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NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

Exercise 8.1

Q1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Ans: 

NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

Given that,
AC = BD
To Prove:  ABCD is a rectangle if the diagonals of a parallelogram are equal.
To show ABCD is a rectangle, we have to prove that one of its interior angles is right-angled.
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
BC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD [SSS congruency]
∠A = ∠B [Corresponding parts of Congruent Triangles]
also,
∠A+∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Therefore, ABCD is a rectangle.
Hence Proved.


Q2. Show that the diagonals of a square are equal and bisect each other at right angles.
Ans: Given: A square is given.
To Prove: The diagonals of a square are the same and bisect each other at 90
Proof: Consider ABCD to be a square.
NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

Consider the diagonals AC and BD intersect each other at a point O.
We have to show that the diagonals of a square are equal and bisect each other at right angles,
AC  =  BD, OA  =  OC, OB  =  OD .
In ΔABC and ΔDCB,
AB = DC (Sides of the square are equal)
∠ABC = ∠DCB (All the interior angles are of the value 90o)
BC = CB (Common side)
∴ ΔABC ≅ ΔDCB (By SAS congruency)
∴ AC= DB (By CPCT)
Hence, the diagonals of a square are equal in length.
In ΔAOB and ΔCOD,
∠AOB = ∠COD (Vertically opposite angles)
∠ABO = ∠CDO (Alternate interior angles)
AB = CD (Sides of a square are always equal)
∴ ΔAOB ≅ ΔCOD (By AAS congruence rule)
∴ AO = CO and OB = OD (By CPCT)
As a result, the diagonals of a square are bisected.
In ΔAOB and ΔCOB,
Because we already established that diagonals intersect each other,
AO = CO
AB = CB (Sides of a square are equal)
BO = BO (Common)
∴ ΔAOB ≅ ΔCOB (By SSS congruency)
∴ ∠AOB = ∠COB (By CPCT)
However, ∠AOB + ∠COB = 180° (Linear pair)
∠AOB +∠AOB = 180°
2 x ∠AOB = 180°
∠AOB = 90°

As a result, the diagonals of a square are at right angles to each other.


Q3. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig.). Show that
NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

Show that
(i) It is bisecting C also, 
(ii) ABCD is a rhombus

Ans: Given: Diagonal AC of a parallelogram ABCD is bisecting A
(i) ABCD is a parallelogram.
DAC = BCA (Alternate interior angles) ... (1)
And BAC = DCA (Alternate interior angles) ... (2)
However, it is given that AC is bisecting A
DAC = BAC ... (3)
Putting the values of eq (1) and eq (2) in eq (3)
BCA = DCA
Hence, AC is bisecting C
(ii) From Equation (4), we obtain
DAC = DCA
DA = DC (Side opposite to equal angles are equal)
However, DA = BC and AB = CD (Opposite sides of a parallelogram)
AB = BC = CD = DA
As a result, ABCD is a rhombus.


Q4. ABCD is a rectangle in which diagonal AC bisects A as well as C . Show that:
(i) ABCD is a square 
(ii) Diagonal BD bisects 
B as well as D.
Ans: Given: ABCD is a rectangle where the diagonal AC bisects A as well as C.
NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

(i) It is given that ABCD is a square.
A = C
⇒ 1/2 A = 1/2 C (AC bisects A and C)
DAC = 1/2 ∠DCA
CD = DA (Sides that are opposite to the equal angles are also equal)
Also, DA = BC and AB = CD (Opposite sides of the rectangle are same)
AB = BC = CD = DA
ABCD is a rectangle with equal sides on all sides.
Hence, ABCD is a square.
(ii) Let us now join BD.
In ΔBCD,
BC = CD (Sides of a square are equal to each other)
CDB = CBD (Angles opposite to equal sides are equal)
However, CDB = ABD (Alternate interior angles for AB || CD)
CBD = ABD
BD bisects B.
Also, CBD = ADB (Alternate interior angles for BC || AD)
CDB = ABD
BD bisects D and B.


Q5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure).
NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals
Show that: 
(i) ΔAPD ≅ ΔCQB 

(ii) AP = CQ 

(iii) ΔAQB ≅ ΔCPD 

(iv) AQ = CP

(v) APCQ is a Parallelogram
Ans: 
(i) In ΔAPD and ΔCQB,
AD = CB (Opposite sides of the parallelogram ABCD)
ADP = CBQ (Alternate interior angles for BC || AD)
DP = BQ (Given)
∴ ΔAPD  ΔCQB (Using SAS congruence rule)
(ii) As we had observed that ΔAPD  ΔCQB,
∴ AP= CQ (CPCT)
(iii) In ΔAQB and ΔCPD,
ABQ = CDP (Alternate interior angles for AB || CD )
AB = CD (Opposite sides of parallelogram ABCD)
BQ = DP (Given)
∴ ΔAQB  ΔCPD (Using SAS congruence rule)
(iv) Since we had observed that ΔAQB  ΔCPD,
∴ AQ = CP (CPCT)
(v) From the results obtained in (ii) and (iv),
AQ = CP and AP = CQ
APCQ is a parallelogram because the opposite sides of the quadrilateral are equal.


