# Important Questions Class 9 Maths Chapter 8

## Important Questions Class 9 Mathematics Chapter 8 – Quadrilaterals

Mathematics is based on application, and its application in the right manner strengthens the conceptual understanding of the topic. The Quadrilaterals Chapter is considered one of the important chapters in Geometry. One can find all the concepts, theories, and NCERT solutions, along with many solved examples, on our Extramarks official website.

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### Important Questions Class 9 Mathematics Chapter 8 – With Solutions

The Extramarks in-house subject experts have collated a comprehensive list of Important Questions Class 9 Mathematics Chapter 8 from various sources. These practical questions and their solutions help students to get a clear understanding of all concepts related to Quadrilaterals.

Given below are a few questions and their answers from our question bank of Important Questions in Class 9 Mathematics Chapter 8.

Answer 1:A quadrilateral is a four-sided polygon having a closed shape. It is a 2-dimensional shape.

The six types of quadrilaterals include

• Rectangle
• Square
• Parallelogram
• Rhombus
• Trapezium
• Kite

Question 2: The diagonals of which quadrilaterals are equal and bisect each other at 90°.

Answer 2: It is a square. The diagonals of a square are equivalent and bisect each other at 90°.

Question 3: Identify the type of quadrilaterals:

(i) The quadrilateral is formed by joining the midpoints of consecutive sides of a quadrilateral whose diagonals are perpendicular.

(ii) The quadrilateral is formed by joining the required midpoints of consecutive sides of a quadrilateral whose diagonals are the same.

(i) The required quadrilateral formed by joining the midpoints of consecutive sides of a quadrilateral whose diagonals are perpendicular is a rectangle.

(ii) A rhombus is a quadrilateral formed by joining the midpoints of consecutive sides whose diagonals are congruent.

Question 4: Find all the angles of a parallelogram if one angle is 80°.

For a required parallelogram ABCD, opposite angles are equivalent.

So, the required angles opposite the given 80° angle will also be 80°.

It is also said that the sum of angles of any quadrilateral = 360°.

So, if ∠A = ∠C = 80° then,

∠A + ∠B + ∠C + ∠D = 360°

Furthermore, ∠B = ∠D

Hence,

80° + ∠B + 80° + ∠D = 360°

∠B +∠ D = 200°

Therefore, ∠B = ∠D = 100°

Here, all angles of the quadrilateral are found, which are:

∠A = 80°

∠B = 100°

∠C = 80°

∠D = 100°

Question 5: In a given rectangle, one diagonal is inclined to one of its sides at 25°. Calculate the acute angle between the two diagonals.

Let ABCD be the required rectangle where AC and BD are the given two diagonals which are intersecting at point O.

Assuming ∠BDC = 25° ( is given)

Here, ∠BDA = 90° – 25° = 65°

Also, ∠DAC = ∠BDA (as diagonals of the necessary rectangle divide the rectangle into two equal right triangles)

Then, ∠BOA = the acute angle between the two diagonals = 180° – 65° – 65° = 50°

Question 6: Is it possible to draw a quadrilateral whose all angles are obtuse?

It is said that the sum of the angles of the required quadrilateral is always 360°. To consider all angles as obtuse, the necessary angles of the quadrilateral will be greater than 360°. So, it is impossible to draw a quadrilateral whose all angles are obtuse.

Question 7: Prove that the angle bisectors of a parallelogram form a rectangle.

LMNO is the required parallelogram in which bisectors of the angles L, M, N, and O intersect at the given points P, Q, R and S to form the necessary quadrilateral PQRS.

LM || NO (The opposite sides of the required parallelogram LMNO)

L + M = 180° (The sum of consecutive interior angles is 180°)

MLS + LMS = 90°

In LMS, MLS + LMS + LSM = 180°

90 + LSM = 180°

LSM = 90°

RSP = 90° (The vertically opposite angles)

SRQ = 90°, RQP = 90° and SPQ = 90°

Therefore, PQRS is the required rectangle.

Question 8: In a given trapezium ABCD, AB∥CD. Measure ∠C and ∠D if ∠A = 55° and ∠B = 70°.

In a given trapezium ABCD, ∠A + ∠D = 180° and ∠B + ∠C = 180°

Then, 55° + ∠D = 180°

Or, ∠D = 125°

Similarly,

70° + ∠C = 180°

∠C = 110°

Question 9: Calculate all the angles of a parallelogram if one of its angles is twice its adjacent angle.

Let the angle of the required parallelogram be “x”.

Then, its adjacent angle will be 2x.

It is said that the opposite angles of a parallelogram are equal.

So, all the angles of a necessary parallelogram will be x, 2x, x, as well as 2x

The sum of the required interior angles of a given parallelogram = 360°,

x + 2x + x + 2x = 360°

x = 60°

Therefore, all the necessary angles will be 60°, 120°, 60°, and 120°.

Question 10: Calculate all the angles of a quadrilateral if they are in the ratio 2:5:4:1.

As the necessary angles are in the ratios 2:5:4:1, they can be noted as-

2x, 5x, 4x, and x

Then, as the sum of the required angles of a quadrilateral is 360°,

2x + 5x + 4x + x = 360°

x = 30°

Here, all the angles will be,

2x =2 × 30° = 60°

5x = 5 × 30° = 150°

4x = 4 × 30° = 120°, and

x = 30°

Question 11: The angles of the quadrilateral are in the ratio 3: 5 : 9: 13. Find all the angles of the quadrilateral.

Let the necessary angles of the quadrilateral be 3x, 5x, 9x and 13x.

3x + 5x + 9x + 13x = 360°

[The property angle sum of a quadrilateral]

30x = 360°

x = 360°/ 30

x = 12°

Here, the required angles are as follows

3x = 3 x 12° = 36°

5x = 5 x 12° = 60°

9x = 9 x 12° = 108°

13x = 13 x 12° = 156°

The necessary angles of the quadrilateral are 36°, 60°, 108° and 156°.

Question 12:If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Let ABCD be the required parallelogram such that AC = BD.

In given Triangles ABC and DCB,

Given:-AC = DB

AB = DC [The opposite sides of the required parallelogram]

BC = CB [Taking common]

∴ ∆ABC ≅ ∆DCB [By SSS congruence]

∠ABC = ∠DCB [By Corresponding Parts of Congruent Triangles] …(1)

Here, AB || DC and BC a transversal. [ ∵ ABCD is the required parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we obtain

∠ABC = ∠DCB = 90°

That is, ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is the required rectangle.

Question 13: Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Let ABCD be the required quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.

In the given Triangles AOB and ∆AOD, we obtain

AO = AO [Taking common]

OB = OD [O is the given midpoint of BD]

∠AOB = ∠AOD [Each is 90°]

∴ ∆AQB ≅ ∆AOD [By,SAS congruence]

AB = AD [By Corresponding Parts of Congruent Triangles] ……..(1)

Here, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

∴ From (1), (2), (3) and (4), we obtain

AB = BC = CD = DA,

Therefore, the quadrilateral ABCD is the required rhombus.

Question 14: Show that the diagonals of a square are equal and bisect each other at right angles.

Let ABCD be the required square such that its diagonals AC and BD intersect at O.

(i) To prove that the necessary diagonals are equivalent, we must prove AC = BD.

In given Triangles ABC and BAD, we obtain

AB = BA [Taking common]

BC = AD [The sides of the given square ABCD]

∠ABC = ∠BAD [Each is 90°]

∆ABC ≅ ∆BAD [By SAS congruence]

AC = BD [By Corresponding Parts of Congruent Triangles] …(1)

(ii) AD || BC and AC is the required transversal. [∵ A square is also a parallelogram]

∠1 = ∠3

[The alternate interior angles are equivalent]

Now, ∠2 = ∠4

Here, in ∆OAD and ∆OCB, we obtain

AD = CB [The sides of the given square ABCD]

∠1 = ∠3 [It is proved]

∠2 = ∠4 [It is proved]

∆OAD ≅ ∆OCB [By ASA congruence]

OA = OC as well as OD = OB [By C.P.C.T.]

