Question

A solid cylinder is of height 15cm and diameter 7cm.Two equal conical holes of radius 3cm and height 4cm are cut off, one from each circular end .Find the surface area of the remaining solid

Answer 1

$\bold{ANSWER}$

$\boxed{235.5 cm}$

$\bold{SOLUTION}$

For surface area of solid cylinder,

$\boxed{S A = 2 \times π \times r \times h}$

Given,

height =15 cm

diameter = 7cm

radius = 3.5cm

S.A = 2 × 3.14 ×3.5 ×15

S.A = 329.7 cm

When two cones of

height = 4cm(perpendicular)

radius = 3cm(base)

length (hypotenuse)= ?

By pythagoras theorem,

$\boxed{{hypotenuse}^{2} = {perpendicular}^{2} + {base}^{2}}$

${hypotenuse}^{2} = {4}^{2} + {3}^{2}$

${hypotenuse}^{2} = 16 + 9$

${hypotenuse}^{2} = 25$

$hypotenuse = \sqrt{25}$

hypotenuse =5 cm

Surface area of cone = πrl

Since, two equal cones are cut off

$\boxed{S A= 2 \times π \times r \times l}$

= 2 × 3.14 × 3 ×5

=94.2cm

So,the surface area of remaining solid =

Surface area of cylinder - surface area both the cones

= 329.7 - 94.2

=235.5 cm