Surface Areas & Volume

Mathematics Class Tenth

10th-Math-home

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Formulae & NCERT Exercise 13.1

(1) Cuboid

Volume of cuboid `=(lxxbxxh)` cubic units

Where, `l=` length, `b=` breadth, and `h=` height of the cuboid

Surface Area of cuboid `=2(lb+bh+lh)` square units

Diagonal of cuboid `=sqrt(l^2+b^2+h^2)` units

(2) Cube

Let each edge of a cube be of length `a`, then

Volume of cube `=a^3` cubic unit

Surface area of cube `=6a^2` square unit

Diagonal of cube `= sqrt3 a` unit

(3) Cylinder

If radius of base `=r` and height or length `=h`, then

Volume of cylinder`=(pi r^2h)` cubic unit

Curved surface area (CSA) of cylinder `=2pirh` square unit

Total surface are (TSA) of cylinder `= 2 pi r h + 2 pi r^2` square unit

`=2pir(h+r)` square unit

(4) Cone

If radius of base `=r` and height `=h`, then

Slant height of cone, `l = sqrt(h^2+r^2)` unit

Volume of cone `=(1/3 pi r^2 h)` cubic unit

Curved surface area (CSA) of cone `=(pi r l)` square unit

Total surface area (TSA) of cone `= (pi r l+pi r^2)` square unit

(5) Sphere

If the radius of the sphere is `r`, then

Volume of sphere `= (4/3 pi r^3)` cubic unit

Surface area of sphere `= (4pi r^2)` square unit

(6) Hemisphere

If the radius of a hemisphere is `r`, then

Volume of hemisphere ` = (2/3 pi r^3)` cubic unit

Curved surface area of hemisphere `=(2pi r^2)` square unit

Total surface area (TSA) of hemisphere `=(3pi r^2)` square unit

NCERT Exercise 13.1

Unless stated otherwise, take `pi = 22/7`

Question: 1. 2 cubes each of volume `64 cm^3` are joined end to end. Find the surface area of the resulting cuboid.

Solution:

10 math surface area volume ex 13.1_1

Given, volume of one cube `=64 cm^3`

Therefore, surface area of resulting cuboid = ?

We know that, Volume of a cube `=a^3`

`64 cm^3 = a^3`

`=>a = root3(64 cm^3)`

`=>a = root3(4xx4xx4 cm^3)`

`=>a = 4 cm`

Thus side of one cube `=4cm`

Thus, length of cuboid, `l=4+4 =8cm`

Height of cuboid, `h=4cm` and breadth of cuboid, `b=4cm`

Now, since Surface area of a Cuboid ` = 2(lb+bh+lh)`

`= 2 (8xx4+4xx4+8xx4)`

`=2(32+16+32)`

`=2xx80 = 160 cm^2`

Thus, surface area of the rsulting cuboid `=160 cm^2` Answer

Question: 2 . A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is `14 cm` and the total height of the vessel is `13cm`. Find the inner surface area of the vessel.

Solution:

10 math surface area volume ex 13.1_2

Given, diameter of hemisphere = 14 cm

Therefore, radius of hemisphere `= 14/2 = 7 cm`

Height of vessel = 13 cm

Therefore, height of vessel, i.e. cylindrical portion excluding hemisphere

= Height of vessel - radius of hemisphere

`= 13 cm - 7 cm = 6 cm`

We know that, curved surface area of cylinder `= 2 pi r h`

`= 2 xx 22/7 xx 7 cm xx 6 cm`

` = 2 xx 22 xx cm xx 6 cm`

`= 44 xx 6 cm^2`

Or, Curved surface area of cylinder excluding hemisphere `= 264 cm^2`

Now, we know that curved surface area of hemisphere ` = 2 pi r^2`

`= 2 xx 22/7 xx (7\ cm)^2`

`= 2 xx 22 xx 7\ cm^2`

`= 44 xx 7\ cm^2 `

∴ Curved surface area of hemisphere `= 308\ cm^2`

Thus, Total surface are of given vessel = Curved surface area of cylinder (excluding hemisphere) + Curved surface area of hemisphere

`= 264\ cm^2 + 308\ cm^2`

`= 572\ cm^2`

Thus, total inner surface area of the vessel `= 572\ cm^2` Answer

Question: 3. A toy is in the form of a cone of radius `3.5` cm mounted on a hemisphere of same radius. The total height of the toy is `15.5` cm. Find the total surface area of the toy.