That is, the diagonals AC and BD bisect each other at O. …….(2)

(iii) In the given Triangles OBA and ODA, we obtain

OB = OD [It is proved]

BA = DA [The sides of a given square ABCD]

OA = OA [Taking common]

∴ ∆OBA ≅ ∆ODA [By SSS congruence]

∠AOB = ∠AOD ….[By CPCT] …(3)

∵ ∠AOB and ∠AOD form the required linear pair.

∠AOB + ∠AOD = 180°

∠AOB = ∠AOD = 90°… [By(3)]

AC ⊥ BD…(4)

From (1), (2) and (4), we acquire AC and BD are equivalent and bisect each other at right angles.

Question 15: Show that if the diagonals of a quadrilateral are equivalent and bisect each other at right angles, then it is a necessary square.

Let ABCD be the required quadrilateral such that diagonals AC and BD are equivalent and bisect each other at right angles.

Here, in ∆AOD and ∆AOB, We obtain

∠AOD = ∠AOB [Each angle is 90°]

AO = AO [Taing common]

OD = OB [ ∵ O is the given midpoint of BD]

∴ ∆AOD ≅ ∆AOB [By SAS congruence]

AD = AB [By Corresponding Parts of Congruent Triangles] …(1)

Now, we obtain

AB = BC … (2)

BC = CD …(3)

CD = DA …(4)

From (1), (2), (3) and (4), we acquire

AB = BC = CD = DA

∴ The required quadrilateral ABCD has all sides equivalent.

In the given Triangles AOD and COB, we acquire

Given:-AO = CO

Given:- OD = OB

∠AOD = ∠COB [The vertically opposite angles]

∆AOD ≅ ∆COB [By SAS congruence]

∴∠1 = ∠2 [By C.P.C.T.]

But, they form the given pair of alternate interior angles.

Now, AB || DC

∴ ABCD is the necessary parallelogram.

∴ A given parallelogram having all its sides equivalent is a rhombus.

∴ ABCD is the required rhombus.

Here, in the given Triangles ABC and BAD, we obtain

Given :- AC = BD

BC = AD [It is proved]

AB = BA [Taking common]

∴ ∆ABC ≅ ∆BAD [By SSS congruence]

∴ ∠ABC = ∠BAD [By Corresponding Parts of Congruent Triangles] ……(5)

Here, AD || BC and AB is the given transversal.

∴∠ABC + ∠BAD = 180° .. .(6) [ The co – interior angles]

⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)]

Therefore, rhombus ABCD has one angle equivalent to 90°.

Hence, ABCD is the necessary square.

Question 16: Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

We have a given parallelogram ABCD in which diagonal AC bisects ∠A

∠DAC = ∠BAC

(i) ABCD is the given parallelogram.

∴ AB || DC and AC are the required transversal.

∠1 = ∠3 …(1)

[ The alternate interior angles are equivalent]

Furthermore, BC || AD and AC are the required transversal.

∠2 = ∠4 …(2)

[ The alternate interior angles are equivalent]

Furthermore, ∠1 = ∠2 …(3)

[ ∵ AC bisects ∠A]

From (1), (2) and (3), we obtain

∠3 = ∠4

AC bisects ∠C

(ii) In the required ABC, we obtain

∠1 = ∠4 [From equations (2) and (3)]

BC = AB …(4)

[ ∵ The sides opposite to equivalent angles of a triangle are equivalent]

But, ABCD is the necessary parallelogram. [It is given]

∴ AB = DC …(6)

From equations (4), (5) and (6), we acquire

AB = BC = CD = DA

Thus, ABCD is the necessary rhombus.

Question 17: Quadrilateral ABCD is the required rhombus. Express that diagonal AC bisects ∠A as well as ∠C, and diagonal BD bisects ∠B as well as ∠D.

Since ABCD is a rhombus.

⇒ AB = BC = CD = DA

Also, AB || CD and AD || BC

∠1 = ∠2 …….(1)

[ ∵ The angles opposite to equivalent sides of a triangle are equivalent]

Furthermore, AD || BC and AC are the required transversals.

[ ∵ Each rhombus is a parallelogram]

∠1 = ∠3 …(2)

[ ∵ The alternate interior angles are equivalent]

From equations (1) and (2), we obtain

∠2 = ∠3 …(3)

Since AB || DC and AC are the required transversal.

∠2 = ∠4 …(4)

[ ∵ The alternate interior angles are equivalent]

From equations (1) and (4),

we obtain ∠1 = ∠4

∴ AC bisects ∠C as well as ∠A.

We can say that BD bisects ∠B as well as ∠D.

Question 18: Quadrilateral ABCD is the required rectangle in which diagonal AC bisects ∠A as well as ∠C. Express that

(i) ABCD is the required square

(ii) The required diagonal BD bisects ∠B as well as ∠D.

We have the required rectangle ABCD such that AC bisects ∠A and ∠C.

That is, ∠1 = ∠4 and ∠2 = ∠3 ……..(1)

(i) Since every rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AB || CD and AC is a transversal.

∴∠2 = ∠4 …(2)

[ ∵ Alternate interior angles are equal]

From equations (1) and (2), we obtain

∠3 = ∠4

In the required Triangle ABC, ∠3 = ∠4

AB = BC

[ ∵ The sides opposite to equivalent angles of A are equivalent]

Here, CD = DA

Quadrilateral ABCD is the necessary square.

(ii) Quadrilateral ABCD is a square, and the diagonals of a square bisect the opposite angles.

Here, BD bisects ∠B as well as ∠D.

Question 19: In parallelogram ABCD, two points, P and Q, are taken on diagonal BD such that DP = BQ (see figure). Show that

We have a required parallelogram ABCD, where BD is the diagonal, and points P and Q are such that PD = QB

(i) AD || BC and BD is the required transversal.

∴ ∠ADB = ∠CBD [ ∵ The alternate interior angles are equivalent]

Here, in the required Triangle APD and CQB, we obtain

AD = CB [The opposite sides of a given parallelogram ABCD are equivalent]

Given:- PD = QB

∠ADP = ∠CBQ [It is proved]

∴ ∆APD ≅ ∆CQB [By SAS congruence]

(ii) ∆APD ≅ ∆CQB [It is proved]

AP = CQ [By C.P.C.T.]

(iii) AB || CD and BD are the required transversal.

∴ ∠ABD = ∠CDB

∠ABQ = ∠CDP

Here, in the required triangles AQB and CPD, we obtain

Given:- QB = PD

∠ABQ = ∠CDP [It is proved]

AB = CD [ The opposite sides of the required parallelogram ABCD are equivalent]

∴ ∆AQB = ∆CPD [By SAS congruence]

(iv) ∆AQB = ∆CPD [It is proved]

AQ = CP [By C.P.C.T.]

(v) In the required ∆PCQ,

The opposite sides are equivalent. [It is proved]

∴ ∆PCQ is the necessary parallelogram.

Question 20: ABCD is a parallelogram, and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

Answer 20: (i) In the required triangles APB and CQD, we obtain

∠APB = ∠CQD [Each angle is 90°]

AB = CD [ ∵ The opposite sides of the given parallelogram ABCD are equivalent]

∠ABP = ∠CDQ

[ ∵ The alternate angles are equivalent to AB || CD, and BD is the required transversal]

∴ ∆APB = ∆CQD [By AAS congruence]

(ii) ∆APB ≅ ∆CQD [It is proved]

AP = CQ [By C.P.C.T.]

Question 21: In the given triangles ABC and DEF, AB = DE, AB || DE, BC – EF and BC || EF. The required vertices A, B and C are joined to the given vertices D, E and F, respectively (observe the figure).

