NCERT Class 9 Maths Solutions Chapter - 8 Quadrilaterals, Ex - 8.2

Ex - 8.2

Question 1.  ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:

(i)     SR || AC and SR =  AC             
(ii)    PQ = SR
(iii)    PQRS is a parallelogram.

Solution

(i)  In ADC, S and R are the mid points of sides AD and CD respectively.
     In a triangle the line segment joining the mid points of any two sides of the triangle is

     parallel to the third side and is half of it.
     SR || AC and SR =  AC                ... (1)

(ii)  In ABC, P and Q are mid points of sides AB and BC respectively. So, by using

      mid-point theorem, we have
      PQ || AC and PQ =  AC                ... (2) 
      Now using equations (1) and (2), we have
      PQ || SR and PQ = SR                 ... (3)
       PQ = SR
(iii)  From equations (3), we have
       PQ || SR and PQ = SR        
       Clearly one pair of opposite sides of quadrilateral PQRS is parallel and equal
       Hence, PQRS is a parallelogram.

Question 2.  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

In ABC, P and Q are mid points of sides AB and BC respectively.
     PQ || AC and PQ =  AC         (using mid-point theorem)    ... (1)

    In ADC
    R and S are the mid points of CD and AD respectively
     RS || AC and RS =  AC         (using mid-point theorem)    ... (2)

    From equations (1) and (2), we have
     PQ || RS and PQ = RS

    As in quadrilateral PQRS one pair of opposite sides are equal and parallel to each other, so, it is a parallelogram.
    Let diagonals of rhombus ABCD intersect each other at point O.
    Now in quadrilateral OMQN
    MQ || ON                (  PQ || AC)
    QN || OM                 ( QR || BD)
    So, OMQN is parallelogram
    MQN = NOM
    PQR = NOM

    But, NOM = 90^o     (diagonals of a rhombus are perpendicular to each other)

 PQR = 90^o

    Clearly PQRS is a parallelogram having one of its interior angle as 90�.
    Hence, PQRS is 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 the quadrilateral PQRS is a rhombus

Solution

 Let us join AC and BD
    In ABC
    P and Q are the mid-points of AB and BC respectively
    PQ || AC and PQ =  AC         (mid point theorem)        ... (1)

    Similarly in ADC
    SR || AC and SR =   AC         (mid point theorem)    ...  ... (2)
    Clearly, PQ || SR and PQ = SR
    As in quadrilateral PQRS one pair of opposite sides is equal and parallel to
    each other, so, it is a parallelogram.
     PS || QR and PS = QR        (opposite sides of parallelogram)... (3)

    Now, in BCD, Q and R are mid points of side BC and CD respectively.
     QR || BD and QR = BD         (mid point theorem)        ... (4)

    But diagonals of a rectangle are equal
     AC = BD                   ... ...  (5)

    Now, by using equation (1), (2), (3), (4), (5) we can say that

PQ = QR = SR = PS
    So, PQRS is a rhombus.

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

Solution

By converse of mid-point theorem a line drawn, through the mid point of any side of a triangle and parallel to another side bisects the third side.
Now in ABD
EF || AB and E is mid-point of AD
So, this line will intersect BD at point G and G will be the mid-point of DB.
Now as EF || AB and AB || CD
EF || CD    (Two lines parallel to a same line are parallel to each other)

Now, in BCD, GF || CD and G is the midpoint of line BD. So, by using converse of mid-point theorem, F 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 (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.

Solution

ABCD is a parallelogram
 AB || CD

So, AE || FC
Again AB = CD         (opposite sides of parallelogram ABCD)
     AB = CD
          AE = FC             (E and F are midpoints of side AB and CD)
As in quadrilateral AECF one pair of opposite sides (AE and CF) are parallel and equal to each other. So, AECF is a parallelogram.
AF || EC            (Opposite sides of a parallelogram)
Now, in DQC, F is mid point of side DC and FP || CQ (as AF || EC). So, by using converse of mid-point theorem, we can say that
P is the mid-point of DQ
DP = PQ            ... (1)
Similarly, in APB, E is mid point of side AB and EQ || AP (as AF || EC). So, by using converse of mid-point theorem, we can say that
Q is the mid-point of PB
PQ = QB            ... (2)
From equations (1) and (2), we may say that
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

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

Solution

Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In ABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =   BD             ... (1)
Similarly in BCD
QR || BD and QR =  BD               ... (2)
From equations (1) and (2), we have
SP || QR and SP = QR
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.

Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Question 7.  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

(iii)    CM=MA=AB

Solution

(i)     In ABC
        Given that M is mid point of AB and MD || BC.
        So, D is the mid-point of AC.        (Converse of mid-point     theorem)

(ii)    As DM || CB and AC is a transversal line for them.
        So, MDC + DCB = 180�         (Co-interior angles)
        MDC + 90� = 180�
        MDC = 90�
         MD  AC

(iii)   Join MC

  In AMD and CMD
     AD = CD                         (D is the midpoint of side AC)
     ADM = CDM             (Each 90�)
     DM = DM                         (common)
     AMDCMD        (by SAS congruence rule)

     So, AM = CM                   (by CPCT)
     But AM =  AB              (M is mid point of AB)
     So, CM = MA =  AB 

Ex-8.1

Ex-9.1