Explore topic-wise InterviewSolutions in .

Correct option is: D) 45°

Correct option is: A) 1

Correct option is: A) 115°

i) True 

ii) True 

iii) False 

iv) True 

v) False 

vi) False

∵ ABCD is a parallelogram

∴ AB = DC

and AB || DC …(i)

Reason: Opposite sides of a parallelogram are equal and parallel. 

Again ABEF is a parallelogram

⇒ AB = FE

and AB || FE …(ii)

From (i) and (ii)

DC = FE and DC || FE

Therefore, CDFE is a parallelogram.

Hence proved.

We know that all the sides are equal in a rhombus

Consider △ ABD

We get

AB = AD and

∠ ABD = ∠ ADB

It can be written as

x = y ……. (1)

Consider △ ABC

We get

AB = BC and

∠ CAB = ∠ ACB

We know that

∠ ACB = 40^o

By using the sum property of a triangle

∠ B + ∠ CAB + ∠ ACB = 180^o

By substituting values in the above equation

∠ B + 40^o + 40^o = 180^o

On further calculation

∠ B = 180^o – 40^o – 40^o

By subtraction

∠ B = 180^o – 80^o

So we get

∠ B = 100^o

∠ DBC can be written as

∠ DBC = ∠ B – x^o

By substituting the values

∠ DBC = 100^o – x^o

From the figure we know that ∠ DBC and ∠ ADB are alternate angles

∠ DBC = ∠ ADB = y^o

By substituting the value of ∠ DBC

100^o – x^o = y^o

Consider the equation (1) we know that x = y

100^o – x^o = x^o

On further calculation

2x^o = 100^o

By division

x^o = 50^o

Therefore, x = y = 50^o.

We know that all the sides are equal in a rhombus

Consider △ ABC

We know that AB = BC

It can be written as

∠ CAB = ∠ ACB = x^o

By sum property of a triangle

We get

∠ CAB + ∠ ABC + ∠ ACB = 180^o

By substituting the values in above equation

x + 110^o + x = 180^o

By addition we get

2x + 110^o = 180^o

On further calculation

2x = 180^o – 110^o

By subtraction

2x = 70^o

By division

x = 35^o

Therefore, x = 35^o and y = 35^o.

From the figure we know that the diagonals are equal and bisect at point O.

Consider △ AOB

We get AO = OB

We know that the base angles are equal

∠ OAB = ∠ OBA = 35^o

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180^o

By substituting the values we get

∠ AOB + 35^o + 35^o = 180^o

On further calculation

∠ AOB = 180^o – 35^o – 35^o

By subtraction

∠ AOB = 180^o – 70^o

∠ AOB = 110^o

From the figure we know that the vertically opposite angles are equal

∠ DOC = ∠ AOB = y = 110^o

In △ ABC

We know that ∠ ABC = 90^o

Consider △ OBC

We know that

∠ OBC = x^o = ∠ ABC – ∠ OBA

By substituting the values

∠ OBC = 90^o – 35^o

By subtraction

∠ OBC = 55^o

Therefore, x = 55^o and y = 110^o.

From the figure we know that

∠ A = ∠ C = 62^o

Consider △ BCD

We get

BC = DC

It can be written as

∠ CDB = ∠ DBC = y^o

Using the sum property of triangle

∠ BDC + ∠ DBC + ∠ BCD = 180^o

By substituting the values

y + y + 62^o = 180^o

On further calculation

2y = 180^o – 62^o

By subtraction

2y = 118^o

By division

y = 59^o

We know that the diagonals of a rhombus are perpendicular to each other

Consider △ COD as a right angle triangle

∠ DOC = 90^o

∠ ODC = y = 59^o

It can be written as

∠ DCO + ∠ ODC = 90^o

To find ∠ DCO

∠ DCO = 90^o – ∠ ODC

By substituting the values

∠ DCO = 90^o – 59^o

∠ DCO = x = 31^o

Therefore, x = 31^o and y = 59^o.

Given that □ABCD is a parallelogram. 

□ABEF is a rectangle. 

In ΔAFD and ΔBEC 

AF = BE ( ∵ opp. sides of rectangle □ABEF) 

AD = BC (∵ opp. sides of //gm □ABCD) 

DF = CE (∵ AB = DC = DE + EC , AB = EF = DE + DF) 

∴ ΔAFD ≅ ΔBEC (SSS congruence)

From the figure we know that the diagonals of a rectangle are equal and bisect each other.

