NCERT Answers for Class 9 Mathematics Chapter 8: Exercise 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 = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram
Solution:
Given that, P, Q, R and S are the mid points of quadrilateral ABCD
Theorem 8.9: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
(i) So here, taking ∆ACD
we can see S and R are the mid points of side AD and DC respectively. [Given]
Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)……………………….(1)
(ii) So here, taking ∆ACB
we can see P and Q are the mid points of side AB and BC respectively. [Given]
Hence, PQ || AC and PQ = ½ AC (NCERT Theorem 8.9)…………………………(2)
From (1) and (2) we can say,
PQ = SR
(iii) so from (i) and (ii) we can say that
PQ || AC and SR || AC
so, PQ || SR and PQ = SR
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Hence, PQRS is a parallelogram.
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 that, P, Q, R and S are the mid points of Rhombus ABCD.
Theorem 8.9: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Construction: Join AC and BD
So here, taking ∆ABD
We can see P and S are the mid points of side AB and AD respectively. [Given]
Hence, PS || BD and PS = ½ BD (NCERT Theorem 8.9)……………………….(1)
Similarly, is we take ∆CBD
We can see R and Q are the mid points of side CD and CB respectively. [Given]
Hence, RQ || BD and RQ = ½ BD (NCERT Theorem 8.9)……………………….(2)
So from (1) and (2), we conclude that
PS || RQ and PS = RQ
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Hence, PQRS is a parallelogram.
Now in ∆ACD
We can see S and R are the mid points of side AD and CD respectively. [Given]
Hence, SR || AC and RS = ½ AC (NCERT Theorem 8.9)
from (2) RQ || BD and RQ = ½ BD (NCERT Theorem 8.9)
Hence, OGSH is a parallelogram.
∠HOG = 90° (Diagonal of rhombus intersect at 90°)
So ∠HSG = 90° (opposite angle of a parallelogram are equal)
As, PQRS is a parallelogram having vertices angles equal to 90°.
Hence, PQRS is a Rectangle.
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 that, P, Q, R and S are the mid points of Rectangle ABCD.
Construction: Join AC
Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
So here, taking ∆ACD
we can see S and R are the mid points of side AD and DC respectively. [Given]
Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)……………………….(1)
Now, taking ∆ACB
we can see P and Q are the mid points of side AB and BC respectively. [Given]
Hence, PQ || AC and PQ = ½ AC (NCERT Theorem 8.9)…………………………(2)
So from (1) and (2), we conclude that
SR || PQ and SR = PQ
Hence, PQRS is a parallelogram. (If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram)
Now in ∆QBP and ∆QCR
QC = QB (Q is the mid point of BC)
RC = PB (opposite sides are equal, hence half length is also equal)
∠QCR = ∠QBP (Each 90°)
∆QBP ≅ ∆QCR (By SAS congruency)
QR = QP (By C.P.C.T.)
As PQRS is a parallelogram and having adjacent sides equal
Hence, PQRS is a rhombus
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:
Let O be the point of intersection of lines BD and EF
Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
So here, taking ∆ADB
we can see S is the mid point of side AD and ED || AB [Given]
Hence, OD = ½ BD ………..(NCERT Theorem 8.10)
Now, taking ∆BCD
we can see O is the mid point of side BD and OF || AB [Proved and Given]
Hence, CF = ½ BC…….. (NCERT Theorem 8.10)
Hence proved!!
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:
Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
Given, E and F are the mid points of side AB and CD of parallelogram ABCD.
As, AB || CD and AB = CD (opposite sides of parallelogram)………..(1)
AE = CF (halves of opposite sides of parallelogram)………………………(2)
from (1) and (2)
AECF is a parallelogram
Hence, AF || EC
Now taking ∆APB
we can see E is the mid point of side AB and EF || AP [Given and proved]
Hence, BQ = PQ………..(NCERT Theorem 8.10)………………….(1)
Now taking ∆CQD
we can see F is the mid point of side CD and CQ || FP [Given and proved]
Hence, DP = PQ………..(NCERT Theorem 8.10)………………..(2)
From (1) and (2) we conclude that,
BQ = PQ = DQ
Hence, we can say that line segments AF and EC trisect the diagonal BD
Question 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution:
Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
So here, taking ∆ACD
we can see S and R are the mid points of side AD and DC respectively. [Given]
Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)……………………….(1)
Now, taking ∆ACB
we can see P and Q are the mid points of side AB and BC respectively. [Given]
Hence, PQ || AC and PQ = ½ AC (NCERT Theorem 8.9)…………………………(2)
So from (1) and (2), we conclude that
SR || PQ and SR = PQ
Hence, PQRS is a parallelogram. (If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram)
And since the diagonal of parallelogram bisects each other
so, QS and PR bisects each other.
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:
Theorem 8.10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
(i) while taking ∆ABC
we can see M is the mid point of side AB and DM || BC [Given]
This implies, DC= AD ……….. (NCERT Theorem 8.10)
Hence, D is the mid-point of AC.
(ii) As we know MD || BC and AC is transversal
This implies, ∠ACB = ∠ADM = 90°
Hence, MD ⊥AC
(iii) Considering ∆ADM and ∆CDM
AD = CD (D is the mid point of AC (Proved))
∠CDM = ∠ADM (proved, MD ⊥AC)
DM = DM (common)
∆ADM ≅ ∆CDM (By SAS congruency)
CM = AM (By C.P.C.T.)
CM = AM = ½ AB (M is the mid point of AB)
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
In NCERT Class 9 Maths Chapter 8, you will find a concise and comprehensive explanation of the properties of special quadrilaterals, such as Parallelogram, Rhombus, Rectangle, Square, and Trapezium. The solutions provided by the experts at w3wiki cover the essential principles regarding angles, diagonals, and relations specific to each quadrilateral. These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals offer conceptual clarity, enabling students to grasp the topics effectively.
The Class 9 Maths Chapter 8 focuses on the Quadrilateral section. The significance of the properties of parallelograms is thoroughly discussed in the NCERT Solutions for Class 9 Maths Chapter 8. Section 4 of this chapter primarily covers the Mid-point theorem and its wide range of applications. By studying these topics, students can gain a better understanding and strengthen their mathematical skills.
NCERT Class 9 Maths Chapter 8 Quadrilaterals covers variety of topics such as:
Exercises under NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals |
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NCERT Maths Solutions Class 9 Exercise 8.1 – 12 Questions (2 Short Answers, 6 Long Answers, 4 Very Long Answers) |
NCERT Maths Solutions Class 9 Exercise 8.2 – 7 Questions (2 Short Answers, 3 Long Answers, 3 Very Long Answers) |