Three angles of a quadrilateral are respectively equa

CHAPTER – 14

QUADRILATERALS

Exercise 14.1

Page number 14.4

Question 1: Three angles of a quadrilateral are respectively equal to

1100, 500 and 400. Find its fourth angle.

Solution:

Three angles of a quadrilateral are 1100, 500 and 400

Let the fourth angle be ‘x’

We know, sum of all angles of a quadrilateral = 3600

1100 + 500 + 400 + x0 = 3600

⇒ x = 3600 – 2000

⇒x = 1600

Therefore, the required fourth angle is 1600.

Question 2: In a quadrilateral ABCD, the angles A, B, C and D are in

the ratio of 1:2:4:5. Find the measure of each angles of the

quadrilateral.

Solution:

Let the angles of the quadrilaterals are A = x, B = 2x, C = 4x and D = 5x


A + B + C + D = 3600

x + 2x + 4x + 5x = 3600

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12x = 3600

x = 360°

12 = 300

Therefore,

A = x = 300

B = 2x = 600

C = 4x = 1200

D = 5x = 1500

Question 3: The angles of a quadrilateral are in the ratio 3:5:9:13.

Find all the angles of the quadrilateral.

Solution:

The angles of a quadrilateral are 3x, 5x, 9x and 13x respectively.

We know, sum of all interior angles of a quadrilateral = 3600

Therefore, 3x + 5x + 9x + 13x = 3600

30x = 3600

or x = 120

Hence, angles measures are

3x = 3(12) = 360

5x = 5(12) = 600

9x = 9(12) = 1080

13x = 13(12) = 1560

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Question 4: In a quadrilateral ABCD, CO and DO are the bisectors

of ∠C and ∠D respectively. Prove that ∠COD = 𝟏

𝟐 (∠A + ∠B).

Solution:

In ΔDOC,

∠CDO + ∠COD + ∠DCO = 1800 [Angle sum property of a triangle]

or 1

2∠CDA + ∠COD +

1

2∠DCB = 1800

∠COD = 1800 – 1

2(∠CDA + ∠DCB) …... (i)

Also


∠CDA + ∠DCB = 3600 – (∠DAB + ∠CBA) ……(ii)

Substituting (ii) in (i)

∠COD = 1800 – 1

2{3600 – (∠DAB + ∠CBA)}

We can also write, ∠DAB = ∠A and ∠CBA = ∠B

∠COD = 1800 − 1800 + 1

2(∠A + ∠B))

∠COD = 1

2(∠A + ∠B)

Hence Proved.

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Exercise 14.2

Page number 14.18

Question 1: Two opposite angles of a parallelogram are (3x – 2)0 and

(50 – x) 0. Find the measure of each angle of the parallelogram.

Solution:

Given: Two opposite angles of a parallelogram are (3x – 2)0 and (50 –

x) 0.

We know, opposite sides of a parallelogram are equal.

(3x – 2)0 = (50 – x) 0

3x + x = 50 + 2

4x = 52

x = 13

Angle x is 130

Therefore,

(3x – 2) 0 = (3(13) – 2) = 370

(50 – x) 0 = (50 – 13) = 370

Adjacent angles of a parallelogram are supplementary.

x + 37 = 1800

x = 1800 − 370 = 1430

Therefore, required angles are: 370, 1430, 370 and 1430.

Question 2: If an angle of a parallelogram is two-third of its adjacent

angle, find the angles of the parallelogram.

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Solution:

Let the measure of the angle be x. Therefore, measure of the adjacent

angle is 2𝑥

3.

We know, adjacent angle of a parallelogram is supplementary.

x + 2𝑥

3 = 1800

3x + 2x = 5400

5x = 5400

or x = 1080

Measure of second angle is 2𝑥

3=

2(108°)

3 = 720

Similarly measure of 3rd and 4th angles are 1080 and 720

Hence, four angles are 1080, 720, 1080, 720

Question 3: Find the measure of all the angles of a parallelogram, if

one angle is 240 less than twice the smallest angle.

Solution:

Given: One angle of a parallelogram is 240 less than twice the smallest

angle.

Let x be the smallest angle, then

x + 2x – 240 = 1800

3x – 240 = 1800

3x = 1080 + 240

3x = 2040

x = 204°

3 = 680

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So, x = 680

Another angle = 2x – 240 = 2(680) – 240 = 1120

Hence, four angles are 680, 1120, 680, 1120.

Question 4: The perimeter of a parallelogram is 22cm. If the longer

side measures 6.5cm what is the measure of the shorter side?

Solution:

Let x be the shorter side of a parallelogram.

Perimeter = 22 cm

Longer side = 6.5 cm

Perimeter = Sum of all sides = x + 6.5 + 6.5 + x

22 = 2 (x + 6.5)

11 = x + 6.5

or x = 11 – 6.5 = 4.5

Therefore, shorter side of a parallelogram is 4.5 cm

Question 5: In a parallelogram ABCD, ∠D = 135°, determine the

measures of ∠A and ∠B.

Solution:

It is given that ABCD is a parallelogram with ∠D = 135°

We know that the opposite angles of the parallelogram are equal.

Therefore,

∠𝐵 = ∠𝐷

∠𝐵 = 135°

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Also, ∠𝐴 and ∠𝐵 are adjacent angles, which must be supplementary.

Therefore,

∠𝐴 + ∠𝐵 = 180°

∠𝐴 + 135° = 180°

∠𝐴 = 180° − 135°

∠𝐴 = 45°

Hence , ∠𝐴 = 45° and ∠𝐵 = 135°.

Question 6: ABCD is a parallelogram in which ∠A = 70°. Compute

∠B, ∠C and ∠D.

Solution:

It is given that ABCD is a parallelogram with ∠A = 70°

We know that the opposite angles of the parallelogram are equal.

Therefore,

∠𝐶 = ∠𝐴

∠𝐶 = 70°

Also, ∠𝐴 and ∠𝐷 are adjacent angles, which must be supplementary.

Therefore,

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∠𝐴 + ∠𝐷 = 180°

70° + ∠𝐷 = 180°

∠𝐷 = 180° − 70°

∠𝐷 = 110°

Also, ∠𝐵 and ∠𝐷 are opposite angles of a parallelogram.

Therefore,

∠𝐵 = ∠𝐷

∠𝐵 = 110°

Hence, the angles of a parallelogram are 70°, 110°, 70° and 110°.

Question 7: In the given figure, ABCD is a parallelogram in which

∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Solution:

The figure is given as follows:

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It is given that ABCD is a parallelogram.

Thus 𝐴𝐷||𝐵𝐶

And ∠𝐷𝐵𝐶 are ∠𝐴𝐷𝐵 alternate interior opposite angles.

Therefore,

∠𝐴𝐷𝐵 = ∠𝐷𝐵𝐶

∠𝐴𝐷𝐵 = 60° …… (i)

We know that the opposite angles of a parallelogram are equal. Therefore,

∠𝐴 = ∠𝐶

Also, we have ∠𝐴 = 75°

Therefore,

∠𝐶 = 75° …… (ii)

In ∆𝐵𝐶𝐷

By angle sum property of a triangle.