Consider △ AOB

We get

OA = OB

We know that the base angles are equal

∠ OAB = ∠ OBA

By using the sum property of triangle

∠ AOB + ∠ OAB + ∠ OBA = 180^o

By substituting the values

110^o + ∠ OAB + ∠ OBA = 180^o

We know that ∠ OAB = ∠ OBA

So we get

2 ∠ OAB = 180^o – 110^o

By subtraction

2 ∠ OAB = 70^o

By division

∠ OAB = 35^o

We know that AB || CD and AC is a transversal

From the figure we know that ∠ DCA and ∠ CAB are alternate angles

∠ DCA = ∠ CAB = y^o = 35^o

Consider △ ABC

We know that

∠ ACB + ∠ CAB = 90^o

So we get

∠ ACB = 90^o – ∠ CAB

By substituting the values in above equation

∠ ACB = 90^o – 35^o

By subtraction

∠ ACB = x = 55^o

Therefore, x = 55^o and y = 35^o.

It is given that ABCD is a parallelogram

So we know that AD || BC

From the figure we know that ∠ DAM and ∠ AMB are alternate angles

So we get

∠ DAM = ∠ AMB

We know that ∠ BAM = ∠ DAM

It can be written as

∠ BAM = ∠ AMB

From the figure we know that the sides opposite to equal angles are equal

So we get

BM = AB

We know that the opposite sides of a parallelogram are equal

AB = CD

So we can write it as

BM = AB = CD ……. (1)

We know that M is the midpoint of the line BC

So we get

BM = ½ BC

We know that BC = AD

We get

BM = ½ AD

Based on equation (1)

CD = ½ AD

By cross multiplication

AD = 2CD.

Therefore, it is proved that AD = 2CD.

(i) We know that opposite angles are equal in a parallelogram.

So we get

∠ C = ∠ A = 60^o

We know that the sum of all the angles in a parallelogram is 360^o

It can be written as

∠ A + ∠ B + ∠ C + ∠ D = 360^o

So we get

∠ B + ∠ D = 360^o – (∠ A + ∠ C)

By substituting values in the above equation we get

∠ B + ∠ D = 360^o – (60^o + 60^o)

On further calculation we get

∠ B + ∠ D = 360^o – 120^o

By subtraction

∠ B + ∠ D = 240^o

We know that ∠ B = ∠ B

So the above equation becomes

∠ B + ∠ B = 240^o

2 ∠ B = 240^o

By division

∠ B = ∠ D = 120^o

We know that AB || DP and AP is a transversal

From the figure we know that ∠ APD and ∠ PAD are alternate angles

∠ APD = ∠ PAD = 60^o/2

∠ APD = ∠ PAD = 30^o ……. (1)

We know that AB || PC and BP is a transversal

So we get

∠ ABP = ∠ CPB = ∠ B/2

i.e. ∠ ABP = ∠ CPB = 120^o/2

We get ∠ ABP = ∠ CPB = 60^o …….. (2)

We know that DPC is a straight line

It can be written as

∠ APD + ∠ APB + ∠ CPB = 180^o

By substituting the values we get

30^o + ∠ APB + 60^o = 180^o

On further calculation

∠ APB = 180^o – 30^o – 60^o

By subtraction

∠ APB = 180^o – 90^o

∠ APB = 90^o

Therefore, it is proved that ∠ APB = 90^o

(ii) From equation (1) we know that

∠ APD = 30^o

We know that

∠ DAP = 60^o/2

By division

∠ DAP = 30^o

So we get

∠ APD = ∠ DAP ……… (3)

From the figure we know that the sides of an isosceles triangle are equal

So we get

DP = AD

We know that

∠ CPB = 60^o and ∠ C = 60^o

By sum property of triangle

We get

∠ C + ∠ CPB + ∠ PBC = 180^o

By substituting the values in the above equation

60^o + 60^o + ∠ PBC = 180^o

On further calculation we get

∠ PBC = 180^o – 60^o – 60^o

By subtraction

∠ PBC = 180^o – 120^o

∠ PBC = 60^o

We know that all the sides of an equilateral triangle are equal

So we get

PB = PC = BC ………. (4)

Therefore, it is proved that AD = AP and PB = PC = BC.

