The radius and height of a cone are in the ratio 4:3 . The area of the base is 154 cm^2 . Find the area of the curved surface .

Solution:  Let $r$ be the radius and $h$ be the height of the cone . it is given that

$\frac{r}{h}= \frac{4}{3}$

Let $r= 4x$   and $h= 3x$

Now , Area of the base $= 154$  $cm^2$

$\Rightarrow$                $\pi r^2=154$

$\Rightarrow$                $\frac{22}{7}\times (4x)^2=154$

$\Rightarrow$               $\frac{22}{7}\times 16 x^2=154$

$\Rightarrow$                           $x^2= \frac{154\times 7}{22\times 16}=\frac{49}{16}\Rightarrow x= \sqrt{\frac{49}{16}}=\frac{7}{4}cm$

$\therefore$                         $r=4x=4\times \frac{7}{4}cm ,$       and      $h= 3 \times \frac{7}{4}=\frac{21}{4}cm$

Let the slant height be $l$ cm . Then,

$l^2=r^2+h^2$

$\Rightarrow$                    $l=\sqrt{r^2+h^2}=\sqrt{7^2+ (\frac{21}{4})^2}cm = \sqrt{49+\frac{441}{16}}=\frac{35}{4}cm$

$\therefore$              curved surface area

$=\pi rl = \frac{22}{7}\times7\times \frac{35}{4}cm^2=192.5cm^2$