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Option 3 : 352 cm^{2}

⇒ Formula of curved surface area of a cone = πrl

where r is the radius and l is the slant height of cone.

⇒ r = √(l^{2} – h^{2})

⇒ r = √(256 – 207)

⇒ r = √49 = 7 cm

∴ Curved surface area of the cone = (22/7) × 7 × 16 = 352 cmOption 2 : 4r

**Calculation:**

Let the height of cone be h

Volume of cone = 1/3 × π × r^{2} × h

Volume of sphere = 4/3 × π × r^{3}

According to the question,

⇒ 1/3 × π × r^{2} × h = 4/3 × π × r^{3}

Option 1 : 1237.5

Height of the conical tent h = 36 m

As we know

Curved surface area of conical tent = πrl

Circumference of the base of a conical tent = 2 πr

2 πr = 66

⇒ 2 × (22/7) × r = 66

⇒ r = 66 × (7/22) × (1/2)

⇒ r = 21/2

As we know,

l^{2} = r^{2} + h^{2}

⇒ l^{2} = (21/2)^{2} + 36^{2}

⇒ l^{2} = 441/4 + 1296

⇒ l^{2} = (441 + 5184)/4 = 5625/4

⇒ l = √5625/4

⇒ l = 75/2

∴ Curved surface area of conical tent = πrl = (22/7) × (21/2) × (75/2) = 1237.5 mOption 2 : 68.75

Radius of base tent = r and area of its base is 38.5 cm^{2} = πr^{2}

Then r = √(12.25) = 3.5

Also conical volume = 1/3 π r^{2 }h=154

h = 12 and

l = √ h^{2} + r^{2}

l = √12^{2} + 3.5^{2}

l = 12.5

Curved surface area = πrl = 137.50 cm^{2}

Curved surface area of conical tent = Area of canvas = length × width = 137.50 cm^{2}

Option 1 : 156π cm^{2}

h = 5 cm, r = 12 cm

Let the lateral height be l,

⇒ l^{2} = r^{2} + h^{2}

⇒ l^{2 }= 144 + 25 = 169

⇒ l = 13 cm

⇒ The curved surface area of the cone = π × r × l

⇒ The curved surface area of the cone = π × 12 × 13 = 156 π cm^{2}

∴ The curved surface area of the cone is 156 π cm^{2}.

Option 2 : 2/√5

As we know,

Volume of hemisphere = (2/3)πr^{3}

Volume of cone = (1/3)πr^{2}h

According to the question

(1/3)πr^{2}h = (2/3)πr^{3}

h = 2r

As we know,

l^{2} = r^{2} + h^{2}

⇒ l^{2} = r^{2} + (2r)^{2}

⇒ l^{2} = 5r^{2}

⇒ l = √5 r

Curved surface area of hemisphere = 2πr^{2}

Curved surface area of cone = πrl = πr × (√5r) = √5r^{2}

Option 3 : 550 cm^{2}

The volume of a right circular cone \(= \frac{1}{3}\pi {r^2}h\)

\(\Rightarrow 1232 = \frac{1}{3} × \frac{{22}}{7} × {r^2} × 24\)

\(\Rightarrow \frac{{1232 × 21}}{{22 × 24}} = {r^2}\)

⇒ r^{2} = 49

⇒ r = 7 cm

⇒Lateral height = √(r^{2} + h^{2})

⇒ Lateral height (l) = √(7^{2} + 24^{2}) = √625 = 25

⇒ Surface area = π r l = 22/7 × 7 × 25

∴ Curved surface area is 550 cmOption 1 : 144π

Given that r = 6 cm h = 12 cm

Volume of cone = \(\frac{1}{3}\pi {r^2}h\)

\(= \frac{1}{3}\pi {6^2}12 = 144\pi\) cm^{3}

Option 2 : 5.8 cm

Let the height of the upper section of the cone be h.

So, the height of the cone = (H) = (14 + h) cm

The radius of the cone = (R) = 14 cm

According to the question:

(1/3)πR^{2}H = 4928

(1/3) × (22/7) × 14 × 14 × (14 + h) = 4928

(616/3) × (14 + h) = 4928

14 + h = 24

h = 10 cm

The height of the cone = 14 + 10 = 24 cm

Now, the radius of the upper section of the cone:

(The height of upper section × the radius of cone)/The height of the cone

= (h × R)/H

(10 × 14)/24 = 5.8 cm (approx.)Option 1 : 616√10 cm^{2}

Let the radius of small cone be r.

According to the question:

Volume of large cone = 4 × Volume of small cones

(1/3)π14^{2} × 21 = 4 × (1/3)πr^{2} × 21

r = 7 cm

Slant height of small cones = √(7^{2} + 21^{2}) = 7√10 cm

Option 2 : 70

The curved surface area of cone = πrl where r is the radius and l is the slant height.

