Question

The slant height of a frustum of a coneis 4cm and the perimetre of itd circular ends are 18cm and 6cm.find the curved surface area of the frustum?

Answer 1

Slant height (l) = 4 cm

Circumference of bigger end = 18 cm

2πr1=18

$2 \times \frac{22}{7} \times r1 = 18$

$r1 = \frac{18 \times 7}{22 \times 2}$


$r1 = \frac{126}{44}$

R1 = 2.86 cm

Circumference of smaller end = 6 cm

2πr = 6cm

$2 \times \frac{22}{7} \times r = 6$



$r = \frac{6 \times 7}{2 \times 22}$


r=42/44

r= 0.95 cm

CSA of frustum of cone = π(r1+r) × l

$= \frac{22}{7} \times (2.86 + 0.95) \times 4$


$= \frac{22}{7} \times 3.81 \times 4$


$= \frac{335.28}{7}$


=47.89 cm^2

Or

=47.9 cm^2

$\bf{Hope\:it\:helps}$