therefore

∠ LQM=90°

or

∠ PQR=90°

Now

PQRS is a parallelogram with one angle as right angle. Hence PQRS is a rectangle

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

The figure is shown as below

To Prove: quadrilateral PQRS is a rhombus

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR

PQ ||SR

Hence PQSR is a parallelogram

Now in Δ BCD

Q and R are mid points of sides BC and DC

By Mid point theorem

QR= 1/2 BD

Nowe AC = BC

Hence

PQ=SR=QR

Now a parallelogram whose adjacent sides are equal is a rhombus.

Hence proved

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 . Show that F is the mid-point of BC.

Let G be the point on BD where EF intersect

Now in Δ ABD

E is the mid-point of AD

EG ||AB

By converse of Mid-point theorem

G is the mid-point of BD

Now in Δ BDC

G is the mid-point of BD

FG || BC

By converse of Mid-point theorem

G is the mid-point of BC

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

AB=CD ( Opposite side of parallelogram)

Since E and F are mid-points of AB and DC respectively.

AE = 1/2 AB and CF = 1/2 DC

Therefore

AE = CF and AE || CF.

One pair of opposite sides is parallel and equal

Hence AECF is a parallelogram.

AF ||EC

In Δ BAP

E is the mid-point of AB

Now AF ||EC

So, EQ || AP

Now Converse of mid-point theorem

Q is mid-point of PB

PQ = QB ---(X)

Similarly, in ΔDQC, P is the mid-point of DQ

DP = PQ ---(Y)

From (X) and (Y), we have DP = PQ = QB

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

To Prove : EG and FH bisect each other.

Construction : Join EF, FG, GH, HE and AC

Proof:

In Δ ABC

E and F are mid-points of AB and BC respectively.

By Mid point theorem

EF = 1/2 AC and EF || AC ----(i)

In Δ ADC

H and G are mid-points of AD and CD respectively.

By Mid point theorem

HG = 1/2 AC and HG || AC ----(ii)

From (i) and (ii), we get

EF = HG

and EF || HG

Hence EFGH is a parallelogram

Now We know that Diagonals of the parallelogram bisects each other

Here EG and FH are diagonals of the parallelogram EFGH.

Hence Proved

ABC is a triangle right angles at C. A line through the mid -point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

- D is mid- point of AC
- MD ⊥ AC
- $CM = MA = \frac {1}{2} AB$

Construction : Join CM.

Proof:

(i) In Δ ABC

M is the mid-point of AB.

BC || DM

By converse of Mid-point theorem

D is the mid-point of AC

(ii)

Now BC || DM

Hence Corresponding angles]

∠ ADM = ∠ACB

Now ∠ACB=90°

Hence

∠ ADM =90°

Hence MD ⊥ AC

(iii) In Δ ADM and CMD

AD=DC

∠ ADM = ∠ CDM = 90°

DM is common

Therefore by SAS congruence

&Delta ADM ≅ &Delta CDM

Now by CPCT

CM=MA

Since M is mid-point of AB,

MA = 1/2 AB

Hence, $CM = MA = \frac {1}{2} AB$

- NCERT solutions for class 9 maths chapter 8 Quadrilaterals Exercise 8.2 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
- This chapter 8 has total 2 Exercise 8.1 and 8.2. This is the Last exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

**Notes****Assignments & NCERT Solutions**

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