Explore topic-wise InterviewSolutions in .

1. The vertices of the pentagon are points A, B, C, D and E.

2. The sides of the pentagon are segments AB, BC, CD, DE and EA.

3. The angles of the pentagon are ∠ABC, ∠BCD, ∠CDE, ∠DEA and ∠EAB.

4. The players shown in the above figure form a pentagon. The players are standing on the vertices.

The folds are called diagonals of the quadrilateral.

Two sides of a quadrilateral, which have a common end point, are called consecutive sides. In the diagram,

AB and BC is one pair of consecutive sides.

BC, CD; CD, DA; and DA, AB are the other three pairs of consecutive sides.

Two angles, which do not include a side in their intersection, are called the opposite angles of a quadrilateral.

Two angles of a quadrilateral, which include a side in their intersection, are called consecutive angles.

We can state the defining property of a parallelogram as follows:

"If a quadrilateral is a parallelogram, then its opposite sides are equal".

Converse

"If both pairs of opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram".

The converse statement stated above is a necessary condition for a quadrilateral to be a parallelogram. Similarly, we may formulate the following two other conditions for a quadrilateral to be a parallelogram.

• "If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram".

• "If either pair of opposite sides of a quadrilateral are equal and parallel, the quadrilateral is a parallelogram".

As diagonals of the parallelogram ABCD are equal, it is a rectangle. Therefore, ∠ABC = 90º

This statement is false, because diagonals of a rhombus are perpendicular but not equal to each other.

Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.

As the sum of all interior angles of a quadrilateral is 360º,

∴ 3x + 5x + 9x + 13x = 360º

30x = 360º

x = 12º

Hence, the angles are

3x = 3 × 12 = 36º

5x = 5 × 12 = 60º

9x = 9 × 12 = 108º

13x = 13 × 12 = 156º

Data: ABCD is a parallelogram. Its diagonals are AC and BD. They meet at ‘O’. 

To Prove: i) AO = OC BO = OD 

ii) ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°. 

Proof: In ∆ABC and ∆ABD, 

BC = AD (diagonals of square is equal) 

∠ABC = ∠BAD = 90° 

(angles of square) 

AB is common. 

∴ ∆ABC ≅ ∆ABD (SAS Postulate.) 

In ∆AOB and ∆COD, 

AB = DC (sides of square) 

∠OAB = ∠OCD (alternate angles) 

∠OBA = ∠ODC (alternate angles) 

∴ ∆AOB ≅ ∆COD (ASA Postulate) 

AO = OC 

BO = OD …………….. (i) 

Similarly, ∆AOB ≅ ∆BOC. 

∴ ∠AOB = ∠BOC = 90° 

Now, ∆COD ≅ ∆AOD, then 

∴ ∠COD = ∠DOA = 90° 

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90° …………. (ii) 

from (i) and (ii) 

∴ Sides of a square are equal and bisect at right angles.

Data : ABCD is a parallelogram and diagonals AC and BD bisects at right angles at O’. 

To Prove: ABCD is a rhombus. 

Proof: Here, AC and BD bisect each other at right angles. 

∴ AO = OC 

BO = OD 

and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° 

If sides are euqal to each other, then ABCD is said to be a rhombus. 

Now, ∆AOD and ∆COD, AO = OC (Data) 

∠AOD = ∠COD = 90° (Data) OD is common. 

∴ ∆AOD ≅ ∆COD (SAS Postulate) 

∴ AD = CD …………… (i) 

Similarly, ∆AOD = ∆AOB 

AD = AB ………… (ii) 

∆AOB ≅ ∆COB 

∴ AB = BC ……….. (iii) 

∆COB ≅ ∆COD 

∴ BC = CD ……………. (iv) 

From (i), (ii), (iii) and (iv), 

AB = BC = CD = AD 

All 4 sides of parallelogram ABCD are equal, then it is rhombus.

Data : Diagonals of a parallelogram are equal. 

To Prove : ABCD is a rectangle. 

Proof: Now ABCD is a parallelogram and diagonal AC = Diagonal BD (Data) 

In ∆ABC and ∆ABD, 

BC = AD (Opposite sides of quadrilateral) 

AC = BD (Data) AB common. 

∴ ∆ABC ≅ ∆ABD (SSS postulate) 

∠ABC = ∠BAD But, 

∠ABC + ∠BAD = 180° 

∠ABC + ∠ABC = 180° 

2 ∠ABC = 180° 

∴ ∠ABC = 90° 

If an angle of a parallelogram is right angle, it is called rectangle. 

