Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

EXERCISE 8.1

Write the correct answer in each of the following:

1. Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is
(a) 90º (b) 95º (c) 105º (d) 120º
Sol. Fourth angle of the quadrilateral
= 360° – (75 ° + 90° + 75°)
= 360° – 240°
= 120°
Hence, (d) is the correct answer.
2. A diagonal of a rectangle is inclined to one side of the rectangle at 25°.
The acute angle between the diagonals is
(a) 55° (b) 50° (c) 40° (d) 25°
Sol. ABCD is a rectangle in which
diagonal AC is inclined to one
side AB of the rectangle at an
angle of 25°.
Now, AC = BD
[∵ Diagonals of a
rectangle are equal]
1 1
∴ AC = BD
2 2
⇒ OA = OB
In ∆AOB, we have OA = OB
∴ ∠OBA = ∠OAB = 25°
∴ ∠AOB = 180° – (25° + 25°) = 130°
∠ AOB and ∠AOD form angles of a linear pair.
∴ ∠AOB + ∠AOD = 180°
⇒ ∠ AOD = 180° – 130° = 50°
Hence, the acute angle between the diagonals is 50°.
Therefore, (b) is the correct answer.
3. ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is
(a) 40° (b) 45° (c) 50° (d) 60°

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

Sol. ABCD is a rhombus such that

∠ACB = 40°.
We know that diagonals of a rhombus
bisect each other at right angles.
In right ∆BOC, we have
∠OBC = 180° – (∠BOC + ∠BCO)
= 180° – (90° + 40°) = 50°
∴ ∠DBC = ∠OBC = 50°
Now,
∠ADB = ∠DBC [Alt. int. ∠s]
∴ ∠ADB = 50° [∵ ∠DBC = 50°]
Hence, (c) is the correct answer.
4. The quadrilateral formed by joining the mid-points of the sides of a
quadrilateral PQRS, taken in a order, is a rectangle, if
(a) PQRS is a rectangle
(b) PQRS is a parallelogram
(c) diagonals of PQRS are perpendicular
(d) diagonals of PQRS are equal
Sol. If diagonals of PQRS are perpendicular.
Hence, (c) is the correct answer.
5. The quadrilateral formed by joining the mid-points of the sides of a
quadrilateral PQRS, taken in order, is a rhombus, if
(a) PQRS is a rhombus (b) PQRS is a parallelogram
(c) diagonals of PQRS are perpendicular
(d) diagonals of PQRS are equal
Sol. If diagonals of PQRS are equal.
Hence, (d) is the correct answer.
6. If angles A,B, C and D of the quadrilateral ABCD, taken in a order, are
in the ratio 3: 7: 6: 4, then ABCD is a
(a) rhombus (b) parallelogram
(c) trapezium (d) kite
Sol. As angles A, B, C and D of the quadrilateral ABCD, taken in order, are
in the ratio 3 : 7 : 6 : 4, so let the angles A, B, C and D be 3x, 7x, 6x and
4x.
Now, sum of the angles of a quadrilateral is 360°.
3x + 7x + 6x + 4x = 360°
⇒ 20x = 360° ⇒ x = 360° ÷ 20 = 18°
So, the angles A, B, C and D of quadrilateral ABCD are 3 × 18°,
7 × 18°, 6 × 18° and 4 × 18°

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

i.e., 54°, 126°, 108° and 72°

Now, AD and BC are two lines which are cut by a transversal CD such
that the sum of angles ∠C and ∠D on the same side of transversal is
∠C + ∠D = 108° + 72° = 180°
∴ AD ||| BC
So, ABCD is a quadrilateral in which one pair of opposite sides are
parallel. Hence, ABCD is a trapezium.
Hence, (c) is the correct answer.
7. If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other
at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S,
then PQRS is a D C
(a) rectangle (b) rhombus P

(c) parallelogram S Q
(d) quadrilateral whose
R
opposite angles are
A B
supplementary
Sol. PQRS is a quadrilateral whose opposite angles are supplementary.
Hence, (d) is the correct answer.
8. If APB and CQD are two parallel lines, then the bisectors of the angles
APQ, BPQ, CQP and PQD form
(a) a square (b) a rhombus
(c) a rectangle (d) any other parallelogram

Sol.

PNQM is a rectangle. Hence, (c) is the correct answer.

