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No, all the angles of a quadrilateral cannot be acute angles. As, sum of the angles of a quadrilateral is 360°. So, maximum of three acute angles will be possible.

(i) Four sides of equal length

Rhombus and square are the quadrilaterals that have all four sides of equal length.

(ii) Four right angles

Rectangle and square have all four angles right angles.

Given that E, F, G and H are the midpoints of the sides of quad. ABCD. 

In ΔABC; E, F are the midpoints of the sides AB and BC. 

∴ EF//AC and EF = 1/2 AC

Also in ΔACD; HG // AC

and HG = 1/2 AC

∴ EF // HG and EF = HG 

Now in □EFGH; EF = HG and EF // HG 

∴ □EFGH is a parallelogram.

Given that, the ratio of angles of a quadrilateral 

= 1 : 2 : 3 : 4 

Sum of the terms of the ratio 

= 1 +2 + 3 + 4= 10 

Sum of the four interior angles of a quadrilateral = 360° 

∴ The measure of first angle

= 1/10 x 360° = 36°

The measure of second angle

= 2/10 × 360° = 72°

The measure of third angle

= 3/10 × 360° = 108°

The measure of fourth angle

= 4/10 × 360° = 144°

Consider △ AMO and △ CNO

We know that AB || CD

From the figure we know that ∠ MAO and ∠ NCO are alternate angles

It is given that AM = CN

We know that ∠ AOM and ∠ CON are vertically opposite angles

∠ AOM = ∠ CON

By ASA congruence criterion

△ AMO ≅ △ CNO

So we get

AO = CO and MO = NO (c. p. c. t)

Therefore, it is proved that AC and MN bisect each other.

Correct option is (D) 74°

\(\because\) \(PQ\,||\,RS\)

\(\therefore\) \(x^\circ+65^\circ+52^\circ=180^\circ\)

\(\Rightarrow\) \(x^\circ=180^\circ-65^\circ-52^\circ\)

\(\Rightarrow\) \(x^\circ=180^\circ-117^\circ\)

\(\Rightarrow\) \(x^\circ=63^\circ\)

In \(\triangle PQR,\)

\(52^\circ+90^\circ+(3y+5)^\circ=180^\circ\)    (Sum of interior angle in a triangle is \(180^\circ)\)

\(\Rightarrow\) \(3y+147^\circ=180^\circ\)

\(\Rightarrow\) \(3y=180^\circ-147^\circ=33^\circ\)

\(\Rightarrow\) \(y=\frac{33^\circ}3=11^\circ\)

\(\therefore\) \(x+y=63^\circ+11^\circ=74^\circ\)        

Correct option is  D) 74°

Consider △ ABQ and △ CDP

We know that the opposite sides of a parallelogram are equal

AB = CD

So we get ∠ B = ∠ D

We know that

DP = AD – PA

i.e. DP = 2/3 AD

BQ = BC – CQ

i.e. BQ = BC – 1/3 BC

BQ = (3-1)/3 BC

We know that AD = BC

So we get

BQ = 2/3 BC = 2/3 AD

We get BQ = DP

By SAS congruence criterion

△ ABQ ≅ △ CDP

AQ = CP (c. p. c. t)

We know that

PA = 1/3 AD

We know that AD = BC

CQ = 1/3 BC = 1/3 AD

So we get

PA = CQ

∠ QAB = ∠ PCD (c. p. c. t)… (1)

We know that

∠ QAP = ∠ A – ∠ QAB

Consider equation (1)

∠ A = ∠ C

∠ QAP = ∠ C – ∠ PCD

From the figure we know that the alternate interior angles are equal

∠ QAP = ∠ PCQ

So we know that AQ and CP are two parallel lines.

Therefore, it is proved that PAQC is a parallelogram.

We know that ABCD is a parallelogram whose diagonals intersect each other at O

Consider △ AOE and △ COF

We know that ∠ CAE and ∠ DCA are alternate angles

From the figure we know that the diagonals are equal and bisect each other

AO = CO

We know that ∠ AOE and ∠ COF are vertically opposite angles

∠ AOE =∠ COF

By ASA congruence criterion

△ AOE ≅ △ COF

OE = OF (c. p. c. t)

Therefore, it is proved that OE = OF.

