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∠A = ∠C = 70° [Opposite angles] 

∠A +∠B = 180° [Adjacent angles of a rhombus] 

70° + ∠B = 180° 

∠B = 180° – 70° 

∠B = 110° 

∠B = ∠D = 110° [Opposite angles are equal]. 

Angles are 70°, 110°, 70° & 110°

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 = 360°

3x + 5x + 9x + 13x = 360° 

30 x = 360°

x = 12° 

Length of smallest angle = 3x = 3(12) = 36°

Correct option is (C) right angle

Diagonals of a rhombus intersect each other at right angle.

\(\therefore\) Angle between the diagonals of a rhombus is a right angle.

C) right angle

Option : (A)

The opposite sides of a quadrilateral have no common points.

If a transversal line is intersecting two given lines such that the sum of the measures of the angles on the same side of transversal is 180º, then the given two lines will be parallel to each other

Here, ∠NML + ∠MLK = 180°

Hence, NM||LK

As quadrilateral KLMN has a pair of parallel lines, therefore, it is a trapezium

∠P + ∠Q = 180° (Angles on the same side of transversal)

∠P + 130° = 180°

∠P = 50°

∠R + ∠S = 180° (Angles on the same side of transversal)

90° + ∠R = 180°

∠S = 90°

Yes. There is one more method to find the measure of m∠P.

m∠R and m∠Q are given. After finding m∠S, the angle sum property of a quadrilateral can be applied to find m∠P.

Given that, bar(AB) II bar(DC)

∠B + ∠C = 180° (Angles on the same side of transversal)

120º + ∠C = 180°

∠C = 60°

Area of the quadrilateral = \(\frac{1}{2}\) d(h₁ + h₂) = \(\frac{1}{2}\times8\times(4+6)\)

= \(4\times10\) = 40 cm²

Area of the quadrilateral ABCD

= \(\frac{1}{2}\) x AC x (PD + BQ)

= \(\frac{1}{2}\) x 5.5 x (2.5 + 1.5)

= \(\frac{1}{2}\) x 5.5 x 4 = 11 cm²

Area of the quadrilateral ABCD

= \(\frac{1}{2}\) x AC(BX + DY) = \(\frac{1}{2}\) x 8 x (4.5 + 3.5)

= 4 x 8 = 32 cm²

Area of dark triangle = \(\frac{1}{2}bh\) cm²

Area of parallelogram = bh

= 2 × area of triangle 10 cm²

The area will be maximum for a rectangle and the maximum area is 40 cm².

Area of isoceless trapezium = \(\frac{1}{2}\times h\times(a+b)\)

= \(\frac{1}{2}\times4\times(7+3)\) = 2 x 10 = 20 cm²

Area of the trapezium = \(\frac{1}{2}\) (a + b)d

= \(\frac{1}{2}\) x (30 + 10) x 20 = 400 sq.cm

Area = \(\frac{1}{2}\) x DE x (AB + CD)

= \(\frac{1}{2}\) x 4 x (8 + 10) = \(\frac{1}{2}\) x 4 x 18

= 4 x 9 = 36 cm²

Data: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. 

To Prove: 

(i) ∆AAPD ≅ ∆CQB 

(ii) AP = CQ 

(iii) ∆AQB ≅ ∆CPD 

(iv) AQ = CP 

(v) APCQ is a parallelogram. 

Proof: ABCD is a parallelogram. 

(i) In ∆APD and ∆CQB, 

AD = BC (opposite sides) 

∠ADP = ∠CBQ (alternate angles) 

DP = BQ (Data) 

∴ ∆APD ≅ ∆CQB (ASA Postulate) 

(ii) AP = CQ 

(iii) In ∆AQB and ∆CPD, 

AB = CD (opposite sides) 

∠ABQ = ∠CDP (alternate angles) 

BQ = DP (Data) 

∴ ∆AQB ≅ ∆CPD (ASA Postulate) 

(iv) ∴ AQ = CP 

(v) In ∆AQPand 

∆CPQ, AP = CQ (proved) 

AQ = PC PQ is common. 

∴ ∆AQP ≅ ∆CPQ 

∴ ∠APQ = ∠CQP 

These are pair of alternate angles. 

∴AP || PC 

Similarly, AQ || PC 

∴ APCQ is a parallelogram.

(c) 78°

Since the diagonal of a square bisects the angle at the vertices, ∠DBC = 45°. 

∴ ∠ ABQ = 90° – (21° + 45°) 

= 90° – 66° = 24° 

In ΔBAQ, BA = BQ 

⇒ ∠BQA = ∠BAQ         (isos. Δ property)

∴ ∠BAQ = \(\frac{180°-24°}{2}=\frac{156°}{2}\) = 78°.