Express that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ABC ≅ ∆DEF

Answer 21:(i) Given:- AB = DE

AB || DE

That is, ABED is the required quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equivalent length.

∴ Quadrilateral ABED is the required parallelogram.

(ii) Given:- BC = EF

BC || EF

That is, BEFC is the given quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equivalent length.

∴ BEFC is the required parallelogram.

(iii) ABED is the given parallelogram [It is proved]

[ ∵ The opposite sides of the given parallelogram are equivalent and parallel]

Furthermore, BEFC is a parallelogram. [It is proved]

BE || CF and BE = CF …(2)

[ ∵ The opposite sides of a parallelogram are equivalent and parallel]

From (1) and (2), we obtain

That is, In the required quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equivalent length.

∴Quadrilateral ACFD is the required parallelogram.

(v) Quadrilateral ACFD is a parallelogram. [Proved]

So, AC = DF [∵ The opposite sides of the given parallelogram are equivalent]

(vi) In the required triangles ABC and DFF, we obtain

Given:- AB = DE

BC = EF

AC = DE [It is proved in (v) part]

∆ABC ≅ ∆DFF [By SSS congruence]

Question 22: Quadrilateral ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that

(i )∠A=∠B

(ii )∠C=∠D

(iv) diagonal AC = diagonal BD

We have been given a trapezium ABCD where AB || CD and AD = BC.

(i) Producing AB to E and drawing CF || AD.. .(1)

AB || DC

AE || DC

[ ∵ The opposite sides of the required parallelogram are equivalent]

By equations (1) and (2),

BC = CF

Here, in the given triangles BCF, we obtain BC = CF

∠CEB = ∠CBE …(3)

[∵ The angles opposite to equivalent sides of a triangle are equivalent]

Furthermore, ∠ABC + ∠CBE = 180° … (4)

[Linear pair]

and ∠A + ∠CEB = 180° …(5)

[The co-interior angles of the given parallelogram ADCE]

From equations (4) and (5), we acquire

∠ABC + ∠CBE = ∠A + ∠CEB

∠ABC = ∠A [From equations (3)]

∠B = ∠A …(6)

(ii) AB || CD and AD are the required transversal.

∴ ∠A + ∠D = 180° …(7) [The co-interior angles]

Now, ∠B + ∠C = 180° … (8)

From equations (7) and (8), we obtain

∠A + ∠D = ∠B + ∠C

∠C = ∠D [From equation (6)]

(iii) In the required triangles ABC and BAD, we obtain

AB = BA [Common]

∴ ∆ABC = ∆BAD [By SAS congruence]

(iv) Since, ∆ABC = ∆BAD [Proved]

⇒ AC = BD [By C.P.C.T.]

Question 23: ABCD is the required quadrilateral in which the points P, Q, R and S are mid-points of the required sides AB, BC, CD and DA (see figure). AC is diagonal. Express that

(i) SR || AC as well as SR = 1/2 AC

(ii) PQ = SR

(iii) PQRS is the required parallelogram.

(i) In the required Triangle AC,

S is the given midpoint of AD, and R is the given midpoint of CD.

SR = 1/2 AC as well as SR || AC …(1)

[Utilising mid-point theorem]

(ii) In the required Triangle ABC, P is the given midpoint of AB and Q is the midpoint of BC.

PQ = 1/2 AC and PQ || AC …(2)

[By mid-point theorem]

From equations (1) and (2), we obtain

PQ = 1/2 AC = SR as well as PQ || AC || SR

PQ = SR as well as PQ || SR

(iii) In the given quadrilateral PQRS,

PQ = SR as well as PQ || SR [It is proved]

∴ PQRS is the necessary parallelogram.

Question 24: Quadrilateral ABCD is the required rhombus, and P, Q, R and S are the given mid-points of the sides AB, BC, CD and DA, respectively. Express that the quadrilateral PQRS is a rectangle.

We have a required rhombus ABCD, and the points P, Q, R and S are the given mid-points of the given sides AB, BC, CD and DA, respectively. Joining AC.

In the required Triangle, ABC, P and Q are the midpoints of AB and BC, respectively.

∴ PQ = 1/2 AC

PQ is parallel to AC …(1)

[By mid-point theorem]

In the required Triangle ADC, R and S are the midpoints of CD and DA, respectively.

∴ SR = 1/2

AC and SR || AC…(2)

[By mid-point theorem]

From (1) and (2), we obtain

PQ = 1/2

AC = SR and PQ || AC || SR

PQ = SR and PQ || SR

∴ PQRS is the required parallelogram. …….(3)

Here, in triangles ERC and EQC,

∠1 = ∠2

[ ∵ The diagonals of any given rhombus bisect the opposite angles]

CR = CQ [ ∵CD/2 = BC/2]

CE = CE [The common]

∴ ∆ERC ≅ ∆EQC [By SAS congruence]

∠3 = ∠4 …(4) [By Corresponding Parts Of Congruent Triangles]

But ∠3 + ∠4 = 180° ……(5) [Linear pair]

From equations (4) and (5), we obtain

∠3 = ∠4 = 90°

Here, ∠RQP = 180° – ∠b [ The Co-interior angles for PQ || AC and EQ is transversal]

But ∠5 = ∠3

[ ∵ The vertically opposite angles are equivalent]

∴ ∠5 = 90°

∠RQP = 180° – ∠5 = 90°

∴ One angle of the given parallelogram PQRS is 90°.

Thus, PQRS is the necessary rectangle.

Question 25: Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give a reason for your answer.

Answer 25:No, diagonals of a parallelogram are not perpendicular to each other because they only bisect each other.

Question 26:Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral? Why or why not

• No, we know that the sum of all quadrilateral angles is 360°.
• Here, the sum of the angles = 110°+ 80° + 70° + 95° = 355° ≠ 360°
• So, these angles cannot be the angles of a quadrilateral.

Question 27: Diagonals of a rectangle are equal and perpendicular. Is this statement true? Give a reason for your answer.

Answer 27: No, the diagonals of a rectangle are equal but need not be perpendicular.

Question 28:Can all four angles of a quadrilateral be obtuse angles? Give a reason for your answer.

Answer 28: No, all four angles of a quadrilateral cannot be obtuse. As the sum of the angles of a quadrilateral is 360°, it may have a maximum of three obtuse angles.

Question 29: In the figure, ABCD and AEFG are two parallelograms. If ∠C = 55°, then determine ∠F.

We have ABCD and AEFG are two parallelograms and ∠C = 55°. Since ABCD is a parallelogram, the opposite angles of a parallelogram are equal.

∠A = ∠C = 55° …(i)

Also, AEFG is a parallelogram.

∴ ∠A=∠F = 55° [from Eq. (i)]

Question 30: 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°

As we know, the diagonals of a rectangle are equal in length.

therefore AC =BD { diagonals are equal}

1/2 AC = 1/2 BD {Then dividing both sides by 2}

AO = BO { O is midpoint of diagonal}

Thus ∠OBA = ∠OAB

∠OAB =25° {Given}

∠OAB = 25°

∠BOC = ∠OBA + ∠OAB {exterior angle is equal to the sum of two opposite interior angles}

= 25° + 25°  = 50°

Thus the actual angle between the diagonals is 50°.

Hence option B is correct.

Question 31: The quadrilateral formed by joining the midpoints on the sides of a quadrilateral PQRS taken in order is a rectangle. If

(A) PQRS is a rectangle

(B) PQRS is a parallelogram

(C) Diagonals of PQRS are perpendiculars

(D) Diagonals of PQRS are equal

According to the question quadrilateral, ABCD is formed by joining the midpoints of PQRS.

If PQRS is the necessary rectangle

Quadrilateral ABCD is not a rectangle because a rectangle is a four-sided polygon having all the internal angles = 90°

and the required opposite sides are equal in length.

If PQRS is the necessary parallelogram

Quadrilateral ABCD is not a rectangle because a rectangle is a four-sided polygon having all the internal angles = 90°

and the required opposite sides are equivalent in length.