(i) Adjacent sides is nothing but two sides of a quadrilateral having same end

i.e. there are four pairs of adjacent sides. They are

(AB, BC), (BC, CD), (CD, DA) and (DA, AB)

(ii) Two sides of a quadrilateral have same end point are called opposite sides.

There are two opposite sides. They are

(AB, DC), (AD, BC)

(iii) When two angles of quadrilateral shares the common arm it is called ad adjacent angles.

There are four adjacent angles. They are

(∠A, ∠B), (∠B, ∠C), (∠C, ∠D) and (∠D, ∠A)

(iii) When angles of a quadrilateral are not adjacent ten it is called as opposite angles.

There are two pairs of opposite angles. They are,

(∠A, ∠C), (∠B, ∠D)

(iv) A diagonal is a line segment which joins two opposite vertices.

Here there are 2 diagonals are there. They are

(AC, BD)

2x = 3x – 40 … [Pythagoras theorem]

∴ x = 40° 

Correct option is (D) 40°

(i) Consider △ ABC

We know that

AB + BC > AC …… (1)

Consider △ ACD

We know that

AD + CD > AC ……. (2)

By adding both the equations we get

AB + BC + AD + CD > AC + AC

So we get

AB + BC + AD + CD > 2AC …….. (3)

Therefore, it is proved that AB + BC + AD + CD > 2AC.

(ii) Consider △ ABC

We know that

AB + BC > AC

Add CD both sides of the equation

AB + BC + CD > AC + CD …… (4)

Consider △ ACD

We know that

AC + CD > DA ……. (5)

By substituting (5) in (4) we get

AB + BC + CD > DA …….. (6)

(iii) Consider △ ABD and △ BDC

We know that

AB + DA > BD ……. (7)

So we get

BC + CD > BD ……. (8)

By adding (7) and (8) we get

AB + DA + BC + CD > BD + BD

On further calculation

AB + DA + BC + CD > 2BD ……. (9)

By adding equations (9) and (3) we get

AB + DA + BC + CD + AB + BC + AD + CD > 2BD + 2AC

So we get

2 (AB + BC + CD + DA) > 2 (BD + AC)

Dividing by 2

AB + BC + CD + DA > BD + AC

Therefore, it is proved that AB + BC + CD + DA > BD + AC.

Correct option is (B) 80°

Let \(120^\circ,x^\circ,x^\circ,x^\circ\) are angles of quadrilateral.

\(\therefore\) \(120^\circ+x^\circ+x^\circ+x^\circ=360^\circ\)

\(\Rightarrow\) \(3x^\circ=360^\circ-120^\circ=240^\circ\)

\(\Rightarrow\) \(x^\circ=\frac{240^\circ}3=80^\circ\)

The measure of each equal angle is \(80^\circ.\)

Correct option is  B) 80°

Correct option is (D) rhombus

(i) Name a pair of adjacent sides.

Adjacent sides are: AB, BC, CD and DA

(ii) Name a pair of opposite sides.

Adjacent sides are: AB CD and BC DA

(iii) How many pairs of adjacent sides are there?

Four pairs of adjacent sides.

AB BC, BC CD, CD DA and DA AB

(iv) How many pairs of opposite sides are there?

Two pairs of opposite sides.

AB DC and DA BC

(v) Name a pair of adjacent angles.

Four pairs of Adjacent angles are:

DAB ABC, ABC BCA, BCA CDA and CDA DAB

(vi) Name a pair of opposite angles.

Pair of opposite angles are: DAB BCA and ABC CDA

(vii) How many pairs of adjacent angles are there?

Four pairs of adjacent angles. DAB ABC, ABC BCA, BCA CDA and CDA DAB

(viii) How many pairs of opposite angles are there?

Two pairs of opposite angles. DAB BCA and ABC CDA

(i) A quadrilateral has Four sides.

(ii) A quadrilateral has Four angles.

(iii) A quadrilateral has Four vertices, no three of which are collinear.

(iv) A quadrilateral has two diagonals.

(v) The number of pairs of adjacent angles of a quadrilateral is two.

(vi) The number of pairs of opposite angles of a quadrilateral is two.

(vii) The sum of the angles of a quadrilateral is 360°.

(viii) A diagonal of a quadrilateral is a line segment that joins two opposite vertices of the quadrilateral.

