Ncert Solutions For Class 9 Maths Ex 8.2

Ncert Solutions For Class 9 Maths Chapter 8 Ex 8.2

Q1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB,BC,CDandDA. AC is a diagonal. Show that :
(i) SRAC and SR=12AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

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Solution:

(i) In DAC,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SRAC and SR=12AC

(ii) In BAC,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQAC and PQ=12AC
also, SR=12AC
Thus, PQ = SR

(iii) SRAC– from (i)
and, PQAC– from (ii)
SRPQ– from (i) and (ii)
also, PQ = SR
Thus, PQRS is a parallelogram.

 

Q2. 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:

13

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 Prove,
PQRS is a rectangle.
Construction,
Join AC and BD

Proof,
In DRSandBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
SDR=QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, DRSBPQ (by SAS congruence condition).
RS = PQ (by CPCT)    ………… (i)
In QCRandSAP,
RC = PA (Halves of the opposite sides of the rhombus)
RCQ=PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, QCRSAP (by SAS congruence condition).
RQ = SP (by CPCT)    ……(ii)
Now,
In CBD,
R and Q are the mid points of CD and BC respectively.
QRBD
also,
P and S are the mid points of AD and AB respectively.
PSBD
QRPS
Thus, PQRS is a parallelogram.
also, PQR=90
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
Q=90
Thus, PQRS is a rectangle.

 

Q3. 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.

14

Solution:

Given,
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
Construction,
AC and BD are joined.
To Prove,
PQRS is a rhombus.
Proof,
In ABC
P and Q are the mid-points of AB and BC respectively
Thus, PQAC and PQ=12AC (Mid point theorem) ……….(i)
In ADC,
SRAC and SR=12AC (Mid point theorem)   ……(ii)
So, PQSR and PQ=SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PSQR and PS=QR  (Opposite sides of parallelogram)  ….(iii)
Now,
In BCD,
Q and R are mid points of side BC and CD respectively.
Thus, QRBDandQR=12BD (Mid point theorem)   ……(iv)
AC = BD (Diagonals of a rectangle are equal)   ……… (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
So, PQRS is a rhombus.

 

Q4. ABCD is a trapezium in which ABDC, 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.

 15
Solution:

Given,
ABCD is a trapezium in which ABDC, BD is a diagonal and E is the mid-point of AD.
To prove,
F is the mid-point of BC.
Proof,
BD intersected EF at G.
In BAD,
E is the mid point of AD and also EGAB.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In BDC,
G is the mid point of BD and also GFABDC.
Thus, F is the mid point of BC (Converse of mid point theorem)


Q5. 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.

16 

Solution:

Given
ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,

ABCD is a parallelogram
Therefore, ABCD

also, AEFC

Now,
AB=CD  (Opposite sides of parallelogram ABCD)
12AB=12CD
AE=FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AFEC (Opposite sides of a parallelogram)
Now,
In DQC,
F is mid point of side DC and FPCQ (as AFEC).
P is the mid-point of DQ (Converse of mid-point theorem)
DP=PQ   ………(i)
Similarly,
In APB,
E is mid point of side AB and EQAP (as AFEC).
Q is the mid-point of PB (Converse of mid-point theorem)
PQ=QB    …….(ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

 

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

Solution:

17

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.
Now,
In ACD,
R and S are the mid points of CD and DA respectively.
Thus, SRAC.
Similarly we can show that,
PQAC
PSBD
QRBD
Thus, PQRS is parallelogram.
PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

 

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) MDAC
(iii) CM=MA=12AB

18

 

Solution:
(i) In ACB,
M is the mid point of AB and MDBC
Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) ACB=ADM (Corresponding angles)
also, ACB=90
Thus, ADM=90and MDAC

(iii)  In AMDandCMD,
AD = CD (D is the midpoint of side AC)
ADM=CDM (Each 90°)
DM = DM (common)
Thus, AMDCMD (by SAS congruence condition).
AM = CM (by CPCT)
also, AM=12AB (as M is mid point of AB)
Hence, CM=AM=12AB.

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