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(A) 14 π

Circumference/ (Area of circle) = 2/7

∴ 2πr/πr² = 2/7

∴ r = 7

∴ Circumference = 2πr 

= 2 x π x 7  = 14π

Given: Radius (r) = 18 cm, 

Measure of the arc (θ) = 80° 

To find: Length of the arc. 

Length of arc = θ/360 × 2πr 

= 8/360 × 2 × 3.14 × 18 

= 2/9 × 2 × 3.14 × 18 

= 2 × 2 × 3.14 × 2 = 25.12 cm 

∴ The length of the arc is 25.12 cm.

Given: Radius (r) = 3.5 cm,

length of arc (l) = 2.2 cm 

To find: Area of the sector. 

Area of sector = (l x r)/2

= (2.2 x 3.5)/2

= 1.1 × 3.5 = 3.85 cm² 

∴ The area of the sector is 3.85 cm².

Given: radius (r) = 7 cm, 

m(arc MBN) = θ = 60° 

i. Area of circle = πr² 

= 22/7 × (7)² 

= 22 × 7

= 154 cm² 

∴ The area of the circle is 154 cm² 

ii. Central angle (θ) = ∠MON = 60° 

Area of sector = θ/360 x πr²

∴ A(O – MBN) = 60/360 × 22/7 × (7)² 

= 1/6 × 22 × 7 

= 25.67 cm² 

= 25.7 cm² 

∴ A(O-MBN) = 25.7 cm²

iii. Area of major sector = area of circle – area of minor sector 

∴ A(O-MCN) = Area of circle – A(O-MBN) 

= 154 – 25.7 

∴ A(O-MCN) 

= 128.3 cm²

Area of sector \(=\frac{l \times r}{2}\)

\(=\frac{2.2 \times 3.5}{2}\)

= 1.1 × 3.5 = 3.85 cm²

∴ The area of the sector is 3.85 cm².

Given: ∠ROQ = ∠MON = 60°, 

radius (r) = OR = 7 cm, radius (R) = OM = 21 cm 

To find: Lengths of arc RXQ and arc MYN. 

i. Length of arc RXQ = θ/360 x 2πr

= 60/2 × 2 × 22/7 × 7 

= 1/6 × 2 × 22 

= 7.33 cm 

ii. Length of arc MYN = θ/360 x 2πR 

= 60/2 × 2 × 22/7 × 21 

= 1/6 × 2 × 22 × 3 

= 22 cm 

∴ The lengths of arc RXQ and arc MYN are 7.33 cm and 22 cm respectively.

Given: For the sphere, diameter (d) = 6 cm 

To find: Volume of the sphere. 

Radius (r) = d/2 = 6/2 = 3 cm 

Volume of sphere = 4/3πr² 

= 4/3 × 3.14 × (3)³ 

= 4 × 3.14 × 3 × 3 

= 113.04 cm³ 

∴ The volume of the sphere is 113.04 cm³ .

Let the radius of the bigger circle be R and that of smaller circle be r. 

OA, OB, OC and OD are the radii of the bigger circle. 

∴ OA = OB = OC = OD = R 

PQ = PA = r 

OQ + BQ = OB … [B – Q – O] 

OQ = OB – BQ = R – 9 

OE + DE = OD ….[D – E – O] 

OE = OD – DE = [R – 5] 

As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle, 

OQ × OA = OE × OF 

∴ (R – 9) × R = (R – 5) × (R – 5) …[∵ OE = OF] 

∴ R² – 9R = R² – 10R + 25 

∴ -9R + 10R = 25 

∴ R = [25units] 

AQ = AB – BQ = 2r ….[B-Q-A] 

∴ 2r = 50 – 9 = 41 

∴ r = 41/2 = 20.5 units

Distance covered by Bhavana in one round = Perimeter of the square field 

= 4 × side of square field = 4 × 80 m = 320 m 

Distance covered in 10 rounds = (320 × 10)m = 3200 m 

Distance covered by Sushmita in one round 

= Perimeter of the rectangular field 

= 2 × (length + breadth) 

= 2 × (150 + 60)m = 2 × 210 = 420 m 

Distance covered in 8 rounds = 420 × 8 = 3360 m 

Hence, Sushmita has covered 160 m more than the distance coverd by Bhavna.

Distance covered by Bhavana in one round = Perimeter of the square field = 4 × side of square field = 4 × 80m = 320m 

Distance covered in 10 rounds = (320 × 10)m = 3200m 

Distance covered by Sushmita in one round 

= Perimeter of the rectangular field 

= 2 × (length + breadth) 

= 2 × (150 + 60)m 

= 2 × 210 = 420m 

Distance covered in 8 rounds = 420 × 8 = 3360m 

Hence, Sushmita has covered 160m more than the distance coverd by Bhavna.