9. The figure obtained by joining the mid-points of the sides of a rhombus,
taken in order, is
(a) a rhombus (b) a rectangle
(c) a square (d) any parallelogram
Sol. The figure will be a rectangle.
Hence, (b) is the correct answer.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

10. D and E are the mid-points of the sides AB and AC of ∆ABC and O is
any point on side BC. O is joined to A. If P and Q are the mid-points of
OB and OC respectively, then DEQP is
(a) a square (b) a rectangle
(c) a rhombus (d) a parallelogram
Sol. Since the line segment
joining the mid-points
of any two sides of a
triangle is parallel to
the third side and is
half of it, so

1
∴ DE = BC and DE || BC
2
1
Similarly, DP = AO and DP || AO
2
1
and EQ = AO and EQ || AO
2
1
∴ DP = EQ [∵ Each = AO]
2
and DP || EQ [∵ DP || AO and EQ || AO]
Now, DEQP is a quadrilateral in which one pair of its opposite sides is
equal and parallel.
Therefore, quadrilateral DEQP is a parallelogram.
Hence, (d) is the correct answer.
11. The figure formed by joining the mid-points of the sides of a quadrilateral
ABCD, taken in order, is a square only, if
(a) ABCD is a rhombus (b) diagonals of ABCD are equal
(c) diagonals of ABCD are equal and perpendicular
(d) diagonals of ABCD are perpendicular.
Sol. If diagonals of ABCD are equal and perpendicular.
Hence, (c) is the correct answer.
12. The diagonals AC and BD of a parallelogram ABCD intersect each
other at the point O, if ∠DAC = 32° and ∠AOB = 70º then ∠DBC is
equal to
(a) 24° (b) 86° (c) 38° (d) 32°

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

Sol. AD is parallel to BC and AC cuts

them,
∠DAC = ∠ACB [Alt. int. ∠s]
∠DAC = 32° [Given]
∴ ∠ACB = 32°
Now, in ∆BOC, CO is
produced to A
∴ Ext. ∠BOA =∠OCB + ∠OBC [By exterior angle theorem]
⇒ 70° =32° + ∠OBC
∴ ∠OBC =70° – 32° = 38°
Hence, ∠DBC = 38°.
Therefore, (c) is the correct answer.
13. Which of the following is not true for a parallelogram?
(a) Opposite sides are equal.
(b) Opposite angles are equal.
(c) Opposite angles are bisected by the diagonals.
(d) Diagonals bisect each other.
Sol. Opposite angles are bisected by the diagonals. This is not true for a
parallelogram.
Hence, (c) is the correct answer.
14. D and E are the mid-points of the sides AB and AC respectively of
∆ABC. DE is produced to F. To prove that CF is equal and parallel to
DA, we need an additional information which is
(a) ∠DAE = ∠EFC A
(b) AE = EF
(c) DE = EF D E F
(d) ∠ADE = ∠ECF
Sol. We need DE = EF.
Hence, (c) is the correct answer. B C

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

EXERCISE 8.2

1. Diagonals AC and BD of a parallelogram ABCD intersect each other at

O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.
Sol. We know that diagonals of a parallelogram bisect each other.
∴ AC = 2 × OA = 2 × 3 cm = 6 cm
and BD = 2OD = 2 × 2 cm = 4 cm
Hence, lengths of AC and BD are 6 cm and 4 cm respectively.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

2. Diagonals of a parallelogram are perpendicular to each other. Is this

statement true? Give reason for your answer.
Sol. This statement not true. Diagonals of a parallelogram bisect each other.
3. Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral?
Why or why not?
Sol. Sum of these angles 110° + 80° + 70° + 95° = 355°
But, sum of the angles of a quadrilateral is always 360°. Hence,
110°, 80°, 70° and 95° can not be the angles of a quadrilateral.
4. In quadrilateral ABCD, ∠A + ∠D = 180°. What special name can be
given to this quadrilateral?
Sol. In quadrilateral ABCD, ∠A + ∠D = 180° i.e., the sum of two consecutive
angles is 180°. So, pair of opposite sides AB and CD are parallel.
Therefore, quadrilateral ABCD is a trapezium. Hence, special name
which can be given to this quadrilateral ABCD is trapezium.
5. All the angles of a quadrilateral are equal. What special name is given
to this quadrilateral?
Sol. All the angles of a quadrilateral are equal. Also, the sum of angles of a
quadrilateral is 360o. Therefore, each angle of a quadrilateral is 90o .
So, the given quadrilateral is a rectangle.
Hence, special name given to this quadrilateral is rectangle.
6. Diagonals of a rectangle are equal and perpendicular. Is this statement
true? Give reason for your answer.
Sol. The given statement is not true. Diagonals of a rectangle need not to be
perpendicular.
7. Can all the four angles of a quadrilateral be obtuse angles? Give reason
for your answer.
Sol. No, because then the sum of four angles of the quadrilateral will be
more than 360° whereas sum of four angles of a quadrilateral is always
equal to 360°.
8. In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are
respectively the mid-points of AB and BC, determine the length of DE.
Sol. In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7cm. D and E are respectively
the mid-points of AB and BC.
1 1
∴ DE= AC = × 7cm = 3.5 cm [Using the mid-point theorem]
2 2
9. In the given figure, it is given that BDEF and FDCE are parallelograms.
Can you say that BD = CD? Why or why not?

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

Sol. BDEF is a parallelogram. A

∴ BD = EF ...(1)
[Opposite sides of a parallelogram
are equal]
FDCE is a parallelogram F E
∴ CD = EF ...(2)
From (1) and (2), we get
BD = CD
10. In the given figure, ABCD and AEFG
B D C
are two parallelograms.