(ix) The sum of the angles of a quadrilateral is four right angles.

(x) The measure of each angle of a convex quadrilateral is less than 180°.

(xi) In a quadrilateral the point of intersection of the diagonals lies in interior of the quadrilateral.

(xii) A point os in the interior of a convex quadrilateral, if it is in the interiors of its two opposite angles.

(xiii) A quadrilateral is convex if for each side, the remaining vertices lie on the same side of the line containing the side.

∠DCB + ∠DCE = 180° [Linear pair] 

∠DCB + 105° = 180° 

∠DCB = 180° – 105° 

∠DCB = 75° 

∠A = ∠DCB = 75° 

∠A =∠C = 75° 

[Opposite angles of a parallelogram] 

∠A +∠B = 180° 

[adjacent angles of a parallelogram] 

75° +∠B = 180° 

∠B = 180° + 75° 

∠B = 105° 

∠B = ∠D = 105° 

[Opposite angles of a parallelogram]

In a parallelogram ABCD , ∠C and ∠D are consecutive interior angles on the same side of the transversal CD. 

So, ∠C + ∠D = 180°

Given, 

∠B = 135^∘ 

∠A + ∠B = 180^∘ 

(supplementary angles of parallelogram) 

∠A = 180^∘ – 135^∘ = 45^∘ 

∠C = ∠A = 45^∘ 

(opposite angles of parallelogram) 

∠D = ∠B = 135^∘

Given: In a parallelogram ABCD, if ∠B = 135° 

Here, ∠A = ∠C, ∠B = ∠D and ∠A + ∠B = 180° 

∠A + 135° = 180° 

∠A = 45° 

Answer: 

∠A = ∠C = 45° 

∠B = ∠D = 135°

Given, 

In a parallelogram ABCD 

∠D= 135^∘ 

∠C + ∠D = 180^∘.. (supplementary angles) 

∠C = 180^∘ - 135^∘ = 45^∘ 

∠C = ∠A and ∠D = ∠B.. (Opposite angles of parallelogram)

∠A = 45^∘ 

∠B = 135^∘

Solution: Angle opposite to y = 180° - 70°=110°
Hence, y = 110°
x = 180° - (110° + 40°) = 30°, (triangle’s angle sum)
z = 30° (Alternate angle of a transversal)

(a) 72^o, 108^o

We know that, sum of adjacent angles of a parallelogram = 180^o

Let us assume two angles be 2x and 3x

Then,

2x + 3x = 180^o

5x = 180^o

x = 180^o/5

x = 36^o

Therefore the two angles are 2x = 2 × 36 = 72^o

3x = 3 × 36 = 108^o

The adjacent angles are in the ratio 2 : 1. 

Let the angles be 2x and x 2x + x = 180° 

[adjacent angles of a parallelogram are supplementary] 

3x = 180° 

 x = \(\frac{180^o}{3}\) = 60° 

2x = 2 × 60° =120° 

∴ The angles of the parallelogram are 60°, 120°. 60° and 120°

Solution: 90°, as they add up to 180°

Sum of adjacent angles = 180°

∠A + ∠B = 180º

2∠A = 180º (∠A = ∠B)

∠A = 90º

∠B = ∠A = 90º

∠C = ∠A = 90º (Opposite angles)

∠D = ∠B = 90º (Opposite angles)

Thus, each angle of the parallelogram measures 90º.

Solution: Opposite angles of a parallelogram are always add upto 180°.
So, 180°= 3x + 2x
Or, 5x = 180°
Or, x = 36°
So, angles are; 36° x 3 = 108°
And 36° x 2 = 72°

Let the measures of two adjacent angles, ∠A and ∠B, of parallelogram ABCD are in the ratio of 3:2. Let ∠A = 3x and ∠B = 2x

We know that the sum of the measures of adjacent angles is 180º for a parallelogram.

∠A + ∠B = 180º

3x + 2x = 180º

5x = 180º

∠A = ∠C = 3x = 108º (Opposite angles)

∠B = ∠D = 2x = 72º (Opposite angles)

Thus, the measures of the angles of the parallelogram are 108º, 72º, 108º, and 72º.

(i) False 

(ii) True 

(iii) False 

(iv) False 

(v) True 

(vi) False 

(vii) False 

(viii) True