(d) trapezium

In Δ BAD, 

∠ BDA=∠BAD= 57°        (isos. Δ property) 

In Δ BDC, 

∠BCD=∠BDC= 66°        (isos. Δ property) 

∴ ∠D = 57° + 66° = 123° and 

∠A +∠D = 57° + 123° = 180° 

Also, ∠D + C = 123° + 66° = 189° 

Hence, by the property that co-int. angles are supplementary ⇒ lines are parallel, we have AB || DC and AD not || BC. 

Hence ABCD is a trapezium.

Option : (D)

The bisectors of any two adjacent angles of a parallelogram intersect at 90°

(i) In ∆ADC,

Q is mid point of AC such that

PQ || AD

∴ P is mid point of DC

= DP = DC

(converse of mid point theorem)

(ii) Similarly, 

R is the mid point of BC

= PR = \(\frac{1}{2}\) BD

PR = \(\frac{1}{2}\) AC

∵ diagonals of rectangle are equal

Proved.

Yes, every rectangle is a parallelogram. But every parallelogram is not a rectangle.

Proof: 

□ABCD is a rectangle.

 ∴ ∠A ≅ ∠C = 90° [Given] 

∠B ≅ ∠D = 90° [Angles of a rectangle] 

∴ Rectangle ABCD is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent] 

(a) Consider a regular polygon having the lowest possible number of sides (i.e., an equilateral triangle). The exterior angle of this triangle will be the maximum exterior angle possible for any regular polygon.

Exterior angle of an equilateral triangle = 360°/3 = 120°

(b) Hence, maximum possible measure of exterior angle for any polygon is 120º. Also, we know that an exterior angle and an interior angle are always in a linear pair. Hence, minimum interior angle = 180º − 120° = 60º

Given: □ABCD is a parallelogram.

 AP = BQ = CR = DS

To prove: □PQRS is a parallelogram.

Proof: 

□ABCD is a parallelogram. [Given] 

∴ ∠B = ∠D ….(i) [Opposite angles of a parallelogram] 

Also, AB = CD [Opposite sides of a parallelogram] 

∴ AP + BP = DR + CR [A-P-B, D-R-C] 

∴ AP + BP = DR + AP [AP = CR] 

∴ BP = DR ….(ii) 

In APBQ and ARDS,

 seg BP ≅ seg DR [From (ii)] 

∠PBQ ≅ ∠RDS [From (i)] 

seg BQ ≅ seg DS [Given]

∴ ∆PBQ ≅ ∆RDS [SAS test] 

∴ seg PQ ≅ seg RS …..(iii) [c.s.c.t] 

Similarly, we can prove that 

∆PAS ≅ ∆RCQ 

∴ seg PS ≅ seg RQ ….(iv) [c.s.c.t]

 From (iii) and (iv), 

□PQRS is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

i. AC = 9 cm [Given] 

Points X and Y are the midpoints of sides AB and BC respectively. [Given] 

∴ XY = 1/2 AC [Midpoint theorem] 

= 1/2 x 9 = 4.5 cm

ii. AB = 5 cm [Given] 

Points Y and Z are the midpoints of sides BC and AC respectively. [Given]

∴ YZ = 1/2 AB [Midpoint theorem] 

=1/2 x 5 = 2.5 cm

iii. BC = 11 cm [Given]

 Points X and Z are the midpoints of sides AB and AC respectively. [Given] 

∴ XZ = 1/2 BC [Midpoint theorem] 

= 1/2 x 11 = 5.5 cm 

l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm

The sum of all exterior angles of all polygons is 360º. Also, in a regular polygon, each exterior angle is of the same measure. Hence, if 360º is a perfect multiple of the given exterior angle, then the given polygon will be possible. 

(a) Exterior angle = 22° 360º is not a perfect multiple of 22º. Hence, such polygon is not possible.

(b) Interior angle = 22°

Exterior angle = 180° − 22° = 158°

Such a polygon is not possible as 360° is not a perfect multiple of 158°.

Given: D and E are the midpoints of side AB and side AC respectively. 

ED = DF 

To prove: □AFBE is a parallelogram.

Proof: 

seg AB and seg EF are the diagonals of □AFBE. 

seg AD ≅ seg DB [Given] 

seg DE ≅ seg DF [Given] 

∴ Diagonals of □AFBE bisect each other. 

∴ □AFBE is a parallelogram. [By test of parallelogram]

Given: ABCD is a rectangle. 

To prove: Rectangle ABCD is a parallelogram. 

Proof: 

ABCD is a rectangle. 

∴ ∠A ≅ ∠C = 90° [Given] 

∠B ≅ ∠D = 90° [Angles of a rectangle] 

∴ Rectangle ABCD is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

Given: ABCD is a parallelogram. P and Q are the midpoints of sides AB and DC respectively. 