If ∠C = 55°, determine ∠F.

Sol. We know that opposite angles of a parallelogram are equal.
In parallelogram ABCD, we have
∠A = ∠C
But, ∠C = 55° [Given]
∴ ∠A = 55°
Now, in parallelogram AEFG, we have
∠F = ∠A = 55°
Hence, ∠F = 55°.
11. Can all the angles of a quadrilateral be acute angles? Give reason for
your answer.
Sol. We know that an acute angle is less than 90°. All the angles of a
quadrilateral cannot be acute angles because then angle sum of a
quadrilateral will be less than a 360°, whereas angle sum of a
quadrilateral is 360°.
12. Can all the angles of a quadrilateral be right angles? Give reason for
your answer.
Sol. Yes, all the angles of a quadrilateral can be right angles. Angle sum of a
quadrilateral will be 360°, which is true.
13. Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35°,
determine ∠B.
Sol. As the diagonals of a quadrilateral ABCD bisect each other, so ABCD
is a parallelogram.
Now, ABCD is a parallelogram
∴ ∠ A + ∠B = 180°
[∵ Adjacent angles of a parallelogram are supplementary]

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

∴ 35° + ∠B = 180°
⇒ ∠ B = 180° – 35° = 145°
14. Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm,
determine CD.
Sol. Since opposite angles of a quadrilateral are equal, so it is parallelogram.
Now, ABCD is a parallelogram, so AB = CD.
[∵ Opposite sides of a parallelogram are equal]
But, AB = 4 cm, therefore CD = 4 cm.
Hence, CD = 4 cm.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

EXERCISE 8.3

1. One angle of a quadrilateral is of 108° and the remaining three angles

are equal. Find each of the three equal angles.
Sol. One angle of a quadrilateral is of 108° and let each of the three remaining
equal angles be x°.
As the sum of the angles of a quadrilateral is 360°.
∴ 108° + x + x + x = 360° ⇒ 3x = 360° – 108° = 252°
∴ x = 252° ÷ 3 = 84°
Hence, each of three angles be 84°.
2. ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45°. Find
angles C and D of the trapezium.
Sol. ABCD is a trapezium in which AB || DC.

Now, AB || DC and AD is a transversal.

∴ ∠A + ∠D = 180°
..
[ . Sum of interior angles on the same side of the transversal is 180°]
⇒ 45° + ∠D = 180°
⇒ ∠D = 180° – 45° = 135°
Similarly, ∠B + ∠C = 180°
⇒ 45° + ∠C = 180°
⇒ ∠C = 180° – 45° = 135°
Hence, ∠A = ∠B = 45° and ∠C = ∠D = 135°.
3. The angle between two altitudes of a parallelogram through the vertex
of an obtuse angle of the parallelogram is 60º. Find the angles of the
parallelogram.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

Sol. In quadrilateral PBQD, D C

1
∠1 + ∠2 + ∠B + ∠3 = 360° 60º
⇒ 60° + 90° + ∠B + 90° = 360°
⇒ ∠B + 240° = 360° 3Q
⇒ ∠B = 360° – 240°
⇒ ∠B = 120° 2
A P B
Now, ∠ADC = ∠B = 120°
[∵ Opposite angles of a parallelogram are equal]
∠A + ∠B = 180°
[∵ Sum of consecutive interior angles is 180°]
⇒ ∠A + 120° = 180°
⇒ ∠A = 180° – 120°
⇒ ∠A = 60°
But, ∠C = ∠A = 60°
[∵ Opposite angles of a parallelogram are equal]
4. ABCD is a rhombus in which altitude from D to side AB bisects AB.
Find the angles of the rhombus.
Sol. In ∆APD and ∆BPD, we have D C
AP = BP [Given]
∠1 = ∠2 [∵ Each equal to 90°]
PD = PD [Common side]
So, by SAS criterion of congruence,
we have 1 2 3
4

∆ APD ≅ ∆BPD A B
∴ ∠A = ∠3 [CPCT]
But, ∠3 = ∠4 [∵ Diagonals bisect opposite
angles of a rhombus]
⇒ ∠A = ∠3 = ∠4 ... (1)
Now, AD || BC
So, ∠A + ∠ABC = 180° [∵ Sum of consecutive interior angles is 180°]
⇒ ∠A + ∠3 + ∠4 = 180°
⇒ ∠A + ∠A + ∠A = 180° [Using (1)]
⇒ 3∠A = 180°
180°
⇒ ∠A = = 60°
3
Now, ∠ABC = ∠3 + ∠4
= 60° + 60°
= 120° [∵ Opposite angles of a rhombus are equal]
∴ ∠ADC = ∠ABC = 120° [Same reason as above]

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

5. E and F are points on diagonal AC of a parallelogram ABCD such that

AE = CF. Show that BFDE is a parallelogram.
Sol. Given : A parallelogram ABCD: D C

