Class IX Math NCERT Solution for Quadrilaterals 8.2

NCERT TEXTBOOK QUESTIONS SOLVED

EXERCISE 8.2 (Page 150)

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.

Sol:   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.

Sol:   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.

Sol:   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.

Sol:   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.

Sol:   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.

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

Sol:   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.

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

           

Sol:   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