∠𝐶𝐷𝐵 + ∠𝐴𝐷𝐵 + ∠𝐶 = 180°

From (i) and (ii), we get:

∠𝐶𝐷𝐵 + 60° + 75° = 180°

∠𝐶𝐷𝐵 + 135° = 180°

∠𝐶𝐷𝐵 = 180° − 135°

∠𝐶𝐷𝐵 = 45°

Hence, the required value for ∠𝐴𝐷𝐵 is 60°

And ∠𝐶𝐷𝐵 is 45°.

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Question 8: Which of the following statements are true (T) and which

are false (F)?

(i) In a parallelogram, the diagonals are equal.

(ii) In a parallelogram, the diagonals bisect each other.

(iii) In a parallelogram, the diagonals intersect each other at right

angles.

(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a

parallelogram.

(v) If all the angles of a quadrilateral are equal, it is a parallelogram.

(vi) If three sides of a quadrilateral are equal, it is a parallelogram.

(vii) If three angles of a quadrilateral are equal, it is a parallelogram.

(viii) If all the sides of a quadrilateral are equal it is a parallelogram.

Solution:

(i) Statement: In a parallelogram, the diagonals are equal.

False

(ii) Statement: In a parallelogram, the diagonals bisect each other.

True

(iii) Statement: In a parallelogram, the diagonals intersect each other at

right angles.

False

(iv) Statement: In any quadrilateral, if a pair of opposite sides is equal, it

is a parallelogram.

False

(v) Statement: If all the angles of a quadrilateral are equal, then it is a

parallelogram.

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True

(vi) Statement: If three sides of a quadrilateral are equal, then it is not

necessarily a parallelogram.

False

(vii) Statement: If three angles of a quadrilateral are equal, then it is no


False

(viii) Statement: If all sides of a quadrilateral are equal, then it is a

parallelogram.

True

Question 9: In the given figure, ABCD is a parallelogram in which

∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove

that AD = DP, PC = BC and DC = 2AD.

Solution:


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Thus,

∠𝐴 + ∠𝐵 = 180°

60° + ∠𝐵 = 180°

∠𝐵 = 180° − 60°

∠𝐵 = 120°

Opposite angles of a parallelogram are equal.

Therefore,

∠𝐷 = ∠𝐵

∠𝐷 = 120°

Also, we have AP as the bisector of

Therefore,

∠𝐷𝐴𝑃 = ∠𝐵𝐴𝑃 …… (i)

Similarly,

∠𝐴𝐵𝑃 = ∠𝑃𝐵𝐴 …… (ii)

We have 𝐷𝐶 || 𝐴𝐵,

∠𝐷𝑃𝐴 = ∠𝑃𝐴𝐵

From (i)

∠𝐷𝑃𝐴 = ∠𝑃𝐴𝐵

Thus, sides opposite to equal angles are equal.

𝐴𝐷 = 𝐷𝑃

Similarly, 𝐷𝐶 || 𝐴𝐵

∠𝐶𝑃𝐵 = ∠𝑃𝐵𝐴

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From (ii)

∠𝐶𝑃𝐵 = ∠𝑃𝐵𝐶

Thus, sides opposite to equal angles are equal.

𝑃𝐶 = 𝐵𝐶

Also,

𝐷𝐶 = 𝐷𝑃 + 𝑃𝐶

𝐷𝐶 = 𝐴𝐷 + 𝐵𝐶

𝐷𝐶 = 𝐴𝐷 + 𝐴𝐷

𝐷𝐶 = 2𝐴𝐷

Question 10: In the given figure, ABCD is a parallelogram and E is

the mid-point of side BC. If DE and AB when produced meet at F,

prove that AF = 2AB.

Solution:

Figure is given as follows:

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𝐶𝐸 = 𝐵𝐸

DE and AB when produced meet at F.

We need to prove that 𝐴𝐹 = 2𝐴𝐵

It is given that 𝐷𝐶|| 𝐴𝐵

Thus, the alternate interior opposite angles must be equal.

∠𝐷𝐶𝐸 = 𝐸𝐵𝐹

In ∆𝐷𝐶𝐸 and ∆𝐵𝐹𝐸, we have

∠𝐷𝐶𝐸 = 𝐸𝐵𝐹 (Proved above)

𝐶𝐸 = 𝐵𝐸 (Given)

∠𝐷𝐸𝐶 = ∠𝐵𝐸𝐹 (Vertically opposite angles)

Therefore,

∆𝐷𝐶𝐸 ≅ ∆𝐵𝐹𝐸 (By ASA Congruency)

By corresponding parts of congruent triangles property, we get

DC = BF …… (i)

It is given that ABCD is a parallelogram. Thus, the opposite sides should

be equal. Therefore,

𝐷𝐶 = 𝐴𝐵 …… (ii)

But,

𝐴𝐹 = 𝐴𝐵 + 𝐵𝐹

From (i), we get:

𝐴𝐹 = 𝐴𝐵 + 𝐷𝐶

From (ii), we get:

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𝐴𝐹 = 𝐴𝐵 + 𝐴𝐵

𝐴𝐹 = 2𝐴𝐵

Hence proved.

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Exercise 14.3

Page number 14.38

Question 1: In a parallelogram ABCD, determine the sum of angles

∠C and ∠D.

Solution:

In a parallelogram ABCD, ∠C and ∠D are consecutive interior angles on

the same side of the transversal CD.

So, ∠C + ∠D = 1800

Question 2: In a parallelogram ABCD, if ∠B = 1350, determine the

measures of its other angles.

Solution:

Given: In a parallelogram ABCD, if ∠B = 1350

Here, ∠A = ∠C, ∠B = ∠D and ∠A + ∠B = 1800

∠A + 1350 = 1800

∠A = 450

Answer:

∠A = ∠C = 450

∠B = ∠D = 1350

Question 3: ABCD is a square. AC and BD intersect at O. State the

measure of ∠AOB.

Solution:

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We know, diagonals of a square bisect each other at right angle.

So, ∠AOB = 900

Question 4: ABCD is a rectangle with ∠ABD = 400. Determine ∠DBC.

Solution:

Each angle of a rectangle = 90o

So, ∠ABC = 900

∠ABD = 400 (given)

Now, ∠ABD + ∠DBC = 900

400 + ∠DBC = 900

or ∠DBC = 500.

Question 5: The sides AB and CD of a parallelogram ABCD are

bisected at E and F. Prove that EBFD is a parallelogram.

Solution:



E is the mid-point of AB

Thus,

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𝐴𝐸 = 𝐵𝐸,

𝐵𝐸 =1

2𝐴𝐵 ...... (i)

Similarly,

𝐷𝐹 = 𝐹𝐶

𝐷𝐹 =1

2𝐶𝐷 ……(ii)

From (i) and (ii)

𝐷𝐹 = 𝐵𝐸

Also, 𝐷𝐹 || 𝐴𝐵

Thus, 𝐷𝐹 || 𝐵𝐸

Therefore, EBFD is a parallelogram.

Question 6: P and Q are the points of trisection of the diagonal BD of

a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also

that AC bisects PQ.

Solution:

Figure can be drawn as follows:

We have P and Q as the points of trisection of the diagonal BD of

parallelogram ABCD.

We need to prove that AC bisects PQ. That is, 𝑂𝑃 = 𝑂𝑄.

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Since diagonals of a parallelogram bisect each other.

Therefore, we get:

𝑂𝐴 = 𝑂𝐶 and 𝑂𝐵 = 𝑂𝐷

P and Q as the points of trisection of the diagonal BD.

