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Since PQ||RS and QR is transversal, 

so x = 39° [Alternate interior angles] 

Again QR||ST and RS is a transversal. 

Therefore, y = x [Alternate interior angles] 

or y = 39°

Given that, ∠1 = 120° and ∠2 = 100°

From the figure ∠1 and ∠5 is a linear pair

∠1 + ∠5 = 180°

∠5 = 180° – 120°

∠5 = 60°

Therefore, ∠5 = 60°

∠2 and ∠6 are corresponding angles

∠2 = ∠6 = 100°

Therefore, ∠6 = 100°

∠6 and ∠3 a linear pair

∠6 + ∠3 = 180°

∠3 = 180° – 100°

∠3 = 80°

Therefore, ∠3 = 80°

By, angles of sum property

∠3 + ∠5 + ∠4 = 180°

∠4 = 180° – 80° – 60°

∠4 = 40°

Therefore, ∠4 = 40°

No, ∠1 and ∠2 are not forming a pair of adjacent angles as they do not have a common vertex.

Since ED||BC and AB is a transversal, 

so ∠QBP = ∠APE [Corresponding angles] 

or ∠QBP = 39° 

Now, AB||EF and BC is a transversal. 

Therefore, ∠FQB = ∠QBP [Alternate interior angles]

or ∠FQB = 39° 

Also, ∠CQF + ∠FQB = 180° [Linear pair] 

So ∠CQF + 39° = 180° 

or ∠CQF = 180° – 39° 

or ∠CQF = 141°

AOC is not a line, because ∠ AOB + ∠ COB = 100° + 82° = 182°, which is not equal to 180°. Similarly, BOD is also not a line.

OF bisects ∠BOD

∠BOF = 35°

∠BOC = ?

∠AOD = ?

∠BOD = ∠BOF = 70°(Therefore, OF bisects ∠BOD)

∠BOD = ∠AOC = 70°(Vertically opposite angle)

∠BOC + ∠AOC = 180°

∠BOC + 70° = 180°

∠BOC = 110°

∠AOD = ∠BOC = 110°(Vertically opposite angle)

(i) True: Since, the angles form a sum of 180°.

(ii) False: Since, the two angles unless are on the line are necessarily equal to 90°.

(iii) False: Since, acute are less than 90° and hence two acute angles cannot give a sum of 180°

(iv) True: Since, sum of angles of linear pair is 180° hence, if both the angles are equal they would measure 90°.

Two parallel lines l and m be cut by a transversal t, forming angles.

It is given that ∠1: ∠2 = 5: 7

Let the measures of angel be 5x and 7x,

Then,

5x + 7x = 180°

12x = 180°

x = 180/12

x = 15°

∴ ∠1 = 5x = 5 × 15 = 75°

∠2 = 7x = 7 × 15 = 105°

We know that,

∠2 + ∠3 = 180° … [∵ linear pair]

105° + ∠3 = 180°

= ∠3 = 180° – 105°

= ∠3 = 75°

∠3 + ∠6 = 180° … [∵ the sum of the consecutive interior angle is 180°]

75° + ∠6 = 180°

∠6 = 180 – 75

∠6 = 105°

Now ∠6 = ∠8 = 105° … [∵ vertically opposite angles are equal]

∴ ∠1= 75°, ∠2 = 105°, ∠3 = 75° and ∠8 = 105°.

(d) Since, ∠AOC and ∠BOC are on the same line AOB and forming linear pair.

∴∠AOC + ∠BOC=180°

Hence, ∠AOC and ∠BOC are supplementary angles.

Since points A, O and B are collinear (Given), therefore AB is a straight line. 

(i) As O is a point on the line AB, therefore ∠AOD + ∠DOC + ∠BOC = 180° 

or, ∠AOD + ∠BOC + 90° = 180° 

or, ∠AOD + ∠BOC = 90° 

So, ∠AOD and ∠BOC are complementary angles. 

(ii) Also, ∠AOC and ∠BOC are supplementary as ∠AOC + ∠BOC = 180°

Given that,

∠AOC + ∠BOE = 70°

And,

∠BOD = 40°

∠BOF = ?