If diagonals of PQRS are perpendicular

Here ABCD is a rectangle because here

∠A = ∠B = ∠C = ∠D =90°

and the opposite sides we equivalent that is AB = DC and AD = BD

If diagonals of PQRS are equivalent

ABCD is not a rectangle. It is a square.

Because here AB = BC = CD = AD and ∠A = ∠B = ∠C = ∠D =90°

We see that if diagonals of PQRS are perpendicular to each other, then ABCD is the required rectangle

option C is correct

(C) diagonals of PQRS are perpendicular

Question 32: If the bisector 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 quadrilateral PQRS is a

(A) The necessary rectangle

(B) The necessary rhombus

(C) The necessary parallelogram

(D)The necessary quadrilateral whose opposite angles are supplementary

Foremost, let’s draw the figure as per the question.

From the above diagram

∠QRS = ∠APB…..(1) (Vertically opposite angles)

In Triangle APB = ∠APB + ∠PAB + ∠ABP = 180°

∠APB = 1/2 ∠A = 1/2 ∠B = 180°

∠ABP = 180° – 1/2 ( ∠A + ∠B )…..(2)

From equation 1 & 2

∠QPS = 180° – 1/2 ( ∠A + ∠B )…..(3)

Similarly, ∠QRS = 180° – 1/2 ( ∠C + ∠D )…..(4)

∠QPS + ∠QRS = 360° – 1/2 ( ∠A + ∠B + ∠C + ∠D)

= 360° – 1/2 (360°)

=360° – 180°

=180°

Therefore, ∠QPS + ∠QPS = 180°

Thus, PQRS is quadrilateral, whose opposite angles are supplementary.

Question 33: Triangle ABC, AB = 5 cm, BC = 8cm and CA =7 cm. If D and E are, respectively, the midpoints of AB and BC determine the length of DE.

Given:- AB = 5cm, BC = 8cm and CA = 7cm and D and E are the necessary midpoints of AB and BC.

To Find: The Length of DE

Using mid-point theorem DE II AC

And DE = 1/2 AC

DE= 1/2 X 7

DE = 7/2 cm

DE = 3.5 cm

Hence, the answer is 3.5 cm

Question 34: 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

Answer 34: (D) any given parallelogram

Let ABCD is the required rhombus, and P, Q, R and S are its midpoints.

In Triangle ABC, R is the midpoint of AB and Q is the midpoint of BC.

AC II RQ and RQ = 1/2 AC …..(1) {using mid-point theorem}

AC II SP and SP = 1/2 AC…..(2)

From equation 1 and 2

SPII RQ and SP = RQ

Similarly, when we take Triangle ABD and Triangle BDC, we get SR II PQ and SR = PQ.

Hence PQRS is a parallelogram

Therefore option (D) is correct.

Question 35: D and E are the mid-points of the AB and AC of DABC, and O is a point on side BC. O is joined to A. If P and Q are the midpoints of OB and respectively, DEQP is

(A) a square (B) a rectangle

(C) a rhombus (D) a parallelogram

By midpoint theorem

DE II BC….(1)

DE = 1/2 (BC)

DE = 1/2 (BP + PO + OQ + QC)

DE = 1/2 (2PO +OQ)

{ P and Q are the given midpoints of OB and OC}

DE = PO + OQ

DE = PQ …..(2)

Now in Triangle AOC,

Q and E are the midpoints of OC and AC.

EQ II AO and EQ = 1/2 AO …..(3) {Using mid-point theorem}

Similarly, in Triangle ABO, PD II AO and PD = 1/2 AO…..(4)

From equation 3 and 4

EQ II PD and EQ = PD

From equation 1 and 3

DE II PQ and DE = PQ

Hence DEQP is a parallelogram.

Therefore option (D) is correct.

Question 36: The figure formed by joining the midpoints of the sides of an 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.

Given:- ABCD is a required quadrilateral, and the points P, Q, R and S are the necessary midpoints of sides of AB, BC, CD and DA. Then, PQRS is a square.

PQ = QR = RS = PS……(1)

PR = SQ

Also, PR = BC and SQ = AB

AB = BC

Therefore, all sides are equivalent.

Therefore, ABCD is either a square or a rhombus.

In Triangle ADB by mid-point theorem

SP II DB

SP = 1/2 DB …..(2)

Similarly, in Triangle ABC, AQ = 1/2 AC …..(3)

From equation 1

PS = PQ

1/2 DB = 1/2 AC

Therefore, ABCD is a square, so diagonals of a quadrilateral are also perpendicular.

Therefore option (C) is correct.

Question 37: The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC= 32° and ∠AOB = 70°, THEN ∠DBC is equal to

(A) 24° (B) 86° (C) 38° (D) 32°

Solution.

Given ∠AOB = 70° and ∠DAC = 32°

∠ACB = 32° {AD II BC and AC is a transversal line}

∠AOB + ∠BOC = 180°

70° + ∠BOC =180°

∠BOC = 180° – 70° = 110°

In Triangle BOC, we have

∠BOC + ∠OCB + ∠CBO = 180°

110° + 32° + ∠CBO = 180°

∠CBO = 180° – 110° -32°

∠CBO =38°

∠DBC = ∠CBD = 38°

Hence option C is correct

Question 38: Which of the following is not true for a parallelogram?

(A)The opposite sides are equal

(B)The opposite angles are equal

(C) The diagonals bisect opposite angles

(D) The necessary diagonals bisect each other.

Answer 38: (C) The opposite angles are bisected by the diagonals

Parallelogram: A parallelogram is a special quadrilateral with equal opposite sides and angles.

Therefore, we can say that the opposite sides of a parallelogram are equal in length (Option A), and the required opposite angles of a parallelogram are equal measure (Option B).

In a parallelogram, diagonals bisect each other (Option D), but opposite angles are not bisected by diagonals (Option C).

Therefore option (C) is not true.

Question 39: D and E are the mid-points of the sides AB and AC respectively of Triangle ABC, DE is produced to F. To prove that CF is equal and parallel to DA, we need a piece of additional information which is

(A) ∠DAE = ∠EFC

(B) AE = EF

(C) DE = EF

Answer 39: (C) DE = EF

Let’s draw the required figure as per the question.

AE = EC {E is the required midpoint of AC}

Let DE = EF

∠AED = ∠FEC {vertically opposite angles}

Triangle ADE ≅ Triangle CFE {by SAS congruence rule}

AD = CF {by CPCT rule}

We need DE = EF.

Hence option C is correct.

Question 40: 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.

Given: ABCD is a parallelogram

OA = 3cm and OD = 2 cm

As we understand that the diagonals of a parallelogram bisect each other

AC = 2AO = 2 x 3 = 6

AC = 6 cm

DB = 2DO = 2 x 2 = 4

DB = 4 cm

Hence the lengths of AC and BD are 6 cm and 4 cm, respectively.

Question 41: Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral? Why or why not?

As we understand that the sum of the angles of a quadrilateral is 360°

Here

110° + 80° + 70° + 95° =355° ≠ 360°

These angles cannot be the given angles of a quadrilateral because the sum of these angles is not equivalent to 360°.

Question 42: In required quadrilateral ABCD ∠A +∠D = 180°. What name can be given to this quadrilateral?

Let the required quadrilateral ABCD is,

It is given that ∠A + ∠D = 180°

We can see that the sum of co-interior angles is 180°, a trapezium property.

Thus, the given quadrilateral ABCD is a trapezium.

Question 43: Diagonal of a rectangle are equal and perpendicular. Is this statement True or False? Give a reason for your answer.

Given that the diagonals of a rectangle are equal and perpendicular.

Rectangle: A rectangle is an equiangular quadrilateral, and all angles are equal.

Therefore, the required diagonals of a rectangle are equivalent but not necessarily perpendicular to each other.