Therefore,

𝐵𝑃 = 𝑃𝑄 and 𝑃𝑄 = 𝑄𝐷

Now, 𝑂𝐵 = 𝑂𝐷 and 𝐵𝑃 = 𝑄𝐷

Thus,

𝑂𝐵 − 𝐵𝑃 = 𝑂𝐷 − 𝑂𝑄

𝑂𝑃 = 𝑂𝑄

AC bisects PQ.

Hence proved.

Question 7: ABCD is a square E, F, G and 𝑯 are points

on AB, BC, CD

and DA respectively. such that AE = BF = CG = DH. Prove

that EFGH

is a square.

Solution:

Square ABCD is given:

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E, F, G and H are the points on AB, BC, CD and DA respectively, such

that:

𝐴𝐸 = 𝐵𝐹 = 𝐶𝐺 = 𝐷𝐻

We need to prove that EFGH is a square.

Say, 𝐴𝐸 = 𝐵𝐹 = 𝐶𝐺 = 𝐷𝐻 = 𝑥

As sides of a square are equal. Then, we can also say that:

𝐵𝐸 = 𝐶𝐹 = 𝐷𝐺 = 𝐴𝐻 = 𝑦

In ∆𝐴𝐸𝐻 and ∆𝐵𝐹𝐸,we have:

𝐴𝐸 = 𝐵𝐹 (Given)

∠𝐴 = ∠𝐵 (Each equal to 90°)

𝐵𝐸 = 𝐴𝐻 (Each equal to y)

By SAS Congruence criteria, we have:

∆𝐴𝐸𝐻 ≅ ∆𝐵𝐹𝐸

Therefore, EH = EF

Similarly, EF= FG, FG= HG and HG= HE

Thus, 𝐻𝐸 = 𝐸𝐹 = 𝐹𝐺 = 𝐻𝐺

Also,

∠1 = ∠2 and ∠3 = ∠4

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But,

∠1 + ∠3 = 90° and ∠2 + ∠4 = 90°

Therefore,

∠1 + ∠3 + ∠2 + ∠4 = 90° + 90°

∠1 + ∠4 + ∠1 + ∠4 = 180°

2(∠1 + ∠4) = 180°

∠1 + ∠4 = 90°

i.e., ∠𝐻𝐸𝐹 = 90°

Similarly,

∠𝐹 = 90°

∠𝐺 = 90°

∠𝐻 = 90°

Thus, EFGH is a square.

Hence proved.

Question 8: ABCD is a rhombus, EABF is a straight line such

that EA = AB = BF. Prove that ED and FC when produced meet at

right angles.

Solution:

Rhombus ABCD is given:

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We have

𝐸𝐴 = 𝐴𝐵 = 𝐵𝐹

We need to prove that ∠𝐸𝐺𝐹 = 90°

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore,

𝑂𝐴 = 𝑂𝐶, 𝑂𝐵 = 𝑂𝐷, ∠𝐴𝑂𝐷 = ∠𝐶𝑂𝐷 = 90°

∠𝐴𝑂𝐵 = ∠𝐶𝑂𝐵 = 90°

In ∆𝐵𝐷𝐸 A and O are the mid-points of BE and BD respectively.

By using mid-point theorem, we get:

𝑂𝐴 || 𝐷𝐸

Therefore,

𝑂𝐶 || 𝐷𝐺

In ∆𝐶𝐹𝐴, A and O are the mid-points of BE and BD respectively.

By using mid-point theorem, we get:

𝑂𝐴 || 𝐷𝐸

Therefore,

𝑂𝐷 || 𝐺𝐸

Thus, in quadrilateral DOCG, we have:

𝑂𝐶 || 𝐷𝐺 and 𝑂𝐷 || 𝐺𝑐

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Therefore, DOCG is a parallelogram.

Thus, opposite angles of a parallelogram should be equal.

∠𝐷𝐺𝐶 = ∠𝐷𝑂𝐶

Also, it is given that

∠𝐷𝑂𝐶 = 90°

Therefore,

∠𝐷𝐺𝐶 = 90°

Or,

∠𝐸𝐺𝐹 = 90°

Hence proved.

Question 9: ABCD is a parallelogram, AD is produced to E so

that DE = DC and EC produced meets AB produced in F. Prove

that BF = BC

Solution:

ABCD is a parallelogram, AD produced to 𝐸 such that 𝐷𝐸 = 𝐷𝐶.

Also, AB produced to F.

We need to prove that 𝐵𝐹 = 𝐵𝐶

In ∆ACE, D and O are the mid-points of AE and AC respectively.

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By using Mid-Point Theorem, we get:

𝐷𝑂 || 𝐸𝐶

Since, BD is a straight line and O lies on AC.

And, C lies on EF

𝑂𝐵 || 𝐶𝐹

Therefore,

𝐴𝐵 = 𝐵𝐹 …… (i)

Also, 𝐴𝐵𝐶𝐷 is a parallelogram with 𝐴𝐵 = 𝐷𝐶.

Thus,

𝐷𝐶 = 𝐵𝐹

In ∆𝐸𝐷𝐶 and ∆𝐶𝐵𝐹,we have:

𝐷𝐶 = 𝐵𝐹

∠𝐸𝐷𝐶 = ∠𝐶𝐵𝐹

∠𝐸𝐶𝐷 = ∠𝐶𝐹𝐵

So, by ASA Congruence criterion, we have:

∆𝐸𝐷𝐶 ≅ ∆𝐶𝐵𝐹

By corresponding parts of congruence triangles property, we get:

𝐷𝐸 = 𝐵𝐶

𝐷𝐶 = 𝐵𝐶

𝐴𝐵 = 𝐵𝐶

From (i) equation, we get:

𝐵𝐹 = 𝐵𝐶

Hence proved.

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Exercise 14.4

Page number 14.55

Question 1: In a ΔABC, D, E and F are, respectively, the mid points

of BC, CA and AB. If the lengths of sides AB, BC and CA are 7 cm, 8

cm and 9 cm, respectively, find the perimeter of ΔDEF.

Solution:

Given: AB = 7 cm, BC = 8 cm, AC = 9 cm

In ∆ABC,

In a ΔABC, D, E and F are, respectively, the mid points of BC, CA and

AB.

According to Midpoint Theorem:

EF = 1

2BC, DF =

1

2 AC and DE =

1

2 AB

Now, Perimeter of ∆DEF = DE + EF + DF

= 1

2 (AB + BC + AC)

= 1

2 (7 + 8 + 9)

= 12

Perimeter of ΔDEF = 12cm

Question 2: In a ΔABC, ∠A = 500, ∠B = 600 and ∠C = 700. Find the

measures of the angles of the triangle formed by joining the mid-

points of the sides of this triangle.

Solution:

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In ΔABC,

D, E and F are mid points of AB, BC and AC respectively.

In a Quadrilateral DECF:

By Mid-point theorem,

DE ∥ AC ⇒ DE = AC

2

And CF = AC

2

⇒ DE = CF

Therefore, DECF is a parallelogram.

∠C = ∠D = 700

[Opposite sides of a parallelogram]

Similarly,

ADEF is a parallelogram, ∠A = ∠E = 500

BEFD is a parallelogram, ∠B = ∠F = 600

Hence, Angles of ΔDEF are: ∠D = 700, ∠E = 500, ∠F = 600.

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Question 3: In a triangle, P, Q and R are the mid-points of sides BC,

CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 cm,

find the perimeter of the quadrilateral ARPQ.