4070 = (22/7) × r × 37

Radius of the cone = 35 cm

Therefore, diameter = 2r = 70 cmOption 3 : 2∛30 cm

Area of base = 64π cm^{2}

⇒ πr^{2 }= 64

⇒ Radius of base = 8 cm

⇒ Height of cone = √(289 – 64) = 15

⇒ Volume of cone = πr^{2}h/3 = (π × 8 × 8 × 15) /3 = 320π

cone is remoulded to obtain a solid sphere,

⇒ Volume of cone = volume of sphere

⇒ 320π = (4/3)π × radius^{3}

Option 1 : Rs. 3300

Given, the height of a right circular cone = 24 cm

And, the radius of base = 7 cm

Since, Lateral height = √(h^{2} + r^{2})

So, Lateral height (l) = √(24^{2} + 7^{2}) = 25 cm

Since, The curved surface area of the cone = πrl

So, The curved surface area of the cone \(= {\rm{}}\frac{{22}}{7}{\rm{}} \times {\rm{}}7{\rm{}} \times {\rm{\;}}25{\rm{}} = {\rm{}}550{\rm{\;c}}{{\rm{m}}^2}\)

⇒ The cost of painting for the curved surface area of the cone = 550 × 6 = Rs. 3300

∴ The cost of the painting is Rs. 3300.

Option 1 : 418

Radii of two circular faces of the frustum of a cone are 3 cm and 2 cm.

Height of the circular frustum of cone H = 21 cm

As we know,

Volume of the frustum of cone = 1/3 × π [R^{2} + r^{2} + Rr] H

⇒ 1/3 × 22/7 × [3^{2} + 2^{2} + 3 × 2] × 21

⇒ 22 × [9 + 4 + 6]

⇒ 22 × 19

⇒ 418Option 1 : 2401 : 1

Let the radius of bases of Cones P and Q are ‘r’ and ‘R’

And the slant height of Cones P and Q are l and L

According to question l/L = 1/7

Given,

Curved surface area of P/curved surface area of Q = 7/1

π rl /π RL = 7/1

rl /RL = 7/1

(r/R) × (l/L) = 7/1

r/R × 1/7 = 7/1

r/R = 49/1

(r/R)^{2 }= 2401/1

On multiplying π in both numerator or in denominator we get,

π r^{2} /π R^{2 }= 2401/1

Option 3 : 352 cm^{2}

⇒ Formula of curved surface area of a cone = πrl

Where r is the radius and l is the slant height of cone.

⇒ r = √(l^{2} - h^{2})

⇒ r = √(256 - 207)

⇒ r = √49 = 7 cm

∴ Curved surface area of the cone = (22/7) × 7 × 16 = 352 cmOption 1 : 5 : 3

Radius of base of cone = BD = 8 cm and Height CD = 15 cm

∴ CB^{2} = 64 + 225 = 289

⇒ CB = 17 cm

We can say that the sphere inside the cone will act as the in circle of ΔABC;

∴ In radius = R = ∆/s

∆ = 1/2 × 15 × 16 = 120

s = (17 + 16 + 17)/2 = 25

∴ In radius = R = 120/25 = 24/5

∴ Ratio of radius of base of cone to the radius of sphere = 8 : (24/5) = 5 : 3

Option 4 : 72.8%

Let r = radius of base and h = height of cone

⇒ Volume of cone = (1/3) × πr^{2}h

Given, Radius and height is increased by 20%

⇒ New radius of base = r + r × (20/100) = 6r/5

⇒ New height of cone = h + h × (20/100) = 6h/5

⇒ New volume of the cone = (1/3) × π × (6r/5)^{2} × (6h/5) = (1/3) × π × (36/25)r^{2} × (6h/5)

⇒ Required percentage = [{(1/3) × π × (36/25)r^{2} × (6h/5)} – {(1/3) × πr^{2}h}] / {(1/3) × πr^{2}h} × 100 = {(216/125) – 1} × 100 = 72.8%

Option 4 : 136π cm^{2}

**Given:**

Diameter of 16 cm and height of 15 cm.

**Formula used:**

Radius(r) = Diameter/2

Curved surface area of cone = πrl

Slant height of cone, l = √(r2 + h2)

**Calculation:**

Radius of cone, r = 16/2 = 8 cm

Height of cone, h = 15 cm

Slant height of cone, l = √(r^{2} + h^{2}) = √(64 + 225) = √289 = 17 cm

From a cone of radius 3 cm and height 5 cm, the top portion is removed such that the radius is 2 cm. Find the volume of the remaining portion.

Option 5 : 33.17 cu cm

The smaller upper cone and the original cone will be similar

Thus according to the property of ratios,

⇒ (Radius of smaller cone)/(Radius of bigger cone ) = (Height of smaller cone) / (Height of the bigger cone)

∴ 2/3 = Height of smaller cone/5

∴ Height of the smaller cone = 10/3

∴ Height of the frustum = 5 – 10/3 = 5/3

Volume of frustum = 1/3 πh(r^{2 }+ rR + R^{2})

Where r = upper radius and R = Lower radius

Required volume = 1/3 π × (5/3) × (2^{2 }+ 6 + 3^{2}) = 33.17 cm^{3}

∴ The required volume is 33.17 cu.cm