9th Class Mathematics Mensuration Solids

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  

 

Answer: (a)  

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  

 

Answer: (c)

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  

 

Answer: (b)

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

 

Answer: (b)        

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

 

Answer: (b)  

 

 

  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

 

Answer: (d)  

 

 

  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

 

Answer: (a)        

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