CBSE 10th Maths Chapter 13 - Surface Areas and Volumes-Exercise-13.1

The diagram is given as:

Given,

The Volume (V) of each cube is = 64 cm$^{3}$

This implies that $a^{3}$ = 64 cm$^{3}$

∴ 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) cm$^{2}$

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

= (2×80) cm$^{2}$ = 160 cm$^{2}$

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

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

$2×(\dfrac{22}{7})×7(6+7) cm^{2}$ =$ 572 cm^{2}$

The diagram is as follows:

Given that the radius of the cone and the hemisphere (r) = 3.5 cm or $\dfrac{7}{2}$ 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

Slant height of the cone($l$)=$\sqrt{h^{2}+r^{2}}$

$\Rightarrow l$=$\sqrt{12^{2}+(\dfrac{7}{2})^{2}}$

$\Rightarrow l$=$\sqrt{144+\dfrac{49}{4}}$

$\Rightarrow l$=$\sqrt{\dfrac{(576+49)}{4}}$

$\Rightarrow l$=$\sqrt{\dfrac{625}{4}}$

$\Rightarrow l$=$\dfrac{25}{2}$

∴ The curved surface area of the cone = $πrl$

($\dfrac{22}{7}$)×($\dfrac{7}{2}$)×($\dfrac{25}{2}$) = $\dfrac{275}{2} cm^{2}$

Also, the curved surface area of the hemisphere = 2πr$^{2}$

2×($\dfrac{22}{7}$)×($\dfrac{7}{2}$)$^{2}$

= 77 cm$^{2}$

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

= ($\dfrac{275}{2}$)+77 cm$^{2}$

= $\dfrac{(275+154)}{2}$ cm$^{2}$

= $\dfrac{429}{2}$ cm$^{2}$

= 214.5cm$^{2}$

So, the total surface area (TSA) of the toy is 214.5cm$^{2}$

It is given that each side of the cube is 7 cm. So, the radius will be $\dfrac{7}{2}$ 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 = $6×(side)^{2}+2πr^{2}-πr^{2}$

= $6×(side)^{2}+πr^{2}$

=$6×(7)^{2}+(\dfrac{22}{7})\times(\dfrac{7}{2})\times(\dfrac{7}{2})$

= $(6×49)+(\dfrac{77}{2})$

= 294+38.5 = 332.5 cm$^{2}$

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

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 = $6 (edge)^{2}+2πr^{2}-πr^{2}$

= $6l^{2} + πr^{2}$

= $6l^{2}+π(l/2)^{2}$

= $6l^{2}+\dfrac{πl^{2}}{4}$

= $\dfrac{l^{2}}{4(24+π)}$ sq. units

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

Here, the diameter of the capsule = 5 mm

∴ Radius = $\dfrac{5}{2}$ = 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×($\dfrac{22}{7})×2.5×2.5$

= $\dfrac{275}{7} mm^{2}$

Now, the surface area of the cylinder = 2πrh

= 2×($\dfrac{22}{7}$)×2.5×9

($\dfrac{22}{7})×45$ = $\dfrac{990}{7} mm^{2}$

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

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

= (2×$\dfrac{275}{7}) \times \dfrac{990}{7}$

= ($\dfrac{550}{7}) + (\dfrac{990}{7})$ = $\dfrac{1540}{7}$ = $220 mm^{2}$

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 = $\dfrac{4}{2}$ = 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)

= ($\dfrac{22}{7}$)×2(2.8+2×2.1)

= ($\dfrac{44}{7}$)(2.8+4.2)

= ($\dfrac{44}{7}$)×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.

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 =$\dfrac{1.4}{2}$ = 0.7

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

Slant height of the cone($l$)=$\sqrt{h^{2}+r^{2}}$

=$\sqrt{(2.4)^{2}+(0.7)^{2}}$

=$\sqrt{5.76+0.49}$

=$\sqrt{6.25}$

=2.5cm

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

= $πrl+(2πrh+πr^{2})$

= πr(l+2h+r)

= ($\dfrac{22}{7}$)× 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$^{2}$

height of the cylinder = 10 cm,

base radius = 3.5 cm

Curved surface area of the cylinder = 2πrh

=$\dfrac{2 \times 22 \times 35 \times 10}{ 7 \times 10}cm^{2}$

= 220 cm$^{2}$

Inner surface area of a hemispherical cavity=2πr$^{2}$

= $ 2 \times \dfrac{22}{7} \times \dfrac{35 \times 35}{10 \times 10}cm^{2}$

= 77 $cm^{2}$

Inner surface area of both hemispherical cavity = 77 $cm^{2}$ + 77 $cm^{2}$ = 154 $cm^{2}$

Total surface area of the solid = Curved surface area of the solid + Inner surface area of both hemispherical ends

= 220 $cm^{2}$ + 154 $cm^{2}$ = 374 $cm^{2}$