[5 marks sum]

Question 15 Marks

A hollow metallic cylindrical tube has an internal radius of 3.5 cm and height 21 cm. The thickness of the metal tube is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place.

Answer

Internal radius of the hollow cylinder $=r=3.5 \mathrm{~cm}$
Height $=\mathrm{h}=21 \mathrm{~cm}$
Thickness of the metal $=0.5 \mathrm{~cm}$
Therefore, Outer radius $=\mathrm{R}=(3.5+0.5) \mathrm{cm}=4 \mathrm{~cm}$
Now, Volume of metal used $=\pi h\left(R^2-r^2\right)$
$
\begin{aligned}
& =\frac{22}{7} \times 21 \times\left(4^2-3.5^2\right) \\
& =\frac{22}{7} \times 21 \times(16-12.25) \\
& =\frac{22}{7} \times 21 \times 3.75 \\
& =247.5 \mathrm{~cm}^3
\end{aligned}
$
Volume of metal used $=247.5 \mathrm{~cm}^3$
Therefore, Volume of cone $=247.5 \mathrm{~cm}^3$ and height $=7 \mathrm{~cm}$ Let $\mathrm{r} 1$ be the radius of cone.
$
\begin{aligned}
& \therefore \text { Volume }=\frac{1}{3} \pi r 1^2 h \\
& \Rightarrow \frac{1}{3} \pi r 1^2 h=247.5 \\
& \Rightarrow \frac{1}{3} \times \frac{22}{7} \times r 1^2 \times 7=247.5 \\
& \Rightarrow r 1^2=\frac{247.5 \times 3 \times 7}{22 \times 7} \\
& \Rightarrow r 1^2=33.75 \\
& \Rightarrow r 1=5.8 \mathrm{~cm}
\end{aligned}
$
Radius of the cone $=5.8 \mathrm{~cm}$

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Question 25 Marks

A buoy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.

Answer

Radius of hemispherical part $(\mathrm{r})=3.5 \mathrm{~m}=\frac{7}{2} \mathrm{~m}$
Therefore, Volume of hemisphere $=\frac{2}{3} \pi r^3$
$
\begin{aligned}
& =\frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \\
& =\frac{539}{6} \mathrm{~m}^3
\end{aligned}
$
Volume of conical part $=\frac{2}{3} \times \frac{539}{6} \mathrm{~m}^3$ (2/3 of hemisphere)
Let height of the cone $=\mathrm{h}$
Then,
$
\begin{aligned}
& \frac{1}{3} \pi r^2 h=\frac{2 \times 539}{3 \times 6} \\
& \Rightarrow \frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h=\frac{2 \times 539}{3 \times 6} \\
& \Rightarrow h=\frac{539 \times 2 \times 2 \times 7 \times 3}{3 \times 6 \times 22 \times 7 \times 7} \\
& \Rightarrow \mathrm{h}=\frac{14}{3} \mathrm{~m}=4 \frac{2}{3} \mathrm{~m}=4.67 \mathrm{~m}
\end{aligned}
$
Height of the cone $=4.67 \mathrm{~m}$
Surface area of buoy $=2 \pi r^2+\pi r l$
But $l=\sqrt{r^2+h^2}$
$
\begin{aligned}
& l=\sqrt{\left(\frac{7}{2}\right)^2+\left(\frac{14}{3}\right)^2} \\
& =\sqrt{\frac{49}{4}+\frac{196}{9}}=\sqrt{\frac{1225}{36}}=\frac{35}{6} \mathrm{~m}
\end{aligned}
$
Therefore, surface area $=$
$
\begin{aligned}
& =\left(2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right)+\left(\frac{22}{7} \times \frac{7}{2} \times \frac{35}{6}\right) \mathrm{m}^2 \\
& =\frac{77}{1}+\frac{385}{6}=\frac{847}{6} \\
& =141.17 \mathrm{~m}^2
\end{aligned}
$

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Question 35 Marks

Find the volume of the right circular cone whose height is $12\ cm$ and slant length is $15\ cm . (\pi = 3.14)$

Answer

Slant length $=I=15 cm$
Height $= h =12 cm$
Radius of the base $=r$
We know ,
$ l^2=h^2+r^2$
$\Rightarrow r^2=l^2-h^2$
$\Rightarrow r=\sqrt{l^2-h^2}$
$\Rightarrow r=\sqrt{15^2-12^2}$
$\Rightarrow r=9 cm $
Radius $=9 cm$
$ \text { Volume }=\frac{1}{3} \times\left(\pi r^2\right) \times h$
$=\frac{1}{3} \times 3.14 \times 9 \times 9 \times 12$
$=1017.36 cm ^3 $
Volume of the cone $=1017.36 cm ^3$

