Quadrilaterals – Exercise 8.2 – (MATHEMATICS) – 9th Class

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Exercise 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are midpoints of the sides AB, BC, CD and DA (sec figure), AC is a diagonal. Show that


(i)  SR\parallel AC    and SR = {1 \over 2}AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

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Sol. (i) Consider triangle ACD,
S and R are mid-points of sides AD and DC respectively.

SR\parallel AC and SR = {1 \over 2}AC                            ….(i)

[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]
(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.

∴  PQ\parallel AC  and   PQ= {1 \over 2}AC                                  ….(ii)
[Reason same as above]

From (i) and (ii),

SR\parallel AC   and PQ\parallel AC     ⇒    SR\parallel PQ     …(iii)

and SR = {1 \over 2} AC and PQ ={1 \over 2} AC   ⇒  SR = PQ.          …(iv)
(iii) SR\parallel PQ   and SR = PQ.                                                    [From (iii) and (iv)]

⇒  PQRS is a parallelogram.


2. ABCD is a rhombus and P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

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Sol. First prove that PQRS is a parallelogram.
(i) Consider triangle ACD,

S and R are mid-points of sides AD and DC respectively.

∴  SR\parallel AC    and    SR = {1 \over 2}AC
[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]
(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.
∴   PQ\parallel AC    and    PQ = {1 \over 2}AC        …(ii) [Reason same as above]
From (i) and (ii),
SR\parallel PQ   and  PQ\parallel AC     ⇒  SR\parallel PQ      …(iii)

and   SR = {1 \over 2}AC and  PQ = {1 \over 2}AC   ⇒   SR = PQ.     …(iv)
(iii) SR\parallel PQ   and    SR = PQ.                                              [From (iii) and (iv)]


⇒  PQRS is a parallelogram.

PQRS is a parallelogram.

As  PX\parallel YO  and   PY\parallel OX is a parallelogram.

⇒     ∠YPX = ∠YOX = 90°      [Diagonals of a rhombus bisect each other and are at right angles.]|

As in parallelogram PQRS,

∠SPQ    is 90°.

∴  PQRS is a rectangle.


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.

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Sol. Construction:
Join AC and BD.
As ABCD is a rectangle.


∴  AC = BD                                                              …(i)

Consider ΔABC, P and Q are midpoints of sides AB and BC respectively.

∴      PQ\parallel AC and PQ = {1 \over 2}AC

Similarly, consider AADC, S and R are mid-points of sides AD and DC respectively.
∴   SR\parallel AC   and   SR = {1 \over 2}AC                              ….(iii)
From (ii) and (iii)

PQ = SR = {1 \over 2}AC                                                                  ……(iv)

Similarly, we can show

PS = QR = {1 \over 2}BD

From (i), (iv) and (v), we have PQ = QR .= RS = SP

∴   PQRS is a rhombus.


4. ABCD is a trapezium in which 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 (see figure). Show that F is the mid-point of BC.

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Sol. Consider  ΔADB, AB\parallel EF  ⇒  AB\parallel EG


⇒    G is mid-point of BD.                                                                                  …(i)

[ A line drawn through mid-point of one side, parallel to other bisects the third side.]

Consider triangle BCD,

AB\parallel CD   and   EF\parallel AB

⇒     EF\parallel CD  ⇒   GF\parallel CD


⇒      F is mid-point of BC.                                 [Reason same as above]


5. 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.


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Sol.  AB = CD \Rightarrow {1 \over 2}AB = {1 \over 2}CD

⇒  AE = CF

As AE = CF and   AE = CF                                                  [ AB = CD]

⇒   AECF is a parallelogram.

⇒   AP\parallel CE                                                                    …(i)

Consider triangle ABP,

E is mid-point of AB and EQ\parallel AP                       [From (i)]

⇒  Q is mid-point of BP [A line segment drawn through mid-point of one side of a triangle and parallel to other, bisects the third side.]

BQ = PQ                                                                                           …(ii)

Similarly, by considering triangle DCQ and proceeding as

above, we can show that

DP = PQ                                                                                            …(iii)

⇒     BQ = PQ = DP                                                                         [From (ii) and (iii)]

⇒     P and Q trisect BD.


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

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Sol.
(i) Consider triangle ACD,

S and R are mid-points of sides AD and DC respectively.

∴   SR\parallel AC    and SR  = {1 \over 2}                       …(i)

[Line segment joining mid-points of two sides of a triangle is parallel to the third and half of it.]
(ii) Consider triangle ABC, P and Q are mid-points of sides AB and BC respectively.

∴  PQ = AC and PQ  = {1 \over 2}AC                                     [Reason same as above]
From (i) and (ii),

SR\parallel AC  and PQ\parallel AC  ⇒   SR\parallel PQ                      …(iii)

and SR = {1 \over 2}AC    and    PQ = {1 \over 2}AC  ⇒   SR = PQ               . …(iv)
(iii) SR\parallel PQ  and SR = PQ.                                    [From (iii) and (iv)]

⇒  PQRS is a parallelogram.


We know that diagonals of a parallelogram bisect each other, i.e., OP = OR and OQ = OS.
Hence, line segments joining midpoints of opposite sides of a quadrilateral bisect each other.


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 = {1 \over 2}  AB.

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Sol.
(i) MD\parallel BC    meets AC at D.

∴  D is mid-point of AC.


[A line through the mid-point of a ID side of a triangle parallel to other bisects and third side.]

(ii) MD\parallel BC  and AC is transversal.

∴   ∠ADM =  ∠ACB.                           [ Corresponding angles ]

⇒   ∠ADM  =  90°                                 [ ∠ACB = 90° ]

⇒    MD  ⊥  AC.

(iii) Consider triangles ADM and CDM,
AD = DC                                                               [ From result (i) ]

MD is common.

∠ADM = ∠CDM                                   [90° each]                   [From result (ii)]

∴  ΔADM = ΔCDM                                              [SAS]

∴   MA = CM = {1 \over 2}AB     [ M is mid-point of AB]


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