A right-angled triangle having base 6.3 , m a | vedclass

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(B) First,convert all units to meters: Base $r = 6.3 \, m$,Height $h = 10 \, cm = 0.1 \, m$.When the triangle is rotated around its height,it forms a

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$233$

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(B) First,convert all units to meters: Base $r = 6.3 \, m$,Height $h = 10 \, cm = 0.1 \, m$.When the triangle is rotated around its height,it forms a cone with radius $r = 6.3 \, m$ and height $h = 0.1 \, m$.Volume of the cone $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} 	imes \frac{22}{7} 	imes (6.3)^2 	imes 0.1 = 4.158 \, m^3$.To find the curved surface area,we first find the slant height $l = \sqrt{r^2 + h^2} = \sqrt{(6.3)^2 + (0.1)^2} = \sqrt{39.69 + 0.01} = \sqrt{39.7} \approx 6.3008 \, m$.Curved Surface Area $= \pi r l = \frac{22}{7} 	imes 6.3 	imes 6.3008 \approx 124.75 \, m^2$.Note: If the height was $10 \, m$ instead of $10 \, cm$,then $l = \sqrt{6.3^2 + 10^2} = 11.81 \, m$. Surface Area $= \frac{22}{7} 	imes 6.3 	imes 11.81 = 233.83 \, m^2$. Given the options,the intended height was $10 \, m$.

$A$ right-angled triangle having base $6.3 \, m$ and height $10 \, cm$ is rotated around its height. Find the volume of the cone thus formed. Also,find the curved surface area (in $m^2$). (in $.83$)

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