The area of the base of a right circular cone is 1.54 sq. metre andhei

1.	The area of the base of a right circular cone is 1.54 sq. metre andheight is 2.4 metre. Find the volume and slant height of the cone.
Answer» SOLUTION : $\underline{\bf{Given\::}}}$ Area of the BASE of a RIGHT circular CONE = 1.54 m² Height of cone (h) = 2.4 m Diagram : $\setlength{\unitlength}{1.8cm}\begin{picture}(16,4)\thicklines \put(8,1){\line(1,0){1}}\put(8,1){\line(1,2){1}}\put(10,1){\line(-1,2){1}}\put(9.03,1.6){\sf{2.4m}}\put(9,1){\line(0,2){2}}\put(8.4,0.83){\sf{r}}\qbezier(8,1)(8.4,1.3)(9,1.3)\qbezier(9,1.3)(9.7,1.3)(10,1)\qbezier(8,1)(8.4,0.6)(9,0.6)\qbezier(9,0.6)(9.7,0.6)(10,1)\end{picture}}$ $\underline{\bf{Explanation\::}}}$ $\longrightarrow\sf{\pi r^{2} =1.54}\\\\\longrightarrow\sf{22/7 \times r^{2} =1.54}\\\\\longrightarrow\sf{r^{2} =\dfrac{1.54\times 7}{22} }\\\\\\\longrightarrow\sf{r^{2}=\dfrac{10.78\times 100}{22\times 100} }\\\\\\\longrightarrow\sf{r^{2}=\cancel{\dfrac{1078}{2200} }}\\\\\\\longrightarrow\sf{r^{2}=0.49}\\\\\longrightarrow\sf{r=\sqrt{0.49}} \\\\\longrightarrow\bf{r=0.7\:m}$ $\underline{\mathcal{VOLUME\:\:OF\:\:CONE\::}}$ $\longrightarrow\sf{Volume\:_{(cone)}=\dfrac{1}{3} \pi r^{2} h}\\\\\\\longrightarrow\sf{Volume\:_{(cone)}=\dfrac{1}{3} \times \dfrac{22}{7} \times (0.7)^{2}\times 2.4}\\\\\\\longrightarrow\sf{Volume\:_{(cone)}=\dfrac{22}{21} \times 0.49 \times 2.4}\\\\\\\longrightarrow\sf{Volume\:_{(cone)}=\cancel{\dfrac{25.872}{21} }}\\\\\longrightarrow\bf{Volume\:_{(cone)}=1.23m^{3}}$ $\underline{\mathcal{SLANT\:\:HEIGHT\:\:OF\:\:CONE\::}}$ $\longrightarrow\sf{Slant\:height\:(l)=\sqrt{r^{2}+h^{2}} }\\\\\longrightarrow\sf{Slant\:height\:(l)=\sqrt{(0.7)^{2}+(2.4)^{2} } }\\\\\longrightarrow\sf{Slant\:height\:(l)=\sqrt{0.49+5.76}} \\\\\longrightarrow\sf{Slant\:height\:(l)=\sqrt{6.25} }\\\\\longrightarrow\bf{Slant\:height\:(l)=2.5m}$

The area of the base of a right circular cone is 1.54 sq. metre andheight is 2.4 metre. Find the volume and slant height of the cone.

Answer»

SOLUTION :

$\underline{\bf{Given\::}}}$

Area of the BASE of a RIGHT circular CONE = 1.54 m²
Height of cone (h) = 2.4 m

Diagram :

$\setlength{\unitlength}{1.8cm}\begin{picture}(16,4)\thicklines \put(8,1){\line(1,0){1}}\put(8,1){\line(1,2){1}}\put(10,1){\line(-1,2){1}}\put(9.03,1.6){\sf{2.4m}}\put(9,1){\line(0,2){2}}\put(8.4,0.83){\sf{r}}\qbezier(8,1)(8.4,1.3)(9,1.3)\qbezier(9,1.3)(9.7,1.3)(10,1)\qbezier(8,1)(8.4,0.6)(9,0.6)\qbezier(9,0.6)(9.7,0.6)(10,1)\end{picture}}$

$\underline{\bf{Explanation\::}}}$

$\longrightarrow\sf{\pi r^{2} =1.54}\\\\\longrightarrow\sf{22/7 \times r^{2} =1.54}\\\\\longrightarrow\sf{r^{2} =\dfrac{1.54\times 7}{22} }\\\\\\\longrightarrow\sf{r^{2}=\dfrac{10.78\times 100}{22\times 100} }\\\\\\\longrightarrow\sf{r^{2}=\cancel{\dfrac{1078}{2200} }}\\\\\\\longrightarrow\sf{r^{2}=0.49}\\\\\longrightarrow\sf{r=\sqrt{0.49}} \\\\\longrightarrow\bf{r=0.7\:m}$

$\underline{\mathcal{VOLUME\:\:OF\:\:CONE\::}}$

$\longrightarrow\sf{Volume\:_{(cone)}=\dfrac{1}{3} \pi r^{2} h}\\\\\\\longrightarrow\sf{Volume\:_{(cone)}=\dfrac{1}{3} \times \dfrac{22}{7} \times (0.7)^{2}\times 2.4}\\\\\\\longrightarrow\sf{Volume\:_{(cone)}=\dfrac{22}{21} \times 0.49 \times 2.4}\\\\\\\longrightarrow\sf{Volume\:_{(cone)}=\cancel{\dfrac{25.872}{21} }}\\\\\longrightarrow\bf{Volume\:_{(cone)}=1.23m^{3}}$

$\underline{\mathcal{SLANT\:\:HEIGHT\:\:OF\:\:CONE\::}}$

$\longrightarrow\sf{Slant\:height\:(l)=\sqrt{r^{2}+h^{2}} }\\\\\longrightarrow\sf{Slant\:height\:(l)=\sqrt{(0.7)^{2}+(2.4)^{2} } }\\\\\longrightarrow\sf{Slant\:height\:(l)=\sqrt{0.49+5.76}} \\\\\longrightarrow\sf{Slant\:height\:(l)=\sqrt{6.25} }\\\\\longrightarrow\bf{Slant\:height\:(l)=2.5m}$