Let us consider a rectangle ABCD

Consider Triangle ACD and Triangle BCD

AC = BD (The opposite sides of a rectangle are equivalent)

∠C = ∠D (90°)

AB = CD (The opposite sides of a rectangle are equivalent)

Triangle ACD ≅ Triangle BCD (SAS congruence)

Hence diagonals are equal.

Similarly, we can prove that Triangle ACB and Triangle BDA are congruent

Now, consider Triangle AOC and Triangle B0D

∠AOC = ∠BOD vertically opposite angles

Triangle AOC and Triangle BOD are also congruent.

But we cannot prove that ∠AOC = ∠BOD = 90°

Hence diagonals don’t need to bisect each other at a right angle, so they are not necessarily perpendicular to each other.

Hence the given statement is False.

Question 44:In Figure ABCD and AEFG are two parallelograms if ∠C = 55° determine ∠F.

Given: ABCD and AEFG are two parallelograms

Given ∠C = 55°

then ∠A is also 55°. {opposite angles of a parallelogram are equal}

Also, AEFG is a parallelogram, then

∠A = ∠F = 55°  {opposite angles of a parallelogram is equal}

Hence,∠F = 55° is the required answer.

Question 45: Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35°, determine ∠B.

If the required diagonals of a quadrilateral ABCD bisect each other, then it is a parallelogram.

The required sum of interior angles between the two parallel lines is 180°; that is,

∠A + ∠B = 180°

∠A = 35° (given)

35° + ∠B = 180°

∠B = 180° – 35°

∠B = 145°

Question 46:Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD.

Given: The required opposite angles of a quadrilateral ABCD are equivalent, and AB = 4

Hence, quadrilateral ABCD is a parallelogram, as we understand that opposite angles are equivalent in a parallelogram.

Furthermore, we understand that opposite sides in a parallelogram are equal.

AB = CD = 4cm

Hence, CD = 4cm

Question 47: One of the angles of a quadrilateral of 108° and the remaining three angles are equal. Find each of the three equal angles.

Given that:

One of those angles of a quadrilateral is 108°, and the remaining three angles are equal.

Let every one of the three equivalent angles be X°.

As we understand that the required sum of angles of a given quadrilateral is 360°

108° +X° + X° + X° =360°

3X° + 108° = 360°

3X° = 360° – 108°

3X° = 252°

X°=252°/3

X°= 84°

Therefore, every one of the three equal angles is 84°

Question 48: ABCD is a trapezium AB II DC and ∠A = ∠B = 45°. Find out the necessary angles C and D of the trapezium.

Answer 48: ∠D = ∠C = 135°

Given: AB II DC and ∠A = ∠E = 45°

If AB II CD and BC are transversal, then the sum of co-interior angles is 180°

∠B +∠C = 180°

45° +∠C = 180°

∠C = 180° – 45°

∠C = 135°

Similarly

∠A + ∠D = 180°

45 + ∠D = 180°

∠D= 180° – 45°

∠D = 135°

Hence angles C and D are 135° each.

Question 49: The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Given that the required angle between the two altitudes of a given parallelogram through the necessary vertex of any obtuse angle of the parallelogram is 60°

This figure shows ∠ADC and ∠ABC are obtuse angles, and DE and DF are altitudes.

In quadrilateral BEDF,∠BED = ∠BFD =90°

∠FBE = 360° – ( ∠FDE + ∠BED + ∠BFD)

= 360° – (60°+ 90° +90°)

= 360° – 240°

= 120°

It is also given that ABCD is a parallelogram

∠A + ∠B = 180° {Sum of interior angle = 180°}

∠A = 180° – B

∠A = 180° – 120°

∠A = 60°

∠C = ∠A = 60

Therefore, the angles of the given parallelogram are 60° and 120°

Question 50: ABCD is a rhombus with altitude from D to side AB bisects AB. Find the angles of the rhombus.

Let the required sides of a rhombus be AB = BC = CD = DA = x

Joining DB

Here ∠DLA = ∠DLB = 90° {DL is perpendicular bisector of AB}

AL = BL = X/2

DL = DL {common side of Triangle ADL and Triangle BDL}

Triangle ALD ≅ Triangle BLD {by SAS congruence}

Triangle ADB is an equilateral triangle.

∠ADB = ∠ABD = ∠A = 60°

Similarly, Triangle DCB is an equilateral triangle

∠BDC = ∠DBC =∠C = 60°

Also, ∠A = ∠C

∠D = ∠B…..(1)

∠A + ∠B + ∠C + ∠D = 360°

60 + B + 60 +B = 360°

2∠B= 240/2 = 120°

Hence the answer is 120°, 120°, 60°, 60°.

Question 51: E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF.

Show that BFDE is a parallelogram.

Given: ABCD is the required parallelogram and AE = CF

To prove: OE = OF

Proof: Join BD and CA

Here OA = OC and OD = OB [ABCD is parallelogram]

OA = OC…..(1)

And AE= CF…..(2) {given}

OA – AE = OC – CF {by subtracting 2 from 1}

OE = OF

Hence, BFDE is a parallelogram.

Question 52: E is the midpoint of the side AD of the trapezium ABCD with AB II CD, A-line through E drawn parallel to AB intersects BC at F. Show that F is the midpoint of BC.

Given: ABCD is trapezium and AB II DC

Constructions: Join AC which intersects EF at O

Proof: In Triangle ADC, E is the midpoint of line AD and OE II CD.

By the required mid-point theorem, O is the given midpoint of AC

∴ In Triangle CBA, as O is the midpoint of AC

OF II AB (mid-point theorem

Furthermore, by the mid-point theorem, F is the midpoint of BC.

Therefore, proved

Question 53: Through A, B and C lines RQ, PR and QP have been drawn, respectively, parallel to the given sides BC, CA and AB of a Triangle ABC as shown in Figure Show that BC= 1/2 QR

Given: Triangle ABC, PQ II AB, PR II AC, and PQ II BC.

To Prove : BC = 1/2 QR

Proof: In quadrilateral BCAR, BR II CA and BC II RA

Hence it is a parallelogram

BC = AR ……(1)

In quadrilateral BCQA, BC II AQ and AB II QC

Hence it is also a parallelogram

BC = AQ …..(2)|

Adding equations 1 and 2, we get

2BC = AR + AQ

2BC = AR+ AQ

2BC = RQ

BC = 1/2 QR

Hence proved

Question 54: D, E and F are the midpoints of the sides BC, CA and AB, respectively, of an equilateral triangle ABC. Show that Triangle is also an equilateral triangle.

D and E are the given midpoints of BC and AC, respectively, so utilising the mid-point theorem:

DE = 1/2 AB …..(1)

E and F are the given midpoints of AC and AB, so utilising the mid-point theorem:

EF = 1/2 BC …..(2)

F and D are the given midpoints of AB and BC, so utilising the mid-point theorem:

FD =1/2 AC …..(3)

It is given that Triangle ABC is an equilateral triangle

AB = BC = CA

1/2 AB = 1/2 BC = 1/2 CA

{dividing by 2}

Using 1, 2 and 3, we get

DE = EF = FD

Hence DEF is an equilateral triangle.

Hence proved

Question 55: Points P and Q have been taken on opposite sides AB and CD, respectively, of the given parallelogram ABCD such that AP = CQ (Figure). Express that AC and PQ bisect each other.

Given: ABCD is the required parallelogram and AP = CQ

To Prove that AC and PQ bisect each other.

Proof :

Here ABCD is a parallelogram

AB II DC

AP II QC

It is given that AP = CQ

Thus APCQ is the required parallelogram.

And we understand that diagonals of a parallelogram bisect each other.

Thus, AC and PQ bisect each other.

Therefore, proved

Question 56: In Figure, P is the midpoint of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. Prove that AD = 2CD

Given: ABCD is a parallelogram, and P is a midpoint of BC such that ∠BAD = ∠DAP.

