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∵ AB || CD || EF (given)

∴ ∠y + 58° = 180°

(interior angle sum)

⇒ ∠y = 180° – 58° = 122°

∴ ∠x = 180° – 122°

= 58° (linear pair)

Now AB || CD

∠x = ∠z = 58° (alt. angles) and ∠y = ∠p= 122°

(corresponding angles)

Hence, x = 58°, y = 122°, z = 58°, and

p = 122°.

∠y = 25° vertically opposite angle

From the figure we can write as

∠x = ∠y are vertically opposite angles

∠x + ∠y + ∠z + 25° = 360°

∠x + ∠z + 25° + 25° = 360°

On rearranging we get,

∠x + ∠z + 50° = 360°

∠x + ∠z = 360° – 50° [∠x = ∠z]

2∠x = 310°

∠x = 155°

And, ∠x = ∠z = 155°

Therefore, ∠x = 155°, ∠y = 25° and ∠z = 155°

∵ AB and CD are perpendicular to l and m

∴∠APQ = 90° and ∠CQM = 90°

∠APQ = ∠CQM = 90°

(Corresponding- angles)

∴ AB || CD

(i) A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.

In Figure (i) Corresponding angles are

∠EGB and ∠GHD

∠HGB and ∠FHD

∠EGA and ∠GHC

∠AGH and ∠CHF

A pair of angles in which one arm of each of the angle is on opposite sides of the transversal and whose other arms include the one segment is called a pair of alternate angles.

The alternate angles are:

∠EGB and ∠CHF

∠HGB and ∠CHG

∠EGA and ∠FHD

∠AGH and ∠GHD

(ii) In Figure (ii)

The alternate angle to ∠d is ∠e.

The alternate angle to ∠g is ∠b.

The corresponding angle to ∠f is ∠c.

The corresponding angle to ∠h is ∠a.

(iii) In Figure (iii)

Angle alternate to ∠PQR is ∠QRA.

Angle corresponding to ∠RQF is ∠ARB.

Angle alternate to ∠POE is ∠ARB.

(iv) In Figure (ii)

Pair of interior angles are

∠a is ∠e.

∠d is ∠f.

Pair of exterior angles are

∠b is ∠h.

∠c is ∠g.

a = 180° – 50° = 130° 

(linear pair of angles)

From the figure ∠y and 160° are vertically opposite angles and hence equal. 

∴ ∠y=1600 

Also ∠x = ∠z and (. vertically opposite angles) 

x + 160° = 180° (Linear pair of angles) 

∴ ∠x = 180° – 160° = 20° 

∠z = 20° (∵ ∠x, ∠z are vertically opposite) 

∴ x = 20°

y = 160° 

z = 20°

False

Sum of interior angles on the same side of a transversal with two parallel lines is 90°.

False

Interior angles on the same side of a transversal with two distinct parallel lines are supplementary angles.

(i) ∠AOC = 52° (given)

⇒ ∠BOD = 52°

(Vertically opposite angles)

(ii) ∠AOC + ∠AOD = 180°

(Linear pair axiom)

⇒ 52° + ∠AOD = 180°

⇒ ∠AOD = 180° – 52°

⇒ ∠AOD = 128°

(iii) (∠AOC, ∠BOD) and (∠AOD, ∠BOC) are

vertically opposite pairs of angles.

(iv) The adjacent supplementary angles of ∠AOC are ∠AOD and ∠BOC.

We have, a : b : c = 4 : 3 : 5

⇒ ∠a = 4x, ∠b = 3x and ∠c = 5x

But ∠a + ∠b + ∠c = 180° (Straight angle)

⇒ 4x + 3x + 5x = 180°

⇒ 12x = 180°

⇒ x = 15°

Angles are 4 x 15° = 60°, 3 x 15° 

= 45° and 5 x 15° = 75°

According to question

(2x + 4)° + (x – 1)° = 180°

(Linear pair axiom)

⇒ 3x + 3 = 180

⇒ 3x = 177

\(x=\frac { 177 }{ 2 }=59\)

∴ Angles are (2 x 59 + 4)° = 122° 

and (59 – 1)° = 58

Correct option is  B) 36°

∠1 = 60° (given)

∠3 = ∠1 = 60°

(vertically opposite angle)

and ∠6 = 120° (given)

∠3 + ∠6 = 60° + 120°

= 180°
i. e. sum of the interior angles on the same side of transversal is 180°.

⇒ m || n.

Correct option is (D) l₄

\(\because\) \(l_4\) intersects \(l_1\) & \(l_2\) at two different points.

\(\therefore\) \(l_4\) is a transversal.

Correct option is  D) l₄  

Correct option is (B) 50°

Since, \(110^\circ\) is an exterior angle of \(\triangle ABC\) whose opposite interior angles are \(60^\circ\) & p.