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Question 45 Marks

Answer

Internal radius of the hollow cylinder $=r=3.5 \mathrm{~cm}$
Height $=\mathrm{h}=21 \mathrm{~cm}$
Thickness of the metal $=0.5 \mathrm{~cm}$
Therefore, Outer radius $=R=(3.5+0.5) \mathrm{cm}=4 \mathrm{~cm}$
Now, Volume of metal used $=\pi h\left(R^2-r^2\right)$
$
\begin{aligned}
& =\frac{22}{7} \times 21 \times\left(4^2-3.5^2\right) \\
& =\frac{22}{7} \times 21 \times(16-12.25) \\
& =\frac{22}{7} \times 21 \times 3.75 \\
& =247.5 \mathrm{~cm}^3
\end{aligned}
$
Volume of metal used $=247.5 \mathrm{~cm}^3$
Therefore, Volume of cone $=247.5 \mathrm{~cm}^3$ and height $=7 \mathrm{~cm}$
Let $\mathrm{r} 1$ be the radius of cone.
$
\begin{aligned}
& \therefore \text { Volume }=\frac{1}{3} \pi r 1^2 h \\
& \Rightarrow \frac{1}{3} \pi r 1^2 h=247.5 \\
& \Rightarrow \frac{1}{3} \times \frac{22}{7} \times r 1^2 \times 7=247.5 \\
& \Rightarrow r 1^2=\frac{247.5 \times 3 \times 7}{22 \times 7} \\
& \Rightarrow r 1^2=33.75 \\
& \Rightarrow r 1=5.8 \mathrm{~cm}
\end{aligned}
$
Radius of the cone $=5.8 \mathrm{~cm}$

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Question 55 Marks

A conical tent with a capacity of $600 m^3$ stands on a circular base of area $160 m^2$ Find in $m^2$ the area of the canvas.

Answer

Area of the circular base $=160 \mathrm{~m}^2$
$
\pi r^2=160
$
$\Rightarrow r=\sqrt{\frac{160 \times 7}{22}}$
$
\Rightarrow r=\sqrt{50.909}=7.134 \mathrm{~m}
$
Therefore, radius $=7.134 \mathrm{~m}$
Capacity or Volume of the tent $=600 \mathrm{~m}^3$
$
\begin{aligned}
& \frac{1}{3} \pi r^2 h=600 \\
& \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 7.13 \times 7.13 \times h=600 \\
& \Rightarrow h=\frac{600 \times 3 \times 7}{7.13 \times 7.13 \times 22} \\
& \Rightarrow h=11.265 \mathrm{~m}
\end{aligned}
$
Therefore, vertical height $=11.265 \mathrm{~m}$
We know slant height $(l)=$
$
\begin{aligned}
& I=\sqrt{r^2+h^2} \\
& \Rightarrow l=\sqrt{7.134^2+11.265^2} \\
& \Rightarrow l=\sqrt{177.624}=13.327
\end{aligned}
$
Therefore, slant height $=13.327 \mathrm{~m}$
The curved surface area $=\pi r l=\frac{22}{7} \times 7.134 \times 13.327=298.9 \mathrm{~m}^2$
Hence, the area of the canvas $=298.9 \mathrm{~m}^2$

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Question 65 Marks

The diameter of a right circular cylinder is $12\ m$ and the slant height is $10\ m.$ Find its curved surface area and the total surface area .

Answer

Diameter of the cylinder $=12 m \Rightarrow$ radius $=6 m$
Curved surface area $=$ circumference of the base $\times$ height
$ =2 \pi r \times h$
$=2 \times \frac{22}{7} \times 6 \times 10$
$=377.14 m ^2 $
Curved Surface area $=377.14 m ^2$
Total surface area $=$ Curved surface area $+(2 \times$ base area $)$
$ =2 \pi r h+2 \pi r^2$
$=2 \pi r(h+r)$
$=2 \times \frac{22}{7} \times 6 \times(10+6)$
$=2 \times \frac{22}{7} \times 6 \times 16$
$=603.42 m ^2 $
Total Surface area $=603.42 m ^2$

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Question 75 Marks

Find the curved surface area , the total surface area and the volume of a cone if its :
Height = 8 cm , diameter = 12 cm