(iii) We know that ∠ DPA = ∠ PAD based on equation (3)

We also know that all the sides are equal in an isosceles triangle

DP = AD

From the figure we know that the opposite sides are equal

So we get

DP = BC

Considering equation (4)

DP = PC

From the figure we know that DP = PC and P is the midpoint of the line DC

So we get

DP = ½ DC

By cross multiplication we get

DC = 2AD

Therefore, it is proved that DC = 2AD.

It is given that ABCD is a parallelogram in which ∠ DAB = 80^o and ∠ DBC = 60^o

We know that opposite angles are equal in parallelogram

So we get

∠ C = ∠ A = 80^o

From the figure we know that AD || BC and BD is a transversal

We know that ∠ ADB and ∠ DBC are alternate angles

So we get

∠ ADB = ∠ DBC = 60^o

Consider △ ABD

Using the sum property of triangle

∠ A + ∠ ADB + ∠ ABD = 180^o

By substituting values in the above equation

80^o + 60^o + ∠ ABD = 180^o

On further calculation

∠ ABD = 180^o – 80^o – 60^o

By subtraction

∠ ABD = 180^o – 140^o

So we get

∠ ABD = 40^o

It can be written as

∠ ABC = ∠ ABD + ∠ DBC

By substituting values we get

∠ ABC = 40^o + 60^o

By addition

∠ ABC = 100^o

We know that the opposite angles are equal in a parallelogram

∠ ADC = ∠ ABC = 100^o

We get

∠ ADC = ∠ CDB + ∠ ADB

On further calculation

∠ CDB = ∠ ADC – ∠ ADB

By substituting values

∠ CDB = 100^o – 60^o

By subtraction

∠ CDB = 40^o

Therefore, ∠ ADB = 60^o and ∠ CDB = 40^o.

In △ ABD

We know that AB = AD

From the figure we know that the base angles are equal

∠ ADB = ∠ ABD

We know that ∠ A = 90^o

It can be written as

∠ ADB + ∠ ABD = 90^o

We know that ∠ ADB = ∠ ABD

So we get

2 ∠ ADB = 90^o

By division

∠ ADB = 45^o

Consider △ OXB

From the figure we know that ∠ XOB and ∠ DOC are vertically opposite angles

∠ XOB = ∠ DOC = 80^o

We also know that

∠ ABD = ∠ XBD = 45^o

We can write it as

Exterior ∠ AXO = ∠ XOB + ∠ XBD

By substituting the values

x^o = 80^o + 45^o

By addition

x^o = 125^o

Therefore, the value of x is 125^o.

(i) From the figure we know that ∠ AOB and ∠ COD are vertically opposite angles

So we get

∠ AOB = ∠ COD = 105^o

Consider △ AOB

By sum property of a triangle

∠ OAB + ∠ AOB + ∠ ABO = 180^o

By substituting the values in above equation

35^o + 105^o + ∠ ABO = 180^o

On further calculation

∠ ABO = 180^o – 35^o – 105^o

By subtraction

∠ ABO = 180^o – 140^o

∠ ABO = 40^o

(ii) We know that AB || DC and BD is a transversal

From the figure we know that ∠ ABD and ∠ CDB are alternate angles

It can be written as

∠ CDO = ∠ CDB = ∠ ABD = ∠ ABO = 40^o

So we get

∠ ODC = 40^o

(iii) We know that AB || CD and AC is a transversal

From the figure we know that ∠ ACB and ∠ DAC are alternate opposite angles

So we get

∠ ACB = ∠ DAC = 40^o

(iv) We know that ∠ B can be written as

∠ B = ∠ CBD + ∠ ABO

So we get

∠ CBD = ∠ B – ∠ ABO

In a parallelogram we know that the sum of all the angles is 360^o

So we get

∠ A + ∠ B + ∠ C + ∠ D = 360^o

It can be written as

2 ∠ A + 2 ∠ B = 360^o

By substituting values in the above equation

2 (40^o + 35^o) + 2 ∠ B = 360^o

On further calculation

2 (75^o) + 2 ∠ B = 360^o

So we get

150^o + 2 ∠ B = 360^o

2 ∠ B = 360^o – 150^o

By subtraction

2 ∠ B = 210^o

By division

∠ B = 105^o

So we get

∠ CBD = ∠ B – ∠ ABO

By substituting values

∠ CBD = 105^o – 40^o

By subtraction

∠ CBD = 65^o

(i) From the figure we know that △ EDC is an equilateral triangle.