\(\therefore\) \(60^\circ+p=110^\circ\)

\(\Rightarrow\) \(p=110^\circ-60^\circ=50^\circ\)

Correct option is  B) 50°

Correct option is  D) n and l

Correct option is (C) 90°

Perpendicular distance is always shorter.

\(\therefore\) At \(90^\circ\) of angle with the edge of the road should he walk so that he walks the shortest distance.

Correct option is  C) 90°

 Correct option is (C) 40° and 140°

Let the angle are 2x and 7x.

\(\therefore\) 2x+7x = \(180^\circ\)   \((\because\) Sum of supplementary angles is \(180^\circ)\)

\(\Rightarrow\) 9x = \(180^\circ\)

\(\Rightarrow\) x = \(\frac{180^\circ}9=20^\circ\)

\(\therefore\) 2x = \(40^\circ\) and 7x = \(140^\circ\)

Hence, both required supplementary angles are \(40^\circ\) and \(140^\circ.\)

C) 40° and 140°

Correct option is (D) MN

A ray has an end point.

\(\overrightarrow{AB}\) (one sided arrow) indicates that A is the point where the ray start and arrow direction represent that B is a point in middle to ray AB.

Here, \(\overrightarrow{PQ}\), \(\overrightarrow{AB}\) & \(\overrightarrow{XY}\) represent a ray but MN does not represent a ray, it represent a line.

Correct option is   A) \(\overrightarrow{PQ}\)

Answer is (D) 50°

Given l || m, n perpendicular l 

To prove: n ⊥ m

Since l || m and n intersects them at G and H respectively 

∠1 = ∠2 [Corresponding angles] 

But, U = 90^o [n ⊥ l]

∠2 = 90^o 

Hence n perpendicular m

Correct option is  C) 25°

i) Angles between legs of a folding cot/scamp cot. 

ii) Angles between legs of a folding chair.

iii) Angles between plates of a scissors.

Correct option is (B) Parallel

Lines parallel to the same line are parallel to each other.

Correct option is  B) Parallel

Correct option is (B) 30

\((x+20)^\circ+3x^\circ+(x+10)^\circ=180^\circ\)

\(\Rightarrow\) \((x+20+3x+x+10)^\circ=180^\circ\)

\(\Rightarrow\) 5x + 30 = 180

\(\Rightarrow\) 5x = 180 - 30 = 150

\(\Rightarrow\) x = \(\frac{150}5\) = 30

Correct option is  B) 30

As we know that sum of the measure of an angle and its complement is equal to 90°

(3x – 20)° + (2x – 40)° = 90°

⇒ 5x – 60° = 90°

⇒ 5x = 150°

⇒ x = 30°

Answer is (B) 30°

Answer is (B) 25°

Correct option is  D) 120°

Correct option is (D) 108°

x is exterior angle whose opposite interior angles are \(68^\circ\;and\;40^\circ.\)

\(\therefore\) x = \(68^\circ+40^\circ\) = \(108^\circ\)

\((\because\) Sum of interior opposite angles is equal to exterior angle)

Correct option is  D) 108°

Correct option is (C) 60°

\(\because\) \(135^\circ\) is exterior angle whose opposite interior angles are a & \(75^\circ.\)

\(\therefore\) \(a+75^\circ=135^\circ\)

\((\because\) The measure of an exterior angle is equal to the sum of its opposite interior angles)

\(\Rightarrow\) \(a=135^\circ-75^\circ=60^\circ\)

Correct option is  C) 60°

Let the number of bottles filled by the machine in five hours be x.

The given information in the form of a table is as follows.

The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain

840/6 = x/5

x = (840 x5)/6 = 700

Thus, 700 bottles will be filled in 5 hours.

If more number of bottles required then, time taken will also be more. Therefore it’s directly proportional.

Let us consider the bottles be x, 420/3 = x/5

3 × x = 420 × 5

3x =2100

x = 2100/3

x = 700

∴ 700 bottles are required.

Given:

Base = 25 cm

Height = 16.8 cm

∴ Area of the parallelogram = Base x Height = 25 cm x 16.8 cm = 420 cm²

Given: 

Base = 25 cm 

Height =16.8 cm 

Area of the parallelogram = Base x Height

= 25 cm x 16.8 cm

= 420 cm²

(i) 0.2 × 6

0.2 × 6 = 1.2

(ii) 8 × 4.6

8 × 4.6 = 36.8

(iii) 2.71 × 5

2.71 × 5 = 13.55

(b) 1

Given exp. = \(\frac{17}{15}\times\)\(\frac{17}{15}+\)\(\frac{2}{15}\times\)\(\frac{2}{15}-\)2 x \(\frac{17}{15}\times\)\(\frac{2}{15}\)

= \(\big(\frac{17}{15}\big)^2\) + \(\big(\frac{2}{15}\big)^2\) - 2 x \(\frac{17}{15}\times\)\(\frac{2}{15}\)

= \(\big(\frac{17}{15}-\frac{2}{15}\big)^2\)          [Using a² + b² - 2ab = (a - b)²]

= \(\big(\frac{15}{15}\big)^2\) = 1² = 1

Given: Total contribution from students= Rs. 1156

Let the number of students be ‘x’

And also let the amount contributed by each student be ‘x’

Therefore, total amount contributed by the students = (x × x) = x²=1156

Shifting the square, we get

√1156 = 2 × 2 × 17 × 17 = 17 × 2 =34

Therefore, number of students is 43 and amount contributed by each student is also 34.

