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