E and F are points on diagonal AC of F

O
parallelogram ABCD such that
E
AE = CF.
To prove : BFDE is a parallelogram. A B
Proof: ABCD is a parallelogram.
∴ OD = OB … (1) [∵ Diagonals of parallelogram
bisect each other]
OA = OC … (2) [Same reason as above]
AE = CF … (3) [Given]
Subtracting (3) from (2), we get
OA – AE = OC – CF
⇒ OE = OF … (4)
∴ BFDE is a parallelogram. [∵ OD = OB and OE = OF]
Hence, proved.
6. E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A
line through E drawn parallel to AB intersect BC at F. Show that F is the
mid-point of BC. [Hint: Join AC]
Sol. Given : A trapezium ABCD in
D C
which AB || DC and E is the mid-
point of the side AB. Also, EF || E O F
AB.
To prove : F is the mid-point of
A B
BC.
Construction : Join AC which intersects EF at O.
Proof : In ∆ADC, E is the mid-point of AD and EF || DC.
[∵ EF || AB and DC || AB ⇒ AB || EF || DC]
∴ O is the mid-point of AC. [Converse of mid-point theorem]
Now, in ∆CAB, O is the mid-point of AC and OF || AB.
⇒ OF bisects BC.
Or, F is the mid-point of BC.
Hence, proved.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

7. Through A, B and C, lines RQ, PR R A Q

and QP have been drawn,
respectively parallel to sides BC, CA
and AB of a ∆ABC as shown in the
B C
1
given figure. Show that BC =QR.
2
Sol. Given : Triangles ABC and PQR in
which AB || QP, BC || RQ and
CA || PR. P

1
To prove : BC = QR
2
Proof : Quadrilateral RBCA is a parallelogram.
[∵ RA || BC and BR || CA]
∴ RA = BC ...(1) [∵ Opposite sides of a parallelogram]
Now, quadrilateral BCQA is a parallelogram.
∴ AQ = BC ...(2)[∵ Opposite sides of a parallelogram]
Adding (1) and (2), we get
RA+ AQ = 2BC
⇒ QR = 2BC
1
⇒ BC = QR
2
Hence, proved.

8. D, E and F are the mid-points of the sides BC, CA and AB, respectively
of an equilateral triangle ABC. Show that ∆DEF is also an equilateral
triangle.
Sol. Given : ∆ABC is an equilateral A
triangle. D, E and F are the mid-
points of the sides BC, CA and
AB, respectively of ∆ABC.
To prove : ∆DEF is an equilateral F E
triangle.
Proof : FE joins mid-points of
sides AB and AC respectively.
1 B D C
∴ F E = BC
2
… (1) [Mid-point theorem]
1
Similarly, D E = BC … (2) [Mid-point theorem]
2

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

1
DF = AC … (3) [Mid-point theorem]
2
But, AB = BC = CA … (4) [Sides of an equilateral ∆ ABC]
From (1), (2), (3) and (4), we have
DE = EF = FD
∴ ∆DEF is an equilateral triangle.
Hence, proved.
9. Points P and Q have been taken on opposite sides AB and CD
respectively of a parallelogram ABCD such that AP = CQ. Show that
AC and PQ bisect each other.
Sol. Points P and Q have been taken on opposite sides AB and CD
respectively of a parallelogram ABCD such that AP = CQ.

In ∆AOP and ∆COQ, we have

AP = CQ [Given]
∠1 = ∠2 [Alt. int ∠s are equal]
∠3 = ∠4 [Vertically opp. ∠s]
∴ ∆AOP ≅ ∆COQ [By AAS Congruence rule]
∴ OA = OC and OP = OQ [CPCT]
Hence, AC and PQ bisect each other.
10. In the given figure, P is the mid-point of side BC of a parallelogram
ABCD such that ∠BAP = ∠DAP. Prove that AD = 2CD.
1
Sol. ∠BAP = ∠DAP = ∠A ...(1)
2

Since ABCD is a parallelogram, we have

∠A + ∠B = 180° ...(2)
[∵ Sum of interior angles on the same side of transversal is 180°]

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

In ∆ABP, we have
∠BAP + ∠B + ∠APB = 180°
1
⇒ θ∠A + 180° – ∠A + ∠APB = 180° [Using (1) and (2)]
2

1
⇒ ∠APB = ∠A ...(3)
2
From (1) and (3), we get
∠BAP = ∠APB
BP = AB ...(4)
[∵ Sides opposite to equal angles are equal]
Since opposite sides of a parallelogram are equal, we have
1 1
AD = BC ⇒ AD = BC
2 2
1
⇒ AD = BP [∵ P is the mid-point of BC]
2
1
⇒ AD = AB [∵ From (4), BP = AB ]
2
Since opposite sides of a parallelogram are equal, we have
1
AD = CD ⇒ AD = 2CD
2
Hence, proved.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