∴ ABCD is a rectangle.

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

Let ∠A+ ∠B : ∠C : ∠D = 3 : 5 : 9 : 13. 

Sum of ratio = 3x + 5x + 9x + 13x = 30x 

Sum of 4 angles of quadrilateral is 360° 

∴ ∠A + ∠B + ∠C + ∠D = 360° 

3x + 5x + 9x + 13x = 360 

30x = 360 ∴ x = = 12° 

∠A = 3x = 3 × 12 = 36° 

∠B = 5x = 5 × 12 = 60° 

∠C = 9x = 9 × 12 = 108° 

∠D = 13x = 13 × 12 = 156°

Since, EAB is a straight line. 

∴ ∠DAE + ∠DAB = 180⁰ 

⇒ 73⁰ + ∠DAB = 180⁰ 

 i.e., ∠DAB = 180⁰ - 73⁰ = 1070 

Since, the sum of the angles of quadrilateral ABCD is 360⁰ 

∴ 107⁰ + 105⁰ + x + 80⁰ = 360⁰ 

⇒ 292⁰ + x = 360⁰ 

⇒ x = 360⁰ - 292⁰ 

⇒ x = 68⁰

Suppose the four angles of the quadrilateral are 3x, 5x, 9x and 13x.

∴ 3x + 5x + 9x + 13x = 360°

(by angle sum property of a quadrilateral)

⇒ 30x = 360°

⇒ x = 12°

Therefore, angles of the quadrilateral are 3 x 12°, 5 x 12°, 9 x 12° and 13 x 12° 

i.e., 36°, 60°, 108° and 156°.

Given the ratio between the angles of the quadrilateral = 3 : 5 : 9 : 13 and 3 + 5 + 9 + 13 = 30 

Since, the sum of the angles of the quadrilateral = 360⁰ 

∴ First angle of it = 3/30 × 360⁰ = 36⁰ , 

 Second angle = 5/30 × 360⁰ = 60⁰ , 

 Third angle = 9/30 × 360⁰ = 108⁰ , 

 And, Fourth angle = 13/30 × 360⁰ = 156⁰ 

∴ The angles of quadrilateral are 360⁰ , 60⁰ , 108⁰ and 156⁰ . 

Given, 

PQRS is a rhombus 

∠SRT = 152° 

We know that diagonals of a rhombus bisects each other at 90° 

Hence, 

∠SOR = 90° = y = 90° 

∠SRT+∠SRO = 180° 

(linear pair of angles) 

= 152°+∠SRO = 180° 

∠SRO = 180° - 152°= 28° 

Now, 

∠SRO = z 

∵SR = SP 

= z = 28°

And, 

∠RSO +∠SRO + y = 180° 

(angle sum property of triangle) 

= ∠RSO + 28° + 90° = 180° 

= ∠RSO = 180°-118° = 62° 

= ∠X = ∠RSO = 62° 

(alternate angles)

Given, 

In parallelogram ABCD 

∠A+∠B = 180° 

(Adjacent angles of a parallelogram are supplementary)

Data: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. 

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

Proof: ABCD is a parallelogram. 

AB || DC and AB = DC 

\(\frac{1}{2}\)AB = \(\frac{1}{2}\)DC 

AE = CF 

and AE || CF (∵ AB || CD) 

In AECF quadrilateral, 

∴ AE || CF and AE = CF. 

∴ AECF is a parallelogram. 

∴ AF || EC 

In ∆DQC, 

PF || QC (∵ AF || EC) 

∴ P is the mid-point of DQ. 

∴ DP = PQ ………….. (i) 

In ∆APB, EQ || AP. 

But E is the mid-point of AB (Data) 

∴ Q is the mid-point of PB. 

∴ PQ = QB …………… (ii) 

From (i) and (ii), DP = PQ = QB 

∴ AF and EC line segments trisects the diagonal BD.

Correct option is (C) 4 right angles

Sum of four interior angles of a quadrilateral is \(360^\circ.\)

\(=4\times90^\circ\) = 4 right angles

Hence, the sum of four angles of a quadrilateral is 4 right angles.

C) 4 right angles

Correct option is: D) Trapezium

Correct option is: D) 4

Correct option is (D) Diagonals bisect each other

In a parallelogram,

(i) Diagonals bisect each other.

(ii) Diagonals may not be equal.

(iii) Opposite angles are equal. Each angle may not be \(90^\circ\)  (or equal)

(iv) Each side may not be equal.