∠BOD = ∠AOC = 40°(Vertically opposite angle)

Given,

∠AOC + ∠BOE = 70°

40° + ∠BOE = 70°

∠BOE = 70° – 40°

= 30°

∠AOC and ∠BOC are linear pair angle

∠AOC + ∠COE + ∠BOE = 180°

∠COE = 180° – 30° – 40°

 = 110°

Therefore,

∠COE = 360° – 110°

= 250°

Given that ∠AOC +∠BOE = 70° 

∠BOD = 40° 

∠AOC = 40° 

(∵ ∠AOC, ∠BOD are vertically opposite angles) 

∴ 40° + ∠BOE = 70° 

⇒ ∠BOE = 70° – 40° = 30° 

Also ∠AOC + ∠COE +∠BOE = 180°

( ∵ AB is a line) 

⇒ 40° + ∠COE + 30° = 180° 

⇒ ∠COE = 180° -70° = 110° 

∴ Reflex ∠COE = 110°

(d) ∠2 + ∠3 = 180^o

We know that, interior opposite angles are equal

∠2 = ∠3

1) The line m intersects other two lines / and n at two distinct points A and C respectively. Hence m is a transversal line. 

2) The line n intersects other two lines / and m at two distinct points B and C respectively. Hence n is a transversal line

Given ∠AOC and ∠COB are linear pair. 

But ∠AOC = ∠AOD + ∠DOC

⇒ ∠AOC + ∠COB = 180° (linear pair) 

⇒ ∠AOD + ∠DOC + ZCOB = 180°

Given ∠AOC and ∠COG are linear pair.

∠AOC + ∠COG = 180° (linear pair) 

But ∠AOC = ∠AOB + ∠BOC 

= ∠1 + ∠2 ∠COG – ∠COD + ∠DOE + ∠EOF + ∠FOG – ∠3 + ∠4 + ∠5 + ∠6 

=> (∠AOB + ∠BOC) + (∠COD + ∠DOE + ∠EOF + ∠FOG) = 180° 

=> ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 180° 

Therefore, the sum of angles at a point on the same side of the line is 180°.

(i) Parallel lines: Lines which don’t intersect any where are called parallel lines. 

(ii) Perpendicular lines: Two lines which are at a right angle to each other are called perpendicular lines. 

(iii) Line segment: it is a terminated line. 

(iv) Radius: The length of the line-segment joining the centre of a circle to any point on its circumference is called its radius.

Given : ∠1 = ∠3 

∠3 = ∠4 

∠2 = ∠4 

∴∠1 = ∠2 

∵Both ∠1 and ∠2 are equal to ∠4. (By Euclid’s axiom things which are equal to same things are equal to one another)

Axiom 5 states that the whole is greater than the part. This axiom is known as a universal truth because it holds true in any field, and not just in the field of mathematics. Let us take two cases − one in the field of mathematics, and one other than that.

Case I

Let t represent a whole quantity and only a, b, c are parts of it.

t = a + b + c

Clearly, t will be greater than all its parts a, b, and c. Therefore, it is rightly said that the whole is greater than the part.

Case II

Let us consider the continent Asia. Then, let us consider a country India which belongs to Asia. India is a part of Asia and it can also be observed that Asia is greater than India. That is why we can say that the whole is greater than the part. This is true for anything in any part of the world and is thus a universal truth.

Each side of the rhombus = 18 cm

base of the rhombus = 18 cm

Distance between two parallel sides = 12 cm

Height = 12 cm

Area of the rhombus = base x height = 18 x 12 = 216 cm² 

Let the two parallel sides and the distance between them be 3x, 4x, 2x cm respectively 

Area = 1/2 (sum of parallel sides) x (distance between parallel sides)

= 1/2 (3x + 4x) x 2x = 175 (given)

