Question

If CSA is $x{\text{m}}^2$, TSA is $y{\text{m}}^2$ and Volume is $z{\text{m}}^3$. What is the value of $x+y+z.$

• A. $12\pi +4\sqrt{40}$
• B. $12\pi +2\sqrt{40}$
• C. $6\pi +4\sqrt{40}$
• D. $10\pi +4\sqrt{40}$

Answer 1

The correct option is A $12\pi +4\sqrt{40}$

For the cone given in the figure,
Radius $=r=2$ m and height $=h=6$ m.
Slant height $=l=\sqrt{r^2+h^2}$
Slant height $=l=\sqrt{2^2+6^2}$
$=\sqrt{4+36}$
$=\sqrt{40}cm$

Curved Surface Area(CSA) $=\pi rl$
Curved Surface Area(CSA) $=\pi \times 2\times \sqrt{40}$
Curved Surface Area(CSA) $=2\sqrt{40}\pi c{m}^2$

Total Surface Area(TSA) $=CSA+\pi r^2$
Total Surface Area(TSA) $=2\sqrt{40}\pi +\pi \left(2^2\right)$
Total Surface Area(TSA) $=2\sqrt{40}\pi +4\pi$


$\text{Volume}=\frac{1}{3}\pi r^{2}h$

$\text{Volume}=\frac{1}{3}\times \pi \times (2{)}^2\times (6)$
$\text{Volume}=8\pi c{m}^3$

Hence, CSA + TSA + Volume $=2\sqrt{40}\pi +2\sqrt{40}\pi +4\pi +8\pi$
$=12\pi +4\sqrt{40}$