Question

hlw mates ^_^

===============

a question ^↓^

•••••••••••••••••••

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out .find the total surface area of the remaining solid to the nearest cm².

{{ chapter - 13, ex.13.1 , question no. 8 }}

=====================

answer fast please

______________

Answer 1

t.s.a of remaining solid nearest =18cm^2

Answer 2

Hlo mate :-

Solution :-
_________________________________________________________________________________________________

Given

Height of cylinder = Height of conical cavity, h = 2.4 cm

Diameter, d = 1.4 cm

Radius, r = 1.4/2= 0.7 cm

Slant height, l = √h² + r² = √(2.4)² + (0.7)² = √5.76 + 0.49 = √6.25 = 2.5 cm

TSA of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

TSA = 2πrh+ πrl+ πr²

= πr [ 2(h) + l + r ]

= 22/7 × 0.7 [ 2 × 2.4 + 2.5 + 0.7 ]

= 17.6 cm²

Hence,
TSA of remaining solid is = 17.6 cm²

:- when rounded off = 18 cm^2 ans

_________________________________________________________________________________________________

Hope it's helpful