Perimeter of a rectangle = 2 (length + breadth) 

Length of the rectangular field = 3 × breadth 

Therefore perimeter of field = 2 (3 × breadth + breadth) 

= 2 ( 4 × breadth) 

= 8 × breadth 

The given perimeter = 800 m 

Therefore 8 × breadth = 800 

Or, breadth = 800 ÷ 8 = 100 m 

So, length = 3 × 100 m = 300 m

Perimeter of a rectangle = 2 (length + breadth)

Length of the rectangular field = 3 × breadth 

Therefore perimeter of field = 2 (3 × breadth + breadth)

= 2 ( 4 × breadth) 

= 8 × breadth 

The given perimeter = 800m 

Therefore 8 × breadth = 800 

Or, breadth = 800 ÷ 8 = 100m 

So, length = 3 × 100m = 300m

Square field

Length of rectangular field = 120 cm

According to question,

⇒ 6(breadth) = 120 cm

⇒ breadth = 120/6 cm = 20 cm

∴ Perimeter of the field = 2(length + breadth)

= 2(120 + 20) cm

= 2(140) cm = 280 cm

The area of Anmol’s chart paper = (90 × 40) cm² = 3600 cm²

And the area of Abhishek’s chart paper = (50 × 70) cm² = 3500 cm²

∴ The chart paper of Anmol will cover more area on the table than that of Abhishek by (3600 – 3500) cm² = 100 cm²

Length of rectangular path = 60 m = 6000 cm

Width of rectangular path = 3 m = 300 cm

Side of a square tile = 25 cm

Number of tiles will be in one row along the width of the path = width of the path/ side of a tile = 300/25 = 12

Width of path = width of the path/ side of a tile = 300/25 = 12

Number of rows = length of the path/ side of a tile = 6000/25 = 240

12, 240, 2880

Area of one square tile with side 15 cm = (15 × 15) cm² = 225 cm²

Since, the wall is covered with 28 square tiles.

∴ Area of the wall = (225 × 28) cm²

= 6300 cm²

Given, r = 3.5 cm 

Surface area of sphere = 4πr²

= 4 × \(\frac{22}{7}\) × 3.5 × 3.5 

= \(\frac{88\times12.25}{7}\) 

= 154 cm²

We know that, 

Volume of cone = \(\frac{1}{3}\)πr²h 

Volume of cylinder = πr²h 

Hence, 

volume of cone : volume of cylinder 

= \(\frac{1}{3}\)πr²h : πr²h 

Hence, volume of cone = \(\frac{1}{3}\) × volume of cylinder.

Given,

Height of cylinder = 21 m

Diameter of cylinder = 6 m

Radius of cylinder = \(\frac{6}{2}\) = 3 m

So,

Curved surface area of cylinder = 2πrh = 2 x \(\frac{22}{7}\) x 3 x 21 = 396 m²

∴ Cost of plastering the inner surface at rate Rs.9.50 per m² = 396 × 9.50 = Rs. 3762

Given,

Dimensions of oil tin are = 26 cm × 26 cm × 45 cm

Then,

Area of tin sheet required for making one oil tin = total surface area of oil tin

= 2(lb x bh x hl) = 2(26 x 26 + 26 x 45 + 45 x 26) = 2(676 + 1170 + 1170)

= 2 x 3016 = 6032 cm²

Area of tin sheet required for 20 oil tins = 20 x 6032 = 120640 cm² = 12.064 m²

So,

Cost of 1 m² tin sheet = Rs. 10

∴cost of 12.064 m² tin sheet = 10 x 12.064 = Rs. 120.64

Given,

Dimensions of rectangular sheet = 30 cm × 0.18 cm

Case (i)

When sheet is rolled along its length:

= 2πr = 30

= r = \(\frac{30}{2π}\)

height of cylinder = 18 cm

Hence, Volume of cylinder thus formed = V₁ = \(\frac{22}{7}\) x \((\frac{30}{2π})^2\) x 18 cm³

Case (ii)

When sheet is rolled along its breadth:

 = 2πr = 18

= r = \(\frac{18}{2π}\)

Height of cylinder = 30 cm

Volume of cylinder = V₂ = \(\frac{22}{7}\) x \((\frac{18}{2π})^2\) x 30 cm³

Hence,

\(\frac{V_1}{V_2}\) = \(\frac{[\frac{22}{7}\times 900\times 4π^2\times 18]}{[\frac{22}{7}\times 324\times 4π^2\times 30]}\) = \(\frac{30}{18}\) = \(\frac{5}{3}\)

Correct option is: A) 7920 m²

Diameter is 2r = 105m

\(\therefore\) Radius = r = \(\frac {105}{2}m\) 

Slant height of the cone is l = 40 m.

Height of cylindrical part is h = 4 m.

\(\therefore\) Total area of canvas required to build the tent.

= Curved surface area of cylinder + curved surface area of cone.

=  \(2 \pi rh + \pi rl\)

= \(2 \times \frac {22}7 \times \frac {105}2 \times 4 + \frac {22}7 \times \frac {105}2 \times 40\)

= 88 \(\times\) 15 + 440 \(\times\)15

= 1320 + 6600

= 7920 \(m^2\)

Hence, area of required canvas is 7920 \(m^2\)

Correct option is: A) 7920 m²

Correct option is: (D) 7

Radius of hemisphere is x cm.

\(\therefore\) Total surface area of the hemisphere = 3 \(\pi x^2\) \(cm^2\)

But given that T.S.A of hemisphere is 147\(\pi\) \(cm^2\).

\(\therefore\) 3\(\pi x^2\)  = 147 \(\pi\)

= \(x^2 = \frac {147}3 = 49 = 7^2\)

\(\therefore\) x = 7

Correct option is: (D) 7