EXERCISE 8.4

1. A square is inscribed in an isosceles right triangle so that the square and

the triangle have one angle common. Show that the vertex of the square
opposite the vertex of the common angle bisects the hypotenuse.
Sol. ABC is an isosceles right triangle with A
AB = BC. A square BFED is inscribed in triangle
ABC so that ∠B = common = 90°.
In ∆ADE and ∆EFC, we have
2
DE = EF ...(1) D 1 E
[∵ Sides of a square are equal]
∠1 + ∠2 = 180º 3 4
B C
[Linear pair axiom] F
⇒ 90o + ∠2 = 180° [∵ Each angle of a square = 90°]
⇒ ∠2 = 90º

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

Similarly, ∠4 = 90°
∴ ∠2 = ∠4 ...(2) [∵ Each = 90°]
Now, AB = BC [Given]
∴ ∠C = ∠A ...(3)
[∵ Angles opp. to equal sides are equal]
From (1), (2) and (3), we get
∆ADE ≅ ∆EFC [By AAS Congruence rule]
Hence, AE = EC [CPCT]
2. In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of
∠A meets DC in E. AE and BC produced meet at F. Find the length of
CF.
Sol. ABCD is a parallelogram, in which AB = 10 cm and AD = 6 cm.
The bisector of ∠A meets DC in E. AE and BC produced meet at F.

∠BAE = ∠EAD ...(1) [∵ AE bisects ∠A]

∠EAD = ∠EFB ...(2) [Alt. ∠s]
⇒ ∠BAE = ∠EFB [From (1) and (2)]
∴ BF = AB [∵ Sides opposite to equal ∠s are equal]
⇒ BF = 10 cm [∵ AB = 10 cm]
⇒ BC + CF = 10 cm ⇒ 6 cm + CF = 10 cm
[∵ BC = AD = 6 cm, opposite sides of a ||gm]
⇒ CF = 10 cm – 6 cm = 4 cm
3. P, Q, R and S are respectively the mid-points of the sides AB, BC, CD
and DA of a quadrilateral ABCD in
R
which AC = BD. Prove that PQRS is D C
a rhombus.
Sol. Given: A quadrilateral ABCD in
which AC = BD and P, Q, R and S S Q
are respectively the mid-points of the
sides AB, BC, CD and DA of
A B
quadrilateral ABCD. P

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

To prove: PQRS is a rhombus.

Proof: In ∆ABC, P and Q are the mid-points of sides AB and BC
respectively.
That is, PQ joins mid-points of sides AB and BC.
∴ PQ || AC ...(1)
1
and PQ = AC ...(2) [Mid-point theorem]
2
In ∆ADC, R and S are the mid-points of CD and AD respectively.
∴ SR || AC ...(3)
1
and SR = AC ...(4) [Mid-point theorem]
2
From (1) and (3), we get From (2) and (4), we get
PQ || SR PQ = SR
⇒ PQRS is a parallelogram.
In ∆DAB, SP joins mid-points of sides DA and AB respectively.
1
∴ SP = BD ...(5) [Mid-point theorem]
2
AC = BD ...(6) [Given]
From equations (2), (5) and (6), we get
SP = PQ
∴ Parallelogram PQRS is a rhombus.
Hence, proved.
D R
4. P, Q, R and S are respectively the C
mid-points of the sides AB, BC, CD
and DA of a quadrilateral ABCD
such that AC ⊥ BD. Prove that S
F
Q
PQRS is a rectangle. 1 E
Sol. Given: A quadrilateral ABCD in G
2
which AC ⊥ BD and P, Q, R and S
are respectively the mid-points of A P B
the sides AB, BC, CD and DA of quadrilateral ABCD.
To prove: PQRS is a rectangle.
Proof: In ∆ABC, P and Q are the mid-points of sides AB and BC
respectively.
That is, PQ joins mid-points of sides AB and BC.
∴ PQ || AC ...(1)
1
and PQ = AC ...(2) [Mid-point theorem]
2
In ∆ADC, R and S are the mid-points of CD and AD respectively.
∴ SR || AC ...(3)

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

1
and SR = AC ...(4) [Mid-point theorem]
2
From (1) and (3), we get From (2) and (4), we get
PQ || SR PQ = SR
⇒ PQRS is a parallelogram.
PQ | AC [Proved above]
⇒ PE || GF
In ∆ABD, PS joins mid-points of sides AB and AD respectively.
∴ PS || BD [Mid-point theorem]
⇒ PG || EF
⇒ PEFG is a parallelogram [∵ PE || GF and PG || EF]
⇒ ∠1 = ∠2 [∵ Opposite angles of a
parallelogram are equal]
But, ∠1 = 90º [∵ AC ⊥ BD]
∴ ∠2 = 90º
⇒ Parallelogram PQRS is a rectangle.
5. P, Q, R and S are respectively the mid-points of sides AB, BC, CD and
DA of quadrilateral ABCD in which D R C
AC = BD and AC ⊥ BD. Prove that PQRS is
a square.
Sol. Given : A quadrilateral ABCD is which
AC = BD and AC ⊥ BD. P, Q, R and S are S Q
respectively the mid-points of sides AB, 1
BC, CD and DA of quadrilateral ABCD.
2
To prove : PQRS is a square.
A B
Proof : Parallelogram PQRS is a rectangle. P
[Same as in Q4]
1
PQ = AC ... (1) [Proved as in Q4]
2
PS joins mid-points of sides AB and AD respectively.
1
BD
PS = … (2) [Mid-point theorem]
2
AC = BD … (3) [Given]
From (1), (2) and (3), we get
PS = PQ
⇒ Rectangle PQRS is a square.
6. A diagonal of a parallelogram bisects one of its angles. Show that it is a
rhombus.
Sol. ABCD is a parallelogram and diagonal AC bisects ∠A. We have to
show that ABCD is a rhombus.
∠1 = ∠2 ...(1) [∵ AC bisects ∠A]