D) Diagonals bisect each other

Correct option is (B) All trapeziums are parallelograms.

(i) & (ii) Every parallelogram is a trapezium.

(Because opposite sides are parallel in a parallelogram)

But every trapezium need not be parallelogram because one pair of opposite sides are parallel in a trapezium but need not the other pair of opposite side be parallel.

(iii) All parallelograms are quadrilateral but all quadrilateral need not be a parallelogram.

(iv) A square is always a rhombus.

B) All trapeziums are parallelograms.

Correct option is (A) Parallelogram

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

A) Parallelogram

Correct option is (C) Congruent triangles

A diagonal of a parallelogram divides it into two congruent triangles.

C) Congruent triangles

The statement “diagonals of a rectangle are equal and perpendicular” is false.

We know that,

Diagonals of a rectangle bisect each other.

Therefore, they are equal but they are not perpendicular.

Hence, the statement is not true.

According to the question,

All the angles of a quadrilateral are equal.

Suppose all the angles of the quadrilateral = x

We know that,

Sum of all angles of a quadrilateral = 360°

⇒ x + x + x + x = 360°

⇒ 4x = 360°

⇒ x = 360°/4

⇒ x = 90°

Hence, the quadrilateral is a rectangle.

According to the question,

In quadrilateral ABCD, ∠A + ∠D = 180º

We know that,

In a trapezium,

Sum of co-interior angles = 180°

Hence, the given quadrilateral is a trapezium.

The angles 110º, 80º, 70º and 95º cannot be the angles of a quadrilateral.

Justification:

We know that,

Sum of all angles of a quadrilateral = 360°

Sum of given angles,

110° + 80° + 70° + 95° = 355° ≠ 360°

Hence, 110°, 80°, 70° and 95° cannot be the angles of a quadrilateral.

Solution: (i) It can be , but not always as you need to look for other criteria as well.
(ii) In a parallelogram opposite sides are always equal, here AD BC, so its not a parallelogram.
(iii) Here opposite angles are not equal, so it is not a parallelogram.

Solution: (a) Triangle is the polygon with minimum number of sides and an equilateral triangle is a regular polygon because all sides are equal in this. We know that each angle of an equilateral triangle measures 60 degree. Hence, 60 degree is the minimum possible value for internal angle of a regular polygon.

(b) Each exterior angle of an equilateral triangle is 120 degree and hence this the maximum possible value of exterior angle of a regular polygon. This can also be proved by another principle; which states that each exterior angle of a regular polygon is equal to 360 divided by number of sides in the polygon. If 360 is divided by 3, we get 120.

Solution: Here, each interior angle = 22°
Hence, each exterior angle = 180° - 22° = 158°
Hence, number of sides = 360°/158° = 2.27

Since, answer is not a whole number, thus, a regular polygon with measure of each interior angle as 22⁰ is not possible. 
>Hence, Answer = no

Measure of each interior angle = 165°

Measure of each exterior angle = 180° − 165° = 15°

The sum of all exterior angles of any polygon is 360º.

Thus, number of sides of the polygon 

= 360°/15° = 24

Solution: 
Since, number of sides of a polygon = 360°/each exterior angle
Hence, number of sides of given polygon = 360°/22° = 16.36

Since, answer is not a whole number, thus, a regular polygon with measure of each exterior angle as 22⁰ is not possible.
Hence, Answer = no

Solution: Here, each interior angle = 165°
Hence, each exterior angle = 180° - 165° = 15°
As, measure of each exterior angle = 15°
So, number of sides = 360°/15° = 24
Hence, number of sides = 24

Solution:We know that number of angles of a polygon = number of sides
And we know that sum of exterior angles of a polygon = 360⁰
So, measure of each angle = 24°
Or, number of exterior angles = 360°/24° = 15
Hence, number of sides = 15

(i) 9 sides

Solution: Since, 9 sides of a polygon has nine angles
And we know that sum of exterior angles of a polygon = 360⁰
So, 9 exterior angles = 360°
Or, 1 exterior angle = 360°/9 = 40°

(ii) 15 sides

Solution: Since, 15 sides of a polygon has 15 angles
And we know that sum of exterior angles of a polygon = 360⁰
So, 15 exterior angles = 360°
Or, 1 exterior angle = 360°/15 = 24°

A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon.
(i) 3 sides
Polygon having three sides is called a triangle.

(ii) 4 sides
Polygon having four sides is called a quadrilateral.

(iii) 6 sides
Polygon having six sides is called a hexagon.