Solution:

In ΔABC,

R and P are mid points of AB and BC

By Mid-Point Theorem

RP ∥ AC ⇒ RP = AC

2

In a quadrilateral, ARPQ

RP ∥ AQ ⇒ RP = AQ [A pair of side is parallel and equal]

Therefore, ARPQ is a parallelogram.

Now, AR = AB

2=

30

2 = 15 cm [AB = 30 cm (Given)]

AR = QP = 15 cm [Opposite sides are equal]

And RP = AC

2=

21

2 = 10.5 cm [AC = 21 cm (Given)]

RP = AQ = 10.5cm [Opposite sides are equal]

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Now,

Perimeter of ARPQ = AR + QP + RP +AQ

= 15 +15 +10.5 +10.5

= 51

Perimeter of quadrilateral ARPQ is 51 cm.

Question 4: In a ΔABC median AD is produced to X such that AD =

DX. Prove that ABXC is a parallelogram.

Solution:

In a quadrilateral ABXC,

AD = DX [Given]

BD = DC [Given]

From figure, Diagonals AX and BC bisect each other.

ABXC is a parallelogram.

Hence Proved.

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Question 5: In a ΔABC, E and F are the mid-points of AC and AB

respectively. The altitude AP to BC intersects FE at Q. Prove that AQ

= QP.

Solution:

In a ΔABC

E and F are mid points of AC and AB (Given)

EF ∥ FE ⇒ EF = 𝐵𝐶

2 and [By mid-point theorem]

In ΔABP

F is the mid-point of AB, again by mid-point theorem

FQ ∥ BP

Q is the mid-point of AP

AQ = QP

Hence Proved.

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Question 6: In a ΔABC, BM and CN are perpendiculars from B and

C respectively on any line passing through A. If L is the mid-point of

BC, prove that ML = NL.

Solution:

Given that,

In ΔBLM and ΔCLN

∠BML = ∠CNL = 900

BL = CL [L is the mid-point of BC]

∠MLB = ∠NLC [Vertically opposite angle]

By ASA criterion:

ΔBLM ≅ ΔCLN

So, LM = LN [By CPCT]

Question 7: In figure, triangle ABC is a right-angled triangle at B.

Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the

sides AB and AC respectively, calculate

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(i) The length of BC (ii) The area of ΔADE.

Solution:

In ΔABC,

∠B = 900 (Given)

AB = 9 cm, AC = 15 cm (Given)

By using Pythagoras theorem

AC2 = AB2 + BC2

⇒152 = 92 + BC2

⇒BC2 = 225 – 81 = 144

or BC = 12

Again,

AD = DB = AB

2=

9

2 = 4.5 cm [D is the mid−point of AB]

D and E are mid-points of AB and AC

DE ∥ BC ⇒ DE = BC

2 [By mid−point theorem]

Now,

Area of ΔADE = 1

2× AD × DE

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= 1

2× 4.5 × 6

= 13.5

Area of ΔADE is 13.5 cm2

Question 8: In figure, M, N and 𝐏 are mid-points of AB, AC and BC

respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate

BC, AB and AC.

Solution:

Given: MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm.

M and N are mid-points of AB and AC

By mid−point theorem, we have

MN∥BC ⇒ MN = BC

2

or BC = 2MN

BC = 6 cm [MN = 3 cm given]

Similarly,

AC = 2MP = 2 (2.5) = 5 cm

AB = 2 NP = 2 (3.5) = 7 cm

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Question 9: In the given figure, AB = AC and CP || BA and AP is the

bisector of exterior ∠CAD of ΔABC. Prove that

(i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.

Solution:

We have the following given figure:

We have 𝐴𝐵 = 𝐴𝐶 and 𝐶𝑃||𝐵𝐴 and AP is the bisector of exterior

angle ∠CAD of ΔABC.

(i) We need to prove that ∠PAC = ∠BCA

In ΔABC,

We have 𝐴𝐵 = 𝐴𝐶 (Given)

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Thus, ∠ABC = ∠BCA (Angles opposite to equal sides are equal)

By angle sum property of a triangle, we get:

∠1 + ∠BCA + ∠ABC = 180°

∠1 + ∠BCA + ∠BCA = 180°

∠1 + 2∠BCA = 180° …… (i)

Now,

∠PAC = ∠PAD (AP is the bisector of exterior angle ∠CAD)

∠1 + ∠PAC + ∠PAD = 180° (Linear Pair)

∠1 + ∠PAC + ∠PAC = 180°

∠1 + 2∠PAC = 180° …… (ii)

From equation (i) and (ii), we get:

∠1 + 2∠BCA = ∠1 + 2∠PAC

∠BCA = ∠PAC

(ii) We need to prove that ABCP is a parallelogram.

We have proved that ∠PAC =∠BCA

This means, AP || BC

Also it is given that CP || BA

We know that a quadrilateral with opposite sides parallel is a

parallelogram.

Therefore, ABCP is a parallelogram.

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Question 10: ABCD is a kite having AB = AD and BC = CD. Prove

that the figure formed by joining the mid-points of the sides, in order,

is a rectangle.

Solution:

ABCD is a kite such that AB = AD and BC = CD

Quadrilateral PQRS is formed by joining the mid-points P, Q, R and S of

sides AB, BC, CD and AD respectively.

We need to prove that Quadrilateral PQRS is a rectangle.

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

Therefore,

PQ || AC and PQ = 1

2AC

Similarly, we have

RS || AC and RS = 1

2AC

Thus,

PQ || RS and PQ = RS

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Therefore, PQRS is a parallelogram.

Now,

AB = AD

1

2 AB =

1

2AD

But, P and S are the mid-points of AB and AD

AP = AS …… (I)

In ∆ABD: P and S are the mid-point of side AB and AD

By mid-point Theorem, we get:

PS || BD

Or,

PM || BO

In ∆ABO, P is the mid-point of side AB and PM || BO

By Using the converse of mid-point theorem, we get:

M is the mid-point of AO

Thus,

PM = MS …… (II)

In ∆APM and ∆ASP, we have:

AM = AM (Common)

AP = AS [From (I)]

PM = MS [From (II)]

By SSS Congruence theorem, we get:

∆APM ≅ ∆ASP

By corresponding parts of congruent triangles property, we get:

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∠𝐴𝑀𝑃 = ∠𝐴𝑀𝑆

But,

∠𝐴𝑀𝑃 + ∠𝐴𝑀𝑆 = 180°

∠𝐴𝑀𝑆 + ∠𝐴𝑀𝑆 = 180°

2∠𝐴𝑀𝑆 = 180°

∠𝐴𝑀𝑆 = 90°

and ∠𝐴𝑀𝑃 = 90°

Therefore,

∠𝐴𝑂𝑁 = 90° (PM || BO, Corresponding angles should be equal)

Or, ∠𝑀𝑂𝑁 = 90°

We have proved that PS || BD

Similarly, PQ || AC.

Then we can say that PM || NO and PN || MO

Therefore, PMNO is a parallelogram with ∠𝑀𝑂𝑁 = 90°

Or, we can say that PMNO is a rectangle.

∠𝑀𝑃𝑁 = 90°

We get:

∠𝑆𝑃𝑄 = 90°

Also, PQRS is a parallelogram.

Therefore, PQRS is a rectangle.

Hence proved.