So we get ∠ EDC = ∠ ECD = 60^o

We know that ABCD is a square

So we get ∠ CDA = ∠ DCB = 90^o

Consider △ EDA

We get

∠ EDA = ∠ EDC + ∠ CDA

By substituting the values in the above equation

∠ EDA = 60^o + 90^o

So we get

∠ EDA = 150^o ……. (1)

Consider △ ECB

We get

∠ ECB = ∠ ECD + ∠ DCB

By substituting the values in the above equation

∠ ECB = 60^o + 90^o

So we get

∠ ECB = 150^o

So we get ∠ EDA = ∠ ECB …… (2)

Consider △ EDA and △ ECB

From the figure we know that ED and EC are the sides of equilateral triangle

So we get

ED = EC

We also know that the sides of square are equal

DA = CB

By SAS congruence criterion

△ EDA ≅ △ ECB

AE = BE (c. p. c. t)

(ii) Consider △ EDA

We know that

ED = DA

From the figure we know that the base angles are equal

∠ DEA = ∠ DAE

Based on equation (1) we get ∠ EDA = 150^o

By angle sum property

∠ EDA + ∠ DAE + ∠ DEA = 180^o

By substituting the values we get

150^o + ∠ DAE + ∠ DEA = 180^o

We know that ∠ DEA = ∠ DAE

So we get

150^o + ∠ DAE + ∠ DAE = 180^o

On further calculation

2 ∠ DAE = 180^o – 150^o

By subtraction

2 ∠ DAE = 30^o

By division

∠ DAE = 15^o

AD = BC (Opposite sides of a parallelogram)

Therefore, DX = BY ( 1/2 AD = 1/2 BC) 

Also, DX || BY (As AD || BC) 

So, XBYD is a parallelogram (A pair of opposite sides equal and parallel) 

i.e., PX || QD 

Therefore, AP = PQ (From ∆AQD where X is mid-point of AD) 

Similarly, from ∆CPB, CQ = PQ (1) 

Thus, AP = PQ = CQ [From (1) and (2)] (2)

Given, 

In a parallelgram ABCD, 

E = mid point of side BC

AD || BC

AD || BE

E is mid point of BC 

So, 

In ΔDEC and ΔBEF 

BE = EC .. (E is the mid point) 

∠DEC = ∠BEF 

∠DCB = ∠FBE 

(vertically opposite angles) 

So, 

ΔDEC ≅ ΔBEF 

DC = FB 

= AB + DC = FB + AB 

= 2AB = AF (proved)

Consider △ ABC

It is given that DE || AB

We know that D is the midpoint of BC

Based on the midpoint theorem we know that

E is the midpoint of AC

So we know that BE is the median of △ ABC drawn through B.

Therefore, it is proved that BE is also a median of △ ABC.

We know that AR || BC and AB || RC

From the figure we know that ABCR is a parallelogram

So we get

AR = BC …….. (1)

We know that AQ || BC and QB || AC

From the figure we know that AQBC is a parallelogram

So we get

QA = BC ……… (2)

By adding both the equations we get

AR + QA = BC + BC

We know that AR + QA = QR

So we get

QR = 2BC

It can be written as

BC = QR/2

BC = ½ QR

In the same way

AB = ½ RP and AC = ½ PQ

Perimeter of △ PQR = PQ + QR + RP

It can be written as

Perimeter of △ PQR = 2AC + 2BC + 2AB

By taking 2 as common

Perimeter of △ PQR = 2 (AC + BC + 2AB)

Perimeter of △ PQR = 2 (Perimeter of △ ABC)

Therefore, it is proved that the perimeter of △ PQR is double the perimeter of △ ABC.

Now let us assume number of sides of a regular polygon be n.

WKT, sum of all exterior angles of all polygons is equal to 360^o.

Form the question it is given that each exterior angle has a measure of 45^o.

Then,

n = 360^o/Exterior angle

n = 360^o/(180^o – 135^o)

n = 360^o/45^o

n = 8

(c) We know that, in a parallelogram, opposite sides are equal.