Proof: Here, ABCD is a parallelogram

AD II BC and AB is transversal

∠A + ∠B = 180 {sum of co-interior angles}

∠B = 180 – ∠B = 180 – ∠A …..(1)

In Triangle ABP,

∠PAB +∠BPA + ∠B = 180 {using angle sum property of triangle}

Putting the values,

1/2 ∠A +∠BPA + (180 – ∠A) = 180

∠BPA – ∠A/2 =0

∠BPA =∠A/2 …..(2)

∠BPA =∠BAP

AB = BP {opposite sides of equal triangle}

2AB = 2BP {multiply both sides by 2}

2AB = 2AB { P is the midpoint of BC}

2CD = AD { ABCD is parallelogram AB = CD and BC = AD}

Hence proved

Question 57: A square is inscribed in an isosceles right triangle, so the square and the Triangle have one angle in common. Show that the square vertex opposite the common angle vertex bisects the hypotenuse.

Given: Now, ABC is an isosceles triangle, and ADEF is the required square inscribed in Triangle ABC.

To prove: CE = BE

Proof: In isosceles Triangle ABC and AB = AC…..(1)

∠A = 90

Subtract equation 2 from 1

AB – AD = AC – AF

BD = CF …..(3)

Now in Triangle BDE and Triangle CFE

BD = CF {from equation 3}

∠CFE = ∠EDF (90 each )

DE = EF {side of a square}

Triangle CFE ≅ Triangle BDE {SAS congruence rule}

CE = BE {by Corresponding Parts of Congruent Triangles}

Thus, vertex E of the square bisects the hypotenuse BC.

Therefore proved

Question 58: In a parallelogram ABCD, AB= 10 cm and AD = 6 cm. The bisector of ∠A meets DC and BC produced meet at F. Find the length of CF.

Given: ABCD is a parallelogram in which AB = 10 cm and AD = 6 cm.

Construction: In parallelogram ABCD, drawing the bisector of ∠A that meets DC in point E.

They are producing AE and BC so that they meet at a given point F.

Furthermore, producing AD to H and joining H and F

Here ABFH is a parallelogram

HF II AB

And ∠AFD = ∠FAB …..(1) {alternate interior angles}

AB = HF (opposite sides of a parallelogram)

AF = AF (Common Side)

Triangle HAF ≅ Triangle FAB…..(2) {SAS Congruence}

Now,∠HAF = ∠FAB…..(3) (EA is the bisector of ∠A)

∠AFD = ∠HAF { using 1 and 3 }

So, HF = AH {Sides opposite to equal angles are equal}

HF = AB = 10cm

AH = HF = 10cm

AD + DH = 10 cm

= 10 – 6

= 4 cm

Because FHDC is a parallelogram

opposite sides are equal

DH = CF = 4cm

Hence the answer is 4 cm

Question 59: P, Q, R and S are, respectively, the mid-points of the AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Verify that PQRS is the necessary rhombus.

Given: ABCD is the required quadrilateral in which the points P, Q, R and S are the given mid-points of sides AB, BC, CD and DA.

To prove : PQRS is a rhombus.

Proof :

To Triangle ADC, S and R are the mid-points of AD and DC, respectively.

By mid-point theorem, we get

SR II AC and SR = 1/2 AC …..(1)

In the given Triangle, ABC, P and Q are the mid-points of AB and BC, respectively.

Then by mid-point theorem, we get

PQII AC and PQ= 1/2 AC …..(2)

From equations 1 and 2, we get

SR = PQ = 1/2 AC..(3)

Similarly, in Triangle BCD, we get

RQ II BD and RQ = 1/2 BD …..(4)

SP II BD and SP = 1/2 BD …..(5)

Using equations 4 and 5, we get

RQ = SP = 1/2 BD

It is given that AC = BD

RQ = SP = 1/2 AC……(6)

Now equate equations 3 and 6 we get

SR = PQ = SP = RQ

It displays that all sides of a quadrilateral PQRS are equivalent.

Thus, PQRS is the necessary rhombus.

Therefore, it Proved.

Question 60: Points P, Q, R, and S are, respectively, the given mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD such that AC ⟂ BD. Verify that PQRS is a rectangle.

Given: ABCD is a quadrilateral where P, Q, R and S are the mid-points of sides AB, BC, CD and DA and AC ⟂ BD. To prove: PQRS is a rectangle

It is given that AC ⟂ BD

∠COD = ∠AOD = ∠AOB = ∠COB = 90

In Triangle ADC, S and R are the mid-points of AD and DC, so by mid-point theorem.

SR II AC and SR = 1/2 AC …..(1)

Similarly, in Triangle ABC,

PQ II AC and PQ = 1/2 AC …..(2)

Using 1 and 2

SR = PQ = 1/2 AC…..(3)

Similarly, SP II RQ & SP = RQ – 1/2 BD ……(4)

OE II FR, OF II ER

∠EOF = ∠ERF = 90

{∠COD = 90 because AC ⟂ BD and opposite angles of a parallelogram are equal}

∠SRQ = ∠ERF = ∠SPQ = 90 {Opposite angles in a parallelogram are equal}

∠SRQ = ∠ERF = ∠SPQ = 90

∠RSP + ∠SRQ + ∠RQP + ∠QRS = 360

90 + 90 + ∠RSP + ∠RQP = 360

∠RSP + ∠RQP = 180

∠RSP = ∠RQP = 90

If all the angles in a parallelogram are, then that parallelogram is a rectangle.

So, PQRS is a rectangle.

Hence Proved

Question 61: P, Q, R and S are, respectively, the midpoints of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⟂ BD. Verify that PQRS is a square.

Given: ABCD is the required parallelogram, and P, Q, R and S are the mid-points of sides AB, BC, CD and AD. Also, AC = BD and AC ⟂ BD.

To prove: PQRS is a square.

Proof: In Triangle ADC, S and R are the mid-points of sides AD and DC by mid-point theorem.

SR II AC and SR = 1/2 AC …..(1)

In the given Triangle, ABC, P and Q are the mid-points of AB and BC, respectively. Therefore, by mid-point theorem.

PQ II AC and PQ = 1/2 AC …..(2)

From equations 1 and 2, we get

PQ II SR and PQ = 1/2 AC …..(3)

Similarly, in Triangle BCD, RQ II BD and R, Q is the midpoints of CD, and CB, respectively. Therefore, by the mid-point theorem.

RQ = 1/2 BD = 1/2 AC {Given BD = AC} …..(4)

And in Triangle ABD, SP II BD and S, P are midpoints of AD, AB respectively. Therefore, by mid-point theorem

SP = 1/2 BD = 1/2 AC

{Given AC = BD…..(5)

From equations 4 and 5, we get

SP = RQ = 1/2 AC …..(6)

From equations 3 and 6, we get

PQ = SR = SP = RQ

Thus, all sides are equal

OE II FR, OF II ER

∠EOF = ∠ERF = 90°

{ ∠COD = 90° because AC ⟂ BD and opposite angles of a parallelogram are equal}

∠SRQ = ∠ERF = ∠SPQ = 90° {Opposite angles in a parallelogram are equal}

∠RSP + ∠SRQ + ∠RQP + ∠QRS = 360°

90° + 90° + ∠RSP + ∠ RQP = 360°

∠RSP + ∠RQP = 180°

∠RSP = ∠RQP = 90°

All the sides are equivalent, and all the interior angles of the quadrilateral are

Hence, PQRS is a square.

Hence Proved

Question 62: If a diagonal of a parallelogram bisects its angles. Show that it is a rhombus.

Let ABCD be a parallelogram, and diagonal AC bisect angles A and ∠CAB = ∠CAD.

To Prove: ABCD is the required rhombus.

Proof: It is given that ABCD is the given parallelogram.

