Question

A semicircular sheet of paper of diameter 28 cm is bent into an open Cup .find the slant height of the cone.

Plss answer as fast as possible.

Answer 1

For the semicircle;

diameter = 28 cm

radius (R) = 28/2 = 14 cm

For the cone;

diameter(semicircle) = circumference(base)

28 = 2πr

r = 28/(2π)

r = 14/π

r = 14 ÷ (22/7)

r = 14 x (7/22)

r = 98/22

r = 49/11 cm

height (h) = radius(semicircle) = R

h = 14 cm

slant height

$l \: = \sqrt{ {{r}^{2} + h}^{2} }$

$l \: = \: \sqrt{ {14}^{2} \: + {(49 \div 11)}^{2} }$

$l \: = \: \sqrt{196 \: + \: (2401 \div 121)}$

$l \: = \: \sqrt{196 \: + \: 19.843}$

$l \: = \: \sqrt{215.843} \: cm$

$l \: = 14.69 \: cm \: approx.$

Hence, slant height of the cone is

$\sqrt{215.843 \: cm}$

or 14.69 cm approx.

(you can leave the answer in square root)

Answer 2

Hence, slant height = radius of the circle = 14 cm


I have given the answer experimentally
because to make u understand.