=> 7x x x = 175

=> 7x² = 175

=> x² = 25

=> x = 5

Height i.e. distance between parallel sides = 2x = 2 x 5 = 10 cm

Ratio in length and breadth = 5 : 4

Area of rectangular field = 3380 m² 

Let length = 5x and breadth = 4x

5x x 4x = 3380

=> 20x²= 3380

x² = 3380/20 = 169 = (13)2

x = 13

Length = 13 x 5 = 65 m 

Breadth =13 x 4 = 52 m

Perimeter = (l + b) = 2 x (65 + 52) m = 2 x 117 = 234 m

Rate of fencing = ₹ 75 per m

Total cost = 234 x 75 = ₹ 17550

Given

Circumference = 44 m

Let radius of circle = r m

Thus, 2πr = 44

⇒ 2 × \(\frac { 22 }{ 7 }\) × r = 44

⇒ r = \(\frac { 44\times 7 }{ 2\times 22 } \)

= 7 m

Area of circle = πr²

= \(\frac { 22 }{ 7 }\) × 7 × 7

= 154 m²

Thus, area of circle is 154 m²

Answer is (A) 22 cm²

Given : Radius of two concentre circle are 4 cm and 3 cm.

Thus, R= 4 cm, r = 3 cm

∴ Area between two concentric circles

= π(R² – r²)

= \(\frac { 22 }{ 7 }\) [(4)² – (3)²]

= \(\frac { 22 }{ 7 }\) × [(4 + 3) (4 – 3)]

= \(\frac { 22 }{ 7 }\) × 7 = 22 cm²

Answer is (B) π (R + r)(R – r)

= πR² – πr²

= π(R² – r²)

= π (R + r)(R – r)

Given: Side of square = 14 cm

Diameter of semi circle (AD = CB) = 14 cm

Radius of semicircle (R) = \(\frac { 14 }{ 2 }\) = 7 cm

∴ Area of square ABCD = (side)²

= 14 × 14 = 196 cm²

Area of semicircle = \(\frac { 1 }{ 2 }\)πr²

\(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 77 cm²

Area of two semicircle =2 × 77 = 154 cm²

∴ Area of shaded part Area of square ABCD – Area of two semicircles

= 196 – 154 = 42 cm²

Thus area of shaded part 42 sq cm

Let each side of the square be x cm. 

Its area = x² = 169 (given)

x = √169 

x = 13 cm

(i) Thus, side of the square = 13 cm

(ii) Again perimeter = 4 (side) = 4 x 13 = 52 cm

Given,

Area of the circle = 301.84 cm²            

We know that,

Area of a Circle = πr² = 301.84 cm²  

(22/7) × r² = 301.84

r² = 13.72 x 7 = 96.04

r = √96.04 = 9.8

So, the radius is = 9.8 cm.

Now, Circumference of a circle = 2πr

= 2 × (22/7) × 9.8 = 61.6 cm

Hence, the circumference of the circle is 61.6 cm.

Supplementary angles are those angles whose sum is 180°. Hence, following are the pairs of supplementary angles:

1.  ∠1, ∠8
2.  ∠2, ∠7
3. ∠3, ∠4
4.  ∠4, ∠5
5. ∠5, ∠6
6. ∠6, ∠3

(80°, 100°), (60°, 120°), (108°, 72°) (140°, 40°). 30°, 150°)

(A) 330°, - 60° 

Here, 405° - (- 675°) = 1080° = 3(360°), 

1230° - (- 930°) = 2160° = 6(360°), 

and 450° - (- 630°) = 1080° = 3(360°)  

∴ the above paris of angles are co-terminal. 

Now, 330° - (- 60°) = 390° 

Since 390° is not a multiple of 360°. 

∴ This pair of angles is not co-terminal.

(C) 40°, 60°, 140°, 120°

Let the measures of the angles of the quadrilateral be 2k, 3k, 7k and 6k in degrees. 