Answer

$
\begin{aligned}
& \text { Height }=8 \mathrm{~cm}, \text { diameter }=12 \mathrm{~cm} \\
& \text { Diameter }=12 \mathrm{~cm} \Rightarrow \mathrm{r}=6 \mathrm{~cm} \\
& \text { Curved surface area }=\left(\pi r \sqrt{h^2+r^2}\right) \\
& =\frac{22}{7} \times 6 \times \sqrt{8^2+6^2} \\
& =\frac{22}{7} \times 6 \times \sqrt{100} \\
& =\frac{22}{7} \times 6 \times 10 \\
& =188.57
\end{aligned}
$
Curved surface area $=188.57 \mathrm{~cm}^2$
Total surface area $=$ area of circular base + Curved surface area
$
\begin{aligned}
& =\pi r^2+\left(\pi r \sqrt{h^2+r^2}\right) \\
& =\frac{22}{7} \times 6 \times 6+188.57 \\
& =113.14+188.57 \\
& =301.71
\end{aligned}
$
Total surface area $=301.71 \mathrm{~cm}^2$
$
\begin{aligned}
& \text { Volume }=\frac{1}{3} \times\left(\pi r^2\right) \times h \\
& =\frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 8 \\
& =301.71
\end{aligned}
$
Volume of the cone $=301.71 \mathrm{~cm}^3$

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Question 85 Marks

Find the curved surface area , the total surface area and the volume of a cone if its :
Height = 16 cm , diameter = 24 cm

Answer

$
\begin{aligned}
& \text { Height }=16 \mathrm{~cm}, \text { diameter }=24 \mathrm{~cm} \\
& \text { diameter }=24 \mathrm{~cm} \Rightarrow r=12 \mathrm{~cm} \\
& \text { Curved surface area }=\left(\pi r \sqrt{h^2+r^2}\right) \\
& =\frac{22}{7} \times 12 \times \sqrt{16^2+12^2} \\
& =\frac{22}{7} \times 12 \times \sqrt{400} \\
& =\frac{22}{7} \times 12 \times 20 \\
& =754.29
\end{aligned}
$
Curved surface area $=754.29 \mathrm{~cm}^2$
Total surface area $=$ area of circular base + Curved surface area
$
\begin{aligned}
& =\pi r^2+\left(\pi r \sqrt{h^2+r^2}\right) \\
& =\frac{22}{7} \times 12 \times 12+754.29 \\
& =452.57+754.29 \\
& =1206.86
\end{aligned}
$
Total surface area $=1206.86 \mathrm{~cm}^2$
$
\begin{aligned}
& \text { Volume }=\frac{1}{3} \times\left(\pi r^2\right) \times h \\
& =\frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 16 \\
& =2413.71
\end{aligned}
$
Volume of the cone $=2413.71 \mathrm{~cm}^3$

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Question 95 Marks

Find the curved surface area , the total surface area and the volume of a cone if its :
Height = 15 cm , radius = 8 cm

Answer

$
\begin{aligned}
& \text { Height }=15 \mathrm{~cm}, \text { radius }=8 \mathrm{~cm} \\
& \text { Curved surface area }=\left(\pi r \sqrt{h^2+r^2}\right) \\
& =\frac{22}{7} \times 8 \times \sqrt{15^2+8^2} \\
& =\frac{22}{7} \times 8 \times \sqrt{289} \\
& =\frac{22}{7} \times 8 \times 17 \\
& =427.43
\end{aligned}
$
Curved surface area $=427.43 \mathrm{~cm}^2$
Total surface area $=$ area of circular base + curved surface area
$
\begin{aligned}
& =\pi r^2+\left(\pi r \sqrt{h^2+r^2}\right) \\
& =\frac{22}{7} \times 8 \times 8+427.43 \\
& =201.14+427.43 \\
& =628.57
\end{aligned}
$
Total surface area $=628.57 \mathrm{~cm}^2$
$
\begin{aligned}
& \text { Volume }=\frac{1}{3} \times\left(\pi r^2\right) \times h \\
& =\frac{1}{3} \times \frac{22}{7} \times 8 \times 8 \times 15 \\
& =1005.71
\end{aligned}
$
Volume of the cone $=1005.71 \mathrm{~cm}^3$

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Question 105 Marks

Find the curved surface area , the total surface area and the volume of a cone if its :
Height $= 12\ cm ,$ radius $= 5\ cm$

Answer

$\text { Height }=12 cm , \text { radius }=5 cm$
$\text { Curved surface area }=\left(\pi r \sqrt{h^2+r^2}\right)$
$=\frac{22}{7} \times 5 \times \sqrt{12^2+5^2}$
$=\frac{22}{7} \times 5 \times \sqrt{169}$
$=\frac{22}{7} \times 5 \times 13$
$=204.29$
Curved surface area $=204.29 cm ^2$
Total surface area $=$ area of circular base + curved surface area
$ =\pi r^2+\left(\pi r \sqrt{h^2+r^2}\right)$
$=\frac{22}{7} \times 5 \times 5+204.29$
$=78.57+204.29$
$=282.86 $
Total surface area $=282.86 cm ^2$
$ \text { Volume }=\frac{1}{3} \times\left(\pi r^2\right) \times h$
$=\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12$
$=314.29 $
Volume of the cone $=314.29 cm ^3$

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Mathematics [5 marks sum]

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