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Question 11: Let ABC be an isosceles triangle in

which AB = AC. If D, E, F be the mid-points of the sides

BC, CA and AB respectively, show that the

segment AD and EF bisect each other at right angles.

Solution:

∆ABC, an isosceles triangle is given with D, E and F as the mid-points

of BC, CA and AB respectively as shown below:

We need to prove that the segment AD and EF bisect each other at right

angle.

Let’s join DF and DE.

In ∆ABC, D and E are the mid-points of BC and AC respectively.

Theorem states, the line segment joining the mid-points of any two sides

of a triangle is parallel to the third side and equal to half of it.

Therefore, we get: DE || AB or DE || AF

Similarly, we can get DE || AE

Therefore, AEDF is a parallelogram

We know that opposite sides of a parallelogram are equal.

DF = AE and DE = AF

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Also, from the theorem above we get AF = 1

2𝐴𝐵

Thus, DE = 1

2𝐴𝐵

Similarly, DF = 1

2𝐴𝐶

It is given that ∆ABC, an isosceles triangle

Thus, AB = AC

Therefore, DE = DF

Also, AE = AF

Then, AEDF is a rhombus.

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore, M is the mid-point of EF and AM ⊥ BC

Hence proved.

Question 12: Show that the line segments joining the mid-points of

the opposite sides of a quadrilateral bisect each other.

Solution:

Figure can be drawn as:

Let ABCD be a quadrilateral such that P, Q, R and S are the mid-points of

side AB, BC, CD and DA respectively.

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Therefore,

PQ || AC and PQ = 1

2𝐴𝐶

Similarly, we have

RS || AC and RS = 1

2𝐴𝐶

Thus,



Since, diagonals of a parallelogram bisect each other.

Therefore, PR and QS bisect each other.

Hence proved.

Question 13: Fill in the blanks to make the following statements

correct:

(i) The triangle formed by joining the mid-points of the sides of an

isosceles triangle is........

(ii) The triangle formed by joining the mid-points of the sides of a

right triangle is ........

(iii) The figure formed by joining the mid-points of consecutive sides

of a quadrilateral is ..........

Solution:

(i) The triangle formed by joining the mid-points of the sides of an

isosceles triangle is isosceles.

Explanation:

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Figure can be drawn as: A

∆ABC, an isosceles triangle is given.

F and E are the mid-points of AB and AC respectively.

Therefore,

EF = 1

2𝐵𝐶 …… (I)

Similarly,

𝐷𝐸 =1

2𝐴𝐵 …… (II)

And

𝐹𝐷 =1

2𝐴𝐶 …… (III)

Now, ∆ABC is an isosceles triangle.

AB = AC

1

2𝐴𝐵 =

1

2𝐴𝐶

From equation (II) and (III), we get:

𝐷𝐸 = 𝐹𝐷

Therefore, in ∆ABC two sides are equal.

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Therefore, it is an isosceles triangle.

(ii) The triangle formed by joining the mid-points of the sides of a right

triangle is right triangle.

Explanation:

Figure can be drawn as: A

∆ABC right angle at B is given.

∠𝐵 = 90°

F and E are the mid-points of AB and AC respectively.

Therefore,

EF || BC …… (I)

Similarly,

DE || AB …… (II)

And

DF || CA …… (III)

Now, DE || AB and transversal CB and CA intersect them at D and E

respectively.

Therefore,

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∠𝐶𝐷𝐸 = ∠𝐵

and ∠𝐶𝐸𝐷 = ∠𝐴

Similarly, EF || BC

Therefore,

∠𝐴𝐸𝐹 = ∠𝐶

and ∠𝐴𝐹𝐸 = ∠𝐵

Similarly, DF || CA

Therefore,

∠𝐵𝐷𝐹 = ∠𝐶

∠𝐵𝐹𝐷 = ∠𝐴

Now AC is a straight line.

∠𝐴𝐸𝐹 + ∠𝐷𝐸𝐹 + ∠𝐶𝐸𝐷 = 180°

∠𝐶 + ∠𝐹𝐷𝐸 + ∠𝐴 = 180°

∠𝐹𝐷𝐸 + (∠𝐶 + ∠𝐴) = 180°

Now, by angle sum property of ∆ABC, we get:

∠𝐶 + ∠𝐴 = 180° − ∠𝐵

Therefore,

∠𝐹𝐷𝐸 + 180° − ∠𝐵 = 180°

∠𝐹𝐷𝐸 = ∠𝐵

But, ∠𝐵 = 90°

Then we have:

∠𝐹𝐷𝐸 = 90°

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(iii) The figure formed by joining the mid-points of the consecutive sides

of a quadrilateral is parallelogram.

Explanation:

Figure can be drawn as:

Let ABCD be a quadrilateral such that P, Q, R and S are the mid-points of

side AB, BC, CD and DA respectively.


Therefore,

PQ || AC and PQ = 1

2AC

Similarly, we have

RS || AC and RS = 1

2AC

Thus,



Question 14: ABC is a triangle and through A, B, C lines are drawn

parallel to BC, CA and AB respectively intersecting at P, Q and R.

Prove that the perimeter of ΔPQR is double the perimeter of ΔABC.

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Solution:

ABCQ and ARBC are parallelograms.

Therefore, BC = AQ and BC = AR

⇒AQ = AR

⇒A is the mid-point of QR

Similarly, B and C are the mid-points of PR and PQ respectively.

By mid−point theorem, we have

AB = PQ

2, BC =

QR

2 and CA =

PR

2

or PQ = 2AB, QR = 2BC and PR = 2CA

⇒ PQ + QR + RP = 2(AB + BC + CA)

⇒ Perimeter of ΔPQR = 2(Perimeter of ΔABC)

Hence proved.

Question 15: In figure, BE ⊥ AC, AD is any line from A to BC

intersecting BE in H. P, Q and 𝐑 are respectively the mid-points of

AH, AB and BC. Prove that ∠PQR = 900.

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Solution:

BE⊥AC and P, Q and R are respectively mid-point of AH, AB and BC.

(Given)

In ΔABC, Q and R are mid-points of AB and BC respectively.

By Mid-point theorem:

QR ∥ AC …. (i)

In ΔABH, Q and P are the mid-points of AB and AH respectively

QP ∥ BH …. (ii)

But, BE⊥AC

From (i) and (ii) we have,

QP ⊥ QR

⇒∠PQR = 900

Hence Proved.

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Question 16: ABC is a triangle, D is a point on AB such

that AD = 1

4 AB and E is a point on AC such that AE =

1

4 AC. Prove

that DE = 1

4 BC.

Solution:

∆ABC is given with D a point on AB such that AD = 1

4AB.

Also, E is point on AC such that AE = 1

4AC.

We need to prove that DE = 1

4BC

Let P and Q be the mid-points of AB and AC respectively.

It is given that

AD = 1

4AB and AE =

1

4AC

But, we have taken P and Q as the mid-points of AB and AC respectively.

Therefore, D and E are the mid-points of AP and AQ respectively.

In ∆ABC, P and Q are the mid-points of AB and AC respectively.



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Therefore, we get PQ || BC and PQ = 1

2𝐵𝐶 …… (i)

In ∆APQ, D and E are the mid-points of AP and AQ respectively.

Therefore, we get DE || PQ and DE = 1

2𝑃𝑄 …… (ii)


DE = 1

2𝐵𝐶

Hence proved.

