Quadrilaterals
Chapter 8
Exercise 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 Fig 8.29). AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
(i) In ∆ACD,
∴ S is the mid-point of DA and
R is the mid-point of DC.
Hene, SR || AC and SR = ½
(ii) In ∆ABC,
P is the mid-point of AB and Q is the mid-point of BC.
PQ = ½
From (1) and (2), we get
PQ = ½
⇒ PQ = SR and PQ || SR
(iii) In a quadrilateral PQRS,
PQ = SR and PQ || SR [Proved]
∴ PQRS is a parallelogram.
EX 8.2 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:
EX 8.2 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:
In ∆ABC, P is a midpoint of AB and Q is a midpoint of BC.
PQ = ½
In ∆ACD, S is a midpoint of AD and R is a midpoint of CD.
SR =½AC and SR || AC …(2)
From (1) and (2),
PQ = SR and PQ || SR
∴ PQRS is a parallelogram.
Now, in ∆BCD, Q is a midpoint of BC and R is a midpoint of CD.
QR = ½AD …….(3) [mid-point theorem]
AC = BD………(4) [∵ Diagonals of a rectangle are equal]
From equation (1),(3) and (4)
⇒ PQ = QR
A parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.
EX 8.2 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 Fig. 8.30). Show that F is the mid-point of BC.
Solution:
In ∆ABD, E is the mid-point of AD and EG || AB
G is the mid-point of BD.[converse of the mid-point theorem]
Again In ∆BCD, G is the midpoint of BD and FG || DC.
F is the mid-point of BC. [converse of the mid-point theorem]
EX 8.2 QUESTION 5.
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
Since the opposite sides of a parallelogram are parallel and equal.
∴ AB || DC
⇒ AE || FC …(1)
and AB = DC
⇒½ AB = ½ CD
⇒ AE = CF …(2)
From (1) and (2),
AE || PC and AE = PC
∴ ∆ECF is a parallelogram.
Now, in ∆DQC, F is the mid-point of DC and FP || CQ
⇒ DP = PQ …(3) [By converse of mid-point theorem]
Similarly, In ∆BAP, E is the mid-point of AB and EQ || AP
⇒ BQ = PQ …(4) [By converse of mid-point theorem]
∴ From (3) and (4), we have
DP = PQ = BQ
So, the line segments AF and EC trisect the diagonal BD.
EX 8.2 QUESTION 6.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution:
Let ABCD be a quadrilateral and P, Q, R and S are the midpoints of AB, BC, CD and DA respectively.
In ∆ABC, P is a mid-point AB and Q is a mid-point BC.
∴ PQ || AC and PQ = ½
Similarly, RS || AC and RS = ½
From equation (1) and (2),
PQ || RS, PQ = RS
∴ PQRS is a parallelogram and diagonals of a parallelogram bisect each other
PR and SQ bisect each other.
EX 8.2 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 = ½ AB
Solution:
(i) In ∆ABC,
M is the mid-point of AB. [Given]
DM || BC , [Given]
D is the mid-point of AC. [Converse of mid-point theorem,]
(ii) ∠AMD = ∠ABC [Corresponding angles are equal] As
∠ADM = 90° [Given]
⇒ MD ⊥AC.
(iii) In ∆AMD and ∆CMD,
∠AMD = ∠CMD [Each equal to 90°]
DM = DM [Common]
∴ ∆ADM ≅ ∆CDM [SAS congruency]
⇒ AM = CM [C.P.C.T.] .. .(1)
AM = ½
From (1) and (2),
CM = AM = ½