Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Surface Areas and Volumes

Given

The Volume of each cube is = 64 cm^{3}

let's side of the cube =a

a^{3} = 64 cm^{3}

∴ a = 4 cm

the side of the cube = a = 4 cm

Also, when we joined end to end of each cube then the length and breadth of the resulting cuboid will 8 cm and 4 cm.

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

= 2(8 × 4 + 4 × 4 + 4 × 8) cm^{2}

= 2(32+16+32) cm^{2}

= (2×80) cm^{2} = 160 cm^{2}

The diameter of the hemisphere = D = 14 cm

The radius of the hemisphere = r = 7 cm

the height of the cylinder h = 13 - 7 = 6 cm

the radius of the hollow hemisphere = 7 cm

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

(2πrh+2πr^{2}) cm^{2} = 2πr(h+r) cm^{2}

2×(22/7)×7(6+7) cm^{2} = 572 cm^{2}

Given that radius of the cone (r) = 3.5 cm

r= 7/2 cm

The total height of the toy = 15.5 cm.

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

Total surface area of the toy =curved surface area of cone +curved surface area of the hemisphere = πrl + 2πr^{2}

=(22/7)×(7/2)×(25/2) + 2×(22/7)×(7/2)^{2}

= (275/2)+77 cm^{2}

= (275+154)/2 cm^{2 }

= 429/2 cm^{2} = 214.5cm^{2}

So, the total surface area of the toy is 214.5cm^{2}

the side of cube = Diameter of hemisphere

So side of cube is 7 cm. and the radius will be 7/2 cm.

We know,

The total surface area of solid = surface area of cubical block +curved surface area of hemisphere – Area of base of hemisphere

∴ TSA of solid = 6×(side)^{2}+2πr^{2}-πr^{2}

= 6×(side)^{2}+πr^{2}

= 6×(7)^{2}+(22/7)×(7/2)×(7/2)

= (6×49)+(77/2)

= 294+38.5 = 332.5 cm^{2}

So, the surface area of the solid is 332.5 cm^{2}

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

So, the radius of hemisphere = l/2

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

TSA of remaining solid = 6 (edge)^{2}+2πr^{2} - πr^{2}

= 6l^{2} + πr^{2}

= 6l^{2} + π(l/2)^{2}

= 6l^{2} + πl^{2}/4

= l^{2}/4(24+π) sq. units

Two hemisphere and one cylinder are included

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 = 2πr^{2} = 2×(22/7)×2.5×2.5

= 275/7 mm^{2}

Now, the surface area of the cylinder = 2πrh

= 2×(22/7)×2.5×9

(22/7)×45 = 990/7 mm^{2}

Thus, the required surface area of medicine capsule

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

= (2×275/7) × 990/7

(550/7) + (990/7) = 1540/7 = 220 mm^{2}

we know that

Diameter = 4 m

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 tent = surface area of cone + surface area of cylinder

= πrl+2πrh

= πr(l+2h)

= (22/7)×2(2.8+2×2.1)

= (44/7)(2.8+4.2)

= (44/7)×7 = 44 m^{2}

∴ The cost of the canvas of the tent at the rate of ₹500 per m^{2 }will be

= Surface area × cost per m^{2}

44×500 = ₹22000

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

same height and same diameter is hollowed out. Find the total surface area of the

remaining solid to the nearest cm^{2}.

we know that

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 Total surface area of remaining solid = surface area of conical cavity + Total surface area of the cylinder

= πrl+(2πrh+πr^{2})

= πr(l+2h+r)

= (22/7)× 0.7(2.5+4.8+0.7)

= 2.2×8 = 17.6 cm^{2}

So, the total surface area of the remaining solid is 17.6 cm^{2}