Question 17: In the given figure, ABCD is a parallelogram in

which P is the mid-point of DC and Q is a point on AC such

that CQ = 𝟏

𝟒AC. If PQ produced meets BC at R, prove that R is a mid-

point of BC.

Solution:


ABCD is a parallelogram, where P is the mid-point of DC and Q is a point

on AC such that

CQ = 1

4AC.

PQ produced meets BC at R.

We need to prove that R is a mid-point of BC.

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Let us join BD to meet AC at O.


Therefore, 𝑂𝐶 =1

2𝐴𝐶

(Because diagonals of a parallelogram bisect each other)

Also, CQ = 1

4AC.

Therefore, CQ = 1

4OC

In ∆DCO, P and Q are the mid-points of CD and OC respectively.



Therefore, we get: PQ || DO

Also, in ∆COB, Q is the mid-point of OC and QR || OB

Therefore, R is a mid-point of BC.

Hence proved.

Question 18: In the given figure, ABCD and PQRC are rectangles

and Q is the mid-point of AC. Prove that

(i) DP = PC

(ii) PR = 𝟏

𝟐 AC

Solution:

Rectangles ABCD and PQRC are given as follows:

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Q is the mid-point of AC.

In ∆ACD, Q is the mid-point of AC such that PQ || AD

Using the converse of mid-point theorem, we get:

P is the mid-point of DC

That is;

DP = PC

Similarly, R is the mid-point of BC.

Now, in ∆BCD, P and R are the mid-points of DC and BC respectively.

Then, by mid-point theorem, we get:

PR = 1

2BD

Now, diagonals of a rectangle are equal.

Therefore, putting BD = AC, we get:

PR = 1

2AC

Hence Proved.

Question 19: ABCD is a parallelogram, E and F are the mid-points

of AB and CD respectively. GH is any line intersecting AD, EF and B

C at G, P and H respectively. Prove that GP = PH.

Solution:

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ABCD is a parallelogram with E and F as the mid-points

of AB and CD respectively.

We need to prove that GP = PH

Since E and F are the mid-points of AB and CD respectively.

Therefore,

BE = 1

2AB, AE = BE

And

DF = 1

2CD, DE = CF

Also, ABCD is a parallelogram. Therefore, the opposite sides should be

equal.

Thus,

AB = CD

1

2AB =

1

2CD

BE = CF

Also, BE || CF (Because AB || CD)

Therefore, BEFC is a parallelogram

Then, BE || EF and BE = PH …… (i)

Now, BE || EF

Thus, AD || EF (Because BE || AD as ABCD is a parallelogram)

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We get,

AEFD is a parallelogram

Then, we get:

AE = GP …… (ii)

But, E is the mid-point of AB.

Therefore,

AE = BE

Using (i) and (ii), we get:

GP = PH

Hence proved.

Question 20: RM and CN are perpendiculars to a line passing

through the vertex A of a triangle ABC. IF L is the mid-point of BC,

prove that LM = LN.

Solution:

In ∆ ABC BM and CN are perpendiculars on any line passing through A.

Also.

BL = LC

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We need to prove that ML = NL

From point L let us draw LP ⊥ AN

It is given that BM ⊥ AN, LP ⊥ AN and CN ⊥ AN

Therefore,

BM || LP || CN

Since, L is the mid-points of BC,

Therefore, intercepts made by these parallel lines on MN will also be

equal

Thus,

MP = NP

Now in ∆LMN,

MP = NP

And LP ⊥ AN. Thus, perpendicular bisects the opposite sides.

Therefore, ∆LMN is isosceles.

Hence ML = NL

Hence proved.

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VERY SHORT ANSWER TYPES QUESTION (VSAQs)

Page number 14.62

Question 1: In a parallelogram ABCD, write the sum of angles A and

B.

Solution:

In parallelogram ABCD, Adjacent angles of a parallelogram are

supplementary.

Therefore, ∠A + ∠B = 1800

Question 2: In a parallelogram ABCD, if ∠D = 1150, then write the

measure of ∠A.

Solution:

In a parallelogram ABCD,

∠D = 1150 (Given)

Since, ∠A and ∠D are adjacent angles of parallelogram.

We know, Adjacent angles of a parallelogram are supplementary.

∠A + ∠D = 1800

∠A = 1800 – 1150 = 650

Measure of ∠A is 650.

Question 3: PQRS is a square such that PR and SQ intersect at O.

State the measure of ∠POQ.

Solution:

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PQRS is a square such that PR and SQ intersect at O. (Given)

We know, diagonals of a square bisects each other at 90 degrees.

So, ∠POQ = 900

Question 4: If PQRS is a square, then write the measure of ∠SRP.

Solution:

PQRS is a square.

⇒ All side are equal, and each angle is 90o degrees and diagonals bisect

the angles.

So, ∠SRP = 1

2 (90 o) = 45o

Question 5: If ABCD is a rhombus with ∠ABC = 56o, find the measure

of ∠ACD.

Solution:

In a rhombus ABCD,

∠ABC = 56 o

So, ∠BCD = 2 (∠ACD)

(Diagonals of a rhombus bisect the interior angles)

or ∠ACD = 1

2 (∠BCD) …..(1)

We know, consecutive angles of a rhombus are supplementary.

∠BCD + ∠ABC = 180o

∠BCD = 180o – 56o = 124o

Equation (1) ⇒ ∠ACD = 1

2× 124o = 62o

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Question 6: The perimeter of a parallelogram is 22 cm. If the longer

side measure 6.5 cm, what is the measure of shorter side?

Solution:

Perimeter of a parallelogram = 22 cm. (Given)

Longer side = 6.5 cm

Let x be the shorter side.

Perimeter = 2x + 2 × 6.5

22 = 2x + 13

2x = 22 – 13 = 9

or x = 4.5

Measure of shorter side is 4.5 cm.

Question 7: If the angles of a quadrilateral are in the ratio 3:5:9:13,

then find the measure of the smallest angle.

Solution:

Angles of a quadrilateral are in the ratio 3: 5: 9: 13 (Given)

Let the sides are 3x, 5x, 9x, 13x

We know, sum of all the angles of a quadrilateral = 360o

3x + 5x + 9x + 13x = 360 o

30 x = 360 o

x = 12 o

Length of smallest angle = 3x = 3(12) = 36o.

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Question 8: In a parallelogram ABCD, if ∠A = (3x − 20) °, ∠B = (y +

15) °, ∠C = (x + 40) °, then find the values of x and y.

Solution:

In parallelogram ABCD, ∠A and ∠C are opposite angles.

We know that in a parallelogram, the opposite angles are equal.

Therefore,

∠C = ∠A

We have ∠A = (3x − 20) ° and ∠C = (x + 40) °

Therefore,

𝑥 + 40° = 3𝑥 − 20°

𝑥 − 3𝑥 = −40° − 20°

−2𝑥 = −60°

𝑥 = 30°

Therefore,

∠𝐴 = (3𝑥 − 20)°

∠𝐴 = [3(3𝑥 − 20)]°

∠𝐴 = 70°

Similarly,

∠𝐶 = 70°

Also, ∠𝐵 = (𝑦 + 15)°

Therefore,

∠𝐷 = ∠𝐵

∠𝐷 = (𝑦 + 15)°

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By angle sum property of a quadrilateral, we have:

∠𝐴 + ∠𝐵 + ∠𝐶 + ∠𝐷 = 360°

70° + (𝑦 + 15)° + 70° + (𝑦 + 15)° = 360°

140° + 2(𝑦 + 15)° = 360°

2(𝑦 + 15)° = 360° − 140°

2(𝑦 + 15)° = 220°

(𝑦 + 15)° = 110°

𝑦 = 95°

Hence the required values for x and y are 30° and 95° respectively.

