NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 9 |
| Subject | Maths |
| Chapter | Chapter 8 |
| Chapter Name | Quadrilaterals |
| Exercise | Ex 8.2 |
| Number of Questions Solved | 7 |
| Category | NCERT Solutions |
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 the mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that: D
(i) SR || AC and SR = \(\frac { 1 }{ 2 } \) AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution.
Given : P,Q,R, and S are the mid-points of the sides
AB, BC, CD, and DA, respectively.
AP = BP
BQ = CQ
CR = DR
and AS = DS
To show (i) SR || AC and SR =\(\frac { 1 }{ 2 } \) AC 2
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Proof:
(i) In Δ ADC, we have S is mid-point of AD and R is mid-point of DC.
We know that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
∴ SR || AC …….(i)
SR =\(\frac { 1 }{ 2 } \) AC ………….(ii)
(ii) Similarly, in Δ ABC, we have P is mid-point of AB and Q is the mid-point of BC.
∴ PQ || AC …………..(iii)
PQ = \(\frac { 1 }{ 2 } \) AC…………..(iv)
(iii) From eqs. (i) and (iii), we get
PQ || SI and from Eq. (v), PQ = SR
Since a pair of opposite sides of a quadrilateral PQRS is equal and parallel.
So, 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.
Given: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD, and DA, respectively.
To show: Quadrilateral PQRS is a rectangle.
Proof: By mid-point theorem.
In Δ ADC, we have S and R are the mid-points of DA and CD, respectively.
∴ SR || AC
SR = \(\frac { 1 }{ 2 } \) AC …………..(i)
In Δ ABC, we have P and Q are the mid-points of AB and BC, respectively.
∴ PQ || AC and PQ = \(\frac { 1 }{ 2 } \) AC …..(ii)
From eqs. (i) and (ii), we get PQ || SR and PQ =SR = \(\frac { 1 }{ 2 } \) 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.
Given, ABCD is a rectangle.
∴ ∠A=∠B = ∠C =∠D = 90° and AD = BC, AB = CD
Also, given P,Q,R and S are mid-points of AB, BC, CD and DA respectively.
∴ PQ || AC and PQ = \(\frac { 1 }{ 2 } \) AC ……(i)
AP = BP (Given)
AS= BQ (Given)
∠A= ∠B (Given)
∴ Δ ASP ≅ Δ BQP (By SAS)
∴ SP = PQ (By CPCT) …(iv)
SD = CQ (Given)
DR = RC (Given)
∠C = ∠D (Given)
Δ RDS = Δ RCQ (By SAS)
SR = RQ (By CPCT) …(v)
From Eqs. (iii), (iv) and (v), it is clear that quadrilateral 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 fig.). 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 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 Δ ABD, we have EP || AB [∵ EF || AB] and E is mid-point of AD.
So, by the converse of mid-point theorem.
We get, P is mid-point of BD.
Similarly, in ABCD, we have PF || CD [∵ EF || AB and AB || CD]
and P is mid-point of BD.
So, by the converse of mid-point theorem, F is 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 fig.). Show that the line segments AF and EC trisect the diagonal BD.
Solution.
Given: ABCD is a parallelogram and E, 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]
⇒ AF || FC and \(\frac { 1 }{ 2 } \) AP = \(\frac { 1 }{ 2 } \) DC
⇒ AF || FC and AF = 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 || FC
⇒ AP || FQ and FP || CQ
Since opposite sides of a parallelogram are parallel.
In Δ BAP, 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 Δ DQC, 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.
Let ABCD be a quadrilateral and P, Q, R and S be the D mid-points of the sides AB, BC, CD, and DA, respectively.
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 Δ ADC, S and R are 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 = \(\frac { 1 }{ 2 } \)AC …(i) [by mid-point theorem]
Similarly, in Δ ABC, P and Q are mid-points of AB and BC, respectively.
PQ || AC and PQ = \(\frac { 1 }{ 2 } \)AC [by mid-point theorem]
From Eqs. (i) and (ii), we get
PQ || SP and PQ = SR = \(\frac { 1 }{ 2 } \)AC
So, PQRS is a parallelogram whose diagonals are SQ and also, 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.
[Since a pair of opposite sides of a quadrilaterals PQRS is equal and parallel.]
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
CM = MA = \(\frac { 1 }{ 2 } \)AB
Solution.
Given, ABC is a right angled triangle.
∠C = 90° and M is the mid-point of AB.
Also, DM || BC
(i) In Δ ABC, BC || MD and Mis mid-point of AB.
∴ D is the mid-point of AC. (By converse of mid-point theorem)
Since, MD || BC and CD is transversal. .
∠ADM = ∠ACB (Corresponding angles)
But ∠ACB = 90°
∠ADM = 90° => MD ⊥ AC
Now, in AADM and A CDM, we have
DM = MD
AD = CD
∠ADM = ∠MDC
AADM = ACDM
CM = AM
Also, M is mid-point of AB.
AM = BM = \(\frac { 1 }{ 2 } \) AB
From Eq. (i) and (ii), we get
CM = AM = \(\frac { 1 }{ 2 } \) AB
Hence proved.
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