# Ncert Solutions For Class 9 Maths Ex 8.2

## Ncert Solutions For Class 9 Maths Chapter 8 Ex 8.2

Q1. ABCD$ABCD$ is a quadrilateral in which P, Q, R and S are mid-points of the sides AB,BC,CDandDA$AB,BC,CD\; and \;DA$. AC is a diagonal. Show that :
(i) SRAC$SR\parallel AC$ and SR=12AC$SR=\frac{1}{2}AC$
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:

(i) In DAC$\bigtriangleup DAC$,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SRAC$SR\parallel AC$ and SR=12AC$SR=\frac{1}{2}AC$

(ii) In BAC$\bigtriangleup BAC$,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQAC$PQ\parallel AC$ and PQ=12AC$PQ=\frac{1}{2} AC$
also, SR=12AC$SR=\frac{1}{2} AC$
Thus, PQ = SR

(iii) SRAC$SR\parallel AC$– from (i)
and, PQAC$PQ\parallel AC$– from (ii)
SRPQ$\Rightarrow SR \parallel PQ$– 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:

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$\bigtriangleup DRS \;\;and\;\;\bigtriangleup BPQ$,
DS = BQ (Halves of the opposite sides of the rhombus)
SDR=QBP$\angle SDR=\angle QBP$ (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, DRSBPQ$\bigtriangleup DRS \cong \bigtriangleup BPQ$ (by SAS congruence condition).
RS = PQ (by CPCT)    ………… (i)
In QCRandSAP$\bigtriangleup QCR \;\; and \;\; \bigtriangleup SAP$,
RC = PA (Halves of the opposite sides of the rhombus)
RCQ=PAS$\angle RCQ=\angle PAS$ (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, QCRSAP$\bigtriangleup QCR \cong \bigtriangleup SAP$ (by SAS congruence condition).
RQ = SP (by CPCT)    ……(ii)
Now,
In CBD$\bigtriangleup CBD$,
R and Q are the mid points of CD and BC respectively.
QRBD$\Rightarrow QR \parallel BD$
also,
P and S are the mid points of AD and AB respectively.
PSBD$\Rightarrow PS \parallel BD$
QRPS$\Rightarrow QR \parallel PS$
Thus, PQRS is a parallelogram.
also, PQR=90$\angle PQR=90^{\circ}$
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
Q=90$\angle Q=90^{\circ}$
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.

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$\bigtriangleup ABC$
P and Q are the mid-points of AB and BC respectively
Thus, PQAC$PQ\parallel AC$ and PQ=12AC$PQ= \frac{1}{2}AC$ (Mid point theorem) ……….(i)
In ADC$\bigtriangleup ADC$,
SRAC$SR\parallel AC$ and SR=12AC$SR=\frac{1}{2}AC$ (Mid point theorem)   ……(ii)
So, PQSR$PQ\parallel SR$ and PQ=SR$PQ= SR$
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PSQR$PS\parallel QR$ and PS=QR$PS= QR$  (Opposite sides of parallelogram)  ….(iii)
Now,
In BCD$\bigtriangleup BCD$,
Q and R are mid points of side BC and CD respectively.
Thus, QRBDandQR=12BD$QR\parallel BD\;\; and\;\; QR=\frac{1}{2}BD$ (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$AB\parallel DC$, 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.

Solution:

Given,
ABCD is a trapezium in which ABDC$AB\parallel DC$, 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$\bigtriangleup BAD$,
E is the mid point of AD and also EGAB$EG\parallel AB$.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In BDC$\bigtriangleup BDC$,
G is the mid point of BD and also GFABDC$GF\parallel AB\parallel DC$.
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.

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$AB\parallel CD$

also, AEFC$AE\parallel FC$

Now,
AB=CD$AB=CD$  (Opposite sides of parallelogram ABCD)
12AB=12CD$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$
AE=FC$\Rightarrow AE=FC$ (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$AF\parallel EC$ (Opposite sides of a parallelogram)
Now,
In DQC$\bigtriangleup DQC$,
F is mid point of side DC and FPCQ$FP\parallel CQ$ (as AFEC$AF\parallel EC$).
P is the mid-point of DQ (Converse of mid-point theorem)
DP=PQ$\Rightarrow DP=PQ$   ………(i)
Similarly,
In APB$\bigtriangleup APB$,
E is mid point of side AB and EQAP$EQ\parallel AP$ (as AFEC$AF\parallel EC$).
Q is the mid-point of PB (Converse of mid-point theorem)
PQ=QB$\Rightarrow PQ=QB$    …….(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:

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$\bigtriangleup ACD$,
R and S are the mid points of CD and DA respectively.
Thus, SRAC$SR\parallel AC$.
Similarly we can show that,
PQAC$PQ\parallel AC$
PSBD$PS\parallel BD$
QRBD$QR\parallel BD$
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$MD\perp AC$
(iii) CM=MA=12AB$CM=MA=\frac{1}{2}AB$

Solution:
(i) In ACB$\bigtriangleup ACB$,
M is the mid point of AB and MDBC$MD\parallel BC$
Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) ACB=ADM$\angle ACB=\angle ADM$ (Corresponding angles)
also, ACB=90$\angle ACB=90^{\circ}$
Thus, ADM=90$\angle ADM=90^{\circ}$and MDAC$MD\perp AC$

(iii)  In AMDandCMD$\bigtriangleup AMD\;\; and\;\; \bigtriangleup CMD$,
AD = CD (D is the midpoint of side AC)
ADM=CDM$\angle ADM=\angle CDM$ (Each 90°)
DM = DM (common)
Thus, AMDCMD$\bigtriangleup AMD \cong \bigtriangleup CMD$ (by SAS congruence condition).
AM = CM (by CPCT)
also, AM=12AB$AM=\frac{1}{2}AB$ (as M is mid point of AB)
Hence, CM=AM=12AB$CM=AM=\frac{1}{2}AB$.