Now, 2k + 3k + 7k + 6k = 360°

 ….[ ∵ the sum of angles of a quadrilateral is 360°] 

⇒ 18k = 360°

⇒ k = 20°

The measures of the angles of quadrilateral are 

2k = 2 x 20° = 40° 

3k = 3 x 20° = 60° 

7k = 7 x 20° = 140°

6k = 6 x 20° = 120°

(B) 30°, 60°, 90°

The angles of a triangle are in A.P. and ratio of the number of degrees in the least to the number of radians in the greatest is 60 : π. The angles of the triangle in degrees are 30°, 60°, 90°

(B) 35π cm

S = r\(\theta\) = 20 x 7π/4 = 35π cm

(D)  \(\frac{\pi^c}6\),\(\frac{5\pi^c}{12}\) ,\(\frac{2\pi^c}3\),\(\frac{3\pi^c}4\)

Let the measure of the angles be 2k, 5k, 8k and 9k in radians. 

∴ 2k + 5k + 8k + 9k = 2π

 ….[∵ the sum of the measures of the angles of a quadrilateral is 2π^c]

⇒ 24k = 2π

⇒ k = π/12

∴ Measures of angles of the quadrilateral are

\(\frac{\pi^c}6\),\(\frac{5\pi^c}{12}\) ,\(\frac{2\pi^c}3\),\(\frac{3\pi^c}4\)

 (A) \(\frac{\pi^c}4\) 

Number of sides = 8 

∴ Exterior angle = \(\frac{360°}8\) 

= 45 x \(\frac{\pi^c}{180}\)

= \(\frac{\pi^c}4\)

(B) \(\frac{2\pi^c}5\) 

Number of sides = 5

Exterior angle = \(\frac{360°}{no \,of \,sides}\)

∴ Exterior angle = \(\frac{360°}{5}\) = \((\frac{360}{5}\times\frac{\pi}{180})^c\)

= \(\frac{2\pi^c}5\)

 Increase in Kelvin will be 25K. 

 The quantity of heat given to body depends on: 

(i) The mass of the body, 

(ii) The rise (or fall) in the temperature of the body, and 

(iii) Nature of the material of the body.

Given: 5 rounds = 2πr(5),

t = 2minutes = 120s

To find: Radius (r)

Formula: a_cp = ω²r

Calculation: From formula,

a_cp = ω²r

∴ π² = \(\frac{V^2}{r}\)

But v = \(\frac{2\pi r(5)}{t}=\frac{10\pi r}{t}\)

∴ π² = \(\frac{100 \pi^2 r^2}{rt^2}\)

∴ r = \(\frac{120\times120}{100}\) =144m

The radius of the track is 144 m.

We have H = ms Δθ or Δθ = (H/ms). 

It follows that if s is more, Δθ will be small (for given values of H and m). Thus, for a given body, its specific heat capacity determines the change in temperature produced by a given quantity of heat. It is thus like mass in mechanics which determines the change in velocity (or the acceleration) produced by a given force. It is quite appropriate, therefore, to regard specific heat capacity as a measure of thermal inertia. 

The mass and specific heat capacity are two factors on which the heat absorbed or given out by a body depends.

(i) Water is used as an effective coolant in radiator of a car. 

(ii) Hot water bottles are used for fomentation. 

Specific latent heat of vaporization is the quantity of heat required to convert unit mass of a substance from liquid to vapour state without change of temperature.

Specific heat capacity of land is 5 times less as compared to water. Thus, air above land becomes hot and light and rises up resulting in drop in pressure of land mass during day time. Thus, cool air from sea starts blowing towards and forming sea breeze. 

During night, land as well as sea radiates heat energy. However, temperature of land falls faster than sea water, because of high specific heat capacity of sea water. So, at night the temperature of sea water is more than land. Warm air above the sea rises and cold air from land starts blowing towards sea resulting in land breeze. 

(i) Inter molecular space changes. 

(ii) Intermolecular space increases.

Due to the rise in sea level the buildings and roads in the coastal areas will get flooded as a result population living in that area will suffer damage from hurricanes and tropical storms.

Heat capacity of substance B is more than substance A.

During the state change heat supplied increases potential energy of molecules as distance between molecules increases work is done by heat supplied against attractive force.

1g steam at 100°C contains more heat than 1g water at 100°C. The reason is that lg steam at 100°C, when converts to 1 g water at 100°C, liberates 2260 J heat.