## Solids

Category : 9th Class

### Solids

The objects having definite shape and size are called solids. A solid occupies a definite space.

Cuboid

Solids like matchbox, chalk box, a tile, a book an almirah, a room etc. are in the

Shape of a cuboid

Formulae

For cuboid of length = I, breath = b and height = h, we have:

(i) Volume $=(l\times b\times h)$

(ii) Total surface area $=2(lb\times bh\times lh)$

(iii) Lateral surface area $=[2(l+b)\times h]$

Cube

Solids like ice cubes, sugar cubes, dice etc. are the shape of cube Formula for a cube having each edge = a units, we have:

(i) Volume$={{a}^{3}}$

(ii) Total surface area $=6{{a}^{2}}$

(iii) Lateral surface area $=4{{a}^{2}}$

Cylinder

Solids like measuring jar, circular pencils, circular pipes, road rollers, gas cylinder, are said to have a cylindrical shape. Formula for a cylinder of base radius = r & height (or length) = h, we have

(i) Volume$=\pi {{r}^{2}}h$

(ii) Curved surface area $=2\pi rh$

(iii) Total surface area $=(2\pi rh+2\pi {{r}^{2}})=2\pi r(h+r)$

Hollow Cylinder

Solids like a hollow cylinder having external radius = R, internal radius = r & height = h then , we have

(i) Volume of material = (external volume) - (internal volume) $=(\pi {{R}^{2}}h-\pi {{r}^{2}}h)=\pi h({{R}^{2}}-{{r}^{2}})$

(ii) Curved surface area of hollow cylinder = (external surface area) - (internal surface area) $=(2\pi Rh-2\pi rh)=2\pi h(R-r)$

(iii)   Total surface area of hollow cylinder = (curved surface area) + 2$\times$(area of the base ring) $=2\pi h(R-r)+2(\pi {{R}^{2}}-\pi {{r}^{2}})$ $=2\pi (R-r)(R+r+h)$

Cone

Consider a cone in which base radius = r, height = h & slant height$l=\sqrt{{{h}^{2}}+{{r}^{2}}}$ then we have

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

(ii) Curved surface area of the cone $=\pi rl$

(iii) Total surface area of the cone = (curved surface area) + (area of the base) $\pi rl+\pi {{r}^{2}}=$ $\pi r(l+r)$

Sphere

Objects like a football, a cricket ball, etc. are said to have the shape of the a sphere. For a sphere of radius r, we have

(i) Volume of the sphere $=\left( \frac{4}{3}\pi {{r}^{3}} \right)$

(ii) Surface area of the sphere $(4\pi {{r}^{2}})$

Hemisphere

A plane through the centre of a sphere cuts it into two equal parts, each part is called hemisphere. For a hemisphere of radius r, we have:

(i) Volume of the hemisphere $=\frac{2}{3}\pi {{r}^{3}}$

(ii) Curved surface area of the hemisphere $=(2\pi {{r}^{2}})$

(iii) Total surface area of the hemisphere $=(3\pi {{r}^{2}})$

• 3 is the number of spatial dimensions needed mathematically describe a solid.
• $\pi =$3.14159265358979323846264338327 95088........
• The word "zero" was coined by the Italian mathematician Leonardo Pisano
• $\sqrt{2}$ is also called Pythagoras constant.
• 3 is the only prime 1 less than a perfect square.

• Volume of a cylinder $=\pi {{r}^{2}}h$
• Curved surface area cylinder $=2\pi rh$
• Total surface area cylinder $=2\pi r(h+r)$
• Volume of the cone $=\frac{1}{3}\pi {{r}^{2}}h$
• Volume of the sphere $=\left( \frac{4}{3}\pi {{r}^{3}} \right)$
• Surface area of the sphere $=(4\pi {{r}^{2}})$
• volume of the hemisphere $=\frac{2}{3}\pi {{r}^{2}}$
• Curved surface area of the hemisphere $=(2\pi {{r}^{2}})$
• Total surface area of the hemisphere $=(3\pi {{r}^{2}})$

The inner diameter of a glass is 7 cm and it has a raised portion at the bottom in the shape of a hemisphere as shown in the figure. If the height of the glass is 16 cm then find the apparent capacity and the actual capacity of the glass.

(a) $\text{616 c}{{\text{m}}^{\text{3}}},\text{ 526}.\text{17 c}{{\text{m}}^{\text{3}}}$

(b) $~\text{616 c}{{\text{m}}^{\text{3}}},\text{ 1527}.\text{95 c}{{\text{m}}^{\text{3}}}$

(c) $\text{616 c}{{\text{m}}^{\text{3}}},\text{ 886}.\text{16 c}{{\text{m}}^{\text{3}}}$

(d) $\text{616 c}{{\text{m}}^{\text{3}}},\text{ 9}.\text{58 c}{{\text{m}}^{\text{3}}}$

(e) None of these

Explanation:

In the given figure

radius of the glass$=\frac{7}{2}$cm & height = 16 cm

Apparent capacity of the glass$=\pi {{r}^{2}}h$

$\text{=}\left( \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times \text{l6} \right)\text{c}{{\text{m}}^{\text{3}}}\text{=616 c}{{\text{m}}^{\text{3}}}$

Volume of the hemisphere at the bottom

$\text{=}\frac{2}{3}\pi {{r}^{3}}\left( \frac{2}{3}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2} \right)\text{c}{{\text{m}}^{\text{3}}}\text{=}\frac{539}{6}\text{c}{{\text{m}}^{\text{3}}}=89.83\,c{{m}^{3}}$

A actual capacity of the glass = (volume of the glass) - (volume of the hemisphere) = (616 - $\text{89}.\text{83})\text{c}{{\text{m}}^{\text{3}}}\text{ 526}.\text{17c}{{\text{m}}^{\text{3}}}$

A wooden toy is in the shape of a cone surmounted on a cylinder as shown in figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that on the cylinder part of radius 4 cm. Find the surface area of the toy.

