NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2
NCERT Solutions for Class
9 Maths Chapter 8 Quadrilaterals Ex 8.2 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 8 Quadrilaterals Ex 8.2.
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2
Ex 8.2 Class 9 Maths 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 figure). AC is a
diagonal. Show that
(i)
SR || AC and SR = ½ AC
(ii)
PQ = SR
(iii)
PQRS is a parallelogram.
Solution:
(i) In ∆ACD, we have
∴ S is the mid-point of AD and R is the mid-point of CD.
SR = ½ AC and SR || AC ….…(1) [By mid-point theorem]
(ii) In ∆ABC, P is the mid-point of AB and Q is the mid-point of BC.
PQ = ½ AC and PQ || AC ….…(2) [By mid-point theorem]
From (1) and (2), we get
⇒ PQ = SR and PQ || SR
(iii) In a quadrilateral PQRS,
PQ = SR and PQ || SR [Proved
in (ii)]
∴ PQRS is a parallelogram.
Ex 8.2 Class 9 Maths 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.
To prove: Quadrilateral PQRS is a rectangle.
Proof: By mid-point theorem.
In ΔADC, we have S and R are the mid-points of AD and CD, respectively.
∴ SR || AC and SR = ½ AC …………..(i)
In ΔABC, we have P and Q are the mid-points of AB and BC, respectively.
∴ PQ || AC and PQ = ½ AC ……..…..(ii)
From eqs. (i) and (ii), we get PQ || SR and PQ = SR = ½ AC
Since, a pair of opposite sides of a quadrilaterals PQRS is equal and parallel.
So, PQRS is a parallelogram.
We know that diagonals of a rhombus bisect
each other at right angles.
∠COD = ∠EOF = 90°
Now, in ABCD, R and Q are the mid-points of CD
and BC, respectively.
RQ || DB [by mid-point
theorem]
⇒ RE || OF
Also, SR || AC [from eq. (i)]
⇒ FR || OE
So, OERF is a parallelogram.
∴ ∠ERF = ∠EOF = 90°
[Opposite angles of a parallelogram are equal]
Thus, PQRS is a parallelogram with ∠R = 90°
Hence, PQRS is a rectangle.
Hence proved.
Ex 8.2 Class 9 Maths 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 ΔASP and ΔBQP,
AP = BP (Given)
AS = BQ (Given)
∠A = ∠B (Each 90°)
∴ ΔASP ≅ ΔBQP (By SAS
congruency rule)
∴ SP = PQ (By CPCT) ……(i)
In ΔPBQ and ΔRCQ,
PB = RC (Given)
BQ = CQ (Given)
∠B = ∠C (Each 90°)
∴ ΔPBQ ≅ ΔRCQ (By SAS
congruency rule)
∴ PQ = RQ (By CPCT) ……(ii)
Similarly, we can prove that RQ = SR and SR = SP (iii)
From eqs. (i), (ii) and (iii), we
get SP = PQ = QR = RS.
Therefore, PQRS is a rhombus.
Ex 8.2 Class 9 Maths 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 figure). Show that F is the mid-point of BC.
Solution.
Given: ABCD is a trapezium in which AB || DC, BD is a diagonal and
E is the mid-point of AD and a line is drawn through E parallel to AB
intersecting BC at F such that EF || AB.
To prove: F is the mid-point of BC.
Proof: Let EF intersects BD at P.
In ΔABD, we have EP || AB [∵ EF || AB] and E is the mid-point of AD.
So, by the converse of mid-point theorem, P is
the mid-point of BD.
Similarly, in ΔBCD, we have PF || CD [∵ EF || AB and AB || CD]
and P is the mid-point of BD.
So, by the converse of mid-point
theorem, F is the mid-point of BC.
Ex 8.2 Class 9 Maths Question
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.
Solution.
Given: ABCD is a parallelogram, and E and F are the mid-points of sides AB and CD, respectively.
To prove: Line segments AF and EC trisect the diagonal BD.
Proof: Since ABCD is a parallelogram.
∴ AB || DC and AB = DC [opposite sides of a parallelogram]
⇒ AE || FC and ½ AB = ½ DC
⇒ AE || FC and AE = FC [∵ E and F are the mid-points of AB and CD]
Since a pair of opposite sides of a
quadrilaterals AECF is equal and parallel.
So, AECF is a parallelogram.
Then, AF || EC
⇒ AP || EQ and FP || CQ [Since opposite sides of a parallelogram
are parallel]
In ΔBAP, E is the mid-point of AB and EQ || AP,
so Q is the mid-point of BP.
[by the converse of mid-point theorem]
∴ BQ = PQ ……(i)
Again, in ΔDQC, F is the mid-point of DC and FP
|| CQ.
So, P is the mid-point of
DQ. [by the converse of
mid-point theorem]
∴ QP = DP ……(ii)
From eqs. (i) and (ii), we get BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD.
Hence proved.
Ex 8.2 Class 9 Maths Question
6.
Show that the line segments joining
the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution.
i.e., AS = DS, AP = BP, BQ = CQ and CR = DR
We have to show that PR and SQ bisect each other,
i.e., SO = OQ and PO = OR
Now, in ΔADC, S and R are the mid-points of AD and CD, respectively.
We know that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.
∴ SR || AC and SR = ½ AC ……(i) [by mid-point theorem]
Similarly, in ΔABC, P and Q are mid-points of AB and BC, respectively.
PQ || AC and PQ = ½ AC ……(ii) [by mid-point theorem]
From eqs. (i) and (ii), we get
PQ || SR and PQ = SR = ½ AC
So, PQRS is a parallelogram whose diagonals are SQ
and PR.
We know that diagonals of a parallelogram bisect
each other.
So, SQ and PR bisect each other at O, i.e., SO = OQ
and PO = OR.
Hence proved.
Ex
8.2 Class 9 Maths 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.
∠C = 90° and M is the mid-point of AB. Also, DM || BC
(i) In ΔABC, BC || MD and M is the mid-point
of AB.
∴ D is the mid-point of AC. (By converse of mid-point theorem)
(ii)
Since MD || BC and CD is
transversal.
.
∠ADM = ∠ACB (Corresponding
angles)
But ∠ACB = 90°
⇒ ∠ADM = 90°
⇒ MD ⊥ AC
(iii)
Now, in ΔADM and ΔCDM, we have
DM = MD
(Common)
AD = CD
(D is the mid-point of AC)
∠ADM = ∠CDM (Each 90°)
Therefore, ΔADM ≅ ΔCDM
∴ CM = AM …….(i)
Also, M is the mid-point of AB.
AM = BM = ½ AB
…….(ii)
From eqs. (i) and (ii), we get
CM = AM = ½ AB
Hence proved.
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