AB II CD and AC is transversal

∠CAB = ∠ACD {alternate interior angles} …. (1)

Also, AD II BC and AC is a transversal

∠CAD = ∠ACB {alternate interior angle} …. (2)

Now it is given that ∠CAB = ∠CAD

So from equations (1) and (2)

∠ACD = ∠ACB…. (3)

Furthermore, ∠A = ∠C {The opposite angles of a parallelogram}

∠A/2 = ∠C /2 {dividing by 2}

∠DAC = ∠DCA {from 1 and 2}

CD = AD {sides opposite of equal angles in are equal}

AB = CD and AD = BC { opposite sides of a parallelogram}

Hence,

AB = BC = CD = AD

Hence all sides are equivalent, so ABCD is the necessary rhombus.

Therefore Proved.

Question 63: P and Q are the required mid-points of the opposite sides AB and CD of the given parallelogram ABCD. AQ intersects DP at S, and BQ intersects CP at R. Express that PRQS is the necessary parallelogram.

Given: ABCD is a parallelogram, and P and Q are the mid-points of AB and CD.

To Prove: PRQS is a parallelogram.

Proof: Since ABCD is a parallelogram, AB II CD

AP II QC

Also, AB = DC …..(1)

1/2 AB = 1/2 DC {dividing by 2}

AP = QC

{P and Q are the required mid-points of AB and DC}

As, AP = QC and AP II QC

Thus APCQ is a parallelogram

AQ II PC or SQ II PR …..(2)

AB II CD or BP II DQ

AB = DC

1/2 AB = 1/2 DC {dividing both sides by 2}

BP = QD {P and Q are the mid-points of AB and DC}

BP II QD and BQ II PD

So, BPDQ is a parallelogram

PD II BQ or PS II QR …..(3)

From equation 2 and 3

SQ II RP and PS II QR

So, PRQS is a parallelogram.

Hence Proved

Question 64: In the figure, AB II DE, AB = DE, AC II DF and AC = DF. Prove that BC II EF and BC= EF.

Given: AB II DE and AC II DF

Also, AB = DE and AC = DF

To prove: BC II EF and BC = EF

Proof: AB II DE and AB = DE

AC II DF and AC = DF

AC II FD and AC = FD …..(1)

Thus ACFD is a parallelogram

AB II DE and AB = DE …..(3)

Thus ABED is a parallelogram

From equation 2 and 4

AD = BE = CF and AD II CF II BE …..(5)

In quadrilateral BCFE, BE = CF and BE II CF {from equation 5}

Therefore, BCFE is a parallelogram BC = EF and BC II EF

Hence Proved.

Question 65: E is the midpoint of a median AD of Triangle ABC, and BE is produced to meet AC at F. Express that AF = 1/3 AC

Given: fIn Triangle ABC, AD is a median, and E is the midpoint of AD

To Prove AF = 1/3 AC.

Construction: Draw DP II EF and Triangle ABC as given

Proof: In Triangle ADP, E is the mid-point of AD and EF II DB

So, F is the mid-point of AP {converse of midpoint theorem}

In Triangle FBC, D is the midpoint of BC and DP II BF

So, P is the mid-point of FC {converse of midpoint theorem}

Thus, AF = FP = PC

AF + FP + PC = AC = 3AF

AF = 1/3 AC

Hence Proved

Question 66:Show that the quadrilateral formed by joining the midpoints of the consecutive square is also a square. Verify that all the sides are equivalent and diagonals are equivalent

Given: In the required square ABCD, P, Q, R and S are the given mid-points of AB, BC, CD and DA.

To Prove: PQRS is the given square

Construction: Joining AC and BD

Proof: Here, ABCD is a square

AB = BC = CD = AD

P, Q, R, and S are the given mid-points of AB, BC, CD and DA.

SR II AC

SR = 1/2 AC{using mid-point theorem} …..(1)

In Triangle ABC,

PQ II AC,

PQ = 1/2 AC{using mid-point theorem} …..(2)

From equations 1 and 2

SR II PQ and SR = PQ = 1/2 AC …..(3)

Similarly,

SP II BD and RQ

SP = 1/2 BD and RQ = 1/2 BD {using mid-point theorem}

SP = RQ = 1/2 BD

Since the diagonals of the required square bisect each other at right angles.

AC = BD

SP = RQ = 1/2 AC…..(4)

From equations 3 and 4

SP = PQ = SP = RQ {All sides are equivalent}

OE II FR and OF II ER

∠EOF = ∠ERF + 90

∠SRQ = ∠ERF = ∠SPQ = 90 {Opposite angles in a parallelogram are equal}

∠RSP + ∠SRQ + ∠RQP + ∠QRS = 360

90 + 90 + ∠RSP + ∠RQP = 360

∠RSP + ∠RQP = 180

∠RSP = ∠RQP = 90

Since all the given sides are equivalent and angles are also equivalent, PQRS is a square.

Hence Proved

Question 67: E and F are, respectively, the midpoints of the parallel sides AD and BC of a trapezium ABCD. Prove that EF II AB and EF = 1/2 (AB + CD)

Given: ABCD is the required trapezium in which AB II CD, E and F are the mid-points or sides of AD and BC.

Constructions: Joining BE and producing it to meet CD at G.

Drawing BOD, which intersects EF at O

To Prove: EF II AB and EF = 1/2 (AB + CD)

Proof: In Triangle GCB, E and F are respectively the mid-points of BG and BC, then by mid-point theorem.

EF II GC

But, GC II AB or CD II AB {given}

EF II AB

In Triangle ADB, AB II EO and E are the midpoint of AD. Then by the mid-point theorem, O is the midpoint of BD.

EO = 1/2 AB ……(1)

In Triangle BDC, OF II CD and O is the midpoint of BD

OF = 1/2 CD …..(2)

Adding 1 and 2, we get

EO + OF = 1/2 AB + 1/2 CD

EF = 1/2 (AB + CD)

Hence Proved

Question 68: Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle

Given: Let ABCD be the required parallelogram, and sides AP, BR, CQ, and DS are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively.

To Prove: Quadrilateral PQRS is the given rectangle.

Proof: Since ABCD is a parallelogram

Then DC II AB and DA is transversal.

∠A + ∠B = 180° {sum of co-interior angles of a parallelogram}

1/2 ∠A + 1/2 ∠B = 90° {Dividing both sides by 2}

∠APD = 90° {sum of all the angles of a triangle is 180°}

∠QRS = 90°

Similarly,

∠RBC + ∠RCB = 90°

∠BRC= 90°

∠QRS = 90°

Similarly,

∠QAB + ∠QBA = 90°

∠AQB = 90° {sum of all the angles of a triangle is 180°}

∠RQP = 90° { vertically opposite angles}

Similarly,

∠SDC + ∠SCD =90°

∠DSC = 90° {sum of all the angles of a triangle is 180°}

∠RSP = 90° { The vertically opposite angles}

Thus PQRS is a required quadrilateral whose all angles are 90°

Hence PQRS is the necessary rectangle.

Hence proved

Question 69: P and Q are the required points on opposite sides AD and BC of a parallelogram ABCD such that side PQ passes through the point of intersection O of its diagonals AC and BD. Verify that PQ is bisected at O

Given: ABCD is the required parallelogram whose diagonals bisect each other at O.

To Prove: PQ is the given bisected at O.

Proof: In Triangle ODP and Triangle OBQ

∠BOQ = ∠POD {Vertically opposite angles}

∠OBQ = ∠ODP {interior angles}

OB = OD {given}

Triangle ODP ≅ Triangle OBD {by ASA congruence}

OP = OQ {by CPCT rule}

So, PQ is bisected at O

Question 70: ABCD is a rectangle in which diagonal BD bisects ∠B; show that ABCD is a square

Given: In the required rectangle ABCD, the diagonal BD bisects B

To Prove: ABCD is the given square

Construction: Joining AC

Proof:

Given that ABCD is a rectangle. So all angles are equal to 90

Now, BD bisects ∠B

∠DBA = ∠CBD

Also,

∠DBA + ∠CBD = 90°

So,

2∠DBA = 90°

∠DBA = 45°

In Triangle ABD,

∠ABD + ∠BDA + ∠DAB = 180°

(Angle sum property)

45° + ∠BDA = 45°

∠BDA = 45°

In Triangle ABD,

AD = AB (The sides opposite to equivalent angles in a triangle are equivalent)

Similarly, we can prove that BC = CD

Thus, AB = BC = CD = DA,

Therefore, ABCD is a square.