Q6. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure).
NCERT Solutions for Class 9 Maths Chapter 8 - QuadrilateralsShow that 
(i) ΔAPB ≅ ΔCQD 
(ii) AP = CQ
Ans: 
(i) In ΔAPB and ΔCQD,
APB = CQD (Each 90°)
AB = CD (The opposite sides of a parallelogram ABCD)
ABP = CDQ (Alternate interior angles for AB || CD)
ΔAPB ΔCQD (By AAS congruency)
(ii) By using
ΔAPB ΔCQD , we obtain
AP = CQ (By CPCT)


Q7. ABCD is a trapezium in which AB || CD and AD = BC (see the given figure).
NCERT Solutions for Class 9 Maths Chapter 8 - QuadrilateralsShow that 
(i) ∠A = ∠B 
(ii) ∠C = ∠D 
(iii) 
ΔABC  ΔBAD 
(iv) diagonal AC = diagonal BD
(Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.)

Ans: Let us extend AB by drawing a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram.
(i) AD = CE (Opposite sides of parallelogram AECD)
However, AD = BC (Given)
Therefore, BC = CE
∠CEB = ∠CBE (Angle opposite to the equal sides are also equal)
Considering parallel lines AD and CE.
AE is the transversal line for them (Angles on a same side of transversal)
(Using the relation ∠CEB = ∠CBE) ... (1)
However,  ∠CBE +  ∠CBA  = 180° (Linear pair angles) ... (2)
From Equations (1) and (2), we obtain ∠A = ∠B
(ii) AB || CD
Also, ∠C + ∠B = 180° (Angles on a same side of a transversal)
∴ ∠A + ∠D = ∠C + ∠B
However, ∠A = ∠B (Using the result obtained in (i))
∴ ∠C = ∠D
(iii) In ΔABC and ΔBAD,
AB = BA (Common side)
BC = AD (Given)
∠B = ∠A (Proved before)
ΔABC ΔBAD (SAS congruence rule)
(iv) We had seen that, ΔABC ΔBAD
∴ AC= BD (By CPCT)

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 the given figure). AC is diagonal. Show that:
(i) SR || AC and SR = (1/2) AC
(ii) PQ = SR 
(iii) PQRS is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

Ans: Given: ABCD is a quadrilateral
To prove: (i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
(i) In ΔADC, S and R are the mid-points of sides AD and CD respectively.
In a triangle, the line segment connecting the midpoints of any two sides is parallel to and half of the third side.
∴ SR || AC and SR = 1/2 AC ... (1)
(ii) In ΔABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using midpoint theorem,
PQ || AC and  PQ = 1/2 AC ... (2)
Using Equations (1) and (2), we obtain
PQ || SR and PQ = SR
(iii) From Equation (3), we obtained
PQ || SR and PQ = SR
Clearly, one pair of quadrilateral PQRS opposing sides is parallel and equal. PQRS is thus 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.
Ans: 
NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

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 Prove: PQRS is a rectangle.
Construction:
Join AC and BD.
Proof:
In ΔDRS and ΔBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
ΔDRS ≅ ΔBPQ [SAS congruency]
RS = PQ [CPCT]———————- (i)
In ΔQCR and ΔSAP,
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
ΔQCR ≅ ΔSAP [SAS congruency]
RQ = SP [CPCT]———————- (ii)
Now,
In ΔCDB,
R and Q are the mid points of CD and BC, respectively.
⇒ QR || BD
also,
P and S are the mid points of AD and AB, respectively.
⇒ PS || BD
⇒ QR || PS
PQRS is a parallelogram.
also, ∠PQR = 90°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
PQRS is a rectangle.


Q3. 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.
Ans: 

NCERT Solutions for Class 9 Maths Chapter 8 - QuadrilateralsGiven: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
To prove: The quadrilateral PQRS is a rhombus.
Proof: Let us join AC and BD.
In ΔABC, P and Q are the mid-points of AB and BC respectively.
∴ PQ || AC and PQ = 1/2 AC (Mid-point theorem) ... (1)
Similarly, in ΔADC , SR ||  AR, SR = 1/2 AC (Mid-point theorem) ... (2)
Clearly,  PQ || SR and  PQ = SR
It is a parallelogram because one pair of opposing sides of quadrilateral PQRS is equal and parallel to each other.
∴ PS || QR , PS = QR (Opposite sides of parallelogram) ... (3)
In ΔBCD, Q and R are the mid-points of side BC and CD respectively.
∴ QR || BD, QR = 1/2 BD (Mid-point theorem) ... (4)
Also, the diagonals of a rectangle are equal.
∴ AC = BD ... (5)
By using Equations (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
So, 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 the given figure). Show that F is the mid-point of BC.

NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

Ans: Given: 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.
To prove: F is the mid-point of BC.
Proof: Let EF intersect DB at G.
We know that a line traced through the mid-point of any side of a triangle and parallel to another side bisects the third side by the reverse of the mid-point theorem.
In  ΔABD, EF || AB and E is the mid-point of AD.
Hence, G will be the mid-point of DB.
As  EF || AB, AB || CD,
∴ EF || CD (Two lines parallel to the same line are parallel)
In  ΔBCD, GF || CD and G is the mid-point of line BD. 
So, by using the converse of mid-point theorem, 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 the given figure). Show that the line segments AF and EC trisect the diagonal BD.
Ans: Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively to prove: The line segments AF and EC trisect the diagonal BD.

NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

To Prove: The line segments AF and EC trisect the diagonal BD. 

Proof: ABCD is a parallelogram.
AB || CD
And hence, AE || FC
Again, AB = CD (Opposite sides of parallelogram ABCD)
1/2 AB = 1/2 CD
AE = FC (E and F are mid-points of side AB and CD)
In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and the same as each other. So, AECF is a parallelogram.
∴ AF || EC (Opposite sides of a parallelogram)
In  ΔDQC, F is the mid-point of side DC and  FP || CQ (as  AF || EC ). 
So, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.
∴ DP= PQ ... (1)
Similarly, in ΔAPB , E is the mid-point of side AB and EQ || AP (as  AF || EC ).
As a result, the reverse of the mid-point theorem may be used to say that Q is the mid-point of PB.
∴ PQ = QB ... (2)
From Equations (1) and (2),
DP = PQ= BQ
Hence, the line segments AF and EC trisect the diagonal BD.


Q6. 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 = 1/2 AB
NCERT Solutions for Class 9 Maths Chapter 8 - QuadrilateralsAns: Given: 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.
(i) In ΔABC,
It is given that M is the mid-point of AB and MD || BC.
Therefore, D is the mid-point of AC. (Converse of the mid-point theorem)
(ii) As DM || CB and AC is a transversal line for them, therefore,(Co-interior angles)
(iii) Join MC.
In ΔAMD and  ΔCMD,
AD = CD (D is the mid-point of side AC)
∠ADM = ∠CDM (Each)
DM = DM (Common)
∴ ΔAMD ≅ ΔCMD (By SAS congruence rule)
Therefore,
AM = CM (By CPCT)
However, 
AM = 1/2 AB (M is mid-point of AB)
Therefore, it is said that CM = AM = 1/2 AB.

The document NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

1. What are the properties of a quadrilateral?
Ans.A quadrilateral is a four-sided polygon, and its properties include having four angles that sum up to 360 degrees, opposite sides that may or may not be equal, and the possibility of having parallel sides. Specific types of quadrilaterals, like rectangles and parallelograms, have additional properties such as equal opposite sides and angles.
2. How do you calculate the area of a quadrilateral?
Ans.The area of a quadrilateral can be calculated using various methods depending on the type of quadrilateral. For a rectangle, the area is length multiplied by width. For a trapezium, the area can be found using the formula: Area = 1/2 * (sum of parallel sides) * height. For irregular quadrilaterals, dividing it into triangles or using the formula based on the lengths of the sides and the semi-perimeter may be necessary.
3. What is the difference between a square and a rectangle?
Ans.A square is a special type of rectangle where all four sides are equal in length and all angles are right angles. In contrast, a rectangle has opposite sides that are equal in length, but the lengths of adjacent sides can differ. Therefore, while all squares are rectangles, not all rectangles are squares.
4. Can a quadrilateral have more than one pair of parallel sides?
Ans.Yes, a quadrilateral can have more than one pair of parallel sides. For example, a parallelogram has two pairs of opposite sides that are equal and parallel. A rectangle and a rhombus are also types of parallelograms. In contrast, a trapezium has only one pair of parallel sides.
5. What is the significance of the diagonals in a quadrilateral?
Ans.The diagonals of a quadrilateral are important because they can provide valuable information about the shape's properties. For instance, in a parallelogram, the diagonals bisect each other. In a rectangle, the diagonals are equal in length. Understanding the properties of diagonals helps in solving problems related to the quadrilateral, such as finding area or determining congruence.
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