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Quadrilaterals Exercise 8.2





In this page we have NCERT book Solutions for Class 9th Maths:Quadrilaterals for EXERCISE 8.2 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1.
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
Class 9 Maths NCERT Solutions for Quadrilaterals Exercise 8.2
(i) SR || AC and SR =1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
Solution
(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

Question 2.
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.
Solution
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

Question 3.
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.
Solution
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

Question 4.
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.

Solution
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

Question 5.
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
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

Question 6.
Show that the line segments joining the mid- point of the opposite sides of a quadrilateral bisect each other.
Solution
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

Question 7.
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
  1. D is mid- point of AC
  2. MD ⊥ AC
  3. $CM = MA = \frac {1}{2} AB$
Solution

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

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