Hence proved

Question 71: D, E and F are, respectively, the mid-points of the sides AB, BC and CA of ABC. Verify that by joining these required mid-points D, E and F, the triangles ABC is divided into four given congruent triangles

Given: In Triangle ABC, D, E and F are respectively the mid-points of the sides AB, BC and CA.

To prove: Triangle ABC is divided into four congruent triangles.

Proof: Using the given conditions, we have

AD = BD = 1/2 AB, BE = EC =1/2 BC, AF = CF = 1/2 AC

Using mid-point theorem

EF II AB and EF = 1/2 AB = AD = BD

ED II AC and ED= 1/2 AC = AF = CF

DF II BC and DF = 1/2 BC = BE = EC

In Triangle ADF and Triangle EFD

AF = DE

DF = FD {common side}

Triangle ADF ≅Triangle EFD {by SSS congruence}

Similarly, we can prove that,

Triangle DEF ≅Triangle EDB

Triangle DEF ≅ Triangle CFE

So, Triangle ABC is divided into four congruent triangles

Hence proved

Question 72: Prove that the line is joining the midpoint and that the line diagonals of a trapezium are parallel to the parallel sides of the trapezium.

Given: Let ABCD be the required trapezium in which AB II CD and let M and N be the given midpoints of diagonals AC and BD.

To Prove: MN II AB II CD

Proof: Joining CN and producing it to meet AB at E

In Triangle CDN and Triangle EBN, we have

DN = BN {N is midpoint of BD}

∠DCN = ∠BEN{alternate interior angle}

∠CDN = ∠EBN{alternate interior angles}

Triangle CDN ≅ Triangle EBN

DC = EB and CN = NE {by CPCT}

Thus in Triangle CAE, the points M and N are the mid-points of AC and CE, respectively.

MN II AE {By mid-point theorem}

MN II AB II CD

Hence Proved

Question 73: P is the required mid-point of the given side CD of a parallelogram ABCD. A necessary line through C parallel to PA intersects AB at Q and DA produced at R. Verify that DA = AR and CQ = QR.

Given: In the required parallelogram ABCD, P is the given midpoint of DC

To Prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram

Also, DC = AB and DC II AB

P is the midpoint of DC

DP = PC = 1/2 DC

Now QC II AP and PC II AQ

So APCQ is a parallelogram

AQ= PC = 1/2 DC

…1/2 AB = BQ..(1) { DC = AB}

In Triangle QAR & Triangle BQC

AQ = BQ {from equation 1}

∠QAR = ∠BQR {vertically opposite angles}

∠ARQ = ∠BCQ {alternate angles of transversal}

Triangle QAR ≅ Triangle BQC {AAS congruence}

AR = BC {by Correspond Parts Congruent Triangles}

BC = DA

AR = DA Also, CQ = QR {by Corresponding Parts Congruent Triangles} Hence Proved .

### Benefits of Solving Important Questions Class 9 Mathematics Chapter 8

Mathematics requires consistent practice. Classes 8, 9, and 10 are very important for students to develop a strong fundamental knowledge of Geometry. We recommend students get access to Extramarks comprehensive set of Important Questions Class 9 Mathematics Chapter 8. By regularly solving questions and going through our answer solutions, students will gain good confidence in solving complex problems from the Quadrilaterals chapter.

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Q.1 In the figure given below, ABCD and PQRC are rectangles and Q is the mid – point of AC. Prove that PR = ½ AC.

Marks:2
Ans

In ADC, Q is the mid-point of AC such that PQ || AD.
P is the mid-point of DC.
DP = PC [By converse of mid-point theorem]
Similarly, R is the mid-point of BC.
In BDC
PR = 1/2(BD) [By mid-point theorem]
so, PR=1/2(AC) [BD=AC]

Q.2 Find the four angles P, Q, R and S in the parallelogram PQRS as shown below.

Marks:2
Ans

InQRS,wehave QSR+ SRQ+ RQS=180’3a+6a+3a=180’12a=180’a=15 R=6Ã—15=90Sinceoppositeanglesareequalinaparallelogram.Therefore, R= P=90…1Sincethesumoftheanglesofaquadrilateralis360o.Therefore, P+ Q+ R+ S=360’2 P+ S=360 P+ S=180 S=90ByequationiHence, P= Q= R= S=90

Q.3 ABCD is a rhombus. Show that diagonals AC bisects angle A as well as angle C.

Marks:2
Ans

InABCandADC,wehaveAB=ADµABCDisarhombusBC=CD ABC Disarhombus and, AC=CA Common So,by SSS congruencecriterion, we have ABC‰…ADC BAC= DAC and ACB= ACD’AC bisectangle A as well as angleC.

Q.4 P,Q,R are, respectively, the mid points of sides AB, BC and CA and of a triangle ABC. PR and AQ meet at X. BR and PQ meet at Y. Prove that XY = ¼ AB.

Marks:4
Ans

In triangle ABC

PR || BC and PR = ½ BC ( since P and R are the mid points of AB and BC)

RQ || AB and RQ = ½ AB

PQ|| AC and PQ = ½ AC

PR|| BC and RQ || AB therefore PR || QB and RQ || BP
Therefore PRQB is a parallelogram

RB and PQ are the diagonal of parallelogram.

Therefore Y is mid point of PQ

Similarly we can show X is the mid point of RP

In triangle PRQ

X and Y are the mid points of sides PR and PQ respectively

XY || RQ and XY = ½ RQ

RQ = ½ AB (proved above)

Hence

XY = ¼ AB

Q.5 P is the mid-point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R. prove that AR = 2BC.

Marks:4
Ans

ABCD is a parallelogram

P is a mid point of AB (given)

BR || DP intersecting DC at Q

In triangle ARB

P is mid point of AB

DP|| BR

Therefore D is the mid point of AR (A line passing through the mid point of one side parallel to
other side meets the third side at mid point).

AD = BC (opposite sides of parallelogram are equal)

BC = ½ AR

AR = 2BC

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### 1. How many total chapters will students study in Class 9 Mathematics?

There are 15 chapters in Class 9 Mathematics. The list is given below:

• Chapter 1- Number System
• Chapter 2 –Polynomials
• Chapter 3 – Coordinate Geometry
• Chapter 4 –Linear Equations In Two Variables
• Chapter 5 – Introduction To Euclids Geometry
• Chapter 6 – Lines And Angles
• Chapter 7 –Triangles
• Chapter 9 –Areas Of Parallelograms And Triangles
• Chapter 10 – Circles
• Chapter 11- Constructions
• Chapter 12- Heron’s Formula
• Chapter 13-Surface Area And Volumes
• Chapter 14- Statistics
• Chapter 15- Probability

### 2. What study resources can I get from the Extramarks website?

Extramarks is one of the finest educational platforms as it has its own archive of academic resources, which helps students  ace their exams. You can obtain all the NCERT-related material like NCERT solutions, solved exemplar solutions, NCERT-based mock tests, CBSE revision notes, and Important Questions Class 9 Mathematics Chapter 8 on the Extramarks website. Apart from this, you can acquire comprehensive advice from our subject experts and doubt-clearing sessions after you sign up on our official website for any study resources.

### 3. How can I score excellent marks in chapter 8 - Quadrilaterals?

Students can score excellent marks by solving all the Mathematics Class 9 Chapter 8 Important Questions provided by our well-renowned platform Extramarks. Along with it students should refer to the official NCERT oriented textbooks and exemplar books.