The correct option is: (D) Only II and either I or III

Explanation:

Volume of cylindrical tank = πr²h

So, we need radius and height.

Radius can be found from either statement I or III and height from statement II.

We know that, 

The smallest number that is divisible by each o0f these numbers is their L.C.M 

So, 

L.C.M of 8, 12, 15, 20 = 120 

Resolving into prime factors, we get 

120 = 2 × 2 × 2 × 3 × 5 

So, for making it a perfect square we have to multiply it by 2 × 3 × 5 = 30 

Multiplying the number by 30, we get 

Required number = 120 × 30 

= 3600

(i) 0.0000000000085 = 8.5 × 10⁻¹²

(ii) 0.00000000000942 = 9.42 × 10⁻¹²

(iii) 6020000000000000 = 6.02 × 10¹⁵

(iv) 0.00000000837 = 8.37 × 10⁻⁹

(v) 31860000000 = 3.186 × 10¹⁰

We know that, 

The smallest number that is divisible by each o0f these numbers is their L.C.M 

So, 

L.C.M of 6, 9, 15, 20 = 180 

Resolving into prime factors, we get 

180 = 2 × 2 × 3 × 3 × 5 

So, for making it a perfect square we have to multiply it by 5 

Multiplying the number by 5, we get 

Required number = 180 × 5 

= 900

Given numbers = 8, 12, 15 and 20

The smallest number divisible by 8, 12, 15 and 20 is their L.C.M. i.e. 120

Resolving the L.C.M. as prime factors we get, 120= 2 × 2 × 2 × 3 × 5

To make it perfect square multiply by 2 × 3 × 5=30

Which implies 120 × 30 = 3600

Least square number which is exactly divisible by 6, 9, 15 and 20 is 3600.

Given numbers = 6, 9, 15 and 20

The smallest number divisible by 6, 9, 15 and 20 is their L.C.M. i.e. 180

Resolving the L.C.M. as prime factors we get, 180= 2 × 2 × 3 × 3 × 5

To make it perfect square multiply by 5 then it becomes, 180= 2 × 2 × 3 × 3 × 5 × 5

Which implies 180 × 5 = 900

Least square number which is exactly divisible by 6, 9, 15 and 20 is 900.

The correct option is: (B) 15 cm

Explanation:

Volume of cuboid = 44 x 30 x 15 cm³

This volume is equal to volume of cylinder

.'. 44 x 30 x 15 = π x r² x 28 

.'. r = 15 cm

(i) 1156 

Resolving 1156 into prime factors we get, 

1156 = 2 × 2 × 17 × 17 

Now, 

grouping the factors into pairs of equal factors We get, 

1156 = (2 × 2) × (17 × 17) 

As all factors are paired 

Hence, 1156 is a perfect square  

Again, 

1156 = (2 × 17) × (2 × 17) 

= 34 × 34 

= (34)² 

Thus, 1156 is a square of 34.

(ii) 2025 

Resolving 2025 into prime factors we get, 

2025 = 3 × 3 × 3 × 3 × 5 × 5 

Now, 

grouping the factors into pairs of equal factors We get, 

2025 = (3 × 3) × (3 × 3) × (5 × 5) 

As all factors are paired 

Hence, 2025 is a perfect square 

Again, 

2025 = (3 × 3 × 5) × (3 × 3 × 5) 

= 45 × 45 

= (45)²

Thus, 2025 is a square of 45.

(iii)14641 

Resolving 14641 into prime factors we get, 

14641 = 11 × 11 × 11 × 11 

Now, 

grouping the factors into pairs of equal factors We get, 

14641 = (11 × 11) × (11 × 11) 

As all factors are paired 

Hence, 14641 is a perfect square 

Again, 

14641 = (11 × 11) × (11 × 11) 

= 121 × 121 

= (121)² 

Thus, 14641 is a square of 121.

(iv) 4761 

Resolving 4761 into prime factors 

we get, 

4761 = 3 × 3 × 23 × 23 

Now, 

grouping the factors into pairs of equal factors We get, 

4761 = (3 × 3) × (23 × 23) 

As all factors are paired 

Hence, 4761 is a perfect square 

Again, 

4761 = (3 × 23) × (3 × 23) 

= 69 × 69 

= (69)² 

Thus, 4761 is a square of 69.