(a) $\text{264}\,\text{c}{{\text{m}}^{\text{2}}}$

(b) $\text{25}0\,\text{c}{{\text{m}}^{\text{2}}}$

(c) $\text{322}.\text{14}\,\text{c}{{\text{m}}^{\text{2}}}$

(d) $\text{25}0\,\text{c}{{\text{m}}^{\text{2}}}$

(e) None of these

Explanation:

Area of conical part of toy = (upper area of cone) + (base area of the cone) - (base area of cylinder$=(\pi Rl+\pi {{R}^{2}}-\pi {{r}^{2}})$sq units When

R = Radius of the conical part = 2.5 cm

H = Height of the conical part = 6 cm

L= Slant height of the conical part$=\sqrt{{{R}^{2}}+{{h}^{2}}}$$=\sqrt{{{(2.5)}^{2}}+{{6}^{2}}}$=6.5 cm

R = Radius or cylinder = 2 cm

Area$=\pi (Rl+{{R}^{2}}-{{r}^{2}})$

$=\frac{22}{7}\left\{ 2.5\times 6.5+{{(2.5)}^{2}}-{{2}^{2}} \right\}c{{m}^{2}}=\frac{407}{7}c{{m}^{2}}$$=\text{58}.\text{14 c}{{\text{m}}^{\text{2}}}$

Area of cylindrical part = (curved surface area of the cylinder) + (area of the base of the cylinder) $=2\pi rh+\pi {{r}^{2}}$

When h = height of cylinder = (26 - 6) = 20 cm

$=\pi r(2h+r)\text{c}{{\text{m}}^{\text{2}}}=\left\{ \frac{22}{7}\times 2\times (40+2) \right\}c{{m}^{2}}=\text{264}\,\text{c}{{\text{m}}^{\text{2}}}$

Then total surface area = Area of upper part + Area of lower part

$=\text{58}.\text{14 }{{\text{m}}^{\text{2}}}+\text{264 }{{\text{m}}^{\text{2}}}=\text{322}.\text{14 c}{{\text{m}}^{\text{2}}}$

The base of a triangular field is twice its altitude. If the cost of ploughing the field at Rs. 3.75 per 100 ${{m}^{2}}$ is Rs 13500. Find the altitude and the base of the field

(a) 1240m, 400m

(b) 1200 m, 600 m

(c) 1350m, 200m

(d) 9050 m, 100 m

(e) None of these

Explanation:

The cost of ploughing is Rs 3.75, if area=100${{\text{m}}^{\text{2}}}$

If the cost of ploughing is Rs 1, area$=\left( \frac{100}{3.75} \right){{m}^{2}}$

If the cost of ploughing is Rs 13500, area$=\left( \frac{100}{3.75}\times 13500 \right){{m}^{2}}=360000\,{{m}^{2}}$

area of the field$=\text{36}0000\text{ }{{\text{m}}^{\text{2}}}$

Let the Altitude of the field be$x$meters

Then base$=2x$meters

Area of the field$=\left( \frac{1}{2}\times 2x\times x \right){{m}^{2}}=({{x}^{2}}){{m}^{2}}$

${{x}^{2}}=360000$       $\Rightarrow$$x=\sqrt{360000}=600$

altitude = 600 m, base = 1200 m

Each side of an equilateral triangle measures 10 cm. Calculate

(i) The area of triangle &

(ii) The height of the triangle

Given: $\sqrt{3}=1.732$

(a) 18.65cm

(b) 18.85cm

(c) 8.99cm

(d) 18.66cm

(e) None of these

Explanation

Here, a = 10 cm

(i) Area of the triangle$=\left( \frac{\sqrt{3}}{4}\times {{a}^{2}} \right)$

$=\left( \frac{\sqrt{3}}{4}\times 10\times 10 \right)c{{m}^{2}}=(25\times \sqrt{3})c{{m}^{2}}$

$=(25\times 1.732)c{{m}^{2}}=43.3c{{m}^{2}}$

(ii) Height of the triangle$=\left( \frac{\sqrt{3}}{2}\times a \right)$units

$=\left( \frac{\sqrt{3}}{2}\times 10 \right)cm=(5\times \sqrt{3})cm=(5\times 1.732)cm=8.66\,cm$

The height of an equilateral triangle is 6 cm. Find the area of the triangle, correct to two decimal places take $\sqrt{3}=\text{1}.\text{732}$

(a) $\text{2}0.\text{88}\,\text{c}{{\text{m}}^{\text{2}}}$

(b) $\text{2}0.\text{78}\,\text{c}{{\text{m}}^{\text{2}}}$

(c) $\text{2}0.\text{78}\,\text{c}{{\text{m}}^{\text{2}}}$

(d) $\text{2}0.\text{78}\,\text{c}{{\text{m}}^{\text{4}}}$

(e) None of these

Find the area of an isosceles triangle, each of whose equal sides is 3 cm and base is 24 cm.

(a) 3.42

(b) 3.50

(c) 3.62

(c) 3.30

(d) None of these

The minute hand of a clock $\frac{x}{2}$cm long. Find the area of the face of the clock described by the minute hand in 35 minutes

(a) $\frac{11{{x}^{2}}}{24}$

(b) $\frac{7{{x}^{2}}}{24}$

(c) $\frac{5{{x}^{2}}}{24}$

(d) $\frac{13{{x}^{2}}}{24}$

(e) None of these