Question 9: If measures opposite angles of a parallelogram are (60 −

x) ° and (3x − 4) °, then find the measures of angles of the

parallelogram.

Solution:

Let ABCD be a parallelogram, with ∠𝐴 = (60 − 𝑥)° and ∠𝐶 =(3𝑥 − 4)°.

We know that in a parallelogram, the opposite angles are equal.

Therefore,

∠𝐴 = ∠𝐶

60 − 𝑥 = 3𝑥 − 4

−𝑥 − 3𝑥 = −4 − 60

4𝑥 = −64

𝑥 = 16

Thus, the given angles become

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∠𝐴 = (60 − 𝑥)°

= (60 − 16)°

= 44°

Similarly,

∠𝐶 = 44°

Also, adjacent angles in a parallelogram form the consecutive interior

angles of parallel lines, which must be supplementary.

Therefore,

∠𝐴 + ∠𝐵 = 180°

44° + ∠𝐵 = 180°

∠𝐵 = 180° - 44°

∠𝐵 = 136°

Similarly,

∠𝐷 = ∠𝐵

∠𝐷 = 136°

Thus, the angles of a parallelogram are 44°, 136°, 44° and 136°.

Question 10: In a parallelogram ABCD, the bisector of ∠A also

bisects BC at X. Find AB: AD.

Solution:

Parallelogram ABCD is given as follows:

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We have AX bisects ∠A bisecting BC at X.

That is, BX = CX

We need to find AB: AD

Since, AX is the bisector ∠A

That is,

∠1 = 1

2∠A …… (i)

Also, ABCD is a parallelogram

Therefore, AD || BC and AB intersects them

∠𝐴 + ∠𝐵 = 180°

∠𝐵 = 180° − ∠𝐴 …… (ii)

In ∆ABX, by angle sum property of a triangle:

∠1 + ∠2 + ∠𝐵 = 180°


1

2∠𝐴 + ∠2 + 180° − ∠𝐴 = 180°

∠2 +1

2∠𝐴 = 180°

∠2 +1

2∠𝐴 …… (iii)

From (i) and (iii), we get:

∠1 = ∠2

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Sides opposite to equal angles are equal. Therefore,

BX = AB

2BX = 2AB

As X is the mid-point of BC. Therefore,

BC = 2AB

Also, ABCD is a parallelogram, then, BC = AD

AD = 2AB

Thus,

AB: AD = AB: 2AB

AB: AD = 1: 2

Hence the ratio of AB: AD is 1: 2.

Question 11: In the given figure, PQRS is an isosceles trapezium. Find

x and y.

Solution:

Trapezium is given as follows:

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We know that PQRS is a trapezium with SR || PQ

Therefore,

∠𝑃 + ∠𝑆 = 180°

2𝑥 + 3𝑥 = 180°

5𝑥 = 180°

𝑥 =180°

5

𝑥 = 36°

Hence, the required value for x is 36°.

Question 12: In the given figure, ABCD is a trapezium. Find the

values of x and y.

Solution:


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We know that ABCD is a trapezium with AB || DC

Therefore,

∠𝐴 + ∠𝐷 = 180°

It is given that ∠𝐴 = 𝑥 + 20° and ∠𝐷 = 2𝑥 + 10°.

(𝑥 + 20)° + (2𝑥 + 10)° = 180°

3𝑥 + 30° = 180°

3𝑥 = 180° − 30°

3𝑥 = 150°

𝑥 = 50°

Similarly,

𝑦 + 92° = 180°

𝑦 = 180° − 92°

𝑦 = 88°

Hence, the required values for x and y is 50° and 88° respectively.

Question 13: In the given figure, ABCD and AEFG are two

parallelograms. If ∠C = 58°, find ∠F.

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Solution:

ABCD and AEFG are two parallelograms as shown below:

Since ABCD is a parallelogram, with ∠C = 58°

We know that the opposite angles of a parallelogram are equal.

Therefore,

∠𝐴 = ∠𝐶

∠𝐴 = 58°

Similarly, AEFG is a parallelogram, with ∠𝐴 = 58°

We know that the opposite angles of a parallelogram are equal.

Therefore,

∠𝐹 = ∠𝐶

∠𝐹 = 58°

Hence, the required measure for ∠𝐹 is 58°.

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Question 14: Complete each of the following statements by means of

one of those given in brackets against each:

(i) If one pair of opposite sides are equal and parallel, then the figure

is ........................ (parallelogram, rectangle, trapezium)

(ii) If in a quadrilateral only one pair of opposite sides are parallel,

the quadrilateral is ................ (square, rectangle, trapezium)

(iii) A line drawn from the mid-point of one side of a triangle ..............

another side intersects the third side at its mid-point.

(perpendicular to parallel to, to meet)

(iv) If one angle of a parallelogram is a right angle, then it is

necessarily a ................. (rectangle, square, rhombus)

(v) Consecutive angles of a parallelogram are ...................

(supplementary, complementary)

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it

is necessarily a ............... (rectangle, parallelogram, rhombus)

(vii) If opposite angles of a quadrilateral are equal, then it is

necessarily a .................... (parallelogram, rhombus, rectangle)

(viii) If consecutive sides of a parallelogram are equal, then it is

necessarily a .................. (kite, rhombus, square)

Solution:

(i) If one pair of opposite sides are equal and parallel, then the figure is

parallelogram.

Reason:

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In ∆ABC and ∆CDA,

AB = DC (Given)

AC = AC (Common)

∠𝐵𝐴𝐶 = ∠𝐷𝐶𝐴 (Because AB || CD, Alternate interior angles are equal)

So, by SAS Congruence rule, we have

∆ABC ≅ ∆CDA

Also,

∠𝐵𝐶𝐴 = ∠𝐷𝐴𝐶 (Corresponding parts of congruent triangles are equal)

But, these are alternate interior angles, which are equal.

AD || BC

Thus, AB || CD and AD || BC

Hence, quadrilateral ABCD is parallelogram

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the

quadrilateral is trapezium.

(iii) A line drawn from the mid-point of one side of a triangle parallel to

another side intersects the third side at its mid-point.

Reason:

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This is a theorem.

(iv) If one angle of a parallelogram is a right angle, then it is necessarily

a rectangle.

Reason:

Let ABCD be the given parallelogram.

We have, ∠𝐴 = 90°

In a parallelogram, opposite angles are equal.

Therefore,

∠𝐶 = 90°

Similarly,

∠𝐴 + ∠𝐷 = 180°

90° + ∠𝐷 = 180°

∠𝐷 = 90°

Also, ∠𝐵 = 90°

Thus, a parallelogram with all the angles being right angle and opposite

sides being equal is a rectangle.

(v) Consecutive angles of a parallelogram are supplementary.

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Reason:

Let ABCD be the given parallelogram.

Thus, AB || DC.

Therefore, ∠𝐴 + ∠𝐷 = 180°

Consecutive angles ∠𝐴 and ∠𝐷 are supplementary.

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is


Reason:

ABCD is a quadrilateral in which AB = CD and BC = DA.

We need to show that ABCD is a parallelogram.

