Quadrilateral CLASS-9th (EXERCISE 8.2 ) CH-8

Quadrilateral

CLASS-9th

EXERCISE 8.2 

 

Q1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that.
           (i) SR || AC and 
           (ii) PQ = SR
           (iii) PQRS is a parallelogram.
SOLUTION:   We have Pas the mid-point of AB, Q as the mid-point of BC,
           (i) To prove that  and SR || AC
                R as the mid-point of CD, S as the mid-point of DA, and AC as the diagonal of quadrilateral ABCD.
                In ΔACD, we have
                S as the mid-point of AD,
                R as the mid-point of CD.
                ∵ The line segment joining the mid-point of any two sides of a triangle is parallel to the third side and half of it.
 and SR || AC
           (ii) To prove that PQ = SR.
                  In ΔABC, we have
                  P is the mid-point of AB,
                  Q is the mid-point of BC.
                  From (1) and (2), PQ = SR
           (iii) To prove that PQRS is a parallelogram.
                   In ΔABC, P and Q are the mid-points of AB and BC.
                   In ΔACD, S and R are the mid-points of DA and CD.
                   From (3) and (4), we get
                   ⇒ PQ = SR and PQ || SR
                   i.e. One pair of opposite sides in quadrilateral PARS is equal and parallel.
                   ∴ 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:   We have P as the mid-point of AB, Q as the mid-point of BC, R as the mid-point of CD, S as the mid-point of DS.
                   We have to prove that PQRS is a rectangle. Let us join AC.
                   ∵ In DABC, P and Q are the mid-points of AB and BC.
                   Also in ΔADC, R and S are the mid-points of CD and DA.
                   From (1) and (2), we get
                                  ⇒ PQ = SR and PQ || SR
                   i.e. One pair of opposite sides of quadrilateral PQRS is equal and parallel.
                   ∴ PQRS is a parallelogram.
                   Now, in ΔERC and ΔEQC,
                 ∠1 = ∠2
[ ∵The diagonal of a rhombus bisects the opposite angles]
                 CR = CQ
[Each is equal to  of a side of rhombus]
                 CE = CE
[Common]
                 ΔERC ≌ ΔEQC
[SAS criteria]
                 ⇒∠3 = ∠4
[c.p.c.t.]
                 But ∠3 + ∠4 = 180°
[Linear pair]
                 ⇒ ∠3 = ∠4 = 90°
                 But ∠5 = ∠3
[Vertically opposite angles]
                 ∠5 = 90°
                 PQ || AC ⇒ PQ || EF
                 ∴PQEF is a quadrilateral having a pair of opposite sides parallel and one of the angles is 90°.
                 ∴PQEF is a rectangle.
                 ⇒∠RQP = 90°
                 ∴One angle of parallelogram PQRS is 90°.
                 Thus, PQRS is a rectangle.
Q3.   ABCD is a rectangle 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 rhombus.
SOLUTION:   In a rectangle ABCD, P is the mid-point of AB, Q is the mid-point of BC, R is the mid-point of CD, S is the mid-point of DA AC is the diagonal.
             From (1) and (2), we get
                                       PQ = SR and PQ || SR
             Similarly, by joining BD, we have
                                       PS = QR and PS || QR
             i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.
             ∴ PQRS is a parallelogram.
             Now, in ΔPAS and ΔPBQ,
             ∴Their corresponding parts are equal.
                ⇒
PS = PQ
                Also
PS = QR
[Proved]
                and
PQ = SR
[Proved]
PQ = QR = RS = SP
             i.e. PQRS is a parallelogram having all of its sides equal.
             ⇒PQRS is a rhombus.
Q4.   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 figure). Show that F is the mid point of BC.
SOLUTION:   In trapezium ABCD, AB || DC. E is the mid-point of AD. EF is drawn parallel to AB. We have to prove that F is the mid-point of BC.
          Join BD.
          In ΔDAB,
          ∵ E is the mid-point of AD
          and                    EG || AB
          ∴ Using the converse of mid-point theorem, we get that G is the mid-point BD.
          Again in ΔBDC
          ∵G is the mid-point of BD
[Proved]
          GF || DC
[ ∵AB || DC and EF || AB and GF is a part of EF]
          ∴ Using the convene of the mid-point theorem, we get that F is the mid-point of BC.
Q5.   In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.
SOLUTION:   We have ABCD is a parallelogram such that:
          E is the mid-point of AB and F is the mid-point of CD. Let us join the opposite vertices B and D.
          Since, the opposite sides of a parallelogram are parallel and equal.
          ∴ AB || DC ⇒ AE || FC
…(1)
          Also                    AB = DC
          From (1) and (2), we can say that AECF is quadrilateral having a pair of the opposite sides as parallel and equal.
          ∴ AEFC is a parallelogram.
          ⇒ AE || CF
          Now, in ΔDBC,
         F is the mid-point of DC
[Given]
         and FP || CQ
[∵ AF || CE]
         ⇒ P is the mid-point of DQ
[Converse of mid-point theorem]
         ⇒ DP = PQ
…(3)
         Similarly, in ΔBAP,
                           BQ = PQ
…(4)
         ∴ From (3) and (4), we have
                           DP = PQ = BQ
         ⇒ The line segments AF and EC trisect the diagonal BD.
Quadrilateral CLASS-9th EXERCISE 8.2
Q6.   Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
SOLUTION:   A quadrilateral ABCD such that the mid-points of AB, BC, CD and DA are P, Q, R and S respectively,
           we have to prove that diagonals of PARS are bisected at O.
           Join PQ, QR, RS and SP. Let us also join PR and SQ.
           Now, in AABC, we have P and Q as the mid-points of its sides AB and BC respectively.
           ⇒ PQRS is a quadrilateral having a pair of opposite sides (PQ and RS) as equal and parallel.
           ∴ PQRS is a parallelogram.
           But the diagonals of a parallelogram bisect each other.
           i.e. PR and SQ bisect each other.
           Thus, the line segments joining the mid-points of opposite sides of a quadrilateral ABCD bisect each other.
Quadrilateral CLASS-9th EXERCISE 8.2
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) MD ⊥ AC
SOLUTION:   We have a triangle ABC, such that ∠C = 90°
           M is the mid-point of AB and MD || BC.
           (i) To prove that D is the mid-point of AC.
                In ΔACB, we have
                M as the mid-point of AB. [Given]
                MD || BC [Given]
                ∴ Using the converse of mid-point theorem, D is the mid-point of AC.
          
 (ii) To prove that MD ⊥ AC.     [Given]
                Since, MD || BC  and AC is a transversal.
∠MDA = ∠BCA  [Corresponding angles]
               But
∠BCA = 90° [Given]
                ∴ ∠MDA = 90°
                ⇒ MD ⊥ AC.
          
 (iii) To prove that CM = MA =  AB
                   In ΔADM and ΔCDM, we have
                       ∠ADM = ∠CDM                [Each = 90°]
                        MD = MD          [Common]
                       AD = CD     [∵M is the mid-point of AC (Proved)]
                   
                     ΔADM = ΔCDM    [SAS criteria]
                   
                     MA = MC                           [c.p.c.t.] …(1)
                 
                   ∵M is the mid-point AB.       [Given]
                   ∴ MA =  AB                               …(2)
                   From (1) and (2), we have
                  CM = MA =  AB

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