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

∠2 = ∠4
...(2) [∵ Alt. interior angles]
From (1) and (2), we get
∠1 = ∠4
Now, in ∆ABC, we have
∠1 = ∠4 [Proved above]
∴ BC = AB
[∵ Sides opp. to equal ∠s are equal]
Also, AB = DC and AD = BC [∵ Opposite sides of a parallelogram are
equal]
So, ABCD is a parallelogram in which its sides AB = BC = CD = AD.
Hence, ABCD is a rhombus.
7. P and Q are the mid-points of the opposite sides AB and CD of a
parallelogram ABCD. AQ intersects Q
D C
DP at S and BQ intersects CP at R.
Show that PQRS is a parallelogram.
Sol. Given : A parallelogram ABCD in S R
which P and Q are the mid-points
of the sides AB and CD
respectively. AQ intersects DP at S
and BQ intersects CP at R.
To prove : PRQS is a parallelogram. A P B
Proof : DC || AB [∵ Opposite sides of a parallelogram are parallel]
⇒ AP || QC
DC = AB [∵ Opposite sides of a parallelogram are equal]
1 1
⇒ DC = AB
2 2
⇒ QC = AP [∵ P is mid-point of AB and Q is
mid-point of CD]
⇒ APCQ is a parallelogram. [∵ AP || QC and QC = AP]
∴ AQ || PC [∵ Opposite sides of a || gm are parallel]
⇒ SQ || PR
Similarly, SP || QR
∴ Quadrilateral PRQS is a parallelogram.
Hence, proved.
8. ABCD is a quadrilateral in which AB || DC and AD = BC. Prove that ∠A
= ∠B and ∠C = ∠D.
Sol. Given : A quadrilateral ABCD in which AB || DC and AD = BC.
To prove : ∠A = ∠B and ∠C = ∠D

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

Construction : Draw DP ⊥ AB and D C

3 4
CQ ⊥ AB.
Proof : In ∆APD and ∆BQC, we
have
∠1 = ∠2
[∵ Each equal to 90°]
AD = BC [Given]
1 2
DP = CQ A P Q B
[Distance between parallel lines]
So, by RHS criterion of congruence, we have
∆APD ≅ ∆BQC
∴ ∠A = ∠B [CPCT]
Now, DC || AB
∠A+ ∠3 = 180° ...(1) [∵ Sum of consecutive interior
angles is 180°]
∠B+ ∠4 = 180° … (2) [Same reason]
From (1) and (2), we get
∠A + ∠3 = ∠B + ∠4
⇒ ∠3 = ∠4 [∵ ∠A = ∠B]
⇒ ∠C = ∠D
Hence, proved.
9. In the given figure, AB || DE, AB = DE, AC || DF and AC = DF. Prove that
BC || EF and BC = EF.

A D
Sol.

E
B

F
C

Given : AB || DE, AB = DE, AC || DF and AC = DF

To prove : BC || EF and BC = EF
Proof : AC || DF [Given]
and AC = DF [Given]
∴ ACFD is a parallelogram.
⇒ AD || CF ...(1) [ ∵ Opposite sides of a || gm are parallel]
and AD = CF ...(2) [∵ Opposite sides of a || gm are equal]
Now, AB || DE [Given]
and AB = DE [Given]

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

∴ ABED is a parallelogram.
⇒ AD || BE ...(3) [∵ Opposite sides of a || gm are parallel]
and AD = BE ...(4) [∵ Opposite sides of a || gm are equal]
From (1) and (3), we get
CF || BE
And, from (2) and (4), we get
CF = BE
∴ BCFE is a parallelogram.
⇒ BC || EF [∵ Opposite sides of a || gm are parallel]
and BC = EF [∵ Opposite sides of a || gm are equal]
Hence, proved.
10. E is the mid-point of a median AD of ∆ABC and BE is produced to
1
meet AC at F. Show that AF = AC .
3
Sol. Given : A ∆ABC in which E is A
the mid-point of median AB
and BE is produced to meet
AC at F. F