In ∆ABC and ∆CDA, we have

AC = CA (Common)

CB = AD (Given)

AB = CD (Given)

So, by SSS criterion of congruence, we have

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∆ACB ≅ ∆CAD

By corresponding parts of congruent triangles property.

∠𝐶𝐴𝐵 = ∠𝐴𝐶𝐷 …… (i)

And ∠𝐴𝐶𝐵 = ∠𝐶𝐴𝐷

Now lines AC intersects AB and DC at A and C, such that

∠𝐶𝐴𝐵 = ∠𝐴𝐶𝐷 (From (i))

That is, alternate interior angles are equal.

Therefore, AB || DC.

Similarly, AD || BC.

Therefore, ABCD is a parallelogram.

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily

a parallelogram.

Reason:

ABCD is a quadrilateral in which ∠𝐴 = ∠𝐶 and ∠𝐵 = ∠𝐷.

We need to show that ABCD is a parallelogram.

In quadrilateral ABCD, we have

∠𝐴 = ∠𝐶

∠𝐵 = ∠𝐷

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Therefore,

∠𝐴 + ∠𝐵 = ∠𝐶 + ∠𝐷 …… (i)

Since sum of angles of a quadrilateral is 360°

∠𝐴 + ∠𝐵 + ∠𝐶 + ∠𝐷 = 360°

From equation (i), we get:

(∠𝐴 + ∠𝐵) + (∠𝐴 + ∠𝐵) = 360°

2(∠𝐴 + ∠𝐵) = 360°

∠𝐴 + ∠𝐵 = 180°

Similarly, ∠𝐶 + ∠𝐷 = 180°

Now, line AB intersects AD and BC at A and B respectively

Such that ∠𝐴 + ∠𝐵 = 180°

That is, sum of consecutive interior angles is supplementary.

Therefore, AD || BC.

Similarly, we get AB || DC.

Therefore, ABCD is a parallelogram.

(viii) If consecutive sides of a parallelogram are equal, then it is

necessarily a rhombus.

We have ABCD, a parallelogram with AB = BC.

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Since ABCD is a parallelogram.

Thus, AB = CD

And BC = AD

But, AB = BC

Therefore, all four sides of the parallelogram are equal, then it is a

rhombus.

Question 15: In a quadrilateral ABCD, bisectors of angles A and B

intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.

Solution:

∠AOB = 75o (given)

In a quadrilateral ABCD, bisectors of angles A and B intersect at O, then

∠AOB = 1

2 (∠ADC + ∠ABC)

or ∠AOB = 1

2 (∠D + ∠C)

By substituting given values, we get

75 o = 1

2 (∠D + ∠C)

or ∠C + ∠D = 150 o

Question 16: The diagonals of a rectangle ABCD meet at O. If ∠BOC

= 44o, find ∠OAD.

Solution:

ABCD is a rectangle and

∠BOC = 44o (given)

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∠AOD = ∠BOC (vertically opposite angles)

∠AOD = ∠BOC = 44o

∠OAD = ∠ODA (Angles facing same side)

and OD = OA

Since sum of all the angles of a triangle is 180 o, then

So, ∠OAD = 1

2 (180 o – 44 o) = 68 o

Question 17: If ABCD is a rectangle with ∠BAC = 32o, find the

measure of ∠DBC.

Solution:

ABCD is a rectangle and ∠BAC = 32o (given)

We know, diagonals of a rectangle bisect each other.

AO = BO

∠DBA = ∠BAC = 32o (Angles facing same side)

Each angle of a rectangle = 90 degrees

So, ∠DBC + ∠DBA = 90o

or ∠DBC + 32o = 90o

or ∠DBC = 58o

Question 18: If the bisectors of two adjacent angles A and B of a

quadrilateral ABCD intersect at a point O such that ∠C +

∠D = k ∠AOB, then find the value of k.

Solution:

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The quadrilateral can be drawn as follows:

We have AO and BO as the bisectors of angles ∠𝐴 and ∠𝐵 respectively.

In ∆AOB, we have,

∠AOB + ∠1 +∠2 = 180°

∠AOB = 180° – (∠1 +∠2)

∠AOB = 180° – (1

2∠𝐴 +

1

2∠𝐵)

∠AOB = 180° – 1

2(∠𝐴 + ∠𝐵) …… (I)

By angle sum property of a quadrilateral, we have:

∠𝐴 + ∠𝐵 + ∠𝐶 + ∠𝐷 = 360°

∠𝐴 + ∠𝐵 = 360° − (∠𝐶 + ∠𝐷)

Putting in equation (I):

∠AOB = 180° – 1

2[360° − (∠𝐶 + ∠𝐷)]

∠AOB = 180° – 180° +(∠𝐶+∠𝐷)

2

∠AOB = 1

2(∠𝐶 + ∠𝐷)

(∠𝐶 + ∠𝐷) = 2∠AOB …… (II)

On comparing equation (II) with

(∠𝐶 + ∠𝐷) = 𝑘∠AOB

We get k = 2.

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Hence, the value for k is 2.

Question 19: In the given figure, PQRS is a rhombus in which the

diagonal PR is produced to T. If ∠SRT = 152°, find x, y and z.

Solution:

Rhombus PQRS is given.

Diagonal PR is produced to T.

Also, ∠SRT = 152°.

We know that in a rhombus, the diagonals bisect each other at right angle.

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Therefore,

y = 90°

Now,

∠1 + ∠SRT = 180°

∠1 + 152° = 180°

∠1 = 28°

In ∆SOR, by angle sum property of a triangle, we get:

∠1 + y + ∠OSR = 180°

28° + 90° + ∠OSR = 180°

118°+ ∠OSR = 180°

∠OSR = 62°

Or, ∠OSR = 62° (Because O lies on SQ)

We have, SR || PQ. Thus the alternate interior opposite angles must be

equal.

Therefore,

𝑥 = ∠𝑄𝑆𝑅

𝑥 = 62°

In ∆SPR, we have

Since opposite sides of a rhombus are equal.

Therefore,

PS = SR

Also,

Angles opposite to equal sides are equal.

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Thus,

z = ∠1

But ∠1 = 28°

Thus, z = 28°

Hence the required values for x, y and z are 62°, 90° and 28° respectively.

Question 20: In the given figure, ABCD is a rectangle in which

diagonal AC is produced to E. If ∠ECD = 146°, find ∠AOB.

Solution:

ABCD is a rectangle

With diagonal AC produced to point E.

We have

∠1 + ∠𝐷𝐶𝐸 = 180° (Linear pair)

∠1 + 146° = 180°

∠1 = 34°

We know that the diagonals of a parallelogram bisect each other.

Thus OC = OD

Also, angles opposite to equal sides are equal.

Therefore,

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∠𝑂𝐷𝐶 = 34°

By angle sum property of a triangle

∠𝑂𝐷𝐶 + ∠1 + ∠𝐶𝑂𝐷 = 180°

34° + 34° + ∠𝐶𝑂𝐷 = 180°

68° + ∠𝐶𝑂𝐷 = 180°

∠𝐶𝑂𝐷 = 112°

Also, ∠𝐶𝑂𝐷 and ∠𝐴𝑂𝐵 are vertically opposite angles.

Therefore, ∠𝐴𝑂𝐵 = 112°

Hence, the required measure for ∠𝐴𝑂𝐵 is 112°.

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Three angles of a quadrilateral are respectively equa

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