Question

5. From a right cylindrical solid of radius 7 cm
and height 12 cm, a right cone, half the height
of the cylinder, is removed. The base of the
cone is one of the plane faces of the cylinder,
their centres being
the same. If the
radius of the base of
the cone is half of
the radius of the
base of the cylinder,
find the total surface
area of the solid left.
​

Answer 1

Solution:

Given:

$\implies$ Radius: 7 cm

$\implies$ Height: 12 cm

Now:

Total surface area of solid left = Total surface area of cylinder - Total surface area of cone

We know that:

$\boxed{\sf{T. S. A.\:of\:Cylinder = 2 \pi r(h+r)}}$

Putting the values in the formula,

We get:

$\implies$ 2 × 22/7 × 7 × (12+7)

$\implies$ 44 × 19

$\implies$ 836

T. S. A. of the cone: π × r^2 + π × r × l

By applying pythagorus theorem, we get,

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

$\implies$ I^2= (7/2)^2 + 6^2

$\implies$ I^2 = 49/4 + 36

$\implies$ I^2 =193/4

$\implies$ I = 6.95

Total surface area of cone:

$\implies$ 22/7 × 7/2 × 7/2 + 22/7 × 7/2 × 6.95

$\implies$ (22 × 7) / 4 + 11 × 6.95

$\implies$ 38.5 + 76.45

$\implies$ 114.9

Total surface area of the Solid left:

$\implies$ 836 - 114.9

$\implies$ 721.1 cm

____________________

Answered by: Niki Swar, Goa❤️

Answer 2

Answer:

89+43=836 bhj#Alaskan soonæ