To prove: APCQ is a parallelogram.

Proof: 

AP = (1/2) AB …..(i) [P is the midpoint of side AB] 

QC = (1/2) DC ….(ii) [Q is the midpoint of side CD] 

ABCD is a parallelogram. [Given] 

∴ AB = DC [Opposite sides of a parallelogram] 

∴ (1/2) AB = (1/2) DC [Multiplying both sides by 1/2] 

∴ AP = QC ….(iii) [From (i) and (ii)] 

Also, AB || DC [Opposite angles of a parallelogram]

i.e. AP || QC ….(iv) [A – P – B, D – Q – C] 

From (iii) and (iv),

APCQ is a parallelogram. [A quadrilateral is a parallelogram if its opposite sides is parallel and congruent]

i. AC = 9 cm [Given] 

Points X and Y are the midpoints of sides AB and BC respectively. [Given] 

∴ XY = (1/2) AC [Midpoint tfyeprem] 

= (1/2) x 9 = 4.5 cm 

ii. AB = 5 cm [Given]

Points Y and Z are the midpoints of sides BC and AC respectively. [Given] 

∴ YZ = (1/2) AB [Midpoint theorem] 

= (1/2) x 5 = 2.5 cm 

iii. BC = 11 cm [Given] 

Points X and Z are the midpoints of sides AB and AC respectively. [Given] 

∴ XZ = (1/2) BC [Midpoint theorem] 

= (1/2) x 11 = 5.5 cm 

l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm

Correct option is (C) Parallelogram

In parallelogram

(i) Opposite angles are equal but adjacent sides may not be equal.

(ii) Opposite sides are equal but adjacent sides may not be equal.

C) Parallelogram

In ΔBEC

∠BEC + ∠ECB +∠CBC = 180° [Sum of angles of a triangle is 180°]

90° + 40° + ∠CBC = 180°

∠CBC = 180°-130°

∠CBC =50°

∠B = ∠D = 50° [Opposite angles of a parallelogram are equal]

∠A + ∠B = 180° [Sum of adjacent angles of a triangle is 180°]

∠A + 50° = 180°

∠A = 180°- 50°

∠A = 130°

In ΔDFC

∠DFC + ∠FCD +∠CDF = 180° [Sum of angles of a triangle is 180°]

90° + ∠FCD + 50° = 180°

∠FCD = 180°- 140°

∠FCD =40°

∠A = ∠C = 130° [Opposite angles of a parallelogram are equal]

∠C = ∠FCE +∠BCE + ∠FCD

∠DCF + 40° + 40° = 130°

∠DCF = 130°- 80°

∠DCF = 50°

seg OP = seg OR, and seg OQ = seg OS 

Thus we can conclude that, point O divides the diagonals PR and QS in two equal parts

In parallelogram BDEF

BD = EF ………..(i) [In a parallelogram opposite sides are equal]

In parallelogram DCEF

DC = EF ………..(ii) [In a parallelogram opposite sides are equal]

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

BD = EF = DC

Hence, BD = DC Proved

Correct option is: C) 4

In parallelogram ABCD

∠C =∠A = 55° [In a parallelogram opposite angles are equal]

In parallelogram AEFG

∠A =∠F = 55° [In a parallelogram opposite angles are equal]

Given, 

ABCD & AEFG are two parallelograms 

∠C = 58° 

∵ AB || CD, 

AE || FG,

BC || AD, 

EF || AG 

GF produced to H so GH || AB 

= ∠C = ∠H 

(corresponding angles) 

= ∠H = 58° 

AD || BC and GH cuts them 

Hence, 

∠F = ∠H 

(corresponding angles) 

∠F = 58°.

Sum of angles of a quadrilateral is 360°

In the quadrilateral MPNO

∠NOP = 45°, ∠OMP = ∠PNO = 90°,

Let angle ∠MPN is x°

∠NOP + ∠OMP + ∠PNO + ∠MPN = 360°

45° + 90° + 90° + x° =360°

x° = 360° - 225°

x° = 135°

Therefore ∠MPN is 135°

Quadrilaterals are figures enclosed by four line segments.

The sum of the four interior angles of a quadrilateral is 360°.

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

∠A = ∠C = 55° ...(i)

Also, AEFG is a parallelogram.

∴ ∠A=∠F = 55° [from Eq. (i)]

Yes, all the angles of a quadrilateral can be right angles. In this case, the quadrilateral becomes rectangle or square.

(i) bisect each other

In a Parallelogram, rectangle, rhombus and square diagonals bisect each other.

(ii) are perpendicular bisector of each other

In a Rhombus and square diagonals are perpendicular bisector of each other

(iii) are equal.

In a square and rectangle diagonals are of equal length.