But according to the question, adjacent sides are also equal.

Thus, the parallelogram in which all the sides are equal is known as rhombus.

(a) Square

These are the properties of a square, i.e. in a square, all sides, diagonals and angles are equal.

The sum of interior angles of a polygon of n sides is 2n – 4 right angles.

The name of three-sided regular polygon is an equilateral triangle.

equilateral triangle, as polygon is regular, i.e. length of each side is same.

∠AOB = 75° (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° = 1/2 (∠D + ∠C) 

or ∠C + ∠D = 150°

Ratio of angles of quadrilateral = 2 : 4 : 5 : 7 

Let the four angles be 2x, 4x, 5x and 7x. 

Sum of them = 2x + 4x + 5x + 7x = 360° 

⇒ 18x = 360° ⇒ x = 20° 

∴ Angles in a quadrilateral = 40°, 80°, 100°, 140°

angles

In a regular polygon, all sides are equal and all angles are equal.

We cannot construct △ABD. So, if we start first from △ABD, 

it is impossible to construct □ ABCD.

 [∵ The length of  \(\overline{AD}\) is not given]

Correct option is (B) 135°

ABCD is a cyclic quadrilateral.

\(\angle A\;\&\;\angle C\) are opposite angles in cyclic quadrilateral ABCD

\(\therefore\) \(\angle A+\angle C\) \(=180^\circ\)    \((\because\) Sum of opposite angles in a cyclic quadrilateral is \(180^\circ)\)

\(\Rightarrow\) \(\angle C\) \(=180^\circ-\angle A\)

\(=180^\circ-45^\circ\)    \((\because\) \(\angle A=45^\circ)\)

= \(135^\circ\)

Correct option is  B) 135°

False

By definition of a regular polygon, we know that, a polygon is regular, if all sides and all angles are equal.

True

Since the sum of exterior angles of a hexagon is 360° and the sum of interior angles of a hexagon is 720°, i.e. double the sum of exterior angles.

(a) 40° 

From the given figure TS ∥ BE and also BS is transversal line.

By the rule of alternate interior angles, ∠EBS = ∠BST = 40^o

Then, ∠y = 90^o … [∵diagonal bisect at 90^o]

Consider triangle TSO,

By the rule of exterior angle property of triangle

∠STO + ∠TSO = ∠SOE

x + 40^o = 90^o

x = 90^o – 40^o

x = 50^o

So, the value of y – x is = 90^o – 40^o = 50^o

True

As the sum of interior angles of a triangle is 180° and the sum of exterior angles is 360°, i.e. double the sum of interior angles.

(d) Each exterior angle = ((n – 2) × 180^o)/n)

We know that, (a) and (b) are the formulae to find the measure of each exterior angle, when number of sides and measure of an interior angle respectively are given and (c) is the formula to find number of sides of polygon when exterior angle is given.

Hence, the formula given in option (d) is not true for an exterior angle of a regular polygon with n sides.

The closed curve which is also a polygon is figure (a). Because there is no line segments intersect each other.

Let the shorter side be x

Perimeter = x + 6.5 + 6.5 + x

22 = 2(x+6.5)

11 = x + 6.5

x = 11 - 6.5 = 4.5 cm

Shorter side = 4.5 cm.

in the fig AB = 25 m and FE = 15 m 

∴ DF + EC = 25 – 15= 10m 

But DF = EC 

∴DF = EC = 5m also AD = BC = 5m 

Area of rectangle ∆BCD = AB x BC = 1 x b = 25 x 5 = 125. m² 

Area of ∆ADF = \(\frac{1}{2}\)base × height 

= \(\frac{1}{2}\)5 × 5 = \(\frac{25}{2}\)m² 

∆ADF ≅ ABCE [data] 

Area of ∆BCE = \(\frac{25}{2}\)m² 

Sum of Area of the two triangles 

= \(\frac{25}{2}\) + \(\frac{25}{2}\) = \(\frac{50}{2}\) = 25m²

^{\(\frac{Area\; of\; 2\; triangles}{Area\; of\; rectangles}\) = \(\frac{25}{125} = \frac{1}{5}\)}

∴ \(\frac{1}{5}\) of the field is occupied by the flower bed.