1 E
To prove : AF = AC G
3
Construction: Draw DG || BF
intersecting AC at G.
Proof : In ∆ADG, E is the mid- B D
C
point of AD and EF || DG.
∴ AF = FG … (1) [Converse of mid-point theorem]
In ∆FBC, D is the mid-point of BC and DG || BF.
∴ FG = GC … (2) [Converse of mid-point theorem]
From equations (1) and (2), we get
AF = FG = GC … (3)
But, AC = AF + FG + GC
⇒ AC = AF + AF + AF [Using (3)]
⇒ AC = 3 AF
1
⇒ AF = AC
3
Hence, proved.
11. Show that the quadrilateral formed by joining the mid-points of the
consecutive sides of a square is also a square.
Sol. Given: A square ABCD in which P, Q, R, S are the mid-points of sides
AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

To prove: PQRS is a square. D R C

Construction: Join AC and BD.
Proof: In ∆ABC, P and Q are the mid-points
of sides AB and BC respectively. O
S Q
1
∴ PQ|| AC and PQ = AC ...(1)
2
In ∆ADC, R and S are the mid-points of CD A B
and AD respectively. P

1
∴ RS || AC and RS = AC ...(2)
2
From eqs. (1) and (2), we get
PQ || RS and PQ = RS ...(3)
Thus, in quadrilateral PQRS one pair of opposite sides are equal and
parallel.
Hence, PQRS is a parallelogram.
Since ABCD is a square.
∴ AB = BC = CD = DA
1 1 1 1
⇒ AB = CD and AB = BC
2 2 2 2
⇒ PB = RC and BQ = CQ
Thus, in ∆s PBQ and RCQ, we have
PB = RC
BQ = CQ [⇒ PB = CR and BQ = CQ]
and, ∠PBQ = ∠RCQ [∵ Each equal to 90°]
So, by SAS criterion of congruence, we have
∆PBQ ≅ ∆RCQ
⇒ PQ = QR [CPCT] ...(4)
From (3) and (4), we have
PQ = QR = RS
But, PQRS is a parallelogram
∴ QR = PS
So, PQ = QR = RS = PS ...(5)
Now, PQ || AC [From (1)]
⇒ PM || NO ...(6)
Since P and S are the mid-points of AB and AD respectively.
PS || BD
⇒ PN || MO ...(7)
Thus, in quadrilateral PMON, we have
PM || NO [From (6)]
and PN || MO [From (7)]

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

So, PMON is a parallelogram.

⇒ ∠MPN = ∠MON
⇒ ∠MPN = ∠BOA [∵ ∠MON = ∠BOA]
⇒ ∠MPN = 90° [∵ Diagonals of square are ⊥
∴ AC ⊥ BD ⇒ ∠BOA = 90°]
⇒ ∠QPS = 90°
Thus, PQRS is a quadrilateral such that PQ = QR = RS = SP and
∠QPS = 90°,
Hence, PQRS is a square.
12. E and F are respectively the mid-points of the non-parallel sides AD and
1
BC of a trapezium ABCD. Prove that EF || AB and EF = (AB + CD) .
2
[Hint: Join BE and produce it to meet CD produced at G]
D C
Sol.
1
D F
2
3
A G
B
Given : A trapezium ABCD in which E and F are respectively the
mid-points of the non-parallel sides AD and BC.
1
To prove : EF || AB and EF = (AB + CD)
2
Construction: Join DF and produce it to intersect AB produced at G.
Proof : In ∆CFD and ∆BFG, we have
DC || AB
∴ ∠C = ∠3 [Alternate interior angles]
CF = BF [Given]
∠1 = ∠2 [Vertically opposite angles]
So, by ASA criterion of congruence, we have
∆CFD ≅ ∆BFG
∴ CD = BG [CPCT]
EF joins mid-points of sides AD and GD respectively
∴ EF || AG [∵ Mid-point theorem]
⇒ EF || AB
1
So, EF = AG [Mid-point theorem]
2
1
⇒ EF = (AB + BG)
2
1
⇒ EF = (AB + CD) [∵ CD = BG]
2
Hence, proved.

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

13. Prove that the quadrilateral formed by the bisectors of the angles of a
parallelogram is a rectangle.
D C
Sol. Given : A parallelogram ABCD in which
R
bisectors of angles A,B,C,D intersect at
P,Q,R,S to form a quadrilateral PQRS. S Q
To prove : PQRS is a rectangle.
Proof : Since ABCD is a parallelogram. P
Therefore, AB || DC A B
Now, AB || DC and transversal AD intersects them at D and A respectively.
Therefore,
∠A + ∠D = 180°[∵ Sum of consecutive interior angles is 180°]
1 1
⇒ ∠A + ∠D = 90°
2 2
⇒ ∠DAS + ∠ADS = 90° ...(1)
[∵ DS and AS are bisectors of ∠A and ∠D respectively]
But, in ∆DAS, we have
∠DAS + ∠ASD + ∠ADS = 180°
[∵Sum of the angles of a triangle is 180°]
⇒∠90° + ∠ASD = 180° [Using (1)]
⇒ ∠ASD = 90°
⇒ ∠PSR = 90° [∵∠ASD and ∠PSR are vertically opposite
angles ∴ ∠PSR = ∠ASD]
Similarly, we can prove that
∠SRQ = 90°, ∠RQP = 90° and ∠SPQ = 90°
Hence, PQRS is a rectangle.
14. P and Q are points on opposite sides AD and BC of a parallelogram
ABCD such that PQ passes through the point of intersection O of its
diagonals AC and BD . Show that PQ is bisected at O.
Sol. ABCD is a parallelogram. Its diagonals AC and BD bisect each other
at O. PQ passes through the point of intersection O of its diagonal AC
and BD.