Solution:

10 math surface area volume ex 13.1_3

Given, height of the toy = 15.5 cm

Radius of the toy, i.e. radius of cone `(r)= 3.5` cm

Radius of hemisphere `(r)= 3.5` cm

Therefore, height of cone, `h =` total height - radius of hemisphere

`=>h = 15.5 ? 3.5 = 12\ cm`

Now, slant height of cone, `l = sqrt(h^2+r^2)`

`:. l = sqrt((12cm)^2+(3.5 cm)^2)`

`= sqrt (144 cm^2 + 12.25 cm^2)`

`= sqrt (156.25\ cm^2)`

`=> l = 12.5\ cm`

Now, curved surface area of cone `= pi r l` sq unit

`= 22/7 xx 3.5\ cm xx 12.5\ cm`

`= 22 xx 0.5\ cm xx 12.5\ cm`

`=137.5\ cm^2`

Thus, Curved surface area of cone `=137.5\ cm^2`

Now, Curved surface area of hemisphere `=2 pi r^2`

`= 2xx 22/7 xx (3.5\ cm)^2`

`= 2xx 22/7 xx 12.25\ cm^2`

`= 77\ cm^2`

Thus Curved surface area of hemisphere `=77\ cm^2`

Therefore, total surface area of the toy = curved surface area of cone + curved surface area oh hemisphere

`= 137.5\ cm^2+ 77\ cm^2`

Thus total surface area of toy `= 214.5\ cm^2` Answer

Question: 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

10 math surface area volume ex 13.1_4

Given, side of cubical block = 7 cm

Greatest (maximum) diameter of hemisphere = ?

And Surface area of solid =?

Since hemisphere is surmounted on the cubical block, thus greatest (maximum) diameter of the hemisphere = side of the cubical block = 7 cm

Now, Surface area of solid = Surface area of cubical block + curved surface area of hemisphere - surface area of base of the hemisphere

Surface area of cubical block `= 6 a^2`

Where `a` = side of cube

`= 6 xx (7\ cm)^2`

`=6 xx 49\ cm^2`

∴ surface area of cubical block `= 294\ cm^2`

Now, curved surface area of hemisphere `= 2pi r^2` sq. unit

Here, diameter of hemisphere `= 7 cm`

∴ Radius, `r = 7/2=3.5\ cm`

∴ Curved surface area of given hemisphere `= 2xx22/7 xx (3.5\ cm)^2`

` = 2xx 22/7 xx 3.5\ cm xx 3.5\ cm`

`= 77\ cm^2`

Now, surface area of base of the hemisphere `= pi r^2`

`=22/7 xx (3.5\ cm)^2`

`= 22/7 xx 3.5\ cm xx 3.5\ cm`

`=38.5\ cm^2`

∴ Surface area of solid = Surface area of cubical block + curved surface area of hemisphere - surface area of base of the hemisphere

` = 294\ cm^2 + 77\ cm^2 - 38.5\ cm^2`

`= 332.5\ cm^2`

∴ Surface area of given solid `= 332.5\ cm^2` Answer

Question: 5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter `l` of the hemisphere is equal to the edge of the cube. Determine the surface area of remaining solid.

Solution:

Given, edge of the cube `= l`

And, diameter of the hemisphere ` = l`

∴ radius of hemisphere `=l/2`

Now, curved surface area of hemisphere `= 2 pi r^2`

` = 2 xx pi (l/2)^2`

`= 2 pi\ l^2/4`

`=(pi\ l^2)/2`

Surface Area of base of hemisphere ` = pi r^2`

` = pi (l/2)^2`

`=(pi\ l^2)/4`

Surface area of cube `=6a^2`

`= 6xxl^2`

`=6l^2`

Now, Total surface area of remaining solid

=Surface area of solid + curved surfaced area of hemisphere -Surface Area of base of hemisphere

`=6l^2 + (pi\ l^2)/2 - (pi\ l^2)/4`

`=6l^2 + (pi (2l^2-l^2))/4`

`=6l^2 + (pil^2)/4`

`=(24l^2+pil^2)/4`

`=1/4(24+pi)l^2` unit²

∴ Total surface area of remaining solid `=1/4(24+pi)l^2` unit² Answer

Question: 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

10 math surface area volume ex 13.1_5

Solution:

Given, diameter of capsule `=5mm`

∴ radius of hemisphere `=5/2 = 2.5 mm`

Now, given length of the capsule `= 14 mm`

Therefore, length of cylinder excluding the two hemisphere

= total length of capsule - `2xx` radius of one hemisphere

`= 14 mm - 2 xx 2.5 mm`

`= 14 mm - 5 mm = 9 mm`

∴ length or height of cylinder, `h = 9mm`

And, radius of cylinder `= 2.5 mm`

Now, curved surface area of cylinder `= 2 pi r h`

`=2 pi xx 2.5 xx 9` mm²

`= 45 pi mm^2`

And, curved surface area of hemisphere ` = 2 pi r^2`

` = 2 pi (2.5)^2 mm^2`

`= 2 pi xx 6.25\ mm^2`

∴ curved surface area of 2 (two) hemisphere ` = 2 xx 2 pi xx 6.25\ mm^2`

`=4 pi xx 6.25\ mm^2`

` = 25 pi\ mm^2`

∴ Total surface area of given capsule

= surface area of cylinder + curved surface area of two hemisphere

`= 45 pi mm^2 + 25 pi\ mm^2`

` = pi (45+25) mm^2`

`=22/7 xx 70\ mm^2`

`= 220\ mm^2`

Thus, surface area of given capsule `=220\ mm^2` Answer

Question: 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are `2.1\ m` and `4\ m` respectively, and the slant height of the top is `2.8\ m`, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs `500` per `m^2`. (Note the base of the tent will not be covered with canvas)

Solution:

10 math surface area volume ex 13.1_6

Given, height of the cylinder of tent `=2.1 m`

And diameter of tent ` = 4 m`

∴ radius of tent `= 4/2 = 2 m`

And Slant height of conical top (cone) `=2.8m`

∴ total surface area of canvas =?

Cost of canvas @ Rs `500//m^2 =?`

Curved surface area of cylindrical portion of tent `= 2 pi r h`

`= 2 xx pi xx 2\ mxx 2.1\ m`

`=8.4 pi\ m^2`

Curved surface area of cone (top of tent) `= pi r l`

`=pi xx 2\ m xx 2.8\ m`

`= 5.6 pi\ m^2`

∴ total surface area of tent = surface area of cylindrical part + curved surface area of conical top

`= 8.4 pi\ m^2 + 5.6 pi \ m^2`

`= pi(8.4+5.6)m^2`

`=22/7 xx 14\ m^2`

`= 44\ m^2`

Now, given cost of `1m^2` of canvas `=` Rs `500`

∴ cost `44\ m^2` of canvas `=` Rs `500 xx 44 =22000`

Thus, area of canvas `= 44\ m^2` And cost of canvas `=` Rs `22000` Answer.

Question: 8. From a solid cylinder whose height is `2.4\ cm` and diameter `1.4\ cm`, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest `cm^2`.

Solution:

10 math surface area volume ex 13.1_7

Given, height of cylinder `h=2.4\ cm`

Diameter of cylinder `=1.4\ cm`

∴ radius, `r = 0.7\ cm`

Total Surface area of remaining solid =?

Now, curved surface area of cylinder `= 2 pi r h`

`=2 pi xx 0.7 xx 2.4\ cm^2`

`= 3.36 pi \ cm^2`

Area of base of the cylinder `=pi r^2`

`= pi (0.7\ cm)^2`

`= 0.49 pi\ cm^2`

Slant height of cone ` l= sqrt(h^2+r^2)`

`=sqrt((2.4\ cm)^2+(0.7\ cm)^2)`

`=sqrt(5.76 cm^2+0.49 cm^2)`

`=sqrt(6.25 cm^2)`

`=2.5 cm`

Thus, curved surface area of cone `= pi r l`

`= pi xx0.7 cm xx 2.5 cm`

`= 1.75 pi cm^2`

Now, total surface area of remaining solid = curved surface area of cylinder + surface area of base of cylinder + curved surface area of cone

`= 3.36 pi cm^2 + 0.49 pi\ cm^2 + 1.75 pi\ cm^2`

`=pi(3.36+0.49+1.75) cm^2`

`=22/7 xx 5.6 cm^2`

`= 17.6 cm^2`

Thus, total surface area of remaining solid `~~ 18 cm^2` Answer

Question: 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base of radius 3.5 cm, find the total surface area of the article.

10 math surface area volume ex 13.1_8