Chapter 12: Surface Areas and Volumes

The diagram is given as:

Given,

The Volume (V) of each cube is = 64 cm³

This implies that a³ = 64 cm³

∴ a = 4 cm

Now, the side of the cube = a = 4 cm

Also, the length and breadth of the resulting cuboid will be 4 cm each, while its height will be 8 cm.

So, the surface area of the cuboid = 2(lb+bh+lh)

= 2(8×4+4×4+4×8) cm²

= 2(32+16+32) cm²

= (2×80) cm² = 160 cm²

The diagram is as follows:

Now, the given parameters are:

The diameter of the hemisphere = D = 14 cm

The radius of the hemisphere = r = 7 cm

Also, the height of the cylinder = h = (13-7) = 6 cm

And the radius of the hollow hemisphere = 7 cm

Now, the inner surface area of the vessel = CSA of the cylindrical part + CSA of the hemispherical part

(2πrh+2πr²) cm² = 2πr(h+r) cm²

2×(22/7)×7(6+7) cm² = 572 cm²

The diagram is as follows:

Given that the radius of the cone and the hemisphere (r) = 3.5 cm or 7/2 cm

The total height of the toy is given as 15.5 cm.

So, the height of the cone (h) = 15.5-3.5 = 12 cm

∴ The curved surface area of the cone = πrl

(22/7)×(7/2)×(25/2) = 275/2 cm²

Also, the curved surface area of the hemisphere = 2πr²

2×(22/7)×(7/2)²

= 77 cm²

Now, the Total surface area of the toy = CSA of the cone + CSA of the hemisphere

= (275/2)+77 cm²

= (275+154)/2 cm²

= 429/2 cm² = 214.5cm²

So, the total surface area (TSA) of the toy is 214.5cm²

It is given that each side of the cube is 7 cm. So, the radius will be 7/2 cm.

We know,

The total surface area of solid (TSA) = surface area of the cubical block + CSA of the hemisphere – Area of the base of the hemisphere

∴ TSA of solid = 6×(side)²+2πr²-πr²

= 6×(side)²+πr²

= 6×(7)²+(22/7)×(7/2)×(7/2)

= (6×49)+(77/2)

= 294+38.5 = 332.5 cm²

So, the surface area of the solid is 332.5 cm²

The diagram is as follows:

Now, the diameter of the hemisphere = Edge of the cube = l

So, the radius of the hemisphere = l/2

∴ The total surface area of solid = surface area of cube + CSA of the hemisphere – Area of the base of the hemisphere

The surface area of the remaining solid = 6 (edge)²+2πr²-πr²

= 6l² + πr²

= 6l²+π(l/2)²

= 6l²+πl²/4

= l²/4(24+π) sq. units

Two hemispheres and one cylinder are shown in the figure given below.

Here, the diameter of the capsule = 5 mm

∴ Radius = 5/2 = 2.5 mm

Now, the length of the capsule = 14 mm

So, the length of the cylinder = 14-(2.5+2.5) = 9 mm

∴ The surface area of a hemisphere = 2πr² = 2×(22/7)×2.5×2.5

= 275/7 mm²

Now, the surface area of the cylinder = 2πrh

= 2×(22/7)×2.5×9

(22/7)×45 = 990/7 mm²

Thus, the required surface area of the medicine capsule will be

= 2×surface area of hemisphere + surface area of the cylinder

= (2×275/7) × 990/7

= (550/7) + (990/7) = 1540/7 = 220 mm²

It is known that a tent is a combination of a cylinder and a cone.

From the question, we know that

Diameter = 4 m

The slant height of the cone (l) = 2.8 m

Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m

Height of the cylinder (h) = 2.1 m

So, the required surface area of the tent = surface area of the cone + surface area of the cylinder

= πrl+2πrh

= πr(l+2h)

= (22/7)×2(2.8+2×2.1)

= (44/7)(2.8+4.2)

= (44/7)×7 = 44 m²

∴ The cost of the canvas of the tent at the rate of ₹500 per m² will be

= Surface area × cost per m²

44×500 = ₹22000

So, Rs. 22000 will be the total cost of the canvas.

The diagram for the question is as follows:

From the question, we know the following:

The diameter of the cylinder = diameter of conical cavity = 1.4 cm

So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7

Also, the height of the cylinder = height of the conical cavity = 2.4 cm

Now, the TSA of the remaining solid = surface area of conical cavity + TSA of the cylinder

= πrl+(2πrh+πr²)

= πr(l+2h+r)

= (22/7)× 0.7(2.5+4.8+0.7)

= 2.2×8 = 17.6 cm²

So, the total surface area of the remaining solid is 17.6 cm²