In ∆AOP and ∆COQ, we have

∠3 = ∠4 [Alternate int. ∠s]

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

OA = O C [Diagonals of a || gm bisect each other]

∠1 = ∠2 [Vertically opposite angles]
∴ ∆AOP ≅ ∆COQ [By ASA Congruence rule]
So, OP = O Q [CPCT]
Hence, PQ is bisected at O.
15. ABCD is a rectangle in which
diagonal BD bisects ∠B. Show
that ABCD is a square.
Sol. Given : A rectangle ABCD in
which diagonal BD bisects ∠B.
To prove : ABCD is a square.
Proof : DC || AB
[∵ Opposite sides of a rectangle
are parallel]
⇒ ∠4 = ∠1 ...(1) [Alternate interior angles]
Similarly, ∠3 = ∠2 ...(2) [Alternate interior angles]
and ∠1 = ∠2 ...(3) [Given]
From equations (1), (2) and (3), we get
∠3 = ∠4
In ∆BDA and ∆BDC, we have
∠1 = ∠2 [Given]
BD = BD [Common side]
∠3 = ∠4 [Proved above]
So, by ASA criterion of congruence, we have
∆BDA ≅ ∆BDC
∴ AB = BC [CPCT]
So, ABCD is a square.
Hence, proved.
16. D, E and F are respectively the mid-points of the sides AB, BC and CA of
a triangle ABC. Prove that by joining these mid-points D, E and F, the
triangles ABC is divided into four congruent triangles.
Sol. Given : A ∆ABC and ∆DEF which is A
formed by joining the mid-points D, E and
F of the sides AB, BC and CA of ∆ABC.
To prove : ∆DEF ≅ ∆EDB ≅ ∆CFE ≅ ∆FAD
Proof : DF joins mid-points of sides AB D F

and AC respectively of ∆ABC.

∴ DF || BC [Mid-point theorem]
⇒ DF || BE
B E C
Similarly, EF || BD

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

So, quadrilateral BEFD is a parallelogram.

⇒ ∆DEF ≅ ∆EDB ...(1)[∵ A diagonal of a parallelogram divides
it into two congruent triangles]
Similarly, ∆DEF ≅ ∆CFE ...(2)
and ∆DEF ≅ ∆FAD ...(3)
From equations (1), (2) and (3), we get
∆DEF ≅ ∆EDB ≅ ∆CFE ≅ ∆FAD
Hence, proved.
17. Prove that the line joining the mid-points of the diagonals of a trapezium
is parallel to the parallel sides of the trapezium.
D C
Sol. 1

3
E F
4

2
A B G

Given : A trapezium ABCD in which E and F are the mid-points of sides

AD and BC respectively.
To prove : EF || AB || DC
Construction : Join DF and produce it to intersect AB produced at G.
Proof : In ∆DCF and ∆GBF, we have
∠1 = ∠2 [Alternate interior angles because DC || BG]
∠3 = ∠4 [Vertically opposite angles]
CF = BF [∵ F is the mid-point of BC]
So, by AAS criterion of congruence, we have
∆DCF ≅ ∆GBF
∴ DF = GF [CPCT]
In ∆DAG, EF joins mid-points of sides DA and DG respectively.
∴ EF || AG [Mid-point theorem]
⇒ EF || AB
But, AB || DC [Given]
∴ EF || AB || DC
Hence, proved.
18. P is the mid-point of the side CD of a parallelogram ABCD. A line
through C parallel to PA intersects AB at Q and DA produced at R.
Prove that DA = AR and CQ = QR.
Sol. ABCD is a parallelogram. P is the mid-point of CD. CR which intersects
AB at Q is parallel to AP.
In ∆DCR , P is the mid-point of CD and AP || CR,

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Chapter 8 - Quadrilaterals NCERT Exemplar - Class 09

∴ A is the mid-point of DR, i.e., D P C

AD = AR.
[∵ The line drawn through the
A 2 4
mid-point of one side of a 1 B
3
triangle parallel to another side
intersects the third side at its
mid-point.] R
In ∆ARQ and ∆BCQ, we have
AR = BC [∵ AD = AR (proved above) and AD = BC]
∠1 = ∠2 [Vertically opposite angles]
∠3 = ∠4 [Alt. ∠s]
∴ ∆ARQ ≅ ∆BCQ [By AAS Congruence rule]
CQ =QR [CPCT]
Hence, DA = AR and CQ = QR is proved.

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